This course introduces algebraic geometry through the classical language of polynomial equations, affine and projective varieties, and the geometric meaning of their algebraic invariants. It asks how solution sets of polynomial systems are organized, how topology and algebra interact through the Zariski topology and coordinate rings, and how geometric objects can be studied using regular functions, rational maps, and local descriptions. The emphasis is on building a working dictionary between geometry and commutative algebra in a setting where the fundamental objects are concrete and explicit. The chapters develop this dictionary step by step. The early chapters establish affine algebraic sets, the Nullstellensatz, and regular maps, then move to localization and rational functions as the local tools needed to analyze geometry on smaller open pieces. From there the course passes to projective space, homogeneous equations, projective closure, and elimination, which provide the natural compact setting for classical geometry and its embeddings. Later chapters study dimension and degree, tangent spaces and singularities, and hypersurface sections mostly in classical language: a fiber first means an inverse image of a point, a projective variety first means a zero set of homogeneous polynomials in projective space, and families of linear subspaces are introduced through explicit coordinates. Some quoted theorem cards in the later chapters are included as optional modern reference points rather than as part of the classical working level of the course. When those cards mention schemes, sheaves, Grassmannians, Cartier divisors, regular local rings, or scheme-theoretic length, they are recording the standard contemporary form of a result whose surrounding discussion is classical. A reader may use the card as a reference label and continue with the point-set, coordinate-ring, and explicit-equation interpretation developed in the main text. # Introduction This course studies geometric objects described by polynomial equations over an [algebraically closed field](/page/Algebraically%20Closed%20Field) $k$. Its guiding question is how much geometry can be recovered from commutative algebra, and how algebraic operations on polynomial rings appear as geometric operations on solution sets. The first half of the course builds the affine dictionary; the second half moves to projective space, where compactness-like phenomena and intersections at infinity enter naturally. A scope convention will be used throughout. Unless a passage explicitly says otherwise, all varieties in this note are classical affine or projective varieties over $k$, meaning reduced algebraic sets cut out by polynomial or [homogeneous polynomial](/page/Homogeneous%20Polynomial) equations. If a map $f:X\to Y$ is discussed, the fiber over a point $y\in Y$ means the ordinary set-theoretic inverse image $f^{-1}(y)$, with its equations inherited from the defining equations of $X$ and the coordinate equations of $y$. Quoted theorem cards that use modern notation should be read as optional reference cards at a higher level of generality. In that notation, $\operatorname{Spec} A$ denotes the affine scheme attached to a ring $A$, $\operatorname{Proj} S$ denotes the projective scheme attached to a graded ring $S$, $\mathcal O_X(d)$ denotes the standard twisting sheaf on a projective variety, and $H^0(X,\mathcal O_X(d))$ denotes its space of global degree-$d$ sections. A scheme-theoretic fiber records the equations of a fiber with possible multiplicity or nilpotent structure; a reduced induced scheme forgets that extra nilpotent information; a Grassmannian parametrizes linear subspaces of a fixed dimension; a Cartier divisor is the modern way to package hypersurface-like codimension-one data; and the length of a zero-dimensional intersection is the algebraic multiplicity counted by its local coordinate ring. These explanations are a reading aid for the cited statements, not a prerequisite list for the course. The surrounding arguments continue to use the classical point-set and coordinate-ring language developed in the course, and any modern card can be skipped without losing the main line of exposition. ## The Central Problem The starting point is a familiar observation from linear algebra and calculus: equations cut out shapes. Algebraic geometry asks what happens when the equations are polynomial, the ambient space is $k^n$, and we insist that the resulting theory remember both geometry and algebra. [explanation: Polynomial Equations as Geometry] Let $k$ be an algebraically closed field. A finite family of polynomials $f_1,\dots,f_r \in k[x_1,\dots,x_n]$ determines a subset of affine space by imposing the simultaneous equations \begin{align*} f_1(a)=\cdots=f_r(a)=0. \end{align*} The geometric object is the common zero set, while the algebraic object is the ideal $(f_1,\dots,f_r)$ generated by the equations. The course is organised around making this passage reversible in the correct sense: geometry determines functions on the space, and functions determine the equations that vanish on it. [/explanation] This reversal is not literal unless the algebra is adjusted. Different sets of equations may define the same subset, and nilpotent information in a [quotient ring](/page/Quotient%20Ring) is invisible to classical point sets. The Nullstellensatz will identify radical ideals as the correct algebraic objects for classical affine varieties. [example: Two Equations With the Same Classical Points] Over a field $k$, the two ideals $(x)$ and $(x^2)$ in $k[x]$ have the same vanishing set in $\mathbb A_k^1$. Indeed, for $a \in k$, \begin{align*} a \in V(x) \text{ if and only if } a=0. \end{align*} Also, \begin{align*} a \in V(x^2) \text{ if and only if } a^2=0. \end{align*} Since a field has no nonzero nilpotent elements, $a^2=0$ implies $a=0$, so \begin{align*} V(x)=V(x^2)=\{0\}. \end{align*} The quotient rings remember different algebra. In $k[x]/(x)$, the class of $x$ is zero because $x \in (x)$. In $k[x]/(x^2)$, the class $\overline{x}$ is not zero: if $\overline{x}=0$, then $x \in (x^2)$, so $x=x^2g(x)$ for some $g(x)\in k[x]$, which is impossible because the right side has degree at least $2$ unless it is zero. But $\overline{x}$ is nilpotent, since \begin{align*} \overline{x}^2=\overline{x^2}=0. \end{align*} Thus $(x)$ and $(x^2)$ define the same classical point set but give different quotient rings; the classical dictionary keeps the radical ideal $(x)$ and deliberately forgets the nilpotent thickening encoded by $k[x]/(x^2)$. Later scheme theory is designed to retain such nilpotent information, but that language is not part of the working framework of this course. [/example] ## Affine Space and Coordinate Rings Once a subset of affine space has been cut out by equations, the next problem is to decide which functions on it should count as polynomial functions. Restricting polynomials from the ambient affine space gives the right class, but two ambient polynomials give the same function on the subset precisely when their difference vanishes there. [definition: Affine Space] Let $k$ be a field and let $n \in \mathbb N$. Affine $n$-space over $k$ is the set \begin{align*} \mathbb A_k^n := k^n. \end{align*} Its elements are written $a=(a_1,\dots,a_n)$. [/definition] The notation treats $k^n$ as a space of points rather than as a [vector space](/page/Vector%20Space) with a preferred origin. Having named the ambient space, the next task is to name the subsets produced by polynomial equations inside it. [definition: Vanishing Set] Let \begin{align*} V:\mathcal P(k[x_1,\dots,x_n]) \to \mathcal P(\mathbb A_k^n) \end{align*} be the operation \begin{align*} S \mapsto V(S) \end{align*} that sends a subset $S \subset k[x_1,\dots,x_n]$ to its vanishing set. For such an $S$, define \begin{align*} V(S) := \{a \in \mathbb A_k^n : f(a)=0 \text{ for all } f \in S\}. \end{align*} [/definition] The operation $S \mapsto V(S)$ is a map from the power set of $k[x_1,\dots,x_n]$ to the power set of $\mathbb A_k^n$. The set $S$ may be infinite, but the [Hilbert Basis Theorem](/theorems/860) will imply that over a Noetherian [polynomial ring](/page/Polynomial%20Ring) every such vanishing set is already defined by finitely many equations. Subsets that arise in this way receive a name, because they are the basic geometric objects of the affine part of the course. [definition: Affine Algebraic Set] A subset $X \subset \mathbb A_k^n$ is an affine algebraic set if there exists a subset $S \subset k[x_1,\dots,x_n]$ such that \begin{align*} X=V(S). \end{align*} [/definition] To compare two different defining families for the same affine algebraic set, we need to move in the reverse direction. Starting from a subset $X$, the relevant algebraic data is not a chosen list of equations but the complete collection of polynomial equations that vanish at every point of $X$. This construction records exactly when two ambient polynomials define the same function on $X$, and it will become the kernel used to form the coordinate ring. [definition: Ideal of a Subset] Let \begin{align*} I:\mathcal P(\mathbb A_k^n) \to \{\text{ideals of } k[x_1,\dots,x_n]\} \end{align*} be the operation \begin{align*} X \mapsto I(X) \end{align*} that sends a subset $X \subset \mathbb A_k^n$ to its ideal. For such an $X$, define \begin{align*} I(X) := \{f \in k[x_1,\dots,x_n] : f(a)=0 \text{ for all } a \in X\}. \end{align*} [/definition] The operation $X \mapsto I(X)$ is a map from the power set of $\mathbb A_k^n$ to the set of ideals of $k[x_1,\dots,x_n]$. The assignments $S \mapsto V(S)$ and $X \mapsto I(X)$ reverse inclusions: more equations give fewer points, and more points impose more vanishing conditions. This motivates the coordinate ring, which turns the slogan "polynomial functions on $X$" into a quotient of the ambient polynomial ring by the equations that vanish on $X$. [definition: Coordinate Ring] Let $X \subset \mathbb A_k^n$ be an affine algebraic set. The coordinate ring of $X$ is \begin{align*} k[X] := k[x_1,\dots,x_n]/I(X). \end{align*} [/definition] Elements of $k[X]$ are polynomial functions on $X$. This quotient is the algebraic object attached to $X$, and later regular maps $X \to Y$ will be described by $k$-algebra homomorphisms $k[Y] \to k[X]$ in the opposite direction. [example: Coordinate Ring of the Coordinate Axes] Let $X=V(xy)\subset \mathbb A_k^2$. A point $(a,b)\in \mathbb A_k^2$ lies in $X$ exactly when $ab=0$, so \begin{align*} X=\{(a,0):a\in k\}\cup \{(0,b):b\in k\}. \end{align*} We compute its coordinate ring by identifying $I(X)$. Since $xy$ vanishes at every point of $X$, we have $(xy)\subset I(X)$. Conversely, let $f\in I(X)$ and write \begin{align*} f(x,y)=\sum_{i,j} c_{ij}x^iy^j. \end{align*} For every $a\in k$, the point $(a,0)$ lies in $X$, so \begin{align*} 0=f(a,0)=\sum_i c_{i0}a^i. \end{align*} Because $k$ is algebraically closed, hence infinite, a polynomial in one variable that vanishes at every $a\in k$ has all coefficients zero; therefore $c_{i0}=0$ for every $i$. Similarly, evaluating on the other axis gives \begin{align*} 0=f(0,b)=\sum_j c_{0j}b^j \end{align*} for every $b\in k$, so $c_{0j}=0$ for every $j$. Thus every nonzero term of $f$ is divisible by both $x$ and $y$, and hence \begin{align*} f=xy\sum_{i,j\ge 1}c_{ij}x^{i-1}y^{j-1}. \end{align*} So $I(X)\subset (xy)$, and therefore $I(X)=(xy)$. By the definition of the coordinate ring, \begin{align*} k[X]=k[x,y]/I(X)=k[x,y]/(xy). \end{align*} In this quotient, the classes of $x$ and $y$ are both nonzero: $x\notin (xy)$ and $y\notin (xy)$ because neither monomial is divisible by $xy$. But their product is zero, since \begin{align*} \overline{x}\,\overline{y}=\overline{xy}=0. \end{align*} Thus the two coordinate functions survive separately on the two axes, while their product records that every point of $X$ lies on at least one of the axes. [/example] ## Topology From Polynomial Equations A geometric theory needs a notion of [closed set](/page/Closed%20Set), continuity, density, and irreducibility. In algebraic geometry these are not imported from Euclidean topology; they are defined by polynomial equations themselves. [definition: Zariski Closed Set] A subset $X \subset \mathbb A_k^n$ is Zariski closed if there exists a subset $S \subset k[x_1,\dots,x_n]$ such that \begin{align*} X=V(S). \end{align*} [/definition] This definition proposes the closed sets, but a topology requires stability under the operations that closed sets must satisfy. Polynomial equations have exactly the permanence properties needed here: arbitrary intersections correspond to combining all equations from the closed sets involved, the whole affine space is cut out by the zero polynomial, the empty set is cut out by the unit ideal, and finite unions are encoded by products of equations. Thus these sets form the closed sets of the Zariski topology. This verification is only about the formal closed-set axioms; it does not say that the Zariski topology resembles the Euclidean or metric topology. Its hypotheses are doing different jobs. The ambient affine-space hypothesis supplies a polynomial ring $k[x_1,\dots,x_n]$ whose equations can be evaluated at points of $\mathbb A_k^n$; without a specified ambient polynomial ring, the expression $V(S)$ has no defined meaning. The polynomial-equation hypothesis is what makes finite unions work: products of equations encode the logical condition "vanish on the first set or on the second set." For instance, if closed sets in $\mathbb A_k^1$ were instead declared to be solution sets of linear equations only, then $\{0\}$ and $\{1\}$ would be closed but $\{0,1\}$ would not be cut out by linear equations. Arbitrary unions do not have the same product encoding: in $\mathbb A_k^1$, each singleton $\{a\}$ is Zariski closed, but an infinite proper subset such as $\mathbb N \subset \mathbb A_k^1$ when $k=\mathbb C$ is not Zariski closed, since a nonzero polynomial has only finitely many roots. This is why closed sets are required to be stable under finite unions rather than arbitrary unions. The topology makes irreducibility a central geometric notion, because a closed set that cannot be expressed as a union of two smaller closed sets behaves like a single geometric piece. Algebraically, irreducibility corresponds to primeness of the ideal defining a variety, so decomposing a space into irreducible pieces mirrors decomposing an ideal into prime data. [example: The Zariski Topology on the Affine Line] Let $k$ be algebraically closed, and identify $\mathbb A_k^1$ with $k$. We show that the proper Zariski closed subsets are exactly the finite subsets of $k$. If $V(S)\ne k$, then there is some $b\in k$ with $b\notin V(S)$, so for some $f\in S$ we have $f(b)\ne 0$. Thus $f$ is a nonzero polynomial and \begin{align*} V(S)\subseteq V(f)=\{a\in k:f(a)=0\}. \end{align*} A nonzero polynomial in one variable has only finitely many roots: each root $a$ gives a factor $(x-a)$, and the degree drops by $1$ after factoring; this process cannot occur more times than $\deg(f)$. Hence $V(S)$ is finite. Conversely, let $F=\{a_1,\dots,a_m\}\subset k$ be finite. If $m=0$, then $F=V(1)$. If $m\ge 1$, set \begin{align*} g(x)=\prod_{i=1}^m (x-a_i). \end{align*} For $a\in k$, the product $g(a)$ is zero exactly when one factor $a-a_i$ is zero, since $k$ is a field. Therefore \begin{align*} V(g)=\{a_1,\dots,a_m\}=F. \end{align*} So the Zariski closed subsets of $\mathbb A_k^1$ are $k$ itself and the finite subsets. It follows that the nonempty Zariski open subsets are complements of finite subsets. If $U=k\setminus F$ and $W=k\setminus G$ with $F,G$ finite, then \begin{align*} U\cap W=k\setminus(F\cup G). \end{align*} The set $F\cup G$ is finite, while an algebraically closed field is infinite: if $k=\{a_1,\dots,a_m\}$ were finite, the polynomial $1+\prod_{i=1}^m(x-a_i)$ would have value $1$ at every element of $k$, contradicting algebraic closedness. Thus $k\setminus(F\cup G)$ is nonempty, so any two nonempty Zariski open subsets of $\mathbb A_k^1$ meet. [/example] ## The Role of the Nullstellensatz The main obstruction in the affine dictionary is that ideals and point sets do not match perfectly without hypotheses. The Nullstellensatz supplies the bridge, but in two related forms that should be kept separate. The weak form says that, over an algebraically closed field $k$, maximal ideals of $k[x_1,\dots,x_n]$ are exactly the evaluation ideals \begin{align*} (x_1-a_1,\dots,x_n-a_n) \end{align*} for points $a=(a_1,\dots,a_n)\in \mathbb A_k^n$. Thus closed points of affine space can be read algebraically from maximal ideals. The stronger ideal-theoretic form says that for an ideal $I\subset k[x_1,\dots,x_n]$, the ideal of all polynomials vanishing on the common zero set $V(I)$ is $\sqrt I$, the radical of $I$. The algebraic closedness hypothesis is essential: over $\mathbb R$, the maximal ideal $(x^2+1) \trianglelefteq \mathbb R[x]$ has quotient $\mathbb C$, so it is maximal but is not the ideal of evaluation at any real point. Thus over a non-algebraically closed field, maximal ideals may represent algebraic points defined only after a [field extension](/page/Field%20Extension), not actual points of $\mathbb A_k^n$. Forward in the course, the weak form is the first guarantee that algebraic quotients of the coordinate ring produce geometric points: it will be used whenever closed points are identified with evaluation homomorphisms and when affine varieties are compared through their coordinate rings. The weak form still identifies only closed points, so it does not describe which arbitrary ideals have the same zero set or how nilpotent elements disappear from classical geometry. The stronger ideal-theoretic form supplies that missing part: for an arbitrary ideal, the equations vanishing on its common zero set recover the radical of the ideal rather than the ideal itself. This is the point at which nilpotent and redundant algebraic information disappears from the classical affine variety. The algebraically closed hypothesis cannot be removed without changing that stronger statement. Over $\mathbb R$, the ideal $(x^2+1)$ has empty real vanishing set, so \begin{align*} I(V(x^2+1)) &= I(\varnothing)=\mathbb R[x],& \sqrt{(x^2+1)} &= (x^2+1). \end{align*} The polynomial-ring hypothesis is also part of the content: the proof uses finite generation over $k$ through [Zariski's lemma](/theorems/2872) and the Rabinowitsch trick in the larger polynomial ring $k[x_1,\dots,x_n,t]$. For an arbitrary $k$-algebra there need not be enough classical points for the same ideal-point dictionary to hold. The theorem also has a classical limitation: it recovers only the radical of an ideal, so it deliberately forgets nilpotent thickening such as the difference between $(x)$ and $(x^2)$. This loss is a feature of classical varieties rather than an accident, and it is the reason reduced finitely generated $k$-algebras, not all quotient rings, match affine algebraic sets. Forward in the course, the Nullstellensatz is the main computational dictionary: to understand a zero set, compute or compare radicals; to understand polynomial functions on that set, pass to the reduced coordinate ring $k[x_1,\dots,x_n]/I(X)$. [example: A Radical Ideal and a Non-Radical Ideal] Let $k$ be a field. First consider $(xy)\subset k[x,y]$. A point $(a,b)\in \mathbb A_k^2$ lies in $V(xy)$ exactly when \begin{align*} ab=0. \end{align*} Since $k$ is a field, $ab=0$ means $a=0$ or $b=0$, so \begin{align*} V(xy)=\{(a,0):a\in k\}\cup\{(0,b):b\in k\}. \end{align*} The ideal $(xy)$ is radical. Indeed, $(x)$ and $(y)$ are prime because \begin{align*} k[x,y]/(x)\cong k[y] \end{align*} and \begin{align*} k[x,y]/(y)\cong k[x], \end{align*} and both $k[x]$ and $k[y]$ are integral domains. Also, \begin{align*} (x)\cap(y)=(xy). \end{align*} To see this, write $f=\sum c_{ij}x^iy^j$. If $f\in (x)\cap(y)$, then every nonzero term of $f$ is divisible by $x$ and by $y$, so every nonzero term has $i\ge 1$ and $j\ge 1$, hence $f\in (xy)$. The reverse inclusion $(xy)\subset (x)\cap(y)$ is immediate. Now if $f^m\in (xy)=(x)\cap(y)$, then $f^m\in (x)$ and $f^m\in (y)$. Since $(x)$ and $(y)$ are prime, $f\in (x)$ and $f\in (y)$, so $f\in (x)\cap(y)=(xy)$. Therefore $\sqrt{(xy)}=(xy)$. Now consider $(x^2,y)$. A point $(a,b)$ lies in $V(x^2,y)$ exactly when \begin{align*} a^2=0 \end{align*} and \begin{align*} b=0. \end{align*} A field has no nonzero nilpotent elements, so $a^2=0$ implies $a=0$. Hence \begin{align*} V(x^2,y)=\{(0,0)\}. \end{align*} Its radical is $(x,y)$. Since $x^2\in (x^2,y)$ and $y\in (x^2,y)$, we have \begin{align*} x\in \sqrt{(x^2,y)} \end{align*} and \begin{align*} y\in \sqrt{(x^2,y)}. \end{align*} Thus $(x,y)\subset \sqrt{(x^2,y)}$. Conversely, $(x^2,y)\subset (x,y)$, and $(x,y)$ is radical because \begin{align*} k[x,y]/(x,y)\cong k \end{align*} is reduced. Therefore \begin{align*} \sqrt{(x^2,y)}\subset \sqrt{(x,y)}=(x,y). \end{align*} Combining the two inclusions gives \begin{align*} \sqrt{(x^2,y)}=(x,y). \end{align*} Thus $(xy)$ already equals its radical and records the two coordinate axes, while $(x^2,y)$ has the same classical point set as $(x,y)$ but contains extra nilpotent information: in $k[x,y]/(x^2,y)$, the class $\overline{x}$ is nonzero and satisfies $\overline{x}^2=0$. This is exactly the kind of information forgotten by the classical point-set dictionary described by *Hilbert Nullstellensatz*. [/example] ## Maps, Local Geometry, and Projective Completion After affine varieties and their coordinate rings are established, the course turns to maps and local structure. The guiding problem is to understand a variety not only as a set of points, but through the functions on it, the behaviour near each point, and the way it sits inside a larger compact-like projective space. [explanation: Regular Maps and Contravariance] A regular function on an affine variety is represented by an element of its coordinate ring. A regular map $\varphi:X \to Y$ is determined by pulling back regular functions on $Y$ to regular functions on $X$, giving a $k$-algebra homomorphism \begin{align*} \varphi^*:k[Y] \to k[X]. \end{align*} The direction reverses because functions are evaluated after applying the map: a function on $Y$ becomes a function on $X$ by composition with $\varphi$. [/explanation] This contravariance is a recurring theme. It also prepares the later scheme-theoretic viewpoint, where spaces are often reconstructed from rings rather than placed inside a fixed affine space. [definition: Zariski Tangent Space] Let $X \subset \mathbb A_k^n$ be an affine algebraic set and let $a \in X$. The Zariski tangent space to $X$ at $a$ is \begin{align*} T_aX := \left\{v \in k^n : \sum_{i=1}^n \frac{\partial f}{\partial x_i}(a)v_i=0 \text{ for all } f \in I(X)\right\}. \end{align*} [/definition] The use of $I(X)$ is essential: the tangent space of a classical affine algebraic set must depend on the set $X$, not on a non-radical presentation chosen for it. For example, the same point $\{0\}\subset \mathbb A_k^1$ can be written as $V(x)$ or $V(x^2)$, and differentiating only the chosen generator $x^2$ would give the wrong presentation-dependent answer at $0$. The tangent space measures the first-order failure of all equations vanishing on $X$ to change at a point. Smoothness will be detected by comparing the dimension of this vector space with the dimension of the variety near the point. [example: Tangent Space of the Cusp] Let $X=V(y^2-x^3)\subset \mathbb A_k^2$, and assume $\operatorname{char}(k)\ne 2,3$. Write \begin{align*} f(x,y)=y^2-x^3. \end{align*} Since the ideal $(f)$ is prime in $k[x,y]$ and $k$ is algebraically closed, *Hilbert Nullstellensatz* gives \begin{align*} I(X)=I(V(f))=\sqrt{(f)}=(f). \end{align*} Thus it is enough to impose the tangent equation coming from $f$. The partial derivatives are \begin{align*} \frac{\partial f}{\partial x}=-3x^2,\qquad \frac{\partial f}{\partial y}=2y. \end{align*} At the origin $(0,0)$, these derivatives evaluate to \begin{align*} \frac{\partial f}{\partial x}(0,0)=-3\cdot 0^2=0,\qquad \frac{\partial f}{\partial y}(0,0)=2\cdot 0=0. \end{align*} For a tangent vector $v=(u,w)\in k^2$, the tangent equation is therefore \begin{align*} 0\cdot u+0\cdot w=0. \end{align*} This equation holds for every $(u,w)\in k^2$, so \begin{align*} T_{(0,0)}X=k^2. \end{align*} Now let $(a,b)\in X$ with $(a,b)\ne(0,0)$. Since $(a,b)\in X$, we have \begin{align*} b^2-a^3=0. \end{align*} If $b=0$, then $a^3=0$, hence $a=0$ because $k$ is a field, contradicting $(a,b)\ne(0,0)$. Thus $b\ne 0$, so $2b\ne 0$ because $\operatorname{char}(k)\ne 2$. The tangent equation at $(a,b)$ for $v=(u,w)$ is \begin{align*} -3a^2u+2bw=0. \end{align*} Since $2b\ne 0$, this is equivalent to \begin{align*} w=\frac{3a^2}{2b}u. \end{align*} Thus every tangent vector is uniquely determined by the free parameter $u\in k$, so $T_{(a,b)}X$ is a one-dimensional subspace of $k^2$. The cusp therefore has a two-dimensional tangent space at the origin but one-dimensional tangent spaces at its other points, so the origin is the singular point detected by the first-order equations. [/example] The cusp shows that affine equations can encode delicate local behaviour. A different limitation of affine geometry is global: curves that should meet may miss each other by escaping to infinity, so the next ambient space adds those missing directions in a controlled way. [definition: Projective Space] Let $k$ be a field and let $n \in \mathbb N$. Projective $n$-space over $k$ is \begin{align*} \mathbb P_k^n := (k^{n+1}\setminus\{0\})/\sim, \end{align*} where $a \sim b$ if there exists $\lambda \in k^\times$ such that $b=\lambda a$. [/definition] A point of $\mathbb P_k^n$ is written in homogeneous coordinates $[a_0:\cdots:a_n]$. Polynomial equations on projective space must be homogeneous, because scaling coordinates must not change whether a point satisfies the equation. [example: Projective Closure of an Affine Parabola] Consider the affine curve $C=V(y-x^2)\subset \mathbb A_k^2$. To embed this equation into projective space, use homogeneous coordinates $[X:Y:Z]$ and replace $x$ by $X/Z$ and $y$ by $Y/Z$ on the affine chart $Z\ne 0$. The affine equation becomes \begin{align*} \frac{Y}{Z}=\left(\frac{X}{Z}\right)^2. \end{align*} Multiplying both sides by $Z^2$ gives \begin{align*} YZ=X^2. \end{align*} Thus the projective closure is the projective plane curve \begin{align*} \overline C=V(YZ-X^2)\subset \mathbb P_k^2. \end{align*} On the chart $Z\ne 0$, every point can be written as $[X:Y:Z]=[X/Z:Y/Z:1]$. Setting $x=X/Z$ and $y=Y/Z$, the homogeneous equation $YZ=X^2$ becomes \begin{align*} \frac{Y}{Z}=\left(\frac{X}{Z}\right)^2, \end{align*} so the part of $\overline C$ with $Z\ne 0$ is exactly the original affine parabola $y=x^2$. It remains to find the points of $\overline C$ with $Z=0$. Substituting $Z=0$ into $YZ=X^2$ gives \begin{align*} Y\cdot 0=X^2. \end{align*} Hence \begin{align*} X^2=0. \end{align*} Since $k$ is a field, $X^2=0$ implies $X=0$. A projective point cannot have all homogeneous coordinates zero, so $Y\ne 0$. Therefore every point at infinity on $\overline C$ has the form $[0:Y:0]$ with $Y\in k^\times$, and \begin{align*} [0:Y:0]=[0:1:0]. \end{align*} So the projective closure of the affine parabola adds exactly one point at infinity, namely $[0:1:0]$. [/example] ## What This Course Builds Toward The final aim is to make classical varieties flexible enough for computations and conceptual enough to motivate schemes. The objects remain concrete: zero sets, coordinate rings, rational functions, divisors, tangent spaces, projective closures, and intersection numbers. [explanation: From Classical Varieties to Schemes] Classical algebraic geometry works with reduced varieties over an algebraically closed field and studies them through their polynomial functions. This is already rich enough for conics, cubics, singularities, projective embeddings, divisors on curves, and elementary intersection theory. Scheme theory later removes several restrictions at once: fields need not be algebraically closed, nilpotents are retained, and rings themselves become geometric spaces. The present course is therefore the geometric and computational foundation for that generalisation. [/explanation] The course should be read as a sequence of dictionaries. Affine geometry is translated into coordinate rings; topology is translated into ideals and radicals; local geometry is translated into local rings and tangent spaces; projective geometry is translated into homogeneous coordinate rings. Each translation loses its mystery once enough examples have been computed by hand. Having translated geometry into coordinate rings, ideals, radicals, local rings, tangent spaces, and homogeneous data, we now begin using that dictionary in the simplest setting. The first chapter fixes the affine side of the correspondence, where polynomial equations define sets and vanishing ideals recover the algebra behind them. Once that order-reversing relation is in place, the Zariski topology becomes the natural language for the rest of the course. # 1. Affine Algebraic Sets and the Zariski Topology This chapter begins the course by fixing the first half of the algebra–geometry dictionary. Polynomial equations in affine space define geometric sets, while subsets of affine space determine ideals of polynomials vanishing on them. The central theme is that these two operations reverse inclusions, and that radicals, irreducibility, and Noetherianity explain why finite algebraic data controls apparently infinite geometric objects. ## Polynomial Equations in Affine Space The first question is how to turn a system of polynomial equations into a geometric object. Throughout this chapter, let $k$ be an algebraically closed field unless stated otherwise, and let $k[x_1,\dots,x_n]$ be the polynomial ring in $n$ variables. Affine $n$-space is the set of all $n$-tuples of scalars over $k$, and its subsets cut out by polynomials are the first objects of study. [definition: Affine Space] The affine $n$-space over $k$ is the set \begin{align*} \mathbb A_k^n := k^n. \end{align*} [/definition] Affine space is only a set at this stage; its geometry will come from declaring polynomial equations to be closed conditions. The basic construction needed next takes any collection of polynomials, finite or infinite, and asks for their common zero set. [definition: Vanishing Locus] Let $S \subset k[x_1,\dots,x_n]$. The vanishing locus of $S$ is \begin{align*} V(S) := \{a \in \mathbb A_k^n : f(a)=0 \text{ for all } f \in S\}. \end{align*} [/definition] Since all equations in $S$ must vanish simultaneously, enlarging $S$ imposes more conditions and usually makes the zero set smaller. Before developing this reversal formally, it is useful to see that familiar plane curves already arise from single polynomial equations. [example: Lines and Conics in the Affine Plane] In $\mathbb A_k^2=k^2$, fix $m,b\in k$ and consider the polynomial $f(x,y)=y-mx-b$. For a point $(a,c)\in k^2$, the definition of vanishing locus gives \begin{align*} (a,c)\in V(y-mx-b)\iff c-ma-b=0. \end{align*} Adding $ma+b$ to both sides gives \begin{align*} c-ma-b=0\iff c=ma+b. \end{align*} Thus $V(y-mx-b)$ is exactly the affine line with slope $m$ and intercept $b$. For the polynomial $g(x,y)=x^2+y^2-1$, its zero set is \begin{align*} V(x^2+y^2-1)=\{(a,c)\in k^2:a^2+c^2-1=0\}. \end{align*} Equivalently, \begin{align*} (a,c)\in V(x^2+y^2-1)\iff a^2+c^2=1. \end{align*} If $\operatorname{char}(k)\ne 2$, this is an affine plane conic, because the equation has total degree $2$ and contains the quadratic terms $x^2$ and $y^2$. Over an algebraically closed field, this describes the set of solutions in $k^2$; it does not carry the Euclidean notions of distance, angle, boundedness, or compactness. The same displayed equation therefore has a different geometric feel over $\mathbb R$ than over an algebraically closed field: over $\mathbb R$ it resembles a circle, while in affine algebraic geometry it is a plane algebraic curve defined by one quadratic equation. [/example] The example uses one equation, but systems of equations are better handled by ideals because addition and multiplication by arbitrary polynomials preserve common vanishing. This raises a basic well-definedness problem for algebraic geometry: if a set of equations is replaced by all equations algebraically forced by it, should the solution set remain unchanged? In the affine coordinate ring $k[x_1,\dots,x_n]$, the answer is immediate from the definition of an ideal. If $I=(S)$ is generated by $S$ and a point $p\in\mathbb A_k^n$ makes every polynomial in $S$ vanish, then every finite combination $\sum_i h_i f_i$ with $f_i\in S$ also vanishes at $p$, because evaluation at $p$ respects addition and multiplication. Thus $V(S)=V((S))$. This affine observation says that the geometry cannot see the difference between a list of equations and all polynomial consequences of that list. The restriction to consequences is essential: if we enlarge $S$ by equations not forced by $S$, the zero set may shrink, for instance $V(0)=\mathbb A_k^1$ but $V(x)=\{0\}$. In computations, this means one should first pass to the ideal of equations and then use ideal operations rather than manipulating arbitrary lists. Because the zero set only depends on the generated ideal, we can write $V(I)$ for an ideal $I\trianglelefteq k[x_1,\dots,x_n]$. This motivates giving a name to the subsets obtained from ideals, since they are the affine geometric objects before imposing irreducibility. [definition: Affine Algebraic Set] A subset $X\subset \mathbb A_k^n$ is an affine algebraic set if there exists an ideal $I\trianglelefteq k[x_1,\dots,x_n]$ such that \begin{align*} X=V(I). \end{align*} [/definition] Different ideals can define the same algebraic set. For instance, powers of an equation do not change its zero set, so the geometry loses nilpotent information unless a later construction remembers it; reducible examples also show that one equation can encode several pieces. [example: Coordinate Axes] In $\mathbb A_k^2=k^2$, consider the polynomial $xy\in k[x,y]$. For a point $(a,b)\in k^2$, the definition of vanishing locus gives \begin{align*} (a,b)\in V(xy)\iff ab=0. \end{align*} Since $k$ is a field, it has no zero divisors, so \begin{align*} ab=0\iff a=0\text{ or }b=0. \end{align*} The condition $a=0$ is exactly $(a,b)\in V(x)$, and the condition $b=0$ is exactly $(a,b)\in V(y)$. Therefore, for every $(a,b)\in k^2$, \begin{align*} (a,b)\in V(xy)\iff (a,b)\in V(x)\cup V(y). \end{align*} Hence \begin{align*} V(xy)=V(x)\cup V(y). \end{align*} Thus one polynomial equation can define a union of simpler algebraic pieces: here the two coordinate axes. [/example] The reverse construction starts with a subset of affine space and asks which polynomials vanish on it. This is the algebraic shadow needed to compare subsets by equations rather than by points. The practical problem is that a subset may be given parametrically, by a picture, or by a finite list of points; $I(X)$ turns that geometric information into equations. [definition: Ideal of a Subset] The vanishing ideal assignment sends a subset $X\subset \mathbb A_k^n$ to \begin{align*} I(X):=\{f\in k[x_1,\dots,x_n]: f(a)=0 \text{ for all } a\in X\}\subset k[x_1,\dots,x_n]. \end{align*} [/definition] The set $I(X)$ should be compatible with the algebra of the polynomial ring, since the next chapter will form quotient rings by these objects. This compatibility is not just terminology: if two polynomials vanish on $X$, then their sum should vanish on $X$, and multiplying a vanishing polynomial by any other polynomial should still give a vanishing polynomial. These are exactly the closure properties needed for $I(X)$ to serve as an ideal of equations. [quotetheorem:9404] [citeproof:9404] The verification that $I(X)$ is an ideal matters because quotient rings by these ideals will later become coordinate rings. The condition is also fragile in the right way: the set of polynomials nonzero on $X$ is usually not closed under addition, so vanishing is the algebraically stable condition. In computations, to find $I(X)$ for a finite set $X=\{a_1,\dots,a_r\}$, one imposes the linear conditions $f(a_i)=0$ on the coefficients of a polynomial of bounded degree and then recognises the resulting ideal. Now that both directions of the dictionary have been defined, the first structural question is how containment behaves under them. The reversal of containment is needed throughout the course because geometric inclusions will repeatedly be translated into algebraic inclusions in the opposite direction. [quotetheorem:9405] [citeproof:9405] The order reversal is the first place where the dictionary becomes genuinely contravariant. The hypotheses cannot be reversed: from $V(J)\subset V(I)$ one cannot yet conclude $I\subset J$, since $I=(x)$ and $J=(x^2)$ have the same vanishing locus in $\mathbb A_k^1$ but neither direction captures equality of ideals in the naive way. This failure is not a nuisance but a signal that radicals will be unavoidable. The correspondence can now be composed in both possible orders. The next theorem is needed to identify what information is lost when we pass from equations to points and back, or from points to equations and back. [quotetheorem:9406] [citeproof:9406] These containments are closure operations rather than inverse identities. The second containment says that $V(I(X))$ is the smallest algebraic set one can hope to recover from $X$, so a non-closed subset is automatically enlarged; for example, an infinite subset of $\mathbb A_k^1$ has closure all of $\mathbb A_k^1$. The first containment can be strict: in $k[x]$, the ideals $(x)$ and $(x^2)$ define the same point $\{0\}$, and $x\in I(V(x^2))$ although $x\notin (x^2)$. This motivates a refinement of which ideals classical geometry can see. Since vanishing cannot distinguish a polynomial from a power of that polynomial, the correct algebraic objects for ordinary affine algebraic sets must ignore nilpotent thickening. [definition: Radical Ideal] An ideal $I\trianglelefteq R$ in a commutative ring $R$ is radical if, whenever $f^m\in I$ for some $m\in \mathbb N$, then $f\in I$. [/definition] A given ideal need not be radical, so we need a canonical way to pass from an ideal to the radical ideal with the same underlying vanishing behaviour. This motivates the radical of an ideal. [definition: Radical of an Ideal] Let $I\trianglelefteq R$. The radical of $I$ is \begin{align*} \sqrt I := \{f\in R : f^m\in I \text{ for some } m\in\mathbb N\}. \end{align*} [/definition] The radical is not just formal algebra here; it is forced by the geometry of zero sets. The following elementary containment is the part of the Nullstellensatz visible before the full theorem. [quotetheorem:9407] [citeproof:9407] This theorem is only the easy direction of the radical story: powers vanish wherever their bases vanish. The reverse inclusion is much deeper and uses the algebraic closedness of $k$; over a non-algebraically closed field, equations can vanish for accidental field-theoretic reasons, such as $x^2+1$ having no real zero on $\mathbb A_{\mathbb R}^1$. The converse inclusion over algebraically closed fields is the strong Nullstellensatz, which is the main theorem of the next chapter. For now, the important point is conceptual: affine algebraic sets forget the difference between $I$ and $\sqrt I$, and singular examples show that even a single reduced equation can have delicate local geometry. [example: A Cusp] Let \begin{align*} F(x,y)=y^2-x^3 \end{align*} and let $C=V(F)\subset \mathbb A_k^2$. Assume $\operatorname{char}(k)\ne 2,3$. The formal partial derivatives are \begin{align*} \frac{\partial F}{\partial x}=-3x^2 \end{align*} and \begin{align*} \frac{\partial F}{\partial y}=2y. \end{align*} At the origin, \begin{align*} F(0,0)=0^2-0^3=0, \end{align*} so $(0,0)\in C$. Also \begin{align*} \frac{\partial F}{\partial x}(0,0)=-3\cdot 0^2=0 \end{align*} and \begin{align*} \frac{\partial F}{\partial y}(0,0)=2\cdot 0=0. \end{align*} Thus $(0,0)$ is a singular point of the plane curve $C$. Now take a point $(a,b)\in C$ at which both partial derivatives vanish. Then \begin{align*} -3a^2=0 \end{align*} and \begin{align*} 2b=0. \end{align*} Since $\operatorname{char}(k)\ne 3$, the scalar $-3$ is nonzero, so $a^2=0$, hence $a=0$ because $k$ is a field. Since $\operatorname{char}(k)\ne 2$, the scalar $2$ is nonzero, so $b=0$. Therefore the only point of $C$ where both partial derivatives vanish is $(0,0)$; at every other point of $C$, at least one of $-3x^2$ and $2y$ is nonzero. The defining polynomial is nevertheless squarefree. If a nonconstant polynomial $h$ satisfied $h^2\mid F$, then differentiating $F=h^2q$ would show that $h$ divides both partial derivatives of $F$. Hence $h\mid -3x^2$ and $h\mid 2y$. Because $-3$ and $2$ are units in $k$, this gives $h\mid x^2$ and $h\mid y$. But $x$ and $y$ have no common nonconstant divisor in $k[x,y]$, so $h$ must be a unit, a contradiction. Thus $F$ has no repeated factor, while its zero set still has the singular point $(0,0)$; reducedness of one equation does not imply smoothness of the curve it defines. [/example] ## The Zariski Topology and Irreducibility Once algebraic sets have been defined, the next problem is to decide which topology they naturally determine. In ordinary Euclidean geometry, closed sets are defined by limits. In algebraic geometry, closed sets are defined by polynomial equations, so the topology is built to make every vanishing locus closed. [definition: Zariski Closed and Open Sets] A subset $F\subset \mathbb A_k^n$ is Zariski closed if there exists an ideal $I\trianglelefteq k[x_1,\dots,x_n]$ such that $F=V(I)$. A subset $U\subset \mathbb A_k^n$ is Zariski open if $\mathbb A_k^n\setminus U$ is Zariski closed. [/definition] A definition of closed sets becomes a topology only if it is stable under the closure operations required of closed subsets. For algebraic sets this is a real algebraic question: arbitrary intersections should correspond to collecting all equations at once, while finite unions require a way to force a point to satisfy one system or the other. Multiplying equations supplies that finite-union mechanism. [quotetheorem:9408] [citeproof:9408] The product ideal in the finite-union formula is the algebraic reason that closed sets behave differently from linear subspaces: unions of solution sets are still solution sets, but usually require multiplying equations. The finiteness in "finite unions" is essential for a topology; arbitrary unions of algebraic sets need not be algebraic, as the union of infinitely many points in $\mathbb A_k^1$ may be an infinite proper subset and hence not closed. This theorem gives the ambient space a topology, but it is much coarser than the Euclidean topology when $k=\mathbb C$. Zariski open sets are often large, and this largeness is already visible on the affine line. A single nonzero polynomial in one variable has finitely many roots over a field, so proper closed subsets of the affine line are finite. [example: Zariski Topology on the Affine Line] Identify $\mathbb A_k^1$ with $k$, with coordinate $x$. The whole line is closed because \begin{align*} \mathbb A_k^1=V(0). \end{align*} The empty set is closed because \begin{align*} \varnothing=V(1), \end{align*} since $1(a)=1\ne 0$ for every $a\in k$. Now let $F\subset \mathbb A_k^1$ be Zariski closed, so $F=V(I)$ for some ideal $I\trianglelefteq k[x]$. If $I=(0)$, then every point of $k$ annihilates every polynomial in $I$, so $F=\mathbb A_k^1$. If $I\ne (0)$, choose a nonzero polynomial $f\in I$. Since every point of $V(I)$ is a zero of every polynomial in $I$, we have \begin{align*} V(I)\subset V(f). \end{align*} A nonzero one-variable polynomial of degree $d$ has at most $d$ roots over a field, so $V(f)$ is finite, and hence $F=V(I)$ is finite. Conversely, if $F=\{a_1,\dots,a_r\}$ is finite, then \begin{align*} F=V((x-a_1)(x-a_2)\cdots(x-a_r)). \end{align*} Indeed, for $c\in k$, the product $(c-a_1)(c-a_2)\cdots(c-a_r)$ is zero iff one factor $c-a_i$ is zero, which is equivalent to $c=a_i$ for some $i$. Thus the Zariski closed subsets of $\mathbb A_k^1$ are exactly $\mathbb A_k^1$, $\varnothing$, and the finite subsets. Therefore every nonempty Zariski open subset has the form $\mathbb A_k^1\setminus F$ with $F$ finite. If $U=\mathbb A_k^1\setminus F$ and $W=\mathbb A_k^1\setminus G$ are nonempty open subsets, then \begin{align*} U\cap W=\mathbb A_k^1\setminus (F\cup G). \end{align*} The set $F\cup G$ is finite, while $k$ is infinite because an algebraically closed field cannot be finite: if $k=\{a_1,\dots,a_N\}$, then $1+\prod_i(x-a_i)$ would be a nonconstant polynomial with no root in $k$. Hence $U\cap W\ne\varnothing$. As a consequence, any infinite subset $Y\subset \mathbb A_k^1$ is Zariski dense. Its closure cannot be finite, because it contains the infinite set $Y$; the only remaining closed subset containing it is $\mathbb A_k^1$. Equivalently, no nonzero polynomial in $k[x]$ can vanish on all of $Y$, since a nonzero polynomial has only finitely many roots. [/example] The preceding example shows that Zariski topology is not designed to separate nearby points. Instead, it raises the question of when a subset is large enough that its closure is the whole ambient space. [definition: Zariski Dense Subset] Let $X$ be a [topological space](/page/Topological%20Space) and let $Y\subset X$. The subset $Y$ is dense in $X$ if $\overline{Y}=X$, where $\overline{Y}$ denotes the closure of $Y$ in $X$. [/definition] For affine space, density should be expressible through equations because closed sets are defined by equations. The obstruction to density is an equation that vanishes on all of $Y$: such an equation traps $Y$ inside a proper closed subset. Conversely, if $Y$ is not dense, its closure is a proper closed set and should be detected by some nonzero polynomial. The next criterion turns this obstruction into an exact algebraic test. [quotetheorem:9458] [citeproof:9458] The density criterion turns topology into an equation test: to prove density, prove that the only polynomial vanishing on the subset is zero. The algebraically closed hypothesis is not needed for the formal equivalence with $I(Y)=(0)$, but it matters for interpreting closed sets through the full Nullstellensatz later. The criterion is useful because it replaces a topological closure calculation by an algebraic non-vanishing argument, for example when a parametrised set has infinitely many points in every relevant direction. The next structural question is whether an algebraic set is made from smaller closed pieces. Irreducibility captures the idea that a space has not been assembled as a union of two proper closed subsets. [definition: Irreducible Topological Space] A nonempty topological space $X$ is irreducible if, whenever $X=F_1\cup F_2$ with $F_1,F_2\subset X$ closed, then $X=F_1$ or $X=F_2$. [/definition] Irreducibility is the topological analogue of being a single algebraic piece. Reducible examples are important because many polynomial equations factor or define unions. [example: The Axes Are Reducible] For a point $(a,b)\in \mathbb A_k^2$, the definition of vanishing locus gives \begin{align*} (a,b)\in V(xy)\iff ab=0. \end{align*} Since $k$ is a field, it has no zero divisors, so \begin{align*} ab=0\iff a=0\text{ or }b=0. \end{align*} The condition $a=0$ is equivalent to $(a,b)\in V(x)$, and the condition $b=0$ is equivalent to $(a,b)\in V(y)$. Hence, for every $(a,b)\in \mathbb A_k^2$, \begin{align*} (a,b)\in V(xy)\iff (a,b)\in V(x)\cup V(y), \end{align*} so \begin{align*} V(xy)=V(x)\cup V(y). \end{align*} Both pieces are closed in $V(xy)$ because $V(x)$ and $V(y)$ are Zariski closed in $\mathbb A_k^2$, and \begin{align*} V(x)=V(xy)\cap V(x) \end{align*} while \begin{align*} V(y)=V(xy)\cap V(y). \end{align*} They are proper subsets of $V(xy)$: the point $(1,0)$ lies in $V(xy)$ because $1\cdot 0=0$, but $(1,0)\notin V(x)$ because $1\ne 0$; similarly, $(0,1)\in V(xy)$ but $(0,1)\notin V(y)$. Therefore $V(xy)$ is the union of two proper closed subsets, so it is reducible. Geometrically, the equation $xy=0$ records the two coordinate axes meeting at the origin rather than one irreducible curve. [/example] The axes example suggests that reducibility should be detected by algebraic factorization or, more generally, by primality. It also shows why merely looking at the number of defining equations is misleading: $xy=0$ is one equation, but its zero set has two components. The next theorem is needed to turn irreducibility from a topological condition into a ring-theoretic condition. [quotetheorem:2126] [citeproof:2126] This theorem is the main working test for irreducibility in affine algebraic geometry. Its hypotheses are important: the statement uses the full vanishing ideal $I(X)$, not just an arbitrary ideal defining $X$, because non-radical defining ideals such as $(x^2)$ need not be prime even when their zero set is the irreducible point $V(x)$. In practice, one proves irreducibility by identifying $I(X)$ as prime, often by showing that the coordinate ring $k[x_1,\dots,x_n]/I(X)$ is an [integral domain](/page/Integral%20Domain). After irreducibility has been connected to prime ideals, we need its most useful topological consequence. The next theorem explains why deleting a proper closed subset from an irreducible space still leaves a dense open part. [quotetheorem:9409] [citeproof:9409] This density property makes principal open subsets central examples: they are the basic regions where a chosen polynomial is invertible. The nonempty hypothesis cannot be dropped, since the empty [open set](/page/Open%20Set) is never dense unless the whole space is empty. Such open sets are the first examples of geometric domains on which rational functions and regular maps will later be studied. [example: A Principal Open Subset] For $f\in k[x_1,\dots,x_n]$, define \begin{align*} D(f):=\mathbb A_k^n\setminus V(f)=\{a\in \mathbb A_k^n:f(a)\ne 0\}. \end{align*} Since $V(f)$ is Zariski closed, $D(f)$ is Zariski open by definition. Assume $f\ne 0$ and $\mathbb A_k^n$ is irreducible. The set $D(f)$ is nonempty: if $D(f)=\varnothing$, then $V(f)=\mathbb A_k^n$, so $f(a)=0$ for every $a\in k^n$; since $k$ is infinite and a nonzero polynomial over an infinite field cannot vanish on all of $k^n$, this would force $f=0$, contradicting the hypothesis. Therefore $D(f)$ is a nonempty open subset of the irreducible space $\mathbb A_k^n$, so it is dense by *Nonempty Opens in an Irreducible Space Are Dense*. In $\mathbb A_k^2$, take $f=x$. For a point $(a,b)\in k^2$, \begin{align*} (a,b)\in V(x)\iff a=0. \end{align*} Thus $V(x)$ is the $y$-axis. Its complement is \begin{align*} D(x)=\{(a,b)\in k^2:a\ne 0\}. \end{align*} The point $(1,0)$ lies in $D(x)$ because $1\ne 0$, so $D(x)$ is nonempty; hence, if $\mathbb A_k^2$ is irreducible, $D(x)$ is dense. On $D(x)$ the coordinate $x$ never vanishes, so expressions involving division by $x$ are defined there, and the excluded set is only the proper closed subset $V(x)$. [/example] Open and closed sets are not the only subsets that naturally appear. Images of polynomial maps and locally closed pieces often require Boolean combinations of closed sets, leading to constructible subsets. [definition: Constructible Subset] A subset $C$ of a topological space $X$ is constructible if it is a finite union of subsets of the form $U\cap F$, where $U\subset X$ is open and $F\subset X$ is closed. [/definition] Constructible sets are flexible enough to describe many geometric conditions that are neither open nor closed. A punctured closed subset is the simplest model for this behaviour. [example: A Constructible Punctured Line] In $\mathbb A_k^2$, we compute the set $V(y)\cap D(x)$ from the definitions. For a point $(a,b)\in k^2$, \begin{align*} (a,b)\in V(y)\iff b=0. \end{align*} Also, since $D(x)=\mathbb A_k^2\setminus V(x)$, \begin{align*} (a,b)\in D(x)\iff a\ne 0. \end{align*} Therefore \begin{align*} V(y)\cap D(x)=\{(a,0)\in k^2:a\ne 0\}. \end{align*} This is the affine $x$-axis with the origin removed. It is constructible because it has the form $U\cap F$ with $U=D(x)$ open and $F=V(y)$ closed. It is not closed in $\mathbb A_k^2$. Indeed, let $F=V(I)$ be any Zariski closed set containing $V(y)\cap D(x)$. If $h\in I$, then $h(a,0)=0$ for every $a\in k$ with $a\ne 0$. The one-variable polynomial $p(t)=h(t,0)\in k[t]$ has infinitely many roots, so $p=0$. Hence \begin{align*} h(0,0)=p(0)=0. \end{align*} Thus $(0,0)\in V(I)=F$ for every closed set $F$ containing $V(y)\cap D(x)$, so $(0,0)$ lies in its closure. Since $(0,0)\notin V(y)\cap D(x)$, the punctured line is not closed. [/example] ## Noetherian Spaces and Decomposition into Components The next problem is finiteness. Polynomial rings contain infinitely many polynomials, and closed subsets of affine space may be defined using infinitely many equations. Noetherianity is the mechanism that says descending chains of closed sets eventually stop, so every algebraic set decomposes into finitely many irreducible pieces. [definition: Noetherian Topological Space] A topological space $X$ is Noetherian if every descending chain of closed subsets \begin{align*} F_1\supset F_2\supset F_3\supset \cdots \end{align*} is eventually constant. [/definition] This topological definition mirrors the algebraic definition of a Noetherian ring, but we still need a reason why affine spaces satisfy it. The needed algebraic input is [Hilbert's Basis Theorem](/theorems/2904), which transfers finite generation to polynomial rings. [quotetheorem:860] [citeproof:860] Hilbert's theorem is the algebraic engine behind the finiteness of affine geometry. The Noetherian hypothesis on $R$ is essential: polynomial rings over non-Noetherian rings can contain ideals that require infinitely many generators, so the reduction procedure has no finite stopping point. The proof sketch should be read as a finite-generation algorithm in principle, not as an efficient computational method; Gröbner bases later make this kind of reduction effective. Hilbert's theorem immediately suggests a geometric finiteness statement. The next theorem is needed to replace an arbitrary system of polynomial equations by a finite system defining the same algebraic set. [quotetheorem:9410] [citeproof:9410] This theorem does not say that a natural infinite list of equations is useless; it says that, as far as its common zero set is concerned, finitely many algebraic consequences already suffice. The finite generating set is not unique, and replacing it by another generating set can make the geometry easier or harder to read. The result is therefore a finiteness theorem, not a canonical presentation theorem. Finite generation of ideals also controls chains of closed sets, because closed sets correspond to ideals in the opposite direction. The next theorem is the topological finiteness statement needed for decomposition into components. [quotetheorem:9411] [citeproof:9411] This Noetherian property is a topological translation of [Hilbert's Basis Theorem](/theorems/2907). The direction of chains matters: descending closed sets correspond to ascending vanishing ideals, so the algebraic ascending chain condition becomes the geometric descending chain condition. The theorem fails for many familiar non-algebraic topologies, where one can often construct infinitely decreasing closed subsets without stabilization. Affine algebraic sets are closed subspaces of affine space, so the previous theorem should pass from the ambient space to the varieties inside it. The next result supplies that inheritance property. [quotetheorem:9412] [citeproof:9412] The closed-subspace result lets Noetherianity pass from $\mathbb A_k^n$ to every affine algebraic set $X=V(I)$. The closedness of $Y$ is the clean inheritance hypothesis used here; arbitrary subspaces of Noetherian spaces are also Noetherian, but the closed case is all that is needed for affine algebraic sets at this stage. With Noetherianity established for affine algebraic sets, the main structural payoff is finite decomposition. The next theorem is needed to justify speaking about the irreducible pieces of an algebraic set as a finite list rather than as an uncontrolled family. [quotetheorem:9413] [citeproof:9413] The decomposition theorem is the topological reason that algebraic sets can be studied piece by piece. The condition that no $X_i$ is contained in another is necessary for uniqueness, since redundant decompositions can always be made by adding a smaller irreducible closed subset already lying inside a component. In computations, the theorem suggests looking for maximal irreducible closed subsets, and algebraically this becomes the search for minimal prime ideals over the defining ideal. The decomposition theorem produces distinguished maximal pieces, so we need terminology for them. These pieces are the components that later correspond algebraically to minimal prime ideals over the defining ideal. [definition: Irreducible Component] Let $X$ be a Noetherian topological space. An irreducible component of $X$ is a maximal irreducible closed subset of $X$. [/definition] The axes example now has a precise interpretation: its two lines are its irreducible components. This final example ties together reducibility, closed subsets, and finite decomposition. [example: Components of the Coordinate Axes] Let $X=V(xy)\subset \mathbb A_k^2$. For $(a,b)\in k^2$, the definition of vanishing locus gives \begin{align*} (a,b)\in X\iff ab=0. \end{align*} Since $k$ is a field, $ab=0$ holds exactly when $a=0$ or $b=0$. The condition $a=0$ is equivalent to $(a,b)\in V(x)$, and the condition $b=0$ is equivalent to $(a,b)\in V(y)$. Hence \begin{align*} X=V(x)\cup V(y). \end{align*} The set $V(x)$ is irreducible. Indeed, the map $k\to V(x)$ given by $t\mapsto (0,t)$ identifies $V(x)$ with $\mathbb A_k^1$ with its Zariski topology: if $F=V(x)\cap V(I)$ is closed in $V(x)$, then its image in $k$ is the common zero set of the one-variable polynomials $h(0,t)$ with $h\in I$. Since the closed subsets of $\mathbb A_k^1$ are exactly $\mathbb A_k^1$, $\varnothing$, and finite subsets, and since $k$ is infinite, $\mathbb A_k^1$ cannot be the union of two proper closed subsets. Thus $V(x)$ is irreducible. The same argument, using $t\mapsto (t,0)$, shows that $V(y)$ is irreducible. Neither line contains the other. The point $(0,1)$ lies in $V(x)$ because $x(0,1)=0$, but it does not lie in $V(y)$ because $y(0,1)=1\ne 0$. Similarly, $(1,0)\in V(y)$ and $(1,0)\notin V(x)$. Now let $Z\subset X$ be any irreducible closed subset. Since $X=V(x)\cup V(y)$, \begin{align*} Z=(Z\cap V(x))\cup (Z\cap V(y)). \end{align*} Both intersections are closed in $Z$, so irreducibility forces $Z=Z\cap V(x)$ or $Z=Z\cap V(y)$. Therefore every irreducible closed subset of $X$ lies in one of the two axes, and the maximal irreducible closed subsets of $X$ are exactly $V(x)$ and $V(y)$. Algebraically, the same two pieces appear from the factorisation $xy=x\cdot y$. We have $(xy)\subset (x)\cap(y)$ because $xy$ is divisible by both $x$ and $y$. Conversely, if $h\in (x)\cap(y)$, then $h=xp=yq$ for some $p,q\in k[x,y]$. Since $y$ divides $xp$ and $x$ and $y$ have no common nonconstant factor, $y$ divides $p$, so $p=yr$ for some $r\in k[x,y]$. Hence $h=xyr\in (xy)$, and therefore \begin{align*} (xy)=(x)\cap(y). \end{align*} Thus the two irreducible components of the zero set correspond to the two prime ideals $(x)$ and $(y)$ appearing in this intersection. [/example] This chapter ends with the guiding dictionary that will be sharpened by the Nullstellensatz. Algebraic subsets of affine space correspond contravariantly to ideals, but equality on the geometric side identifies ideals with the same radical. The next chapter proves that over an algebraically closed field, radical ideals are exactly the ideals recovered as $I(X)$ for affine algebraic sets $X$. The affine picture now shows that closed sets and ideals match contravariantly, with radicals recording the geometry exactly. What remained open was whether passing from an ideal to its zero set and back loses anything beyond radical information. The Nullstellensatz settles that point and identifies the coordinate rings that actually arise from affine algebraic sets. # 2. Hilbert Nullstellensatz and Coordinate Rings The previous chapter built the Zariski topology from polynomial equations and showed that ideals and closed subsets are related by order-reversing operations. The main unresolved issue was whether this relation loses information: if a polynomial vanishes on all common zeros of an ideal, must it be forced by the ideal itself? This chapter answers that question over an algebraically closed field and then packages affine algebraic sets into coordinate rings. The result is the first major dictionary of the course: geometry of affine varieties is encoded by reduced finitely generated $k$-algebras. Throughout this chapter $k$ is an algebraically closed field unless a different hypothesis is stated, and we continue to write $\mathbb A_k^n$ for affine $n$-space over $k$ with coordinate ring $k[x_1,\dots,x_n]$. ## Maximal Ideals and the Weak Nullstellensatz The first problem is to recognise which ideals correspond to single points. If $a=(a_1,\dots,a_n)\in \mathbb A_k^n$, then evaluation at $a$ defines a homomorphism $k[x_1,\dots,x_n]\to k$, and its kernel is the ideal generated by the coordinate differences $x_i-a_i$. The [weak Nullstellensatz](/theorems/2123) says that, over an algebraically closed field, every maximal ideal arises in this way. [definition: Point Ideal] Let $a=(a_1,\dots,a_n)\in \mathbb A_k^n$. The point ideal of $a$ is \begin{align*} \mathfrak m_a=(x_1-a_1,\dots,x_n-a_n)\trianglelefteq k[x_1,\dots,x_n]. \end{align*} [/definition] The point ideal definition supplies the basic examples of maximal ideals, since evaluation at $a$ has quotient $k[x_1,\dots,x_n]/\mathfrak m_a\cong k$. For the dictionary between points and ideals to work, the possible obstruction is a maximal ideal that behaves algebraically like a single closed point but has no actual point of $\mathbb A_k^n$ attached to it. The weak Nullstellensatz rules out exactly this obstruction over an algebraically closed field by showing that the point ideals are all the maximal ideals. [quotetheorem:2875] [citeproof:3721] The algebraically closed hypothesis is doing real work here. Over $\mathbb R$, the maximal ideal $(x^2+1)\trianglelefteq \mathbb R[x]$ has quotient $\mathbb C$, so it is maximal but is not the ideal of any real point. The theorem also says only something about maximal ideals: it does not classify arbitrary ideals, nor does it distinguish nilpotent information such as $(x)$ from $(x^2)$ by itself. The next question is the geometric existence problem for arbitrary systems of polynomial equations. If $I$ is merely a proper ideal, it need not itself be maximal, so the weak Nullstellensatz does not apply to $I$ directly. The reason maximal ideals still matter is that every proper ideal is contained in a maximal ideal; after the weak Nullstellensatz identifies that maximal ideal with a point, the containment forces all polynomials in $I$ to vanish at that point. [quotetheorem:2123] [citeproof:2123] The theorem converts an algebraic condition, properness of an ideal, into the existence of an actual $k$-rational solution. Algebraic closedness is necessary: over $\mathbb R$, the proper ideal $(x^2+1)$ has no real zero, as the next example records. The theorem does not say that the ideal is determined by its zero set; many different ideals can have the same nonempty vanishing set. That stronger reconstruction is exactly the radical question addressed by the strong Nullstellensatz. [example: No Real Point for a Proper Ideal] Over $\mathbb R$, the ideal $(x^2+1)\trianglelefteq \mathbb R[x]$ is proper: if $1\in(x^2+1)$, then there would be some $q(x)\in\mathbb R[x]$ with $q(x)(x^2+1)=1$. If $q\ne0$, then $\deg(q(x)(x^2+1))=\deg q+2$, which cannot equal $\deg 1=0$; if $q=0$, then the product is $0$, not $1$. Its real vanishing set is empty, because for every $a\in\mathbb R$ we have $a^2\ge0$, hence \begin{align*} a^2+1\ge1\ne0. \end{align*} Over $\mathbb C$, however, \begin{align*} i^2+1=-1+1=0. \end{align*} Also, \begin{align*} (-i)^2+1=i^2+1=-1+1=0. \end{align*} The factorization is visible from expanding the product: \begin{align*} (x-i)(x+i)=x^2+ix-ix-i^2=x^2+1. \end{align*} Thus the same proper ideal that has no real zero has complex zeros $i$ and $-i$, showing that algebraic closedness is essential in the Nullstellensatz. [/example] For a single point the point ideal is maximal, and the weak Nullstellensatz proves the converse. For more complicated closed sets, maximal ideals are replaced by all functions vanishing on that set, and this leads to the stronger form. ## Radical Ideals and the Strong Nullstellensatz The next problem is to determine exactly which polynomials vanish on the common zero set of an ideal. The containment $I\subseteq I(V(I))$ is automatic, but it is usually too small: powers of a polynomial can belong to $I$ even when the polynomial itself is the function that cuts out the same vanishing condition. The strong Nullstellensatz says that this is the only discrepancy. [definition: Radical Ideal] Let $R$ be a commutative ring and let $I\trianglelefteq R$ be an ideal. The radical of $I$ is \begin{align*} \sqrt I=\{f\in R: f^m\in I\text{ for some }m\in\mathbb N\}. \end{align*} The ideal $I$ is radical if $I=\sqrt I$. [/definition] Radicals record the same zero set because $f(a)=0$ iff $f(a)^m=0$. The obstruction to recovering $I$ from its points is precisely that point sets cannot detect nilpotent or multiplicity information: $f^m$ and $f$ impose the same vanishing condition. The strong form of the Nullstellensatz identifies the full ideal of functions forced to vanish on $V(I)$ with this radical replacement. [quotetheorem:2124] [citeproof:2124] The algebraically closed hypothesis again cannot be removed: over $\mathbb R$, the ideal $(x^2+1)$ has empty real vanishing set, so $I(V(x^2+1))=I(\varnothing)=\mathbb R[x]$ rather than $\sqrt{(x^2+1)}=(x^2+1)$. The theorem also does not preserve multiplicity, because $I$ and $\sqrt I$ define the same classical point set even when the quotient by $I$ has nilpotents. What it does provide is the exact replacement for a naive ideal-set correspondence: closed subsets correspond to radical ideals, and the next theorem packages this into the full order-reversing correspondence promised in the previous chapter. [quotetheorem:9414] [citeproof:9414] This correspondence is the algebraic core of affine geometry, but it is a correspondence for classical closed subsets only. The radical condition is forced: $(x)$ and $(x^2)$ cut out the same subset of $A^1$, so no point-set dictionary can distinguish them. It also does not yet describe maps between algebraic sets; that contravariant part of the dictionary will appear when regular maps are introduced. For now, the practical consequence is that geometric operations on closed sets should be translated into algebraic operations on radical ideals. [example: Intersections and Sums of Ideals] In $\mathbb A_k^2$, let $X=V(x)$ and $Y=V(y)$. A point $(a,b)$ lies in $X\cap Y$ exactly when $a=0$ and $b=0$, so \begin{align*} X\cap Y=\{(0,0)\}. \end{align*} Now compute the ideal of this point. If $f\in k[x,y]$, write it as a finite sum \begin{align*} f=\sum_{i,j\ge0} c_{ij}x^iy^j. \end{align*} Evaluating at $(0,0)$ gives $f(0,0)=c_{00}$. Hence $f$ vanishes at $(0,0)$ iff $c_{00}=0$, and then every remaining monomial has $i+j>0$, so it is divisible by either $x$ or $y$. Therefore \begin{align*} I(\{(0,0)\})=(x,y). \end{align*} The sum of the two defining ideals is \begin{align*} (x)+(y)=(x,y). \end{align*} This ideal is radical: if $f^m\in(x,y)$, then evaluating at $(0,0)$ gives $f(0,0)^m=0$, hence $f(0,0)=0$, so $f\in(x,y)$ by the computation above. Thus \begin{align*} I(X\cap Y)=I(\{(0,0)\})=(x,y)=\sqrt{(x)+(y)}. \end{align*} In this case the sum is already radical, but the formula uses the radical because sums of radical ideals need not be radical in general. [/example] Radicals also explain why multiplicity is not visible in classical affine algebraic sets. To remember multiplicity, later theories enlarge the category of geometric objects; in this chapter, the classical dictionary deliberately forgets it. [example: A Nonreduced Quotient Invisible to Points] For $a\in \mathbb A_k^1$, the polynomial $x$ vanishes at $a$ exactly when $a=0$. Also, $x^2$ vanishes at $a$ exactly when $a^2=0$, and since $k$ is a field, $a^2=0$ implies $a=0$. Hence \begin{align*} V(x)=\{0\}=V(x^2). \end{align*} The quotients still have different algebra. Evaluation at $0$ gives a surjective homomorphism $k[x]\to k$, $f\mapsto f(0)$, whose kernel is $(x)$, so $k[x]/(x)\cong k$. In $k[x]/(x^2)$, let $\bar x$ be the class of $x$. Then \begin{align*} \bar x^2=\overline{x^2}=0. \end{align*} But $\bar x\ne0$: if $\bar x=0$, then $x\in(x^2)$, so $x=q(x)x^2$ for some $q(x)\in k[x]$; if $q=0$ this gives $x=0$, and if $q\ne0$ then $\deg(qx^2)=\deg q+2\ne1=\deg x$. Thus $k[x]/(x^2)$ contains a nonzero nilpotent element, while $k[x]/(x)$ does not. The radical calculation is \begin{align*} \sqrt{(x^2)}=(x), \end{align*} because every multiple of $x$ has square in $(x^2)$, and if $f^m\in(x^2)$ then $f(0)^m=0$, hence $f(0)=0$, so $f\in(x)$. By *Strong Nullstellensatz*, the classical vanishing set remembers only this radical ideal, so it cannot detect the nilpotent class $\bar x$ in $k[x]/(x^2)$. [/example] The next step is to attach a ring directly to a closed set, rather than move back and forth through its defining equations. ## Coordinate Rings and Algebraic Functions Given an affine algebraic set $X\subseteq \mathbb A^n$, the guiding question is: what should the ring of polynomial functions on $X$ be? Two polynomials on $\mathbb A^n$ may define the same function on $X$ even if they are not equal as polynomials. The coordinate ring is obtained by quotienting out exactly this ambiguity. [definition: Coordinate Ring] Let $X\subseteq \mathbb A^n$ be an affine algebraic set over $k$. The coordinate ring of $X$ is \begin{align*} k[X]=k[x_1,\dots,x_n]/I(X). \end{align*} [/definition] Elements of $k[X]$ are polynomial functions on $X$, but each element is represented by many possible ambient polynomials. To use these elements as actual functions, one must remember both the residue class and the pointwise rule it induces on $X$; the quotient by $I(X)$ is exactly what makes different representatives give the same rule. [definition: Algebraic Function on an Affine Algebraic Set] Let $X\subseteq \mathbb A^n$ be an affine algebraic set. An algebraic function on $X$ is an element $\bar f\in k[X]$, represented by a polynomial $f\in k[x_1,\dots,x_n]$, together with its induced map \begin{align*} X&\longrightarrow k, & a&\longmapsto f(a). \end{align*} [/definition] The residue class of $f\in k[x_1,\dots,x_n]$ represents the function $a\mapsto f(a)$ on $X$, and two polynomials represent the same function exactly when their difference lies in $I(X)$. This construction produces finitely generated $k$-algebras, but pointwise function algebras have an extra restriction: they should not contain hidden nilpotent functions. If a function has a power equal to zero at every point, then the function itself vanishes at every point. The next result records this obstruction as reducedness of the coordinate ring. [quotetheorem:9415] [citeproof:9415] Reducedness is the algebraic shadow of working with ordinary point sets: if a function has a power equal to zero in $k[X]$, then it vanishes at every point of $X$, hence it is already zero in the coordinate ring. For this particular reducedness conclusion, the reasoning can also be phrased directly from the definition of $I(X)$: if $f^m$ vanishes on $X$, then $f$ vanishes on $X$. Algebraic closedness becomes essential in the reverse direction, where a reduced finitely generated algebra must be realised as the coordinate ring of its set of $k$-points by the strong Nullstellensatz. The theorem does not say that every finitely generated algebra is a coordinate ring; $k[\varepsilon]/(\varepsilon^2)$ is finitely generated but contains a nonzero nilpotent. The next theorem gives the reverse direction precisely after imposing reducedness. [quotetheorem:9416] [citeproof:9416] The reducedness hypothesis is essential: the dual numbers $k[\varepsilon]/(\varepsilon^2)$ are finitely generated, but no classical affine algebraic set has them as its coordinate ring because classical coordinate rings are reduced. Algebraic closedness is equally tied to the use of classical $k$-points. Over a non-algebraically-closed field, a reduced finitely generated algebra can have too few $k$-points for its ideal to be recovered from pointwise vanishing; for example $\mathbb R[x]/(x^2+1)$ is reduced but has no real point model as a classical subset of $A^1$ over $\mathbb R$. The theorem also describes objects before it describes morphisms; the phrase "arrows reversed" anticipates the later fact that a regular map $X\to Y$ induces a $k$-algebra homomorphism $k[Y]\to k[X]$. Within those limits, this is the point where many computations become easier on the algebra side than on the geometric side, as the following examples show. [example: Coordinate Ring of the Parabola] Let $X=V(y-x^2)\subseteq \mathbb A_k^2$. By the definition of the coordinate ring, \begin{align*} k[X]=k[x,y]/I(X). \end{align*} Since the parabola is cut out by the single polynomial $y-x^2$, we compute with \begin{align*} k[X]=k[x,y]/(y-x^2). \end{align*} Let $\bar x,\bar y$ denote the residue classes of $x,y$ in $k[X]$. Define $\varphi:k[t]\to k[X]$ by $\varphi(t)=\bar x$. To build the inverse, start with the homomorphism $\theta:k[x,y]\to k[t]$ given by \begin{align*} \theta(x)=t,\qquad \theta(y)=t^2. \end{align*} Then \begin{align*} \theta(y-x^2)=t^2-t^2=0. \end{align*} Thus $(y-x^2)\subseteq\ker\theta$, so $\theta$ descends to a homomorphism $\psi:k[X]\to k[t]$ with \begin{align*} \psi(\bar x)=t,\qquad \psi(\bar y)=t^2. \end{align*} Now \begin{align*} (\psi\circ\varphi)(t)=\psi(\bar x)=t, \end{align*} so $\psi\circ\varphi=\operatorname{id}_{k[t]}$. For the other composition, the ring $k[X]$ is generated by $\bar x$ and $\bar y$, and \begin{align*} (\varphi\circ\psi)(\bar x)=\varphi(t)=\bar x. \end{align*} Also, \begin{align*} (\varphi\circ\psi)(\bar y)=\varphi(t^2)=\bar x^2. \end{align*} In the quotient $k[x,y]/(y-x^2)$ we have $\bar y-\bar x^2=0$, hence $\bar y=\bar x^2$. Therefore $(\varphi\circ\psi)(\bar y)=\bar y$, so $\varphi\circ\psi=\operatorname{id}_{k[X]}$. Hence \begin{align*} k[X]\cong k[t]. \end{align*} The parabola therefore has the same coordinate ring as affine line, with the isomorphism corresponding to the parametrisation $t\mapsto(t,t^2)$. [/example] The parabola is smooth and parametrised by one coordinate. Singular curves reveal that the coordinate ring also records local algebraic behaviour at special points. [example: Coordinate Ring of the Nodal Cubic] Let $X=V(y^2-x^2(x+1))\subseteq A^2$, with $\operatorname{char}(k)\ne2$. Using the defining equation, its hypersurface coordinate ring is \begin{align*} k[X]=k[x,y]/(y^2-x^2(x+1)). \end{align*} Write \begin{align*} y^2-x^2(x+1)=y^2-x^3-x^2. \end{align*} At the origin, the terms of smallest total degree are the degree-$2$ terms $y^2-x^2$. Since \begin{align*} (y-x)(y+x)=y^2+yx-xy-x^2=y^2-x^2, \end{align*} the initial quadratic part factors as $y^2-x^2=(y-x)(y+x)$. Because $\operatorname{char}(k)\ne2$, the linear forms $y-x$ and $y+x$ are distinct: if they were scalar multiples, then comparing the $y$-coefficients gives the scalar $1$, while comparing the $x$-coefficients would force $-1=1$, contradicting $\operatorname{char}(k)\ne2$. Thus the lowest-degree part has two distinct linear factors, corresponding to the two tangent directions $y=x$ and $y=-x$ at the origin. This is the coordinate-ring trace of the node: the singular point has two different branches meeting there. [/example] The dictionary also clarifies what classical geometry omits. Nonreduced rings are meaningful algebraically, but they do not arise as coordinate rings of classical affine algebraic sets. [example: Dual Numbers Are Not a Classical Coordinate Ring] The algebra $k[\varepsilon]/(\varepsilon^2)$ is generated as a $k$-algebra by the class $\bar\varepsilon$, so it is finitely generated. In this quotient, \begin{align*} \bar\varepsilon^2=\overline{\varepsilon^2}=0. \end{align*} The element $\bar\varepsilon$ is not zero: if $\bar\varepsilon=0$, then $\varepsilon\in(\varepsilon^2)$, so $\varepsilon=q(\varepsilon)\varepsilon^2$ for some $q\in k[\varepsilon]$. If $q=0$, this gives $\varepsilon=0$; if $q\ne0$, then $\deg(q\varepsilon^2)=\deg q+2\ne1=\deg\varepsilon$. Hence $\bar\varepsilon$ is a nonzero nilpotent, and $k[\varepsilon]/(\varepsilon^2)$ is not reduced. The classical zero set of $\varepsilon^2$ in $\mathbb A_k^1$ is just the origin, because $a^2=0$ in the field $k$ implies $a=0$. For the point $\{0\}$, a polynomial $f(t)\in k[t]$ can be written as \begin{align*} f(t)=f(0)+tq(t) \end{align*} for some $q(t)\in k[t]$, so $f$ vanishes at $0$ exactly when $f(t)\in(t)$. Thus \begin{align*} k[\{0\}]=k[t]/(t)\cong k. \end{align*} More generally, every classical affine coordinate ring in this chapter is reduced: if $\bar f^m=0$ in $k[X]$, then $f(a)^m=0$ for every $a\in X$, hence $f(a)=0$ for every $a\in X$, so $\bar f=0$. Therefore the dual numbers cannot be the coordinate ring of a classical affine algebraic set; their nilpotent element records infinitesimal information that the classical point-set dictionary deliberately forgets. [/example] ## Algebraic Consequences of the Dictionary The final question of the chapter is how to read geometric properties from the coordinate ring. At this stage the main lessons concern points, inclusions, products, and irreducibility. These rules will later support regular maps, dimension theory, tangent spaces, and projective constructions. The dictionary says that $k[X]$ is the ring of functions on $X$. The next theorem is needed to show that the points of $X$ themselves can also be recovered from that ring. [quotetheorem:9417] [citeproof:9417] The algebraically closed hypothesis is again part of the statement, not decoration. For instance, the maximal ideal $(x^2+1)$ of $\mathbb R[x]$ would correspond to a closed point with residue field $\mathbb C$, not to an ordinary real point. The theorem also recovers only the classical point set from maximal ideals; it does not see nilpotent thickening in a nonreduced algebra unless one works with a richer geometric category. In the present classical setting, it tells us that the ring remembers the points, and the next question is whether the ring also remembers when the set has one irreducible component. [quotetheorem:9418] [citeproof:9418] This criterion explains why algebraic geometers often call irreducible affine algebraic sets affine varieties. The hypothesis that $X$ is an affine algebraic set over an algebraically closed field keeps the statement inside the classical point-set dictionary: products vanish on $X$ because they vanish at every point, and the strong Nullstellensatz identifies that pointwise condition with membership in $I(X)$. The result does not say that every domain is a coordinate ring without the finite generation and algebraic closedness hypotheses; those assumptions are what let a ring be realised by a classical set of $k$-points. Nor does it analyse how many components a reducible set has beyond detecting the existence of zero divisors. In these notes, when the distinction matters, algebraic set allows reducible objects and variety means irreducible algebraic set, and later dimension theory will refine this domain criterion by studying prime ideals inside the coordinate ring. [example: Two Coordinate Axes] Let $X=V(xy)\subseteq \mathbb A_k^2$. A point $(a,b)$ lies in $X$ exactly when $ab=0$; since $k$ is a field, this is equivalent to $a=0$ or $b=0$. Thus \begin{align*} X=V(x)\cup V(y), \end{align*} the union of the two coordinate axes. We compute $I(X)$. Write \begin{align*} f(x,y)=\sum_{i,j\ge0} c_{ij}x^iy^j. \end{align*} If $f$ vanishes on $X$, then it vanishes on the $x$-axis, so \begin{align*} 0=f(x,0)=\sum_{i\ge0} c_{i0}x^i. \end{align*} Since $k$ is algebraically closed, it is infinite, so a one-variable polynomial over $k$ that vanishes at every element of $k$ is the zero polynomial. Hence $c_{i0}=0$ for every $i$. Similarly, vanishing on the $y$-axis gives \begin{align*} 0=f(0,y)=\sum_{j\ge0} c_{0j}y^j, \end{align*} so $c_{0j}=0$ for every $j$. Therefore all remaining nonzero terms have $i\ge1$ and $j\ge1$, and \begin{align*} f(x,y)=xy\sum_{i,j\ge1} c_{ij}x^{i-1}y^{j-1}. \end{align*} Thus $f\in(xy)$, and the reverse containment is immediate because every multiple of $xy$ vanishes wherever $xy=0$. Hence \begin{align*} I(X)=(xy), \end{align*} so \begin{align*} k[X]=k[x,y]/(xy). \end{align*} Let $\bar x,\bar y$ be the residue classes of $x,y$ in $k[X]$. Their product is \begin{align*} \bar x\bar y=\overline{xy}=0. \end{align*} Both classes are nonzero: if $\bar x=0$, then $x\in(xy)$, so $x=xyq(x,y)$ for some $q\in k[x,y]$, which would make every term on the right divisible by $y$, impossible for the polynomial $x$. The same argument shows $\bar y\ne0$. Thus $k[X]$ has nonzero zero divisors, exactly reflecting that $X$ splits into the two coordinate-axis components. [/example] The chapter's main conclusion is that affine algebraic geometry over an algebraically closed field can be done in either language. Closed subsets of affine space correspond to radical ideals, points correspond to maximal ideals, and affine varieties correspond to reduced finitely generated coordinate rings. The rest of the course builds on this dictionary by translating maps, dimensions, singularities, and projective constructions into algebraic statements about these rings. With the Nullstellensatz in hand, affine varieties are no longer just zero sets but spaces whose geometry is encoded by reduced finitely generated coordinate rings. That makes maps the next natural object to study, since morphisms should be visible as algebra homomorphisms between these rings. The following chapter turns that expectation into the precise language of regular functions and regular maps. # 3. Regular Functions and Regular Maps This chapter is the affine-morphism part of the course. The preceding chapter attached to an affine variety $X \subseteq \mathbb A^n$ its coordinate ring $k[X]$, using the radical ideal correspondence from the Nullstellensatz to identify polynomials with the same values on $X$. We now assume that language and ask how functions and maps should be recognised without depending on a chosen set of equations in affine space. Regular functions are the functions on an affine variety that deserve to be called polynomial from the intrinsic point of view. The main point is that the coordinate ring is not only an invariant of $X$: it is the algebra of regular $k$-valued functions on $X$. This chapter turns that construction into a language for maps: regular maps are precisely those maps whose coordinate functions lie in the coordinate ring, and their algebraic shadow is a pullback homomorphism of coordinate rings. ## Functions Seen Through the Coordinate Ring What does it mean to evaluate a polynomial on a variety when different ambient polynomials can agree on all points of the variety? The coordinate ring answers this by making functions, rather than formulas, the primary object. A regular function on an affine variety is a function obtained from a polynomial after passing to the quotient by the vanishing ideal. [definition: Regular Function On An Affine Variety] Let $X \subseteq \mathbb A^n$ be an affine variety over an algebraically closed field $k$. A function $f: X \to k$ is regular if there exists $F \in k[x_1,\dots,x_n]$ such that $f(p)=F(p)$ for every $p \in X$. [/definition] The polynomial $F$ is not part of the data of the function. Replacing $F$ by $F+G$ with $G \in I(X)$ gives the same function on $X$, so the quotient by $I(X)$ is exactly the algebra that removes this ambiguity. The next question is whether this quotient is merely a convenient notation or whether it gives the whole algebra of regular functions. [quotetheorem:9419] [citeproof:9419] The affine hypothesis matters because the construction uses a single ambient polynomial ring; for instance, on projective space regular functions are not described by evaluating arbitrary homogeneous representatives in the same way. The quotient by $I(X)$ is also essential: on $X=V(y-x^2)$, the ambient polynomials $y$ and $x^2$ are different formulas but define the same function. The theorem does not say that every set-theoretic function $X\to k$ is regular; even on $\mathbb A^1$, most functions are not polynomial functions. It justifies writing $k[X]$ both for the coordinate ring and for the algebra of regular $k$-valued functions on $X$, and it prepares the next step: maps between varieties can be tested by how they transform these regular functions. A first test case is a variety whose defining equation lets us remove one coordinate. [example: Coordinate Ring Of The Parabola] Let $X=V(y-x^2)\subseteq \mathbb A^2$. Its vanishing ideal is $(y-x^2)$, so \begin{align*} k[X]=k[x,y]/(y-x^2). \end{align*} In this quotient, $y-x^2=0$, hence \begin{align*} \overline{y}-\overline{x}^2=\overline{y-x^2}=0. \end{align*} Therefore $\overline{y}=\overline{x}^2$. Now take any class $\overline{F}\in k[X]$, with \begin{align*} F=\sum_{i,j} a_{ij}x^i y^j\in k[x,y]. \end{align*} Using $\overline{y}=\overline{x}^2$, each monomial satisfies \begin{align*} \overline{x^i y^j}=\overline{x}^i\overline{y}^j=\overline{x}^i(\overline{x}^2)^j=\overline{x}^{i+2j}. \end{align*} Thus \begin{align*} \overline{F}=\sum_{i,j}a_{ij}\overline{x}^{i+2j}, \end{align*} so every regular function on $X$ is represented by a polynomial in the single class $\overline{x}$. Define $\Phi:k[t]\to k[X]$ by $\Phi(t)=\overline{x}$. Define $\Psi:k[x,y]\to k[t]$ by $\Psi(x)=t$ and $\Psi(y)=t^2$. Since \begin{align*} \Psi(y-x^2)=t^2-t^2=0, \end{align*} the map $\Psi$ factors through the quotient and gives $\overline{\Psi}:k[X]\to k[t]$. The two composites fix the generators: \begin{align*} \overline{\Psi}(\Phi(t))=\overline{\Psi}(\overline{x})=t, \end{align*} and \begin{align*} \Phi(\overline{\Psi}(\overline{x}))=\Phi(t)=\overline{x}, \end{align*} while \begin{align*} \Phi(\overline{\Psi}(\overline{y}))=\Phi(t^2)=\overline{x}^2=\overline{y}. \end{align*} Hence $\Phi$ and $\overline{\Psi}$ are inverse $k$-algebra homomorphisms, so $k[X]\cong k[t]$. Algebraically, the equation $y=x^2$ removes $y$ as an independent coordinate on the parabola. [/example] The parabola example shows a recurring theme: two varieties may have different embeddings into affine space but the same ring of regular functions. This raises the next problem: how should a map between varieties act on regular functions? ## Regular Maps And Pullbacks A function between varieties should be considered algebraic if all polynomial measurements on the target become polynomial measurements on the source. This reverses direction: a map $\varphi:X\to Y$ should induce a homomorphism from functions on $Y$ to functions on $X$ by composition. [definition: Regular Map Of Affine Varieties] Let $X\subseteq \mathbb A^n$ and $Y\subseteq \mathbb A^m$ be affine varieties. A map $\varphi:X\to Y$ is regular if there exist regular functions $f_1,\dots,f_m\in k[X]$ such that $\varphi(p)=(f_1(p),\dots,f_m(p))$ for every $p\in X$. [/definition] The condition that the image lies in $Y$ is part of the assertion that $\varphi$ is a map to $Y$. If $Y=V(J)$ with $J\trianglelefteq k[y_1,\dots,y_m]$, this means that every polynomial relation defining $Y$ is satisfied after substituting the coordinate functions $f_i$. The next construction records this substitution process as a homomorphism of coordinate rings. [definition: Pullback Homomorphism] Let $\varphi:X\to Y$ be a regular map of affine varieties. The pullback homomorphism of $\varphi$ is the $k$-algebra homomorphism $\varphi^*:k[Y]\to k[X]$ defined by $\varphi^*(g)=g\circ \varphi$. [/definition] Pullback is the algebraic trace of a regular map. It is contravariant: maps of varieties go one way, while homomorphisms of coordinate rings go the other way. The next question is whether every homomorphism of coordinate rings arises in this way from a unique regular map. [quotetheorem:9420] [citeproof:9420] The affine hypotheses are doing real work here: the coordinate functions $\overline{y_1},\dots,\overline{y_m}$ generate $k[Y]$, so a homomorphism out of $k[Y]$ is determined by their images. The condition that the defining relations of $Y$ are respected cannot be dropped; for example, assigning $x\mapsto t$ and $y\mapsto t$ does not define a map $\mathbb A^1\to V(y-x^2)$ because $t-t^2$ need not vanish. The theorem does not say that a bijection of underlying point sets is regular, nor that every algebra homomorphism $k[X]\to k[Y]$ gives a map $X\to Y$ in the same direction; the direction is reversed by pullback. This is the operational form of the affine dictionary, and the parabola gives the cleanest example because the parametrisation also identifies the coordinate rings. [example: Parametrising The Parabola] Let $X=\mathbb A^1$ and $Y=V(y-x^2)\subseteq \mathbb A^2$. Define $\varphi:X\to \mathbb A^2$ by $\varphi(t)=(t,t^2)$. For every $t\in k$, the defining polynomial of $Y$ vanishes at this point because \begin{align*} (y-x^2)(t,t^2)=t^2-t^2=0. \end{align*} Thus $\varphi(t)\in Y$, so $\varphi$ is a map $\mathbb A^1\to Y$. Its coordinate functions are $t$ and $t^2$, both elements of $k[t]=k[\mathbb A^1]$, hence $\varphi$ is regular. Since $k[Y]=k[x,y]/(y-x^2)$, the pullback $\varphi^*:k[Y]\to k[t]$ is determined by the coordinate classes: \begin{align*} \varphi^*(\overline{x})(t)=\overline{x}(\varphi(t))=\overline{x}(t,t^2)=t. \end{align*} Similarly, \begin{align*} \varphi^*(\overline{y})(t)=\overline{y}(\varphi(t))=\overline{y}(t,t^2)=t^2. \end{align*} Equivalently, $\varphi^*(\overline{x})=t$ and $\varphi^*(\overline{y})=t^2$. Define $\theta:k[t]\to k[Y]$ by $\theta(t)=\overline{x}$. Then \begin{align*} (\varphi^*\circ\theta)(t)=\varphi^*(\overline{x})=t, \end{align*} so $\varphi^*\circ\theta=\operatorname{id}_{k[t]}$ because $k[t]$ is generated by $t$. On the generators of $k[Y]$, \begin{align*} (\theta\circ\varphi^*)(\overline{x})=\theta(t)=\overline{x}. \end{align*} Also, using the relation $\overline{y}=\overline{x}^2$ in $k[Y]$, \begin{align*} (\theta\circ\varphi^*)(\overline{y})=\theta(t^2)=\theta(t)^2=\overline{x}^2=\overline{y}. \end{align*} Thus $\theta\circ\varphi^*=\operatorname{id}_{k[Y]}$, since $\overline{x}$ and $\overline{y}$ generate $k[Y]$. Therefore $\varphi^*:k[Y]\to k[t]$ is an isomorphism, and the parametrisation $t\mapsto(t,t^2)$ identifies the regular functions on the parabola with polynomials in one variable. [/example] Not every parametrisation that is bijective on points is an isomorphism of varieties. The cusp gives the standard warning: pointwise inverse maps can fail to be regular because the missing inverse coordinate may not lie in the image of the pullback. [example: The Cuspidal Cubic] Let $C=V(y^2-x^3)\subseteq \mathbb A^2$, and define $\psi:\mathbb A^1\to \mathbb A^2$ by $\psi(t)=(t^2,t^3)$. Its image lies in $C$, since for every $t\in k$, \begin{align*} (y^2-x^3)(t^2,t^3)=(t^3)^2-(t^2)^3=t^6-t^6=0. \end{align*} The coordinate functions $t^2$ and $t^3$ belong to $k[t]=k[\mathbb A^1]$, so $\psi$ is a regular map $\mathbb A^1\to C$. The map is injective. If $\psi(s)=\psi(t)$, then $s^2=t^2$ and $s^3=t^3$. If $t=0$, then $s^2=0$, so $s=0=t$. If $t\ne 0$, then also $s\ne 0$, and dividing the equalities gives \begin{align*} s=\frac{s^3}{s^2}=\frac{t^3}{t^2}=t. \end{align*} It is also surjective onto $C$. Let $(a,b)\in C$, so $b^2=a^3$. If $a=0$, then $b^2=0$, hence $b=0$, and $(a,b)=\psi(0)$. If $a\ne 0$, set $t=b/a$. Then \begin{align*} t^2=\frac{b^2}{a^2}=\frac{a^3}{a^2}=a, \end{align*} and \begin{align*} t^3=t\cdot t^2=\frac{b}{a}\cdot a=b. \end{align*} Thus $(a,b)=\psi(t)$. Now compute the pullback. Since $k[C]=k[x,y]/(y^2-x^3)$, the coordinate classes satisfy \begin{align*} \psi^*(\overline{x})(t)=\overline{x}(\psi(t))=\overline{x}(t^2,t^3)=t^2, \end{align*} and \begin{align*} \psi^*(\overline{y})(t)=\overline{y}(\psi(t))=\overline{y}(t^2,t^3)=t^3. \end{align*} Therefore the image of $\psi^*:k[C]\to k[t]$ is the subalgebra generated by $t^2$ and $t^3$, namely $k[t^2,t^3]$. This subalgebra is proper. Every monomial in $t^2$ and $t^3$ has the form \begin{align*} (t^2)^i(t^3)^j=t^{2i+3j} \end{align*} with $i,j\ge 0$, so every element of $k[t^2,t^3]$ is a finite $k$-linear combination of powers $t^n$ with $n=0$ or $n\ge 2$. No such linear combination has a $t^1$ term, and the monomials $1,t,t^2,\dots$ are linearly independent in $k[t]$. Hence $t\notin k[t^2,t^3]$. The inverse set map would have to send $(t^2,t^3)$ back to $t$, so its pullback would have to produce the regular function $t$ on $\mathbb A^1$. Since $t$ is not in the image of $\psi^*$, the inverse set map is not regular at the cusp. [/example] The cusp separates topology from algebra: a bijective regular map need not be an isomorphism. Isomorphisms are governed by inverse homomorphisms of coordinate rings, not by bijectivity alone. This motivates the next section, where standard geometric constructions are translated into algebraic tests. ## Isomorphisms, Embeddings, Products, And Graphs How do the usual constructions with spaces appear inside the affine dictionary? Isomorphisms become algebra isomorphisms, closed embeddings become quotient maps of coordinate rings, products become tensor products, and graphs turn regularity into a closedness condition. [definition: Isomorphism Of Affine Varieties] Let $X$ and $Y$ be affine varieties. A regular map $\varphi:X\to Y$ is an isomorphism if there exists a regular map $\psi:Y\to X$ such that $\psi\circ\varphi=\operatorname{id}_X$ and $\varphi\circ\psi=\operatorname{id}_Y$. [/definition] The definition is geometric, but the previous section translates it into algebra. The next result states the full affine equivalence of categories, which explains why coordinate rings are not only invariants but complete algebraic models of affine varieties. Here "affine variety" is being used in the broad classical sense of an affine algebraic set; under the stricter convention that varieties are irreducible, the corresponding algebraic category is the category of finitely generated reduced integral $k$-algebras. [quotetheorem:2127] [citeproof:2127] Each hypothesis in this equivalence rules out a specific failure. Algebraic closedness is what lets the Nullstellensatz recover radical ideals from point sets; over $\mathbb R$, the algebra $\mathbb R[x]/(x^2+1)$ has no corresponding real affine variety with the expected points. Reducedness is also necessary, since nilpotents such as the class of $\varepsilon$ in $k[\varepsilon]/(\varepsilon^2)$ cannot be detected by ordinary pointwise functions. The theorem does not say that all geometric spaces are affine, or that non-reduced scheme-theoretic information is visible in classical varieties. It is nevertheless the central organising result for affine varieties, and it foreshadows the scheme-theoretic viewpoint where nilpotents and non-affine gluing are retained rather than discarded. As a concrete application, automorphisms of the affine line reduce to automorphisms of a polynomial algebra in one variable. [example: Automorphisms Of The Affine Line] An automorphism $\varphi:\mathbb A^1\to\mathbb A^1$ gives a $k$-algebra automorphism $\alpha:k[t]\to k[t]$ by pullback. Write $\alpha(t)=f(t)$. Since $\alpha$ is surjective, there is some $g(t)\in k[t]$ with $\alpha(g(t))=t$. Because $\alpha$ fixes scalars and sends $t$ to $f(t)$, applying $\alpha$ to $g(t)$ means substituting $f(t)$ for $t$, so \begin{align*} g(f(t))=t. \end{align*} The polynomial $f$ cannot be constant, since then $g(f(t))$ would also be constant. Thus $\deg f\ge 1$. Similarly, $g$ cannot be constant because $g(f(t))=t$ is nonconstant, so $\deg g\ge 1$. Taking degrees in the identity above gives \begin{align*} \deg(g(f(t)))=(\deg g)(\deg f)=\deg t=1. \end{align*} Since $\deg f$ and $\deg g$ are positive integers, both must be $1$. Therefore \begin{align*} f(t)=at+b \end{align*} for some $a\in k^\times$ and $b\in k$. Conversely, if $f(t)=at+b$ with $a\ne 0$, then the map $t\mapsto at+b$ has inverse $t\mapsto a^{-1}(t-b)$, because \begin{align*} a^{-1}((at+b)-b)=t \end{align*} and \begin{align*} a(a^{-1}(t-b))+b=t. \end{align*} Hence the automorphisms of $\mathbb A^1$ are exactly the affine linear maps $t\mapsto at+b$ with $a\in k^\times$ and $b\in k$. [/example] Automorphisms compare varieties of the same size. The next natural construction is an embedding, where one variety is realised as a closed subvariety of another and functions are obtained by restriction from the ambient variety. [definition: Closed Embedding Of Affine Varieties] A regular map $\iota:X\to Y$ is a closed embedding if $\iota$ is an isomorphism from $X$ onto a closed subvariety of $Y$. [/definition] The quotient nature of restriction gives the algebraic test for such embeddings. A closed subvariety has fewer functions because functions on it are restrictions of functions on the ambient space. What can fail for a general injective regular map is surjectivity on functions: the cuspidal parametrisation $\mathbb A^1\to V(y^2-x^3)$ is bijective on points, but the function $t$ on the source is not the pullback of a regular function on the cusp. The next theorem makes the restriction principle into an iff criterion. [quotetheorem:9421] [citeproof:9421] Surjectivity is the exact algebraic form of saying that no regular functions on $X$ are missing from the ambient variety. If it fails, as for $\mathbb A^1\to V(y^2-x^3)$ with pullback image $k[t^2,t^3]$, the map may still be bijective on points but cannot be a closed embedding. The theorem does not classify all injective regular maps, and it does not say that a homeomorphism onto a closed subset is automatically an isomorphism of varieties. It gives the test needed for later constructions where subvarieties are specified by quotienting coordinate rings. Products are the next construction to translate. The product variety should represent pairs of points and should allow regular maps into it to be the same thing as pairs of regular maps. A naive product of defining equations would be too small: for $X=V(f)$ and $Y=V(g)$, the equation $fg=0$ describes a union-type condition rather than independent membership in $X$ and $Y$. To make the product construction usable, we need its coordinate ring. [definition: Product Of Affine Varieties] Let $X\subseteq \mathbb A^n$ and $Y\subseteq \mathbb A^m$ be affine varieties. Their product is the affine variety $X\times Y\subseteq \mathbb A^{n+m}$ cut out by the equations defining $X$ in the first $n$ coordinates and the equations defining $Y$ in the last $m$ coordinates. [/definition] The definition matches the set-theoretic product, but its usefulness comes from the coordinate ring computation. The possible failure is the appearance of accidental mixed relations between the two coordinate systems. The equations defining $X$ should constrain only the first coordinates, and the equations defining $Y$ should constrain only the last coordinates; no extra relation should mix them unless it follows from those two independent systems. Algebraically, the operation designed to impose exactly this independence over the common base field is the [tensor product](/page/Tensor%20Product). Thus the product construction still needs an algebraic check: its coordinate ring should be obtained by adjoining the two independent coordinate systems over $k$ and imposing only the old relations from each factor. Without that check, maps into $X\times Y$ would not reliably be the same as pairs of maps into $X$ and $Y$. [quotetheorem:9422] [citeproof:9422] The tensor product hypothesis reflects that the two coordinate systems are independent over the same base field $k$; without working over a common base, the expression $k[X]\otimes_k k[Y]$ would not encode scalar multiplication consistently. The algebraically closed hypothesis is also tied to the point-set interpretation: over $\mathbb R$, the variety $V(x^2+1)$ has no real points, while its coordinate ring $\mathbb R[x]/(x^2+1)$ is nonzero, and \begin{align*} \mathbb R[x]/(x^2+1)\otimes_{\mathbb R}\mathbb R[y]/(y^2+1)\cong \mathbb C\otimes_{\mathbb R}\mathbb C. \end{align*} Thus the algebraic tensor product still records base-change information even though the real point-set product is empty. The theorem gives the algebra needed to describe maps into products and, in particular, to encode a map by its graph inside $X\times Y$. The product construction lets us encode a map by its graph inside $X\times Y$. This is useful because it turns the question of whether coordinate functions are regular into a question about whether a certain subset of a product is closed and projects back to the source in the right way. [definition: Graph Of A Map] Let $X$ and $Y$ be affine varieties and let $\varphi:X\to Y$ be a set map. The graph of $\varphi$ is $\Gamma_\varphi=\{(x,y)\in X\times Y: y=\varphi(x)\}$. [/definition] A regular map has polynomial coordinate functions, so its graph is cut out by polynomial equations $y_i=f_i(x)$. Closedness alone is not enough: a closed subset of $X\times Y$ may have several points over a single point of $X$, so it would describe a correspondence rather than a function. Projection being bijective as a set is also too weak unless the inverse is regular, as the cusp example warns. The next theorem gives the precise criterion. [quotetheorem:9423] [citeproof:9423] Both conditions in the graph criterion are necessary. If $\varphi:\mathbb A^1\to\mathbb A^1$ is an arbitrary non-polynomial set map, its graph need not be Zariski closed, so closedness detects the failure of polynomial coordinate functions. If a closed subset of $X\times Y$ projects to $X$ with two points over some $x\in X$, it is not the graph of a function; if the projection is only a bijection of sets but not an isomorphism, it can still hide non-regular inverse coordinate functions. The theorem does not say that every closed subvariety of a product is a graph, only those whose projection to the source is an isomorphism. This formulation will reappear in later chapters when rational maps, projective morphisms, and separatedness-like phenomena are introduced in classical language. The morphism theory completes the affine dictionary by showing how geometric maps are read off from coordinate rings. Once maps are understood globally, the next step is to look locally on open subsets, where functions may be represented by quotients rather than polynomials. Distinguished opens and rational functions provide exactly that local language. # 4. Localization, Distinguished Opens, and Rational Functions These notes develop the first algebraic dictionary for classical affine varieties over an algebraically closed base field $k$. Using the Zariski topology of Chapter 1, the coordinate-ring dictionary of Chapter 2, and the morphism language of Chapter 3, This chapter shifts from global polynomial functions on affine varieties to functions that are regular only after excluding a closed subset. The guiding question is how much geometry can be read from the algebra obtained by inverting one polynomial. Distinguished open sets provide the local pieces of affine varieties, and rational functions record expressions that are meaningful on dense open subsets even when they have poles elsewhere. ## Distinguished Opens and Local Rings The Zariski topology has many open sets, but algebraic computations are controlled by a smaller family. The first problem is to find open neighbourhoods that are both geometrically natural and algebraically explicit. [definition: Distinguished Open Set] Let $X \subset \mathbb A_k^n$ be an affine variety and let $f \in k[X]$. The distinguished open subset determined by $f$ is \begin{align*} D_X(f) := \{p \in X : f(p) \ne 0\}. \end{align*} When the ambient variety is fixed, write $D(f)$ for $D_X(f)$. [/definition] The complement of $D(f)$ is the closed set $V_X(f)$, so distinguished opens are open. Before proving that these opens are sufficient for local work, it is useful to see how their algebra mirrors their set-theoretic behaviour. [example: Basic Distinguished Opens in Affine Space] In $\mathbb A_k^2$ with coordinate ring $k[x,y]$, a point is a pair $p=(a,b)$ with $a,b\in k$. Evaluating the coordinate function $x$ at $p$ gives $x(p)=a$, so \begin{align*} D(x)=\{(a,b)\in \mathbb A_k^2:a\ne 0\}. \end{align*} Its complement is therefore $\{(0,b):b\in k\}$, the $y$-axis. Similarly, evaluating $xy$ at $p=(a,b)$ gives \begin{align*} (xy)(p)=a b. \end{align*} Since $k$ is a field, $ab\ne 0$ holds exactly when $a\ne 0$ and $b\ne 0$. Hence \begin{align*} D(xy)=\{(a,b):ab\ne 0\}=\{(a,b):a\ne 0\text{ and }b\ne 0\}. \end{align*} Also, \begin{align*} D(x)\cap D(y)=\{(a,b):a\ne 0\}\cap \{(a,b):b\ne 0\}=\{(a,b):a\ne 0\text{ and }b\ne 0\}. \end{align*} Thus $D(xy)=D(x)\cap D(y)$, and its complement is the union of the two coordinate axes. The same pointwise argument gives the general identity $D(fg)=D(f)\cap D(g)$: at a point $p$, $(fg)(p)=f(p)g(p)$, and a product in the field $k$ is nonzero exactly when both factors are nonzero. [/example] The example shows that multiplication in the coordinate ring corresponds to intersection of these basic opens. Local arguments need neighbourhoods that are both algebraically controlled and stable under shrinking around a point. Distinguished opens provide such neighbourhoods: they come from single nonvanishing functions, and their multiplicative behaviour makes them compatible with finite intersections. [quotetheorem:9424] [citeproof:9424] The theorem lets us study regularity locally on a controlled collection of open sets. Its limitation is that "basis" does not mean "every open set is distinguished": in $\mathbb A_k^2$, the open set $\mathbb A_k^2\setminus\{(0,0)\}$ is not equal to $D(f)$ for any single $f\in k[x,y]$, since that would make the radical ideal $(x,y)$ the radical of a principal ideal. The affine coordinate-ring hypothesis is what supplies the functions whose nonvanishing cuts out these neighbourhoods; an arbitrary topological space has no comparable algebraic source of basic opens. Since every point of an open set has a distinguished neighbourhood, the algebra of functions on $D(f)$ becomes the next object to identify. [definition: Localization at an Element] Let $A$ be a commutative $k$-algebra and let $f \in A$. The localization of $A$ at the multiplicative set $\{1,f,f^2,\dots\}$ is \begin{align*} A_f := \left\{\frac{a}{f^m}: a \in A,\ m \ge 0\right\}, \end{align*} with the usual fraction [equivalence relation](/page/Equivalence%20Relation). [/definition] The definition formalizes the operation that is geometrically allowed on $D(f)$: powers of $f$ have no zeros there and may be used as denominators. The next theorem is needed to prove that this algebraic localization is not merely a source of examples, but the entire ring of regular functions on the distinguished open. [quotetheorem:9425] [citeproof:9425] This result is the first local form of the coordinate-ring dictionary. Instead of assigning one ring to all of $X$, it assigns $k[X]_f$ to the open region where $f$ has become invertible. The theorem does not say that every open subset has coordinate ring obtained by inverting a single function; $\mathbb A_k^2\setminus\{(0,0)\}$ again shows that many opens are not single distinguished opens. Nor is the reduced affine hypothesis cosmetic: if a nonreduced coordinate algebra such as $k[\varepsilon]/(\varepsilon^2)$ were treated as a ring of ordinary set-valued functions on its one closed point, the nilpotent $\varepsilon$ would define the zero function while remaining nonzero in the algebra. That would destroy the injectivity step, which is why the classical variety setting keeps coordinate rings reduced. The practical test supplied by the theorem is membership in a localization. To decide whether a quotient is regular on $D(f)$, the denominator need not be nonzero everywhere on $X$; it only needs to become invertible after some power of $f$ is allowed in the denominator. [example: The Function One Over X] On $\mathbb A_k^1$, write a point as $p=a\in k$, so the coordinate function satisfies $x(p)=a$. The distinguished open determined by $x$ is \begin{align*} D(x)=\{a\in k:x(a)\ne 0\}=\{a\in k:a\ne 0\}=\mathbb A_k^1\setminus\{0\}. \end{align*} On this open set the denominator $x$ is nowhere zero, so the quotient \begin{align*} a\longmapsto \frac{1}{x(a)}=\frac{1}{a} \end{align*} is represented by the element \begin{align*} \frac{1}{x}\in k[x]_x. \end{align*} This reciprocal does not come from a regular function on all of $\mathbb A_k^1$. If it did, then some polynomial $g\in k[x]$ would satisfy $g(a)=1/a$ for every $a\ne 0$. Multiplying by $a$ gives \begin{align*} a g(a)-1=0 \end{align*} for every $a\ne 0$, so the polynomial $xg-1$ vanishes on the infinite set $k\setminus\{0\}$ and hence is the zero polynomial. Evaluating this alleged identity at $0$ gives \begin{align*} 0\cdot g(0)-1=-1, \end{align*} which is not $0$. Thus $\frac{1}{x}$ is regular on $D(x)$ but not on the whole affine line, showing that regularity on $D(x)$ is measured by membership in the localization $k[x]_x$. [/example] The example shows that regularity depends on the open neighbourhood under consideration. This motivates the definition of the local ring at a point: instead of choosing one distinguished open such as $D(x)$, we collect all fractions that are regular on some neighbourhood of the chosen point. [definition: Local Ring of a Variety at a Point] Let $X$ be an affine variety and let $p \in X$. The local ring of $X$ at $p$ is \begin{align*} \mathcal O_{X,p} := \left\{\frac{a}{b}: a,b \in k[X],\ b(p) \ne 0\right\}. \end{align*} [/definition] The elements of $\mathcal O_{X,p}$ are functions defined on some distinguished neighbourhood of $p$. To use this ring effectively, one must know which fractions are locally invertible and which ones record vanishing at the chosen point. A function with nonzero value at $p$ should become invertible after shrinking the neighbourhood, while a function vanishing at $p$ cannot be inverted locally at $p$. The next result identifies this dichotomy as the local-ring structure with one distinguished maximal ideal. [quotetheorem:9426] [citeproof:9426] This theorem gives a precise meaning to the phrase "near $p$". Regular functions with nonzero value at $p$ are locally invertible, while those vanishing at $p$ form the maximal ideal measuring first-order and higher-order behaviour. The point condition is essential: on $\mathbb A_k^1$, localizing $k[x]$ at all nonzero polynomials gives the field $k(x)$, where $x$ is a unit. Thus the set of fractions whose numerator vanishes at $0$ cannot be a maximal ideal in that larger ring, because it would contain the unit $x$. The theorem also does not assert that every element of the local ring extends to a global regular function; the reciprocal of $x$ at a point with $x(p)\ne 0$ is local but not globally regular on $\mathbb A_k^1$. This distinction is what lets rational functions keep track of generic formulas while local rings remember behaviour at a chosen point. ## Rational Functions and Dense Open Equality Regular functions are too restrictive for many geometric constructions: projection from a point, parametrisation of curves, and ratios of homogeneous coordinates all naturally involve quotients. The next problem is to introduce such quotients without pretending they are defined everywhere. [definition: Rational Function] Let $X$ be an irreducible affine variety. A rational function on $X$ is an element of the fraction field \begin{align*} k(X) := \operatorname{Frac}(k[X]). \end{align*} [/definition] Irreducibility is the reason $k[X]$ is an integral domain, so its fraction field exists. A rational function may be written by different quotients, and a denominator that vanishes in one presentation may disappear after cancellation or after rewriting the same fraction. The obstruction to evaluating a rational function is therefore not tied to a single displayed denominator. We need a definition that records the largest region where the rational function itself, independent of its presentation, is genuinely regular. [definition: Domain of Definition] Let $X$ be an irreducible affine variety and let $\varphi \in k(X)$. The domain of definition of $\varphi$ is the union of all open subsets $U \subset X$ such that the class $\varphi$ is represented on $U$ by an element of the regular function ring $\mathcal O_X(U)$. [/definition] The domain of definition is therefore an open subset of $X$. It separates the rational function from any particular written fraction. Its complement is where every representative has an unavoidable denominator problem, and this motivates naming the local obstruction. [definition: Pole] Let $X$ be an irreducible affine variety and let $\varphi \in k(X)$. A point $p \in X$ is a pole of $\varphi$ if $\varphi$ is not regular at $p$, meaning $\varphi \notin \mathcal O_{X,p}$. [/definition] The local-ring formulation makes poles independent of the chosen quotient. The following example shows why this independence matters: cancellation may remove an apparent singularity. [example: Cancellation and Domain of Definition] On $\mathbb A_k^1$, the coordinate ring is $k[x]$ and the rational function field is $k(x)=\operatorname{Frac}(k[x])$. In $k[x]$ we have \begin{align*} x^2-x=x(x-1). \end{align*} Since $x$ is a nonzero element of the domain $k[x]$, the equality in the fraction field is \begin{align*} \frac{x^2-x}{x}=\frac{x(x-1)}{x}=x-1. \end{align*} Thus the written quotient has denominator $x$, so that particular expression is not defined at $0$, but the rational function itself is the polynomial function $x-1$ and is therefore regular on all of $\mathbb A_k^1$. By contrast, the rational function $1/x$ is regular on $D(x)$ because at each point $a\in D(x)$ the value $x(a)=a$ is nonzero, so the quotient $1/x$ is represented by an element of $k[x]_x$. It is not regular at $0$: if $1/x$ belonged to $\mathcal O_{\mathbb A^1,0}$, then for some $a,b\in k[x]$ with $b(0)\ne 0$ we would have \begin{align*} \frac{1}{x}=\frac{a}{b} \end{align*} in $k(x)$. Clearing denominators gives \begin{align*} b=xa \end{align*} in $k[x]$, and evaluating at $0$ gives $b(0)=0\cdot a(0)=0$, contradicting $b(0)\ne 0$. Hence $1/x$ is regular exactly on $D(x)$ and has a pole at $0$. [/example] The example shows that equality of rational functions must be tested after passing to a common open set, not by comparing a single pair of written formulas. The next theorem is needed to justify that agreement on a nonempty open subset determines a rational function on an irreducible variety. [quotetheorem:9427] [citeproof:9427] The [identity principle](/theorems/3357) is the algebraic-geometric analogue of [analytic continuation](/page/Analytic%20Continuation), but its proof uses irreducibility rather than limits. It lets us define maps and functions by formulas on any convenient dense open subset. Irreducibility cannot be dropped: on a reducible variety with two components, a regular function may vanish on a nonempty open subset of one component without vanishing on the other. The theorem also does not say that agreement at many closed points is enough over an arbitrary finite field; the hypothesis is agreement on a nonempty Zariski open subset. [example: Equality on the Punctured Line] Let $X=\mathbb A_k^1$, so $k[X]=k[x]$, and let $g,h\in k[x]$ agree on $D(x)$. Since \begin{align*} D(x)=\{a\in k:x(a)\ne 0\}=\{a\in k:a\ne 0\}, \end{align*} agreement on $D(x)$ means \begin{align*} g(a)=h(a) \end{align*} for every $a\in k$ with $a\ne 0$. Therefore \begin{align*} (g-h)(a)=g(a)-h(a)=0 \end{align*} for every $a\in D(x)$. The open set $D(x)$ is nonempty, for instance $1\in D(x)$ because $x(1)=1\ne 0$. Since $\mathbb A_k^1$ is irreducible, *Identity Principle on Irreducible Varieties* applies to the regular functions $g$ and $h$, and gives \begin{align*} g=h \end{align*} in $k[x]$. Equivalently, the polynomial $g-h$ vanishes on the nonempty open subset $D(x)$, hence vanishes everywhere on the irreducible variety $\mathbb A_k^1$. This shows that on an irreducible variety, equality on a dense open set is already equality in the coordinate ring. [/example] ## Rational Maps in the Affine Setting Polynomial maps are defined everywhere, but many geometric correspondences are only defined away from exceptional subsets. The next problem is to formalize maps whose coordinate functions are rational functions and whose values are meaningful on a dense open domain. [definition: Rational Map] Let $X \subset \mathbb A_k^n$ and $Y \subset \mathbb A_k^m$ be irreducible affine varieties. A rational map $\Phi: X \dashrightarrow Y$ is an equivalence class of regular maps $\Phi_U: U \to Y$ defined on nonempty open subsets $U \subset X$, where two representatives are equivalent if they agree on a nonempty open subset of their common domain. [/definition] The dashed arrow records that the map may be undefined at some points. The open-set definition is intrinsic, but it is not yet convenient for calculations: a formula for a map into affine space is a tuple of rational functions, and a formula for a map into a closed subvariety must also satisfy the target's defining equations. The obstruction is that arbitrary rational coordinate functions may fail those equations in $k(X)$. [quotetheorem:9428] [citeproof:9428] This coordinate test makes rational maps computable, but it also records a necessary obstruction: arbitrary rational coordinate functions define a rational map to affine space, not automatically to a closed subvariety. For example, the pair $(t,0)$ on $X=\mathbb A_k^1$ does not define a rational map to the conic $Y=V(y_1^2+y_2^2-1)$, because $t^2+0^2-1$ is not zero in $k(t)$. More generally, a pair $(\varphi_1,\varphi_2)$ gives a rational map into this conic only if $\varphi_1^2+\varphi_2^2=1$ in $k(X)$. The affine target hypothesis is also part of the statement: a rational map to projective space is not encoded by an ordinary tuple of functions, since homogeneous coordinates are defined only up to a common nonzero scalar; for instance, $(1,t)$ and $(2,2t)$ describe the same map to $\mathbb P_k^1$ but are different affine tuples. The irreducibility hypothesis is doing work as well: for reducible $X$, there is no single function field $k(X)$ controlling all components, and agreement on a nonempty open subset of one component need not determine behaviour on another component. The theorem also does not specify the largest domain of definition; it guarantees some nonempty open subset where all coordinate functions are regular and the equations of $Y$ hold. This contravariant coordinate viewpoint raises the next question: when should two varieties be considered the same after discarding lower-dimensional exceptional sets? Rational maps already identify formulas that agree on a dense open set, but a one-way rational map is not enough to say that two varieties have the same generic geometry. The right equivalence relation should require rational maps in both directions, with the composites recovering the original points wherever all expressions involved are defined. [definition: Birational Equivalence] Let $X$ and $Y$ be irreducible affine varieties. A rational map $\Phi:X\dashrightarrow Y$ is a birational equivalence if there is a rational map $\Psi:Y\dashrightarrow X$ such that $\Psi\circ\Phi$ and $\Phi\circ\Psi$ agree with the identity maps on nonempty open subsets. [/definition] In that case $X$ and $Y$ are called birational. Birational equivalence treats varieties as the same if they contain isomorphic dense open pieces. The remaining problem is to recognize this equivalence without explicitly finding matching open subsets. Since rational functions are already insensitive to deleting proper closed subsets, the function field is the natural invariant of the generic geometry. In the affine setting used here, the working criterion is that two irreducible affine varieties are birational exactly when their function fields are isomorphic as fields over $k$. One direction comes from pulling rational functions through a birational equivalence; the other direction is read on coordinate generators, whose images become rational functions satisfying the same quotient relations on suitable dense open subsets. This criterion explains the role of rational parametrisations: to parametrize a variety by affine space is to show that its function field is a purely transcendental field. The requirement that the isomorphism fix $k$ is part of birational equivalence over the base field; a field isomorphism that moves the embedded copy of $k$ would compare the varieties only after changing the base-field structure. The irreducibility assumption is essential here because $k[X]$ must be an integral domain before $\operatorname{Frac}(k[X])$ is available; a reducible variety has componentwise generic functions rather than one fraction field of its coordinate ring. The invariant is deliberately insensitive to closed exceptional subsets, so it cannot detect singularities or missing finitely many points. For instance, a smooth conic with a rational point and the affine line may be birational even though they are not isomorphic as affine varieties. The first major illustration is the classical parametrisation of a conic by lines through a known rational point. [example: Stereographic Parametrization of a Conic] Let $C\subset \mathbb A_k^2$ be the conic $x^2+y^2=1$, assume $\operatorname{char}(k)\ne 2$, and take $P=(-1,0)\in C$. A line through $P$ with slope $t$ has equation $y=t(x+1)$. Substituting this equation into $x^2+y^2=1$ gives \begin{align*} x^2+t^2(x+1)^2=1. \end{align*} Expanding the square gives \begin{align*} x^2+t^2(x^2+2x+1)-1=(1+t^2)x^2+2t^2x+(t^2-1). \end{align*} The point $P$ corresponds to $x=-1$, and the polynomial factors as \begin{align*} (1+t^2)x^2+2t^2x+(t^2-1)=(x+1)((1+t^2)x+t^2-1). \end{align*} Thus, away from the known intersection $x=-1$, the second intersection satisfies \begin{align*} (1+t^2)x+t^2-1=0. \end{align*} On the distinguished open $D(1+t^2)\subset \mathbb A_k^1$, this gives \begin{align*} x=\frac{1-t^2}{1+t^2}. \end{align*} Then \begin{align*} x+1=\frac{1-t^2}{1+t^2}+1=\frac{2}{1+t^2}. \end{align*} Since $y=t(x+1)$, the corresponding $y$-coordinate is \begin{align*} y=t\cdot \frac{2}{1+t^2}=\frac{2t}{1+t^2}. \end{align*} These formulas define a rational map $\mathbb A_k^1\dashrightarrow C$: \begin{align*} t\longmapsto \left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right). \end{align*} To verify that the image lies on $C$, compute \begin{align*} \left(\frac{1-t^2}{1+t^2}\right)^2+\left(\frac{2t}{1+t^2}\right)^2=\frac{(1-t^2)^2+4t^2}{(1+t^2)^2}. \end{align*} The numerator expands as \begin{align*} (1-t^2)^2+4t^2=1-2t^2+t^4+4t^2=1+2t^2+t^4=(1+t^2)^2. \end{align*} Therefore the sum is $1$ wherever $1+t^2\ne 0$. On the open subset of $C$ where $x\ne -1$, define \begin{align*} t=\frac{y}{x+1}. \end{align*} For the parametrized point, we already computed $x+1=2/(1+t^2)$, so \begin{align*} \frac{y}{x+1}=\frac{2t/(1+t^2)}{2/(1+t^2)}=t, \end{align*} using $2\ne 0$ because $\operatorname{char}(k)\ne 2$. Conversely, if $(x,y)\in C$ and $x\ne -1$, set $t=y/(x+1)$. Then $y=t(x+1)$, so the same factorization gives \begin{align*} (x+1)((1+t^2)x+t^2-1)=0. \end{align*} Since $x+1\ne 0$, we get $(1+t^2)x+t^2-1=0$. If $1+t^2=0$, this equation would give $-2=0$, impossible because $\operatorname{char}(k)\ne 2$. Hence $1+t^2\ne 0$, and \begin{align*} x=\frac{1-t^2}{1+t^2}. \end{align*} Then $y=t(x+1)=2t/(1+t^2)$ as above. Thus the parametrization and $t=y/(x+1)$ are inverse on their common domains. Since $k$ is algebraically closed, the polynomial $1+t^2$ has a zero, so this parametrization is rational rather than regular on all of $\mathbb A_k^1$. [/example] The conic example shows how a smooth curve with a chosen point can become rational by projecting from that point. Singular curves provide another source of rational parametrisations, where projection from the singular point lowers the intersection multiplicity left to solve. [example: Rational Parametrization of the Nodal Cubic] Let $C\subset \mathbb A_k^2$ be the cubic curve \begin{align*} y^2=x^2(x+1) \end{align*} over a field with $\operatorname{char}(k)\ne 2$. The point $(0,0)$ lies on $C$ because $0^2=0^2(0+1)$, and a line through this point with slope $t$ has equation \begin{align*} y=tx. \end{align*} Substituting $y=tx$ into the defining equation of $C$ gives \begin{align*} (tx)^2=x^2(x+1). \end{align*} Since $(tx)^2=t^2x^2$, this is \begin{align*} t^2x^2=x^2(x+1). \end{align*} Moving all terms to one side and factoring $x^2$ gives \begin{align*} 0=x^2(x+1)-t^2x^2=x^2(x+1-t^2). \end{align*} Thus the intersection with the line consists of the node $x=0$ and the remaining solution \begin{align*} x+1-t^2=0. \end{align*} Solving for $x$ gives \begin{align*} x=t^2-1. \end{align*} Then the line equation gives \begin{align*} y=tx=t(t^2-1). \end{align*} So the parametrizing formula is \begin{align*} t\longmapsto (t^2-1,\ t(t^2-1)). \end{align*} To check that this point lies on $C$, compute the left-hand side: \begin{align*} y^2=\bigl(t(t^2-1)\bigr)^2=t^2(t^2-1)^2. \end{align*} The right-hand side is \begin{align*} x^2(x+1)=(t^2-1)^2\bigl((t^2-1)+1\bigr)=(t^2-1)^2t^2. \end{align*} These two expressions are equal, so the formula gives a regular map $\mathbb A_k^1\to C$. On the open subset $D_C(x)\subset C$, define \begin{align*} t=\frac{y}{x}. \end{align*} If $(x,y)\in D_C(x)$ and $t=y/x$, then $y=tx$. Substituting into the equation of $C$ gives \begin{align*} t^2x^2=x^2(x+1). \end{align*} Since $x\ne 0$, division by $x^2$ gives \begin{align*} t^2=x+1. \end{align*} Hence \begin{align*} x=t^2-1 \end{align*} and then \begin{align*} y=tx=t(t^2-1). \end{align*} Conversely, for a parametrized point with $t^2-1\ne 0$, its $x$-coordinate is nonzero and \begin{align*} \frac{y}{x}=\frac{t(t^2-1)}{t^2-1}=t. \end{align*} Thus the parametrization and the rational function $y/x$ are inverse on the open sets where both expressions are defined, so the nodal cubic is birational to the affine line. [/example] These parametrisations are not merely computational tricks. They show that the conic and the nodal cubic are birational to the affine line, even though their global geometry and singular behaviour differ. Localizing affine varieties reveals that geometry is often controlled by functions defined only where denominators do not vanish. Rational parametrizations then show how familiar curves can be compared birationally even when their global shapes differ. This local viewpoint motivates projective geometry, where one adds the missing points at infinity instead of discarding them. # 5. Projective Space and Homogeneous Geometry After distinguished opens and rational functions showed how affine geometry behaves locally, projective geometry enters because affine varieties still miss intersections that should be present after allowing directions at infinity. The projective viewpoint replaces points of affine space by lines through the origin in one higher-dimensional vector space, so equations must be insensitive to rescaling. This chapter builds the algebraic language for that change: homogeneous coordinates, homogeneous ideals, projective coordinate rings, and affine charts. It assumes the affine theory from the previous chapters, especially affine algebraic sets, ideals, radicals, coordinate rings, and the affine Nullstellensatz. The main payoff is that projective varieties can be studied locally by affine calculations while still retaining global geometric features such as points at infinity. ## From Affine Space to Projective Space What should be added to affine space so that parallel affine lines meet and polynomial equations behave well under change of scale? The guiding construction is to regard a point as a one-dimensional linear subspace of a vector space. This forces coordinates to be meaningful only up to simultaneous nonzero scalar multiplication. [definition: Projective Space] Let $k$ be a field and let $n \ge 0$. Projective $n$-space over $k$ is the set \begin{align*} \mathbb P_k^n := (k^{n+1} \setminus \{0\})/\sim, \end{align*} where $a \sim b$ if there exists $c \in k^\times$ such that $a = cb$. [/definition] A point of $\mathbb P_k^n$ is written $[a_0:\cdots:a_n]$, and the brackets record that only the ratio of the coordinates matters. This is the first place where projective geometry differs from affine geometry: individual coordinates are not functions on projective space, but ratios of coordinates are meaningful wherever the denominator is nonzero. The following example motivates the need for equations that respect this ratio notation. [example: Projective Line] The projective line $\mathbb P_k^1$ consists of equivalence classes $[a_0:a_1]$ with $(a_0,a_1)\ne(0,0)$, where \begin{align*} [a_0:a_1]=[ca_0:ca_1]\quad\text{for every }c\in k^\times. \end{align*} If $a_0\ne 0$, then $a_0^{-1}\in k^\times$, so rescaling by $a_0^{-1}$ gives \begin{align*} [a_0:a_1]=[a_0^{-1}a_0:a_0^{-1}a_1]=[1:a_1/a_0]. \end{align*} Thus every point with first coordinate nonzero has a unique form $[1:t]$, because $[1:t]=[1:t']$ implies $(1,t)=c(1,t')$, hence $c=1$ and $t=t'$. The remaining points have $a_0=0$. Since $(a_0,a_1)\ne(0,0)$, this forces $a_1\ne0$, and rescaling by $a_1^{-1}$ gives \begin{align*} [0:a_1]=[0:a_1^{-1}a_1]=[0:1]. \end{align*} So $\mathbb P_k^1$ is the affine chart $\{[1:t]:t\in k\}$, which is identified with $\mathbb A_k^1$, together with the single extra point $[0:1]$ at infinity. [/example] The example shows why homogeneous coordinates are economical: one notation contains both affine points and points at infinity. The price of this notation is that a projective point has many representatives. An equation in the coordinates is meaningful only if replacing $(a_0,\dots,a_n)$ by a nonzero scalar multiple does not change whether the equation vanishes. This rescaling obstruction singles out polynomials whose monomials all have the same total degree. [definition: Homogeneous Polynomial] Let $S=k[x_0,\dots,x_n]$. A polynomial $F \in S$ is homogeneous of degree $d$ if every monomial appearing in $F$ has total degree $d$. [/definition] If $F$ is homogeneous of degree $d$, then $F(ca)=c^dF(a)$ for all $c \in k$. Hence the condition $F(a)=0$ is well-defined on the projective point $[a]$. This motivates the definition of projective algebraic sets as common zero loci of homogeneous equations. [definition: Projective Algebraic Set] Let $S=k[x_0,\dots,x_n]$ and let $T \subset S$ be a set of homogeneous polynomials. The projective algebraic set defined by $T$ is \begin{align*} V_+(T) := \{[a_0:\cdots:a_n] \in \mathbb P_k^n : F(a_0,\dots,a_n)=0 \text{ for all } F \in T\}. \end{align*} [/definition] The plus sign reminds us that only homogeneous equations are being used. The same notation also works for a homogeneous ideal, since the vanishing set depends on all homogeneous equations contained in the ideal. The next example shows how projective equations add a boundary to an affine chart. [example: The Line at Infinity in the Projective Plane] In $\mathbb P_k^2$ with homogeneous coordinates $[x_0:x_1:x_2]$, the equation $x_0=0$ is homogeneous of degree $1$, so its zero condition is unchanged under rescaling: for $c\in k^\times$, \begin{align*} (cx_0)=0 \quad\Longleftrightarrow\quad c x_0=0 \quad\Longleftrightarrow\quad x_0=0. \end{align*} Thus $V_+(x_0)$ is the set of projective points $[0:a_1:a_2]$ with $(a_1,a_2)\ne(0,0)$. On the chart $U_0=\{x_0\ne0\}$, every point has a representative with first coordinate $1$: if $x_0\ne0$, then \begin{align*} [x_0:x_1:x_2]=[1:x_1/x_0:x_2/x_0]. \end{align*} This gives the affine coordinates \begin{align*} u=x_1/x_0,\qquad v=x_2/x_0. \end{align*} Conversely, each affine point $(u,v)\in\mathbb A_k^2$ gives the projective point $[1:u:v]\in U_0$, and applying the coordinate map gives \begin{align*} [1:u:v]\longmapsto (u,v). \end{align*} The two constructions are inverse because \begin{align*} [1:x_1/x_0:x_2/x_0]=[x_0:x_1:x_2]. \end{align*} A point lies outside this affine chart exactly when $x_0=0$. Therefore $V_+(x_0)$ is precisely the projective boundary missing from the copy of $\mathbb A_k^2$ given by $U_0$, and this projective line is called the line at infinity. [/example] The line at infinity is not mysterious when viewed before quotienting by scalar multiplication: it is made from ordinary affine lines through the origin in $k^3$. To connect projective equations back to the affine theory from earlier chapters, we record all representatives of all projective points at once; this motivates the cone construction. [definition: Affine Cone] Let $X \subset \mathbb P_k^n$ be a projective algebraic set. The affine cone over $X$ is \begin{align*} C(X) := \{a \in k^{n+1} : a=0 \text{ or } [a] \in X\} \subset \mathbb A_k^{n+1}. \end{align*} [/definition] The cone records all representatives of all projective points, together with the origin. It is stable under scalar multiplication, and homogeneous polynomials vanish on $X$ exactly when they vanish on $C(X)$. This gives a practical way to visualise projective equations using affine geometry. [example: A Projective Conic] Let $k$ have characteristic not equal to $2$, and consider the projective conic \begin{align*} X=V_+(x_0^2+x_1^2-x_2^2)\subset \mathbb P_k^2. \end{align*} The affine cone $C(X)\subset \mathbb A_k^3$ consists of the origin together with all triples $(a_0,a_1,a_2)$ representing points of $X$, so it is cut out by the same homogeneous equation \begin{align*} a_0^2+a_1^2-a_2^2=0. \end{align*} On the chart $U_2=\{x_2\ne0\}$, every point has a representative with third coordinate $1$: \begin{align*} [x_0:x_1:x_2]=[x_0/x_2:x_1/x_2:1]. \end{align*} Writing \begin{align*} u=\frac{x_0}{x_2},\qquad v=\frac{x_1}{x_2}, \end{align*} and substituting $x_0=ux_2$ and $x_1=vx_2$ into the homogeneous equation gives \begin{align*} (ux_2)^2+(vx_2)^2-x_2^2=0. \end{align*} Expanding the squares gives \begin{align*} u^2x_2^2+v^2x_2^2-x_2^2=0. \end{align*} Factoring out $x_2^2$ gives \begin{align*} x_2^2(u^2+v^2-1)=0. \end{align*} Since $x_2\ne0$ on $U_2$, also $x_2^2\ne0$, so this is equivalent to \begin{align*} u^2+v^2-1=0. \end{align*} Thus the part of $X$ in the affine chart $U_2$ is the affine conic $u^2+v^2=1$. The points of $X$ not seen in this chart are exactly those with $x_2=0$, and they satisfy \begin{align*} x_0^2+x_1^2=0. \end{align*} These are precisely the points at infinity of the affine conic obtained from the chart $x_2\ne0$. [/example] The cone example also warns us that one affine point has no projective counterpart: the origin. Since all homogeneous coordinates cannot vanish at once in projective space, we need a name for the ideal detecting the deleted origin. [definition: Irrelevant Ideal] In the graded ring $S=k[x_0,\dots,x_n]$, the irrelevant ideal is \begin{align*} S_+ := (x_0,\dots,x_n). \end{align*} [/definition] This ideal cuts out only the origin in affine $(n+1)$-space, and the origin has been removed before passing to projective space. Thus $V_+(S_+)=\varnothing$, a fact that will modify the Nullstellensatz in the projective setting. ## Homogeneous Ideals and the Projective Nullstellensatz Which ideals should encode projective algebraic sets? Since projective equations must be homogeneous, the algebra must remember the grading on $k[x_0,\dots,x_n]$. The correct ideals are those generated degree by degree. [definition: Homogeneous Ideal] Let $S=k[x_0,\dots,x_n]$ with its standard grading $S=\bigoplus_{d\ge 0}S_d$. An ideal $I \trianglelefteq S$ is homogeneous if whenever $F\in I$ and $F=\sum_d F_d$ is its decomposition into homogeneous parts, each $F_d$ lies in $I$. [/definition] Homogeneous ideals are exactly the ideals generated by homogeneous polynomials. This matters because the quotient by such an ideal inherits a grading. This motivates the definition of the projective coordinate ring, which stores the homogeneous forms on a projective variety degree by degree. [definition: Projective Coordinate Ring] Let $X\subset \mathbb P_k^n$ be a projective algebraic set. Its homogeneous vanishing ideal is \begin{align*} I_+(X):=\{F\in k[x_0,\dots,x_n] : \text{every homogeneous component of } F \text{ vanishes on } X\}. \end{align*} The projective coordinate ring of $X$ is the graded ring \begin{align*} S(X):=k[x_0,\dots,x_n]/I_+(X). \end{align*} [/definition] Unlike the affine coordinate ring, this graded ring is not best interpreted as a ring of functions on $X$: a homogeneous element of positive degree changes under rescaling. Instead it packages all homogeneous forms on the cone, and degree-zero ratios on suitable open sets become regular functions on affine charts. The twisted cubic gives a standard test case where a parametrised projective curve is recovered from homogeneous equations. [example: Coordinate Ring of the Twisted Cubic] The twisted cubic is the projective curve \begin{align*} C=\{[s^3:s^2t:st^2:t^3] : [s:t]\in \mathbb P_k^1\}\subset \mathbb P_k^3. \end{align*} Let \begin{align*} \varphi:k[x_0,x_1,x_2,x_3]\longrightarrow k[s,t] \end{align*} be the graded homomorphism defined by \begin{align*} \varphi(x_0)=s^3,\qquad \varphi(x_1)=s^2t,\qquad \varphi(x_2)=st^2,\qquad \varphi(x_3)=t^3. \end{align*} The three quadrics \begin{align*} Q_1=x_0x_2-x_1^2,\qquad Q_2=x_0x_3-x_1x_2,\qquad Q_3=x_1x_3-x_2^2 \end{align*} lie in $\ker(\varphi)$ because \begin{align*} \varphi(Q_1)=s^3\cdot st^2-(s^2t)^2=s^4t^2-s^4t^2=0, \end{align*} \begin{align*} \varphi(Q_2)=s^3\cdot t^3-(s^2t)(st^2)=s^3t^3-s^3t^3=0, \end{align*} and \begin{align*} \varphi(Q_3)=s^2t\cdot t^3-(st^2)^2=s^2t^4-s^2t^4=0. \end{align*} To see that these generate the kernel, reduce monomials modulo the relations \begin{align*} x_1^2\equiv x_0x_2,\qquad x_1x_2\equiv x_0x_3,\qquad x_2^2\equiv x_1x_3. \end{align*} Repeatedly replacing $x_1^2$, $x_1x_2$, and $x_2^2$ shows that every monomial is congruent modulo $(Q_1,Q_2,Q_3)$ to a linear combination of monomials of the forms \begin{align*} x_0^a x_3^b,\qquad x_0^a x_1x_3^b,\qquad x_0^a x_2x_3^b. \end{align*} Their images are respectively \begin{align*} s^{3a}t^{3b},\qquad s^{3a+2}t^{3b+1},\qquad s^{3a+1}t^{3b+2}. \end{align*} These monomials in $k[s,t]$ have distinct exponent pairs, so they are $k$-linearly independent. Hence no nonzero reduced linear combination maps to zero, and therefore \begin{align*} \ker(\varphi)=(x_0x_2-x_1^2,\ x_0x_3-x_1x_2,\ x_1x_3-x_2^2). \end{align*} Thus the projective coordinate ring of the twisted cubic is \begin{align*} k[x_0,x_1,x_2,x_3]/(x_0x_2-x_1^2,\ x_0x_3-x_1x_2,\ x_1x_3-x_2^2). \end{align*} Over an algebraically closed field this is the homogeneous vanishing ideal of the parametrised projective curve, while over an arbitrary field it is the scheme-theoretic image ideal of the parametrisation. The equations say exactly that the four projective coordinates behave like consecutive degree-three monomials $s^3,s^2t,st^2,t^3$. [/example] The affine Nullstellensatz said that radical ideals over an algebraically closed field are exactly vanishing ideals of affine algebraic sets. The projective version is similar, but the irrelevant ideal must be excluded because it has empty projective vanishing set. There is also a subtler failure of the naive dictionary: in $k[x_0,x_1]$, the ideals $(x_0)$ and $(x_0^2,x_0x_1)$ define the same projective zero set in $\mathbb P_k^1$, because the second ideal differs from the first only by information supported at the deleted origin. Saturation is the algebraic operation that removes this invisible information. [quotetheorem:2135] [citeproof:2135] This theorem explains why the projective ideal-variety correspondence is not phrased using all radical homogeneous ideals. The hypothesis that $k$ is algebraically closed is inherited from the affine Nullstellensatz; over a non-algebraically closed field, the zero set over $k$-rational projective points can miss equations detected after extending scalars. The saturation by $S_+$ is also essential: for example, $(x_0)$ and $(x_0^2,x_0x_1)$ have the same projective vanishing set in $\mathbb P_k^1$, but only the first is saturated. The theorem does not say that every homogeneous ideal is determined by its projective zero set; it says that the zero set determines the radical after removing information supported at the deleted origin. This distinction is the reason projective coordinate rings are graded objects rather than ordinary rings of functions, and it prepares the ideal-variety correspondence below. [quotetheorem:9429] [citeproof:9429] The correspondence is the algebraic engine behind projective geometry in the course, but it classifies only reduced projective algebraic sets, not embedded components or nilpotent scheme structure. Each hypothesis rules out a specific failure. Radicality is needed because $(x_0^2)$ and $(x_0)$ define the same point $[0:1]$ in $\mathbb P_k^1$ but have different nilpotent information. Saturation is needed because $(x_0^2,x_0x_1)\subset k[x_0,x_1]$ and $(x_0)$ define the same projective point, while the former still contains information concentrated at the irrelevant ideal. Properness, or the separate empty-set convention, is needed because any ideal whose saturation is the whole ring has no projective points. The algebraically closed hypothesis is also necessary: over $\mathbb R$, the homogeneous ideal $(x_0^2+x_1^2)\subset \mathbb R[x_0,x_1]$ has no real projective zero in $\mathbb P_{\mathbb R}^1$, although after extending scalars to $\mathbb C$ it cuts out two projective points. In later computations, this is why parametrised projective varieties and projective closures must be checked after extending the ground field when needed, taking radicals, and saturating. [example: Empty Vanishing from the Irrelevant Ideal] In $\mathbb P_k^2$, consider the homogeneous ideal \begin{align*} I=(x_0,x_1,x_2)\subset k[x_0,x_1,x_2]. \end{align*} A projective zero of $I$ would be a point $[a_0:a_1:a_2]\in\mathbb P_k^2$ such that each generator vanishes at the chosen representative. Thus it would have to satisfy \begin{align*} x_0(a_0,a_1,a_2)=a_0=0,\qquad x_1(a_0,a_1,a_2)=a_1=0,\qquad x_2(a_0,a_1,a_2)=a_2=0. \end{align*} These three equalities force \begin{align*} (a_0,a_1,a_2)=(0,0,0). \end{align*} But homogeneous coordinates in $\mathbb P_k^2$ are defined only for nonzero triples, so no such projective point exists. Hence \begin{align*} V_+(x_0,x_1,x_2)=\varnothing. \end{align*} The ideal $I$ is also radical: if $f^m\in (x_0,x_1,x_2)$ for some $m\ge 1$, then evaluating at the origin gives $f(0,0,0)^m=0$, so $f(0,0,0)=0$, which means the constant term of $f$ is zero and therefore $f\in (x_0,x_1,x_2)$. Thus \begin{align*} \sqrt{(x_0,x_1,x_2)}=(x_0,x_1,x_2). \end{align*} So radicality alone is not enough in projective geometry: this radical homogeneous ideal defines the empty projective set because it cuts out only the deleted origin in affine $3$-space. [/example] ## Affine Charts and Local Calculations How can computations with projective varieties be made if projective coordinates are only ratios? The answer is to cover projective space by standard affine charts. On each chart one coordinate is declared nonzero and used to normalise the others. [definition: Standard Affine Chart] For $0\le i\le n$, the $i$-th standard affine chart in $\mathbb P_k^n$ is \begin{align*} U_i := \{[x_0:\cdots:x_n]\in \mathbb P_k^n : x_i\ne 0\}. \end{align*} [/definition] The chart $U_i$ is identified with $\mathbb A_k^n$ by setting $x_i=1$ and using the remaining ratios as affine coordinates. The issue is that projective space has no single global normalization: a point with $x_i=0$ cannot be described by ratios using $x_i$ as denominator. Any affine description must therefore be local, and the local pieces must still cover every projective point because at least one homogeneous coordinate is nonzero. The following result packages this local affine structure precisely. [quotetheorem:9430] [citeproof:9430] The theorem justifies the method used throughout projective computations: work chart by chart, then check compatibility on overlaps. Each condition in the chart construction has a role. The requirement $x_i\ne0$ is essential because the ratios \begin{align*} \frac{x_j}{x_i} \end{align*} are undefined on the hyperplane $x_i=0$; for instance, the point $[0:1]\in\mathbb P_k^1$ is invisible in the chart $U_0$ but appears in $U_1$. The exclusion of the zero vector in the definition of projective space is also necessary: if $[0:0]$ were allowed in $\mathbb P_k^1$, it would lie in no standard chart and no normalisation would be available. Finally, the field hypothesis is what lets us divide by a nonzero coordinate; over $\mathbb Z/6\mathbb Z$, for example, $2$ is nonzero but has no inverse, so the same ratio formula does not define an affine chart in this elementary set-theoretic way. The theorem does not identify projective space with a single affine space, nor does it make homogeneous coordinates into global functions. It says instead that projective geometry is locally affine, and the global information lies in how the affine charts overlap. To carry this method out algebraically, we need a formal operation that converts a homogeneous equation into its affine equation on a chosen chart. Without such an operation, a homogeneous equation is not yet an affine equation: it still remembers an arbitrary scale factor until one nonzero coordinate has been fixed. [definition: Dehomogenisation] Let $S=k[x_0,\dots,x_n]$ and fix $0\le i\le n$. The dehomogenisation map on the chart $U_i$ is the $k$-algebra homomorphism \begin{align*} \delta_i:S\longrightarrow k[x_0,\dots,\widehat{x_i},\dots,x_n]. \end{align*} It is determined on generators by $\delta_i(x_i)=1$ and, for $j\ne i$, \begin{align*} \delta_i(x_j)=x_j. \end{align*} For a homogeneous form $F\in S_d$, its dehomogenisation on $U_i$ is $\delta_i(F)$. [/definition] Dehomogenisation turns projective equations into affine equations on a chosen chart. The reverse operation, homogenisation, embeds affine calculations back into projective space by adding a new coordinate and restoring equal total degree. [example: Projective Closure of an Affine Plane Curve] Let $Y\subset \mathbb A_k^2$ be the affine curve with equation $y=x^2$. In the standard chart $U_0\subset \mathbb P_k^2$, write \begin{align*} u=\frac{x_1}{x_0},\qquad v=\frac{x_2}{x_0}. \end{align*} The affine equation $v=u^2$ is equivalently $v-u^2=0$. Substituting the chart coordinates gives \begin{align*} \frac{x_2}{x_0}-\left(\frac{x_1}{x_0}\right)^2=0. \end{align*} Since $x_0\ne0$ on $U_0$, multiplying by $x_0^2$ preserves the zero set on this chart: \begin{align*} x_0^2\left(\frac{x_2}{x_0}-\frac{x_1^2}{x_0^2}\right)=0. \end{align*} Distributing the factor gives \begin{align*} x_0x_2-x_1^2=0. \end{align*} Thus the homogeneous equation cutting out the projective closure is $x_0x_2-x_1^2=0$, so \begin{align*} \overline{Y}=V_+(x_0x_2-x_1^2)\subset \mathbb P_k^2. \end{align*} To find the points added at infinity, impose $x_0=0$ together with the projective equation: \begin{align*} 0\cdot x_2-x_1^2=0. \end{align*} This is \begin{align*} -x_1^2=0. \end{align*} Since $k$ is a field, $x_1^2=0$ implies $x_1=0$. A projective point cannot have all three homogeneous coordinates zero, so $x_2\ne0$, and rescaling by $x_2^{-1}$ gives \begin{align*} [0:0:x_2]=[0:0:1]. \end{align*} Therefore the affine parabola $y=x^2$ acquires exactly one point at infinity in its projective closure, namely $[0:0:1]$. [/example] The same procedure lets us compare different affine descriptions of one projective variety. On chart overlaps, transition functions are rational expressions whose denominators are nonzero on the overlap. [example: Gluing the Charts of the Projective Line] The projective line has two standard affine charts, \begin{align*} U_0=\{[x_0:x_1]\in\mathbb P_k^1:x_0\ne0\} \end{align*} and \begin{align*} U_1=\{[x_0:x_1]\in\mathbb P_k^1:x_1\ne0\}. \end{align*} On $U_0$, normalising by $x_0$ sends \begin{align*} [x_0:x_1]=[1:x_1/x_0], \end{align*} so the affine coordinate is $t=x_1/x_0$. On $U_1$, normalising by $x_1$ sends \begin{align*} [x_0:x_1]=[x_0/x_1:1], \end{align*} so the affine coordinate is $s=x_0/x_1$. A point lies in the overlap $U_0\cap U_1$ exactly when $x_0\ne0$ and $x_1\ne0$. Therefore both $t=x_1/x_0$ and $s=x_0/x_1$ are defined there, and $t\ne0$. Their product is \begin{align*} st=\frac{x_0}{x_1}\cdot\frac{x_1}{x_0}=\frac{x_0x_1}{x_1x_0}=1, \end{align*} because $x_0$ and $x_1$ are nonzero in the field $k$. Hence \begin{align*} s=\frac{1}{t}. \end{align*} Similarly, \begin{align*} t=\frac{1}{s}. \end{align*} Thus the two affine charts are glued only where their affine coordinates are nonzero, and the transition map between them is inversion on the punctured affine line. [/example] For projective varieties, the chart equations are obtained by applying the same dehomogenisation to the defining homogeneous ideal. This reduces many questions about projective varieties to affine algebra, but the global projective object remembers how the affine pieces are glued. [example: Affine Charts of a Projective Conic] Consider $X=V_+(x_0x_2-x_1^2)\subset \mathbb P_k^2$. On the chart $U_0=\{x_0\ne0\}$, define \begin{align*} u=\frac{x_1}{x_0},\qquad v=\frac{x_2}{x_0}. \end{align*} Then $x_1=ux_0$ and $x_2=vx_0$, so substituting into the homogeneous equation gives \begin{align*} x_0x_2-x_1^2=x_0(vx_0)-(ux_0)^2. \end{align*} Expanding the right-hand side gives \begin{align*} x_0(vx_0)-(ux_0)^2=vx_0^2-u^2x_0^2. \end{align*} Factoring out $x_0^2$ gives \begin{align*} vx_0^2-u^2x_0^2=x_0^2(v-u^2). \end{align*} Since $x_0\ne0$ on $U_0$, also $x_0^2\ne0$, so the equation $x_0x_2-x_1^2=0$ is equivalent on this chart to \begin{align*} v-u^2=0. \end{align*} Thus $X\cap U_0$ is the affine parabola $v=u^2$. On the chart $U_2=\{x_2\ne0\}$, define \begin{align*} r=\frac{x_0}{x_2},\qquad q=\frac{x_1}{x_2}. \end{align*} Then $x_0=rx_2$ and $x_1=qx_2$, so \begin{align*} x_0x_2-x_1^2=(rx_2)x_2-(qx_2)^2. \end{align*} Expanding gives \begin{align*} (rx_2)x_2-(qx_2)^2=rx_2^2-q^2x_2^2. \end{align*} Factoring out $x_2^2$ gives \begin{align*} rx_2^2-q^2x_2^2=x_2^2(r-q^2). \end{align*} Since $x_2\ne0$ on $U_2$, this is equivalent to \begin{align*} r-q^2=0. \end{align*} Thus $X\cap U_2$ is the affine parabola $r=q^2$. On the overlap $U_0\cap U_2$, both $x_0$ and $x_2$ are nonzero. Therefore $v=x_2/x_0$ is nonzero and \begin{align*} r=\frac{x_0}{x_2}=\frac{1}{x_2/x_0}=\frac{1}{v}. \end{align*} Similarly, \begin{align*} q=\frac{x_1}{x_2}=\frac{x_1/x_0}{x_2/x_0}=\frac{u}{v}. \end{align*} Using the equation $v=u^2$ on $X\cap U_0$, the overlap also satisfies \begin{align*} q=\frac{u}{u^2}=\frac{1}{u} \end{align*} whenever $u\ne0$. These formulas show explicitly how the two affine parabolas are glued: the nonzero coordinate ratio is inverted, so the two chart descriptions are opposite affine views of the same projective conic. [/example] The chart method is also the bridge to later parts of the course. Dimension, tangent spaces, regular maps, and intersection computations for projective varieties are usually defined or checked locally on affine charts, then assembled into statements independent of the chosen chart. [remark: Why Projective Geometry is Still Algebraic] Projective coordinates are not ordinary functions, but homogeneous polynomials and degree-zero ratios provide enough algebra to recover the geometry. The graded coordinate ring records global homogeneous equations, while affine charts recover the local coordinate rings familiar from affine varieties. This is the classical precursor of the scheme-theoretic habit of studying spaces by local rings and gluing data. [/remark] Projective space completes affine geometry by adding directions at infinity and replacing ordinary coordinates with homogeneous ones. The homogeneous coordinate ring packages global equations, while affine charts recover the local rings already familiar from the affine case. That global-local balance is the starting point for projective closure and elimination. # 6. Projective Closure and Elimination This chapter explains how affine equations are completed into projective equations and how variables are removed from polynomial systems. It uses the affine Nullstellensatz from Chapter 2 and the homogeneous projective dictionary from Chapter 5. The guiding theme is that projective space records limiting directions that affine space forgets, while elimination gives algebraic equations for shadows of varieties under projection. These two operations meet in classical projective geometry: closing an affine curve adds its points at infinity, and projecting a projective variety is controlled by the equations obtained after eliminating the forgotten coordinates. ## Homogenization and Dehomogenization Affine equations live in $k[x_1,\dots,x_n]$, while projective equations must be homogeneous in $k[x_0,x_1,\dots,x_n]$. The first problem is to pass between these two languages without changing the geometry on the affine chart $D_+(x_0)$ where $x_0 \ne 0$. [definition: Homogenization of a Polynomial] For each $d\ge 0$, homogenization with respect to $x_0$ is the map \begin{align*} (-)^h:\{f\in k[x_1,\dots,x_n]:\deg f\le d\}\longrightarrow k[x_0,x_1,\dots,x_n]_d. \end{align*} It sends $f$ to $f^h$, where \begin{align*} f^h(x_0,x_1,\dots,x_n)=x_0^d f(x_1/x_0,\dots,x_n/x_0). \end{align*} [/definition] The formula should be read term by term: a monomial of degree $m$ is multiplied by $x_0^{d-m}$ so that all terms have total degree $d$. This is the algebraic operation that lets the affine equation be evaluated on projective coordinates. [example: Homogenizing an Affine Conic] Let $f=y-x^2\in k[x,y]$, and use projective coordinates $[X:Y:Z]$ on $\mathbb P^2$ so that the affine chart $Z\ne 0$ has affine coordinates $x=X/Z$ and $y=Y/Z$. Since $\deg f=2$, homogenizing with respect to $Z$ means multiplying $f(X/Z,Y/Z)$ by $Z^2$: \begin{align*} f^h(X,Y,Z)=Z^2\left(\frac{Y}{Z}-\left(\frac{X}{Z}\right)^2\right). \end{align*} Expanding the two terms separately gives \begin{align*} Z^2\cdot \frac{Y}{Z}=YZ,\qquad Z^2\cdot \frac{X^2}{Z^2}=X^2. \end{align*} Therefore \begin{align*} f^h=YZ-X^2. \end{align*} On the affine chart $Z=1$, this equation becomes \begin{align*} Y\cdot 1-X^2=Y-X^2. \end{align*} Thus the affine parabola $y=x^2$ is represented in projective space by the conic $V(YZ-X^2)\subset\mathbb P^2$, and setting $Z=1$ recovers exactly the original affine equation. [/example] The example shows how an affine equation enters projective space, but it also raises the reverse problem. To compare a projective equation with the affine chart that produced it, we need an operation that reads homogeneous coordinates as affine ratios. [definition: Dehomogenization of a Polynomial] For each $d\ge 0$, dehomogenization on the affine chart $D_+(x_0)$ is the map \begin{align*} (-)^{\mathrm{deh}}:k[x_0,x_1,\dots,x_n]_d\longrightarrow k[x_1,\dots,x_n]. \end{align*} It sends $F$ to $F^{\mathrm{deh}}$, where \begin{align*} F^{\mathrm{deh}}(x_1,\dots,x_n)=F(1,x_1,\dots,x_n). \end{align*} [/definition] Dehomogenization depends on the chosen chart. On $D_+(x_i)$ one sets $x_i=1$ instead, and the remaining coordinates are ratios relative to $x_i$. The next result measures the failure of homogenization and dehomogenization to be inverse operations: the only lost information is a homogeneous factor supported on the omitted hyperplane. [quotetheorem:9431] [citeproof:9431] The theorem explains why homogeneous factors supported on the hyperplane $x_0=0$ disappear when we enter the affine chart. The hypothesis that $F$ be homogeneous is essential: for a nonhomogeneous polynomial such as $x_0+x_1^2$, setting $x_0=1$ gives $1+x_1^2$, and there is no single projective degree in which dehomogenization can be reversed without making an extra choice. The theorem also does not say that a projective equation is determined by its affine chart; multiplying by a power of $x_0$ changes the projective vanishing along the omitted hyperplane while leaving the chart $D_+(x_0)$ unchanged. For varieties defined by several equations, this point becomes more serious: a generating set may not contain all polynomial consequences needed to control behaviour at infinity. The next definition applies homogenization to the ideal itself, rather than only to a chosen list of equations. [definition: Homogenization of an Ideal] Let $I\trianglelefteq k[x_1,\dots,x_n]$. Its homogenization with respect to $x_0$ is the homogeneous ideal \begin{align*} I^h=(f^h : f\in I)\trianglelefteq k[x_0,x_1,\dots,x_n]. \end{align*} [/definition] Although $I^h$ is defined using all polynomials in $I$, a finite computation is possible when $I$ is generated by a Grobner basis for a degree-compatible order. In this chapter the conceptual definition is the one that matters, since it identifies the projective closure without depending on a chosen generating set. [example: Homogenizing an Ideal Is Not Just Homogenizing Any Generators] Let $I=(x,1-xy)\trianglelefteq k[x,y]$. A point $(a,b)\in\mathbb A^2$ lies in $V(I)$ only if $a=0$ and $1-ab=0$. Substituting $a=0$ gives \begin{align*} 1-ab=1-0\cdot b=1, \end{align*} so there is no such point and $V(I)=\varnothing$. The ideal itself is the whole ring, because \begin{align*} xy+(1-xy)=1. \end{align*} Since $x\in I$, multiplying by $y$ gives $xy\in I$, and since $1-xy\in I$, the displayed sum shows $1\in I$. Therefore $I=(1)$, so its homogenization is \begin{align*} I^h=(1^h)=(1), \end{align*} and $V_+(I^h)=\varnothing$. Now compare this with homogenizing only the two displayed generators in $k[Z,X,Y]$. The generator $x$ has homogenization $X$. The generator $1-xy$ has degree $2$, so \begin{align*} (1-xy)^h=Z^2\left(1-\frac{X}{Z}\frac{Y}{Z}\right)=Z^2-XY. \end{align*} Thus the smaller homogeneous ideal obtained from just those generators is \begin{align*} (X,Z^2-XY). \end{align*} At the projective point $[0:0:1]$, we have $Z=0$, $X=0$, and $Y=1$, so \begin{align*} X=0 \end{align*} and \begin{align*} Z^2-XY=0^2-0\cdot 1=0. \end{align*} Hence $[0:0:1]\in V_+(X,Z^2-XY)$, even though the correct projective vanishing set $V_+(I^h)$ is empty. The extra point appears because the displayed generators do not include the polynomial consequence $1\in I$. [/example] This example motivates using the full ideal, or an algebraic procedure that computes the same ideal, when taking closure. The next section turns the algebra into geometry. ## Projective Closure and Points at Infinity The affine chart $D_+(x_0)\subset\mathbb P^n$ identifies with $\mathbb A^n$ by \begin{align*} (a_1,\dots,a_n)&\longmapsto [1:a_1:\dots:a_n]. \end{align*} The main question is how to describe the smallest projective variety containing a given affine variety inside this chart. [definition: Projective Closure of an Affine Variety] Let $X=V(I)\subset\mathbb A^n$ be an affine variety over an algebraically closed field $k$. The projective closure of $X$ in $\mathbb P^n$ is \begin{align*} \overline{X}^{\mathbb P}=V_+(I^h)\subset\mathbb P^n, \end{align*} where $I^h\trianglelefteq k[x_0,x_1,\dots,x_n]$ is the homogenization of $I$. [/definition] The superscript reminds us that this is closure in projective space, not closure inside affine space. The definition proposes a closed projective variety; the next theorem verifies that it is the topological closure of the affine image and that no finite affine points have been changed. The projective-closure theorem [Projective Closure of an Affine Variety via Homogenisation of the Ideal](/theorems/2141) says that if $X=V(I)\subset\mathbb A_k^n$, then under the chart embedding $\iota_0:\mathbb A_k^n\to\mathbb P_k^n$ given by $\iota_0(y_1,\dots,y_n)=[1:y_1:\cdots:y_n]$, the Zariski closure of $\iota_0(X)$ in $\mathbb P_k^n$ is \begin{align*} \overline{\iota_0(X)}^{\mathbb P_k^n}=V_+(I^h). \end{align*} The theorem makes the added part of the closure computable: after homogenizing, the new points must lie on the hyperplane at infinity. Algebraic closedness is used through the Nullstellensatz; over a non-algebraically closed field, the same equations still define a scheme or variety after base change, but equality with the set-theoretic closure of $k$-rational points can fail. The use of the full ideal is also essential, as the earlier example $I=(x,1-xy)$ shows: homogenizing only the displayed generators creates the spurious point $[0:0:1]$. The theorem does not say that every projective closure is smooth, irreducible, or independent of the affine embedding; it only identifies the smallest projective closed set containing this chosen affine model. This leads to a separate name for the boundary between affine and projective geometry, because these points carry asymptotic information that is invisible in $\mathbb A^n$. [definition: Points at Infinity] Let $X\subset\mathbb A^n$ be an affine variety with projective closure $\overline{X}^{\mathbb P}\subset\mathbb P^n$. The points at infinity of $X$ are the points of \begin{align*} \overline{X}^{\mathbb P}\cap H_\infty, \end{align*} where $H_\infty=V_+(x_0)$. [/definition] These points record limiting directions of unbounded branches of the affine variety. They are often easier to find than to interpret geometrically, so examples are essential. [example: The Parabola Has One Point at Infinity] For $X=V(y-x^2)\subset\mathbb A^2$, homogenizing $y-x^2$ in projective coordinates $[X:Y:Z]$ gives \begin{align*} Z^2\left(\frac{Y}{Z}-\left(\frac{X}{Z}\right)^2\right)=YZ-X^2. \end{align*} Thus the projective closure is \begin{align*} \overline{X}^{\mathbb P}=V_+(YZ-X^2)\subset\mathbb P^2. \end{align*} The line at infinity is $Z=0$, so the points at infinity must satisfy both $Z=0$ and $YZ-X^2=0$. Substituting $Z=0$ into the homogeneous equation gives \begin{align*} Y\cdot 0-X^2=-X^2=0. \end{align*} Since $k$ is a field, $X^2=0$ implies $X=0$. Therefore every point at infinity has the form \begin{align*} [0:Y:0]. \end{align*} A projective point cannot have all coordinates zero, so $Y\ne 0$, and scaling by $Y^{-1}$ gives \begin{align*} [0:Y:0]=[0:1:0]. \end{align*} Hence the parabola has exactly one point at infinity, namely $[0:1:0]$. Along the affine parametrization $x=t$, $y=t^2$, the corresponding projective points are $[t:t^2:1]=[1/t:1:1/t^2]$ for $t\ne 0$, which approach $[0:1:0]$ projectively; this is the vertical direction of the parabola at infinity. [/example] The parabola acquires one point because its highest-degree part is $-x^2$. A different affine curve can acquire several points, or points with multiplicity when the later intersection theory is developed. [example: The Hyperbola $xy=1$] Let $X=V(xy-1)\subset\mathbb A^2$, and use projective coordinates $[X:Y:Z]$ with affine chart $Z\ne 0$, where $x=X/Z$ and $y=Y/Z$. Since $xy-1$ has degree $2$, its homogenization with respect to $Z$ is \begin{align*} Z^2\left(\frac{X}{Z}\frac{Y}{Z}-1\right)=Z^2\cdot\frac{XY}{Z^2}-Z^2=XY-Z^2. \end{align*} Thus the projective closure is \begin{align*} \overline{X}^{\mathbb P}=V_+(XY-Z^2)\subset\mathbb P^2. \end{align*} The line at infinity is $Z=0$, so a point at infinity has the form $[X:Y:0]$ and must satisfy \begin{align*} XY-0^2=XY=0. \end{align*} Because $k$ is a field, $XY=0$ implies $X=0$ or $Y=0$. If $X=0$, then the point is $[0:Y:0]$; since projective coordinates cannot all vanish, $Y\ne 0$, and scaling by $Y^{-1}$ gives $[0:1:0]$. If $Y=0$, then the point is $[X:0:0]$; again $X\ne 0$, and scaling by $X^{-1}$ gives $[1:0:0]$. Hence the points at infinity are exactly \begin{align*} [1:0:0]\quad\text{and}\quad[0:1:0]. \end{align*} They correspond to the two asymptotic directions of the affine hyperbola $xy=1$. [/example] Projective closure turns asymptotes and limiting directions into ordinary projective points. This is one of the main advantages of projective geometry: statements about intersections no longer need separate clauses for finite and infinite behaviour. [remark: Dependence on the Affine Embedding] Projective closure depends on the chosen affine embedding $X\subset\mathbb A^n$. Isomorphic affine varieties can have non-isomorphic projective closures if embedded by different coordinate functions. Later, this dependence is organized by line bundles and projective embeddings, but in this course it is tracked directly through equations. [/remark] The closure operation also prepares the ground for projection. Once a variety lies in projective space, forgetting coordinates becomes a rational or regular geometric operation, and elimination gives the equations of the image. ## Elimination Ideals and Projections Many classical constructions ask for the shadow of a variety after some coordinates are forgotten. Algebraically, this means: given equations in many variables, find all consequences involving only the remaining variables. [definition: Elimination Ideal] Let $I\trianglelefteq k[x_1,\dots,x_n]$, and fix $0\le r\le n$. The $r$-th elimination ideal is \begin{align*} I_r=I\cap k[x_{r+1},\dots,x_n]. \end{align*} [/definition] The ideal $I_r$ contains exactly the polynomial equations in the last $n-r$ variables that are forced by the original system. The geometric question is whether these consequences cut out the image of the projection, or whether additional closure is needed. The next theorem gives the precise affine answer. [quotetheorem:9432] [citeproof:9432] The closure is necessary: projections of algebraic sets need not be closed in the affine Zariski topology, as the hyperbola example below shows by projecting $xy=1$ to the punctured affine line. Algebraic closedness again enters through the Nullstellensatz; without it, elimination still gives equations over $k$, but the equality is most naturally interpreted after passing to an [algebraic closure](/page/Algebraic%20Closure) or in scheme-theoretic language. The theorem also does not claim that $\pi(X)=V(I_r)$, nor that the fibres of $\pi$ have constant size or dimension. Its role is more modest and more useful computationally: it tells us which equations cut out the Zariski closure of the shadow. Projective geometry improves the topological behaviour when the projection is a morphism from a projective variety, because projective varieties are complete; in this course we use the elimination theorem as the affine computational replacement. [example: Projection of a Hyperbola] Let $X=V(xy-1)\subset\mathbb A^2$, and let $\pi:\mathbb A^2\to\mathbb A^1$ be the projection $\pi(x,y)=y$. We compute the elimination ideal $(xy-1)\cap k[y]$. Define a $k$-algebra homomorphism \begin{align*} \varphi:k[x,y]\longrightarrow k(y) \end{align*} by $\varphi(y)=y$ and $\varphi(x)=y^{-1}$. Then \begin{align*} \varphi(xy-1)=y^{-1}y-1=1-1=0. \end{align*} Thus every element of $(xy-1)$ maps to $0$ under $\varphi$. If $g(y)\in (xy-1)\cap k[y]$, then \begin{align*} 0=\varphi(g(y))=g(y) \end{align*} inside the rational function field $k(y)$. Since the inclusion $k[y]\hookrightarrow k(y)$ is injective, this forces $g=0$. Therefore \begin{align*} (xy-1)\cap k[y]=(0). \end{align*} By *Basic Elimination Theorem for Affine Projections*, the Zariski closure of $\pi(X)$ is \begin{align*} V((0))=\mathbb A^1. \end{align*} The actual image is smaller: if $b\in k^\times$, then choosing $a=b^{-1}$ gives \begin{align*} ab-1=b^{-1}b-1=0, \end{align*} so $b\in\pi(X)$. If $b=0$, then for every $a\in k$, \begin{align*} a\cdot 0-1=-1\ne 0, \end{align*} so $0\notin\pi(X)$. Hence \begin{align*} \pi(X)=\mathbb A^1\setminus\{0\}, \end{align*} while its Zariski closure is all of $\mathbb A^1$. [/example] The example displays the affine defect: projection can miss a lower-dimensional set even when its closure is large. For projective varieties, projections are often defined away from a centre, and the missing-set issue is replaced by the question of where the projection is defined. We therefore name the geometric projection operation before computing more images by elimination. [definition: Linear Projection from a Centre] Let $L\subset\mathbb P^n$ be a linear subspace, and let $M\subset\mathbb P^n$ be a complementary linear subspace. The projection from centre $L$ to $M$ is the rational map \begin{align*} \pi_L:\mathbb P^n\dashrightarrow M \end{align*} that sends a point $p\notin L$ to the intersection of $M$ with the linear span of $L$ and $p$. [/definition] In coordinates, this often becomes the operation of forgetting some homogeneous coordinates. The condition that the centre be disjoint from the variety is what prevents the projection from being undefined on the variety. [example: Projecting the Twisted Cubic to a Plane Cubic] Let $C\subset\mathbb P^3$ be the twisted cubic parametrized by \begin{align*} [s:t]\longmapsto [s^3:s^2t:st^2:t^3]. \end{align*} Projecting from $[0:1:0:0]$ forgets the second coordinate, so in the plane with coordinates $[X:Z:W]$ the projected parametrization is \begin{align*} [s:t]\longmapsto [s^3:st^2:t^3]. \end{align*} We first find a homogeneous equation satisfied by every projected point. For $X=s^3$, $Z=st^2$, and $W=t^3$, we have \begin{align*} XW^2=s^3(t^3)^2=s^3t^6. \end{align*} Also, \begin{align*} Z^3=(st^2)^3=s^3t^6. \end{align*} Therefore every projected point satisfies \begin{align*} XW^2-Z^3=0. \end{align*} Conversely, let $[X:Z:W]\in\mathbb P^2$ satisfy $XW^2-Z^3=0$. If $W\ne 0$, scale the point so that $W=1$. The equation becomes \begin{align*} X\cdot 1^2-Z^3=0. \end{align*} Hence $X=Z^3$, so the point is \begin{align*} [X:Z:1]=[Z^3:Z:1]. \end{align*} Taking $[s:t]=[Z:1]$ gives \begin{align*} [s^3:st^2:t^3]=[Z^3:Z\cdot 1^2:1^3]=[Z^3:Z:1]. \end{align*} If $W=0$, then the equation gives \begin{align*} X\cdot 0^2-Z^3=-Z^3=0. \end{align*} Since $k$ is a field, $Z^3=0$ implies $Z=0$. Thus the point has the form $[X:0:0]$, and because projective coordinates cannot all vanish, $X\ne 0$. Scaling by $X^{-1}$ gives \begin{align*} [X:0:0]=[1:0:0]. \end{align*} This point is obtained from $[s:t]=[1:0]$, since \begin{align*} [s^3:st^2:t^3]=[1^3:1\cdot 0^2:0^3]=[1:0:0]. \end{align*} Thus the projected curve is exactly \begin{align*} V_+(XW^2-Z^3)\subset\mathbb P^2. \end{align*} The point $[1:0:0]$ is singular: for $F=XW^2-Z^3$, the partial derivatives are \begin{align*} \frac{\partial F}{\partial X}=W^2,\qquad \frac{\partial F}{\partial Z}=-3Z^2,\qquad \frac{\partial F}{\partial W}=2XW, \end{align*} and all three vanish at $[1:0:0]$. Hence the projection is the cuspidal plane cubic $XW^2=Z^3$, with the cusp at the image of $[1:0]\in\mathbb P^1$. [/example] This example shows the geometric content of elimination: forgetting a coordinate may preserve dimension while changing singularities. The resulting equation is not guessed from the parametrization; it is obtained by eliminating the parameters. ## Resultants and Common Roots of Plane Curves Elimination can be made explicit for two equations in one variable over a coefficient ring. Directly solving the equations fibre by fibre is a poor algebraic procedure: for instance, if $F(X,Y)=Y-X^2$ and $G(X,Y)=Y-aX-b$, solving for $Y$ first hides the fact that the remaining condition is the single polynomial $X^2-aX-b=0$. For plane curves, the classical tool is the resultant, which produces that remaining equation without choosing roots and detects when two polynomials have a common root after specializing the remaining coordinate. [definition: Resultant] Let $A$ be a commutative ring, and fix integers $m,n\ge 1$. The resultant with respect to $T$ is the map \begin{align*} \operatorname{Res}_T:A[T]_{\le m}\times A[T]_{\le n}\longrightarrow A. \end{align*} It sends $(f,g)$ to $\operatorname{Res}_T(f,g)$, where, for $f(T)=a_mT^m+a_{m-1}T^{m-1}+\dots+a_0$ and $g(T)=b_nT^n+b_{n-1}T^{n-1}+\dots+b_0$, the element $\operatorname{Res}_T(f,g)$ is the determinant of the $m+n$ by $m+n$ Sylvester matrix formed from the coefficients of $f$ and $g$. [/definition] The Sylvester matrix records the coefficients of the products $T^{n-1}f,\dots,f,T^{m-1}g,\dots,g$ in the basis $T^{m+n-1},\dots,1$. Its determinant vanishes exactly when these shifted multiples become linearly dependent in the relevant polynomial vector space. The theorem below turns this linear dependence criterion into the geometric statement that the two equations have a common zero. [quotetheorem:9433] [citeproof:9433] The algebraically closed hypothesis is needed for the stated root formulation: over $\mathbb R$, the polynomials $T^2+1$ and $T^2+1$ have vanishing resultant but no common real root, although they do have common roots over $\mathbb C$. The positive-degree hypothesis avoids degenerate conventions for the Sylvester matrix and for constant polynomials; if one polynomial is a nonzero constant, there is no common root even though a separate convention is needed to define the resultant uniformly. The theorem also does not count common roots or record multiplicity; it only detects whether the two polynomials have at least one common root. For plane curves $F(X,Y)=0$ and $G(X,Y)=0$, treating the equations as polynomials in $Y$ gives a polynomial in $X$ whose zeros contain the $X$-coordinates of intersection points. The common-root theorem supplies the fibrewise criterion; the next statement packages it as an elimination theorem for plane curves. [quotetheorem:9434] [citeproof:9434] The degree condition in the converse matters because specialization can move intersections to infinity. For example, with $F=XY-1$ and $G=Y$ as polynomials in $Y$, the resultant is $-1$, but after homogenizing the affine picture one sees that the missing behaviour is controlled by the line at infinity rather than by a finite value of $Y$. A more direct degree-drop phenomenon occurs when the leading coefficient in $Y$ vanishes at $X=a$: the specialized Sylvester matrix no longer represents the same pair of degrees, so a zero of the original resultant may record a root escaping to infinity instead of a finite point $(a,b)$. Thus the theorem gives a reliable finite-fibre criterion only away from degree drops; homogenized resultants avoid part of this issue by treating finite and infinite roots symmetrically. This prepares the later intersection-theoretic viewpoint, where such exceptional fibres are not discarded but counted with multiplicity in projective space. [example: Intersecting a Line and a Parabola by Resultants] Let $F=Y-X^2$ and $G=Y-aX-b$ in $k[X,Y]$. As polynomials in $Y$, both are linear: \begin{align*} F=1\cdot Y+(-X^2). \end{align*} \begin{align*} G=1\cdot Y+(-aX-b). \end{align*} For linear polynomials $\alpha Y+\beta$ and $\gamma Y+\delta$, the Sylvester determinant is $\alpha\delta-\gamma\beta$. Applying this with $\alpha=1$, $\beta=-X^2$, $\gamma=1$, and $\delta=-aX-b$ gives \begin{align*} \operatorname{Res}_Y(F,G)=1\cdot(-aX-b)-1\cdot(-X^2)=X^2-aX-b. \end{align*} Multiplying a resultant by the nonzero scalar $-1$ cuts out the same zero set, so the equivalent eliminated equation is \begin{align*} aX+b-X^2=0. \end{align*} Now fix $c\in k$. A point of intersection with $X$-coordinate $c$ is a number $d\in k$ such that \begin{align*} F(c,d)=d-c^2=0. \end{align*} \begin{align*} G(c,d)=d-ac-b=0. \end{align*} The first equation gives $d=c^2$. Substituting $d=c^2$ into the second equation gives \begin{align*} c^2-ac-b=0. \end{align*} Equivalently, \begin{align*} ac+b-c^2=0. \end{align*} Thus $c$ is the $X$-coordinate of an intersection point exactly when the eliminated equation vanishes at $c$. Conversely, if $c^2-ac-b=0$, set $d=c^2$. Then \begin{align*} F(c,d)=c^2-c^2=0. \end{align*} \begin{align*} G(c,d)=c^2-ac-b=0. \end{align*} So $(c,c^2)$ lies on both the parabola $Y=X^2$ and the line $Y=aX+b$. Therefore the roots of $X^2-aX-b=0$ are exactly the $X$-coordinates of the intersection points. If this quadratic has a repeated root $c$, then \begin{align*} X^2-aX-b=(X-c)^2. \end{align*} Expanding the right-hand side gives \begin{align*} (X-c)^2=X^2-2cX+c^2. \end{align*} Comparing coefficients gives $a=2c$ and $b=-c^2$, so the line is \begin{align*} Y=2cX-c^2. \end{align*} This is the tangent line to $Y=X^2$ at $(c,c^2)$, since its slope is $2c$ and it passes through $(c,c^2)$. [/example] Resultants are therefore both computational and geometric. They compute elimination ideals in basic cases, and they detect when projection collapses intersection data into a discriminant-like equation. ## How the Chapter Fits the Course Projective closure and elimination complete the transition from affine algebraic sets to projective geometry. Homogenization explains how affine equations acquire points at infinity, while elimination explains how projected images inherit equations. In the next part of the course, these tools feed into dimension theory, tangent spaces, and the first systematic intersection computations, where points at infinity and multiplicities become unavoidable rather than exceptional. Once projective space is available, affine varieties can be completed by adding their points at infinity, and elimination can be used to describe images after projection. The resulting equations explain how geometry changes under coordinate forgetfulness, while still retaining the essential projective information. These tools prepare the way for a more systematic study of projective embeddings and classical constructions. # 7. Projective Embeddings and Classical Constructions This chapter studies three classical ways of producing projective varieties: the Veronese construction, the Segre construction, and maps defined by linear systems. The common theme is that projective coordinates may be chosen from homogeneous forms rather than from the original variables themselves. This viewpoint turns conics, rank-one matrices, and families of hypersurfaces into projective geometric objects, and it prepares the ground for the divisor and intersection calculations in Chapters 10 and 11. ## Homogeneous Forms and the Veronese Construction The first problem is how to regard all degree $d$ homogeneous forms in $n+1$ variables at once. A point of projective space gives values for such forms only up to common scalar, so the collection of all degree $d$ monomials should define a new point in a larger projective space. Let $k$ be an algebraically closed field. For $d \ge 1$, the $k$-vector space of homogeneous forms of degree $d$ in $x_0,\dots,x_n$ has basis the monomials $x^\alpha=x_0^{\alpha_0}\cdots x_n^{\alpha_n}$ with $|\alpha|=d$. Its dimension is \begin{align*} N+1=\binom{n+d}{d}. \end{align*} After choosing an ordering of these monomials, their values give homogeneous coordinates on $\mathbb P^N$. [definition: Veronese Map] For integers $n \ge 0$ and $d \ge 1$, the degree $d$ Veronese map is the map \begin{align*} \nu_d:\mathbb P^n \longrightarrow \mathbb P^N, \qquad [x_0:\cdots:x_n] \longmapsto [x^\alpha]_{|\alpha|=d}, \end{align*} where $N+1=\binom{n+d}{d}$ and the coordinates on $\mathbb P^N$ are indexed by multi-indices $\alpha$ with $|\alpha|=d$. [/definition] The definition is well-defined because replacing $(x_0,\dots,x_n)$ by $(\lambda x_0,\dots,\lambda x_n)$ multiplies every coordinate $x^\alpha$ by the same scalar $\lambda^d$. Thus the Veronese map records all degree $d$ measurements of a projective point, while forgetting the common scale. [example: Conics As Points Of Projective Five Space] A plane conic over $k$ is the zero locus in $\mathbb P^2$ of a nonzero homogeneous quadratic form \begin{align*} F(x_0,x_1,x_2)=a_{00}x_0^2+a_{01}x_0x_1+a_{02}x_0x_2+a_{11}x_1^2+a_{12}x_1x_2+a_{22}x_2^2. \end{align*} If $\mu\in k^\times$, then $(\mu F)(p)=0$ exactly when $\mu F(p)=0$, and since $\mu\ne0$ this is equivalent to $F(p)=0$. Thus multiplying all six coefficients by the same nonzero scalar does not change the conic, so conics are parametrised by $\mathbb P^5$ with coordinates $[a_{00}:a_{01}:a_{02}:a_{11}:a_{12}:a_{22}]$. Let $Q=k[x_0,x_1,x_2]_2$, with ordered basis \begin{align*} x_0^2,\ x_0x_1,\ x_0x_2,\ x_1^2,\ x_1x_2,\ x_2^2. \end{align*} For $p=[u_0:u_1:u_2]\in\mathbb P^2$ and a representative $\tilde p=(u_0,u_1,u_2)$, evaluation at $\tilde p$ is the linear functional $\operatorname{ev}_{\tilde p}:Q\to k$ given by $F\mapsto F(\tilde p)$. On the quadratic form above, \begin{align*} \operatorname{ev}_{\tilde p}(F)=a_{00}u_0^2+a_{01}u_0u_1+a_{02}u_0u_2+a_{11}u_1^2+a_{12}u_1u_2+a_{22}u_2^2. \end{align*} Therefore, in the [dual basis](/theorems/414) to the six monomials, the coordinates of $\operatorname{ev}_{\tilde p}$ are \begin{align*} [u_0^2:u_0u_1:u_0u_2:u_1^2:u_1u_2:u_2^2]. \end{align*} If we replace $\tilde p$ by $\lambda\tilde p=(\lambda u_0,\lambda u_1,\lambda u_2)$ with $\lambda\in k^\times$, then each coordinate becomes \begin{align*} (\lambda u_i)(\lambda u_j)=\lambda^2u_iu_j. \end{align*} Thus all six coordinates are multiplied by the same scalar $\lambda^2$, so the projective class $[\operatorname{ev}_{\tilde p}]\in\mathbb P(Q^*)$ depends only on $p$. This class is exactly \begin{align*} \nu_2(p)=[u_0^2:u_0u_1:u_0u_2:u_1^2:u_1u_2:u_2^2], \end{align*} so points of the quadratic Veronese surface record the values of all quadratic monomials at a point of the plane. [/example] The example separates two related projective spaces: one parametrises quadratic forms, and the other contains the image of $\mathbb P^2$ under all quadratic monomials. We now need to know whether this parametrised image is intrinsically a projective variety and whether the parametrisation loses any information about the original point. [quotetheorem:9435] [citeproof:9435] This theorem is often the first substantial example of an embedding that changes the ambient projective space while preserving the original variety. The hypotheses matter in two different ways: $d\ge 1$ prevents the constant degree-zero map, and projective space is essential because homogeneous coordinates make simultaneous scaling harmless. For example, if one tried to use only a proper subset of the degree $d$ monomials, distinct points could become identified or some point could have all chosen coordinates zero. The theorem also does not say that every embedding of $\mathbb P^n$ arises from the complete degree $d$ system; later linear systems will describe projections and subsystems that may lose information. Its equations introduce a recurring pattern: projective parametrisations often have images described by rank or factorisation relations. [example: Rational Normal Curve] For $n=1$, write the target coordinates on $\mathbb P^d$ as $[Y_0:\cdots:Y_d]$, where $Y_i$ corresponds to the monomial $s^{d-i}t^i$. The degree $d$ Veronese map is \begin{align*} \nu_d([s:t])=[s^d:s^{d-1}t:\cdots:st^{d-1}:t^d], \end{align*} and its image is called the rational normal curve of degree $d$. The $2\times2$ minors of the array with first row $Y_0,Y_1,\dots,Y_{d-1}$ and second row $Y_1,Y_2,\dots,Y_d$ are \begin{align*} Y_iY_{j+1}-Y_jY_{i+1}\qquad 0\le i<j\le d-1. \end{align*} Substituting $Y_i=s^{d-i}t^i$ gives \begin{align*} Y_iY_{j+1}=(s^{d-i}t^i)(s^{d-j-1}t^{j+1})=s^{2d-i-j-1}t^{i+j+1}. \end{align*} Similarly, \begin{align*} Y_jY_{i+1}=(s^{d-j}t^j)(s^{d-i-1}t^{i+1})=s^{2d-i-j-1}t^{i+j+1}. \end{align*} The two products have the same exponent of $s$ and the same exponent of $t$, so every minor vanishes on the image: \begin{align*} Y_iY_{j+1}-Y_jY_{i+1}=0. \end{align*} Thus the coordinate sequence of a point on the rational normal curve satisfies the rank-one relations encoded by these minors. For $d=2$, the map is \begin{align*} [s:t]\longmapsto [s^2:st:t^2], \end{align*} and the only minor is \begin{align*} Y_0Y_2-Y_1^2. \end{align*} Indeed, \begin{align*} Y_0Y_2-Y_1^2=(s^2)(t^2)-(st)^2=s^2t^2-s^2t^2=0. \end{align*} So the degree $2$ rational normal curve is the conic $Y_0Y_2-Y_1^2=0$ in $\mathbb P^2$. [/example] The rational normal curve shows how the Veronese construction turns a line into a higher-degree curve. For surfaces, the same construction gives a flexible supply of embedded surfaces whose projections produce many classical examples. [example: Quadratic Veronese Surface And Projection] Write the coordinates on $\mathbb P^5$ as $[Y_{00}:Y_{01}:Y_{02}:Y_{11}:Y_{12}:Y_{22}]$, matching the quadratic monomials \begin{align*} Y_{00}=x_0^2,\quad Y_{01}=x_0x_1,\quad Y_{02}=x_0x_2,\quad Y_{11}=x_1^2,\quad Y_{12}=x_1x_2,\quad Y_{22}=x_2^2. \end{align*} Thus the quadratic Veronese surface is the image of \begin{align*} \nu_2([x_0:x_1:x_2])=[x_0^2:x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2]. \end{align*} For a concrete projection centre disjoint from the surface, take \begin{align*} q=[0:1:0:0:0:0]\in\mathbb P^5. \end{align*} Projection from $q$ is represented on the complement of $q$ by forgetting the coordinate $Y_{01}$: \begin{align*} [Y_{00}:Y_{01}:Y_{02}:Y_{11}:Y_{12}:Y_{22}]\longmapsto [Y_{00}:Y_{02}:Y_{11}:Y_{12}:Y_{22}]. \end{align*} On the Veronese surface this gives \begin{align*} [x_0:x_1:x_2]\longmapsto [x_0^2:x_0x_2:x_1^2:x_1x_2:x_2^2]. \end{align*} These five coordinates cannot all vanish at a point of $\mathbb P^2$: from $x_0^2=0$, $x_1^2=0$, and $x_2^2=0$ we get $x_0=x_1=x_2=0$, which is not a projective point. Hence this projection restricts to a regular map on the whole Veronese surface. If the centre lies on the surface, the failure is visible in coordinates. The point $\nu_2([1:0:0])$ is \begin{align*} [1:0:0:0:0:0]. \end{align*} Projection from this point is represented by forgetting $Y_{00}$, and its restriction to the Veronese surface is \begin{align*} [x_0:x_1:x_2]\longmapsto [x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2]. \end{align*} At $[1:0:0]$ every displayed coordinate is zero, so the formula has no value there; the ambient projection is rational, and its restriction is undefined at the point used as the centre. A disjoint centre can still identify points when it lies on a secant line. Let \begin{align*} A=\nu_2([1:0:0])=[1:0:0:0:0:0] \end{align*} and \begin{align*} B=\nu_2([0:1:0])=[0:0:0:1:0:0]. \end{align*} The point \begin{align*} q=A+B=[1:0:0:1:0:0] \end{align*} lies on the line through $A$ and $B$. It is not on the Veronese surface: if $q=\nu_2([x_0:x_1:x_2])$, then $x_0^2$ and $x_1^2$ are nonzero while $x_0x_1=0$, which is impossible because nonzero $x_0$ and nonzero $x_1$ imply nonzero $x_0x_1$. After quotienting by the line spanned by $q$, the images of $A$ and $B$ become the same projective point because $A+B=q$ gives $\overline A=-\overline B$ in the quotient vector space. Thus the projected surface can acquire identifications exactly from this secant-line incidence. [/example] ## Products and the Segre Construction The second problem is how to place a product of projective spaces inside a single projective space. Since points in each factor are defined up to scalar, the products of their coordinates have exactly the right transformation law: scaling either factor scales all products by a common scalar. [definition: Segre Map] For integers $m,n\ge 0$, the Segre map is \begin{align*} \sigma:\mathbb P^m\times \mathbb P^n\longrightarrow \mathbb P^{(m+1)(n+1)-1}, \qquad ([x_0:\cdots:x_m],[y_0:\cdots:y_n])\longmapsto [x_i y_j]_{0\le i\le m,\ 0\le j\le n}. \end{align*} [/definition] The coordinates of the target may be arranged as an $(m+1)\times(n+1)$ matrix $Z=(Z_{ij})$. This gives an immediate candidate for the equations of the image: if $Z_{ij}=x_i y_j$, then every pair of rows and columns must satisfy a rank-one relation. The next theorem proves that these rank-one equations are not only necessary but also sufficient. [quotetheorem:2145] [citeproof:2145] The case $\mathbb P^1\times\mathbb P^1$ is the basic bridge between products and quadrics. The nonzero projective matrix condition is essential: the zero matrix satisfies all minors but is not a point of projective space, so it cannot represent a pair of projective points. The projective scaling is also part of the uniqueness statement. A rank-one matrix may be written as $xy^\top$, and replacing $x$ by $\lambda x$ and $y$ by $\lambda^{-1}y$ gives the same matrix; after passing to $[x]\in\mathbb P^m$ and $[y]\in\mathbb P^n$, this reciprocal rescaling is exactly what should be forgotten. The boundary cases $m=0$ or $n=0$ fit the theorem: the Segre map becomes the standard identification of a product with a point factor, and there are no $2\times2$ minors to impose. The theorem characterises the image set-theoretically by rank-one equations, but it does not by itself describe all linear sections, projections, or higher-rank determinantal varieties. As a concrete boundary case, allowing rank at most two in a larger matrix would give a different determinantal variety rather than a product of two projective spaces. For $\mathbb P^1\times\mathbb P^1$, the rank-one description gives a smooth quadric surface with two visible families of lines, one from fixing the first factor and one from fixing the second. [example: Rank One Two By Two Matrices] Write coordinates on $\mathbb P^3$ as $[Z_{00}:Z_{01}:Z_{10}:Z_{11}]$, and arrange them as the entries of a $2\times2$ array $Z=(Z_{ij})$. Under the Segre map, the pair $([x_0:x_1],[y_0:y_1])$ has coordinates \begin{align*} [Z_{00}:Z_{01}:Z_{10}:Z_{11}]=[x_0y_0:x_0y_1:x_1y_0:x_1y_1]. \end{align*} Substituting these coordinates into the quadratic equation gives \begin{align*} Z_{00}Z_{11}-Z_{01}Z_{10}=(x_0y_0)(x_1y_1)-(x_0y_1)(x_1y_0). \end{align*} The first product is \begin{align*} (x_0y_0)(x_1y_1)=x_0x_1y_0y_1, \end{align*} and the second product is \begin{align*} (x_0y_1)(x_1y_0)=x_0x_1y_1y_0=x_0x_1y_0y_1. \end{align*} Thus \begin{align*} Z_{00}Z_{11}-Z_{01}Z_{10}=x_0x_1y_0y_1-x_0x_1y_0y_1=0, \end{align*} so every point in the Segre image lies on the quadric $Z_{00}Z_{11}-Z_{01}Z_{10}=0$. Conversely, take a nonzero point $[Z_{00}:Z_{01}:Z_{10}:Z_{11}]\in\mathbb P^3$ satisfying \begin{align*} Z_{00}Z_{11}-Z_{01}Z_{10}=0. \end{align*} On the chart where $Z_{00}\ne0$, this equation says \begin{align*} Z_{00}Z_{11}=Z_{01}Z_{10}. \end{align*} Since multiplying all four homogeneous coordinates by the same nonzero scalar does not change the point, we may multiply by $Z_{00}$ and obtain \begin{align*} [Z_{00}:Z_{01}:Z_{10}:Z_{11}]=[Z_{00}^2:Z_{00}Z_{01}:Z_{00}Z_{10}:Z_{00}Z_{11}]. \end{align*} Using $Z_{00}Z_{11}=Z_{01}Z_{10}$ in the last coordinate gives \begin{align*} [Z_{00}:Z_{01}:Z_{10}:Z_{11}]=[Z_{00}^2:Z_{00}Z_{01}:Z_{00}Z_{10}:Z_{01}Z_{10}]. \end{align*} This is the Segre image of \begin{align*} ([Z_{00}:Z_{10}],[Z_{00}:Z_{01}]), \end{align*} because its four products are $Z_{00}^2$, $Z_{00}Z_{01}$, $Z_{10}Z_{00}$, and $Z_{10}Z_{01}$. If another entry is the chosen nonzero entry instead of $Z_{00}$, the same argument on that chart factors the array using the corresponding nonzero row and column. Hence the quadric $Z_{00}Z_{11}-Z_{01}Z_{10}=0$ is exactly the projectivised space of nonzero rank-one $2\times2$ matrices. [/example] This rank-one description is more than a convenient equation: it explains the geometry of the two rulings. Holding the column factor fixed and varying the row factor gives one line on the quadric, while holding the row factor fixed gives the other. [remark: Bihomogeneous Coordinates] A polynomial on $\mathbb P^m\times\mathbb P^n$ is naturally bihomogeneous: it has one degree in the $x$-variables and another degree in the $y$-variables. Under the Segre embedding, a homogeneous form of degree $e$ in the target coordinates pulls back to a bihomogeneous form of bidegree $(e,e)$, because each Segre coordinate $Z_{ij}=x_i y_j$ has bidegree $(1,1)$. General bidegree $(a,b)$ forms are better regarded as sections of $\mathcal O(a,b)$ on the product; they are not ordinary homogeneous forms in the Segre coordinates unless $a=b$, or unless an additional twist or embedding is introduced. This is the first appearance of multigraded coordinate rings in the course. [/remark] The Segre construction embeds a product by using all products of coordinates of bidegree $(1,1)$. Linear systems generalise this idea by replacing the full list of all monomials with a chosen finite-dimensional family of forms. ## Linear Systems and Rational Maps The third problem is how a family of hypersurfaces determines a map to projective space. If several homogeneous forms have the same degree, then evaluating them at a point gives projective coordinates, except at points where all the forms vanish simultaneously. [definition: Linear System Of Hypersurfaces] Let $V\subseteq k[x_0,\dots,x_n]_d$ be a nonzero finite-dimensional $k$-vector subspace of homogeneous forms of degree $d$. The associated linear system is the projective space $\mathbb P(V)$ parametrising nonzero forms in $V$ up to scalar. [/definition] A point of $\mathbb P(V)$ is a hypersurface from the family, while a basis of $V$ gives coordinates for a map in the opposite direction. The obstruction to obtaining a morphism everywhere is the common zero locus of the chosen forms, so this locus needs its own name before we can state the mapping construction precisely. [definition: Base Locus] Let $V\subseteq k[x_0,\dots,x_n]_d$ be a linear system. Its base locus is \begin{align*} \operatorname{Bs}(V)=\{p\in\mathbb P^n:F(p)=0\text{ for every }F\in V\}. \end{align*} [/definition] The base locus marks exactly the points where every possible coordinate value is zero. Having isolated these bad points, the next problem is to define the projective-coordinate formula on the remaining open set and name the resulting partially defined map. [definition: Rational Map From A Linear System] Let $V\subseteq k[x_0,\dots,x_n]_d$ have basis $F_0,\dots,F_r$. The rational map associated to $V$ is \begin{align*} \varphi_V:\mathbb P^n\dashrightarrow\mathbb P^r, \qquad p\longmapsto [F_0(p):\cdots:F_r(p)], \end{align*} with domain of definition $\mathbb P^n\setminus\operatorname{Bs}(V)$. [/definition] This construction includes the Veronese map as the complete linear system of all degree $d$ forms. It also includes projections, because deleting some coordinates from a projective embedding corresponds to choosing a smaller subspace of forms. The key question is when this rational map is defined on every point of the source. [quotetheorem:9436] [citeproof:9436] The theorem explains why base points are the precise failure of regularity. For instance, the system $V=\operatorname{span}(x_0,x_1)\subset k[x_0,x_1,x_2]_1$ on $\mathbb P^2$ has base point $[0:0:1]$, so the formula $[x_0:x_1]$ cannot define a morphism at that point. The theorem does not say that a base-point-free system is an embedding: it may still identify distinct points or fail to separate tangent directions. When base points are present, the same formula remains meaningful on a dense open subset and is therefore a rational map, while later constructions such as blowing up the base locus explain how such maps can sometimes be resolved. [example: A Pencil Of Plane Conics] Let $F,G\in k[x_0,x_1,x_2]_2$ be linearly independent quadratic forms, and let $V=\operatorname{span}(F,G)$. A point $[a:b]\in\mathbb P^1$ represents the nonzero quadratic form $aF+bG$, so the pencil consists of the conics \begin{align*} aF+bG=0. \end{align*} The associated rational map is \begin{align*} \varphi_V:\mathbb P^2\dashrightarrow\mathbb P^1,\qquad p\longmapsto [F(p):G(p)]. \end{align*} The formula is defined exactly at the points where the two displayed coordinates are not both zero. Hence the base locus is \begin{align*} \operatorname{Bs}(V)=\{p\in\mathbb P^2:F(p)=0\text{ and }G(p)=0\}=V(F,G), \end{align*} the intersection of the two conics $F=0$ and $G=0$. Now take $p\notin V(F,G)$. Then $(F(p),G(p))\ne(0,0)$, so $[F(p):G(p)]$ is a point of $\mathbb P^1$. The corresponding member of the pencil may be written as \begin{align*} G(p)F-F(p)G=0. \end{align*} Evaluating this quadratic form at $p$ gives \begin{align*} \bigl(G(p)F-F(p)G\bigr)(p)=G(p)F(p)-F(p)G(p)=0. \end{align*} Thus every point outside the base locus lies on a unique conic in the pencil, namely the member with parameter $[G(p):-F(p)]$; the rational map records this parameter. [/example] Pencils are the smallest nontrivial linear systems, and they already show the dual role of the construction: the same vector space parametrises hypersurfaces and defines a map from the original projective variety. Higher-dimensional systems give maps to higher-dimensional projective spaces. [example: Projection From The Quadratic Veronese Surface] Let the quadratic Veronese embedding be written in coordinates as \begin{align*} \nu_2([x_0:x_1:x_2])=[x_0^2:x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2]. \end{align*} A linear projection $\pi:\mathbb P^5\dashrightarrow\mathbb P^4$ with centre $q$ may be described by five independent linear forms $L_0,\dots,L_4$ in the six coordinates $Y_{00},Y_{01},Y_{02},Y_{11},Y_{12},Y_{22}$, all vanishing at $q$: \begin{align*} \pi([Y])=[L_0(Y):L_1(Y):L_2(Y):L_3(Y):L_4(Y)]. \end{align*} Pulling these linear forms back along $\nu_2$ gives five quadratic forms \begin{align*} H_i(x_0,x_1,x_2)=L_i(x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2) \end{align*} for $0\le i\le4$. Therefore the composite rational map is \begin{align*} \pi\circ\nu_2:\mathbb P^2\dashrightarrow\mathbb P^4,\qquad [x_0:x_1:x_2]\longmapsto [H_0:H_1:H_2:H_3:H_4], \end{align*} so it is defined by the five-dimensional subspace $\operatorname{span}(H_0,\dots,H_4)\subseteq k[x_0,x_1,x_2]_2$. If the forms $H_0,\dots,H_4$ have no common zero in $\mathbb P^2$, then for every point $p\in\mathbb P^2$ at least one value $H_i(p)$ is nonzero, so $[H_0(p):\cdots:H_4(p)]$ is a well-defined point of $\mathbb P^4$. On the open set $H_i\ne0$, its affine coordinates are \begin{align*} \frac{H_0}{H_i},\dots,\frac{H_{i-1}}{H_i},\frac{H_{i+1}}{H_i},\dots,\frac{H_4}{H_i}, \end{align*} which are regular ratios of homogeneous forms of the same degree on that chart. Thus the projection restricts to a morphism on the Veronese surface. If the projection centre lies on the Veronese surface, say $q=\nu_2(p_0)$, then each $L_i$ vanishes at $q$. Hence \begin{align*} H_i(p_0)=L_i(\nu_2(p_0))=L_i(q)=0 \end{align*} for every $i$. Thus $p_0$ is a base point of the pulled-back linear system, and the formula $[H_0:\cdots:H_4]$ has no value at $p_0$. In this case the projection is only a rational map on $\mathbb P^2$, with indeterminacy exactly where all the pulled-back quadratic coordinates vanish. [/example] The three constructions in this chapter fit into one pattern. Veronese maps use all degree $d$ forms on one projective space, Segre maps use all bidegree $(1,1)$ products on a product, and linear systems use a selected subspace of forms. In each case, projective geometry is controlled by homogeneous coordinates, while the equations of the image come from algebraic relations among the chosen coordinate functions. Projective embeddings show how the same variety can be realized in many different coordinate systems, depending on which homogeneous forms are chosen. Veronese, Segre, and linear-system maps all repackage the same geometry into new projective models. From there, the next question is how to measure the size and complexity of these models numerically. # 8. Dimension and Degree This chapter explains how algebraic geometry measures the size and complexity of a variety. Earlier chapters built the dictionary between varieties and coordinate rings, developed morphisms and local rings, and introduced projective closure. We now attach two numerical invariants to this dictionary: dimension, which measures the length of nested geometric structure, and degree, which measures how many points remain after cutting by enough general linear equations. ## Krull Dimension and Geometric Dimension The first problem is to turn the informal idea that a curve is one-dimensional and a surface is two-dimensional into an invariant that survives changes of coordinates and algebraic presentation. The algebraic side gives a natural candidate: chains of prime ideals in the coordinate ring. Through the variety-ring dictionary, these chains record chains of irreducible closed subsets. [definition: Krull Dimension] Let $R$ be a commutative ring. The Krull dimension of $R$, denoted $\dim R$, is the supremum of all integers $r \ge 0$ for which there exists a strictly increasing chain of prime ideals \begin{align*} \mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_r \end{align*} in $R$. [/definition] This definition counts inclusions of prime ideals rather than generators, so it applies to any coordinate ring without choosing special coordinates. To use it on varieties, we need to transfer the invariant from a ring to the geometric object whose regular functions form that ring. [definition: Dimension of an Affine Variety] Let $X \subset \mathbb A^n$ be an affine variety over an algebraically closed field $k$, with coordinate ring $k[X]=k[x_1,\dots,x_n]/I(X)$. The dimension of $X$ is \begin{align*} \dim X := \dim k[X]. \end{align*} [/definition] This definition is algebraic, but the course also needs a geometric way to read dimension directly from $X$. The obstruction is that a chain of prime ideals lives in the coordinate ring, while geometric dimension should be visible as a chain of closed subvarieties. The bridge is the correspondence between prime ideals in $k[X]$ and irreducible closed subsets of $X$. The following result checks that the ring-theoretic definition is measuring exactly those nested irreducible subvarieties. [quotetheorem:9437] [citeproof:9437] This theorem explains why dimension is a topological invariant for affine varieties with their Zariski topology. The irreducibility hypothesis on the closed subsets is essential: arbitrary closed subsets can be unions of unrelated pieces, so a chain of reducible closed sets would not correspond to a chain of prime ideals. The statement does not yet compute any dimensions; it only changes the language from algebra to geometry. Its value is that lower bounds can now be produced by explicit nested subvarieties, while the matching upper bounds will come from Noether normalization. [example: Affine Space Has Expected Dimension] For $\mathbb A^n$ over $k$, the coordinate ring is $k[\mathbb A^n]=k[x_1,\dots,x_n]$. For each $0\le i\le n$, set $\mathfrak p_i=(x_1,\dots,x_i)$, with $\mathfrak p_0=(0)$. The quotient is \begin{align*} k[x_1,\dots,x_n]/(x_1,\dots,x_i)\cong k[x_{i+1},\dots,x_n], \end{align*} which is an integral domain, so each $\mathfrak p_i$ is prime. The inclusions are strict because $x_{i+1}\in \mathfrak p_{i+1}$, while its image in $k[x_{i+1},\dots,x_n]$ is the nonzero variable $x_{i+1}$, so $x_{i+1}\notin \mathfrak p_i$. Thus \begin{align*} (0) \subsetneq (x_1) \subsetneq (x_1,x_2) \subsetneq \cdots \subsetneq (x_1,\dots,x_n) \end{align*} is a chain of $n$ strict inclusions of prime ideals, and hence $\dim \mathbb A^n\ge n$ by the definition of Krull dimension. For the reverse inequality, take the Noether normalization parameters to be $y_i=x_i$. Then $k[x_1,\dots,x_n]=k[y_1,\dots,y_n]$ is finite over this polynomial subring, indeed free of rank $1$ over itself. By *[Dimension Theorem](/theorems/915) for Affine Varieties*, the dimension is the number of normalization parameters, so $\dim \mathbb A^n\le n$. Combining the two inequalities gives $\dim \mathbb A^n=n$. [/example] The example shows how to find long chains, but it does not explain why no longer chains exist. This is a real obstruction: from the presentation of a quotient such as $k[x,y,z]/(xz-y^2)$, the number of generators and relations alone does not by itself prove the dimension. To prove finite-dimensionality and obtain a general computational method, the course uses Noether normalization: every irreducible affine coordinate ring is finite over a polynomial ring in the right number of variables. [quotetheorem:2939] [citeproof:2939] Noether normalization supplies algebraic parameters, but dimension requires identifying the number of those parameters with the maximum length of prime chains. Without this comparison, $d$ would only measure the size of one convenient polynomial subring, not the geometry of $X$ itself. The next theorem performs that comparison and gives the main dimension formula for irreducible affine varieties. [quotetheorem:9438] [citeproof:9438] This result turns dimension into the number of independent parameters needed to describe a dense part of $X$. Irreducibility matters because the coordinate ring is then a domain, so Noether normalization applies in the clean form above; for a reducible variety, different components may have different parameter counts. The theorem does not say that $X$ is isomorphic to $\mathbb A^d$, only that its coordinate ring is finite over a polynomial ring and therefore that $X$ maps finitely onto affine space of dimension $d$. It is especially effective for hypersurface-like examples, where the coordinate ring has one visible relation among several coordinates. [example: The Quadric Cone in Affine Three-Space] Let $X=V(xz-y^2)\subset \mathbb A^3$, and write its coordinate ring as \begin{align*} k[X]=k[x,y,z]/(xz-y^2). \end{align*} Set $A=k[x,y,z]/(xz-y^2)$. Since $xz-y^2=-(y^2-xz)$ is monic in $y$ up to the unit $-1$, every element of $A$ can be reduced using \begin{align*} y^2=xz \end{align*} to the form $a(x,z)+yb(x,z)$ with $a,b\in k[x,z]$. Thus $A$ is generated as a $k[x,z]$-module by $1$ and $y$. The inclusion $k[x,z]\hookrightarrow A$ is injective: if $a(x,z)$ maps to $0$ in $A$, then $a(x,z)$ lies in the ideal $(y^2-xz)$ inside $k[x,z][y]$. Dividing by the monic polynomial $y^2-xz$ leaves remainder $a(x,z)$, so the remainder can be $0$ only when $a(x,z)=0$. Hence $x$ and $z$ are algebraically independent in $A$, and $A$ is finite over the polynomial subring $k[x,z]$. By *Dimension Theorem for Affine Varieties*, this Noether normalization has two parameters, so \begin{align*} \dim X=2. \end{align*} Geometrically, the equation cuts out a surface in $\mathbb A^3$; its gradient is $(z,-2y,x)$, which vanishes on $X$ only at $(0,0,0)$, the vertex of the quadric cone. [/example] ## Codimension, Hypersurfaces, and Components Once dimension is available, the next question is how dimension changes when we impose additional equations. A single equation should usually lower dimension by one, but it may fail to do so if the equation already vanishes on a component or if embedded algebraic behaviour is present in the ring. The classical theory packages the reliable part of this expectation in codimension and the principal ideal theorem. [definition: Codimension] Let $Y \subset X$ be irreducible varieties. The codimension of $Y$ in $X$ is \begin{align*} \operatorname{codim}_X(Y) := \dim X - \dim Y. \end{align*} [/definition] Codimension measures how many independent conditions are needed to cut out $Y$ inside $X$. This motivates separating the codimension-one case, because zero loci of single nonzero equations are the basic cuts used throughout projective geometry. [definition: Hypersurface] Let $X$ be an irreducible variety. A closed subvariety $Y \subset X$ is a hypersurface in $X$ if every irreducible component of $Y$ has codimension $1$ in $X$. [/definition] The definition is phrased componentwise because a reducible zero set may have several irreducible pieces. The obstruction is that a single equation need not cut anything of codimension one if it is zero, invertible, or allowed to vanish identically on a component. To use hypersurfaces as equation cuts, we need a dimension result saying that a genuinely nonzero nonunit equation in an irreducible coordinate ring has only codimension-one components in its zero locus. [quotetheorem:9439] [citeproof:9439] The theorem is a dimension theorem, not just an equation-counting slogan. The hypotheses exclude the two basic failures: if $f=0$ in $k[X]$, then the zero locus is all of $X$, and if $f$ is a unit, then the zero locus is empty. The domain hypothesis prevents $f$ from vanishing identically on a whole irreducible component; in reducible rings, one equation can contain a component and fail to cut dimension there. The theorem does not control singularities or multiplicities of the hypersurface cut, only the dimensions of its irreducible components. [example: Hypersurfaces in Projective Space] Let $F \in k[x_0,\dots,x_n]$ be a nonzero homogeneous polynomial of degree $m>0$, and let $X=V(F)\subset \mathbb P^n$. Fix a standard affine chart $U_i=\{x_i\ne 0\}\cong \mathbb A^n$, with affine coordinates $t_j=x_j/x_i$ for $j\ne i$. On this chart the equation becomes the dehomogenized polynomial \begin{align*} F_i(t_0,\dots,\widehat{t_i},\dots,t_n)=F(t_0,\dots,t_{i-1},1,t_{i+1},\dots,t_n). \end{align*} Because $F$ is homogeneous, distinct monomials of $F$ remain distinct after setting $x_i=1$: the exponent of $x_i$ is determined by the total degree $m$ and the other exponents. Hence $F_i$ is not the zero polynomial. If $X\cap U_i$ is nonempty, then $F_i$ is not a nonzero constant, since a nonzero constant has empty zero locus. Thus on such a chart, \begin{align*} X\cap U_i=V(F_i)\subset \mathbb A^n \end{align*} is cut out by one nonzero nonunit polynomial. By *Principal Ideal Theorem for Hypersurface Cuts*, every irreducible component of $V(F_i)$ has codimension $1$ in $\mathbb A^n$. Since $\dim \mathbb A^n=n$, each irreducible component of $X\cap U_i$ has dimension \begin{align*} n-1. \end{align*} Now let $Z$ be an irreducible component of $X$. Since the standard charts cover $\mathbb P^n$, choose $i$ with $Z\cap U_i\ne\varnothing$. The set $Z\cap U_i$ is a nonempty open subset of the irreducible variety $Z$, so it has the same dimension as $Z$. It is contained in $X\cap U_i$ and is an irreducible component there near its generic point, so $\dim Z=n-1$. Therefore every irreducible component of the projective hypersurface $V(F)$ has dimension $n-1$. [/example] Projective and affine varieties often appear with several irreducible components, and not all components need have the same dimension. To keep dimension compatible with maximal chains, we measure a reducible variety by its largest irreducible component. [definition: Dimension of a Reducible Variety] Let $X$ be a variety with irreducible components $X_1,\dots,X_s$. The dimension of $X$ is \begin{align*} \dim X := \max_{1\le i\le s} \dim X_i. \end{align*} [/definition] This convention matches chains of irreducible closed subsets: any irreducible chain lies inside a single irreducible component after its first term is enlarged to a maximal irreducible closed subset. It also handles reducible plane curves in the expected way. [example: Coordinate Axes Revisited] Let $X=V(xy)\subset \mathbb A^2$. For a point $(a,b)\in \mathbb A^2$, we have $(a,b)\in V(xy)$ exactly when $ab=0$, which over a field means $a=0$ or $b=0$. Hence \begin{align*} X=V(x)\cup V(y). \end{align*} The coordinate rings of the two pieces are \begin{align*} k[V(x)]=k[x,y]/(x)\cong k[y] \end{align*} and \begin{align*} k[V(y)]=k[x,y]/(y)\cong k[x]. \end{align*} Thus each component is isomorphic to $\mathbb A^1$, and each has dimension $1$. By the definition of dimension for a reducible variety, \begin{align*} \dim X=\max\{\dim V(x),\dim V(y)\}=\max\{1,1\}=1. \end{align*} The origin is \begin{align*} V(x,y)=\{(0,0)\}. \end{align*} Inside the component $V(x)$, it is cut out by the additional equation $y=0$, so its coordinate ring is \begin{align*} k[V(x)]/(y)\cong k[y]/(y)\cong k. \end{align*} Therefore $\dim V(x,y)=0$, and its codimension in $V(x)$ is \begin{align*} \operatorname{codim}_{V(x)}V(x,y)=\dim V(x)-\dim V(x,y)=1-0=1. \end{align*} The same calculation inside $V(y)$ gives codimension $1$ there as well. However, $V(x,y)$ is properly contained in both $V(x)$ and $V(y)$, so it is not an irreducible component of $X$; the irreducible components are the two coordinate axes. [/example] ## Fibers of Dominant Morphisms The next question is how dimension behaves under a morphism. In the classical setting of irreducible varieties, a map $f:X\to Y$ should have fibers of dimension roughly $\dim X-\dim Y$ once its image really fills the target. The word for this condition is dominance, and it is the hypothesis that separates genuine families over $Y$ from maps whose images live in a smaller closed subset. [definition: Dominant Morphism] Let $f:X\to Y$ be a morphism of irreducible varieties. The morphism $f$ is dominant if $f(X)$ is dense in $Y$. [/definition] Dominance is the geometric condition that the map reaches a dense part of the target. With this hypothesis, it becomes meaningful to ask for the dimension of a general fiber, because the target is not artificially larger than the image. Without dominance, the naive difference $\dim X-\dim Y$ can be misleading: a constant map from a curve to a surface has one-dimensional fibers over one point and empty fibers elsewhere, while $1-2$ is not a possible fiber dimension. For this introductory chapter we will use the classical form of the fiber-dimension principle rather than the scheme-theoretic version: after discarding a proper closed subset of the target, the fibers of a dominant morphism have the expected dimension $\dim X-\dim Y$. Special fibers over the discarded closed set may be larger. This is a principle about dimension on a dense open part of the target; it is not a statement that all fibers are isomorphic or smooth. [example: Projection of the Quadric Cone] Let $X=V(xz-y^2)\subset \mathbb A^3$, and let $\pi:X\to \mathbb A^1$ be the projection $\pi(x,y,z)=x$. On coordinate rings, $\pi$ corresponds to the homomorphism \begin{align*} k[t]\longrightarrow k[x,y,z]/(xz-y^2),\quad t\longmapsto x. \end{align*} This map is injective: if $g(x)$ maps to $0$, then $g(x)\in (xz-y^2)\subset k[x,y,z]$; setting $y=0$ and $z=0$ gives $g(x)=0$ in $k[x]$, so $g=0$. Hence the image of $\pi$ is dense in $\mathbb A^1$, so $\pi$ is dominant. For $a\in k$, the fiber over $a$ is obtained by imposing $x=a$ in the equation $xz-y^2=0$. Thus \begin{align*} \pi^{-1}(a)=V(x-a,\;az-y^2)\subset \mathbb A^3. \end{align*} If $a\ne 0$, then $az-y^2=0$ is equivalent to $z=a^{-1}y^2$. The morphism \begin{align*} \mathbb A^1\longrightarrow \pi^{-1}(a),\quad y\longmapsto (a,y,a^{-1}y^2) \end{align*} has inverse given by projection to the $y$-coordinate, since every point of the fiber has the form $(a,y,a^{-1}y^2)$. Therefore $\pi^{-1}(a)\cong \mathbb A^1$ for every $a\ne 0$. For $a=0$, the fiber is \begin{align*} \pi^{-1}(0)=V(x,\;y^2)\subset \mathbb A^3. \end{align*} Its scheme-theoretic coordinate ring is \begin{align*} k[x,y,z]/(xz-y^2,\;x)\cong k[y,z]/(y^2). \end{align*} The element $y$ is nonzero in $k[y,z]/(y^2)$ but satisfies $y^2=0$, so this fiber has nilpotent structure. Its underlying classical reduced variety is obtained by replacing $(y^2)$ with its radical $(y)$: \begin{align*} (k[y,z]/(y^2))_{\mathrm{red}}\cong k[y,z]/(y)\cong k[z]. \end{align*} Thus the reduced fiber over $0$ is the affine line $V(y)\subset \mathbb A^2_{y,z}$, while the scheme-theoretic fiber remembers an extra nilpotent thickening along that line. [/example] ## Degree of Projective Varieties Dimension measures how many parameters a variety has; degree measures how it sits in projective space. The guiding question is: after cutting an irreducible projective variety by enough general hyperplanes to get finitely many points, how many points remain? The answer is the degree. [definition: Degree of a Projective Variety] Let $X\subset \mathbb P^n$ be an irreducible projective variety of dimension $r$. The degree of $X$, denoted $\deg X$, is the number of points in the intersection \begin{align*} X\cap H_1\cap \cdots \cap H_r \end{align*} for general hyperplanes $H_1,\dots,H_r\subset \mathbb P^n$, counted with intersection multiplicity. [/definition] The word general matters: special hyperplanes may pass through singularities, contain components of a reducible slice, or meet with higher tangency. For example, a tangent line to a smooth plane conic meets it at one visible point with multiplicity $2$, while a general line meets it in two distinct points. To make the definition computable, we first treat hypersurfaces, where the count reduces to zeros of one homogeneous polynomial on a line. [quotetheorem:9440] [citeproof:9440] This recovers the familiar fact that a plane conic has degree $2$ and a plane cubic has degree $3$. Nonconstancy of $F$ is needed because a constant homogeneous polynomial defines either the empty set or all of projective space, neither of which is a hypersurface of degree $m$ in this sense. Generality of $L$ is also essential: a line contained in $X$ would give infinitely many intersection points, while a tangent or otherwise special line changes the visible set and must be interpreted with multiplicity. The result does not compute degrees of arbitrary projective varieties, but it gives the model case for later intersection theory and for the [Hilbert polynomial](/theorems/2894) interpretation of degree. [example: Linear Sections of a Quadric Surface] Let $Q=V(x_0x_3-x_1x_2)\subset \mathbb P^3$, and write $F=x_0x_3-x_1x_2$. The polynomial $F$ is homogeneous of degree $2$ and nonconstant, so by *[Degree of a Projective Hypersurface](/theorems/9440)* the projective hypersurface $Q$ has degree $2$. Let $H\subset \mathbb P^3$ be a plane for which $F$ does not vanish identically on $H$. After choosing homogeneous coordinates $[u:v:w]$ on $H\cong \mathbb P^2$, the restriction $F|_H$ is a nonzero homogeneous polynomial of degree $2$, so \begin{align*} Q\cap H=V_H(F|_H) \end{align*} is a plane conic. If $L\subset H$ is a line not contained in this conic, choose homogeneous coordinates $[s:t]$ on $L\cong \mathbb P^1$. Then $(F|_H)|_L$ is a nonzero homogeneous binary quadratic \begin{align*} as^2+bst+ct^2 \end{align*} for some $a,b,c\in k$, not all zero. Since $k$ is algebraically closed, this quadratic has exactly two zeros on $\mathbb P^1$ counted with multiplicity, so \begin{align*} \#(Q\cap H\cap L)=2 \end{align*} with intersection multiplicity. The need for general linear sections is visible from a special plane. If $H_0=V(x_0)$, then on $H_0$ the equation of $Q$ becomes \begin{align*} x_0x_3-x_1x_2=0\cdot x_3-x_1x_2=-x_1x_2. \end{align*} Thus \begin{align*} Q\cap H_0=V(x_0,x_1x_2)=V(x_0,x_1)\cup V(x_0,x_2), \end{align*} because $x_1x_2=0$ holds exactly when $x_1=0$ or $x_2=0$. This special plane section is a union of two lines rather than a smooth conic, while the degree count remains $2$ only after counting the linear section with multiplicity. [/example] Curves in projective space are the first place where degree becomes a practical invariant for embeddings. We now need a standard family of curves whose abstract variety is always $\mathbb P^1$ but whose projective degree changes with the embedding. [definition: Rational Normal Curve] The rational normal curve of degree $n$ in $\mathbb P^n$ is the image of the morphism \begin{align*} \nu_n: \mathbb P^1 \to \mathbb P^n, \quad [s:t] \mapsto [s^n:s^{n-1}t:\cdots:st^{n-1}:t^n]. \end{align*} [/definition] The map records all degree $n$ monomials in two variables. The point still to check is that this parametrized curve has the expected intersection count with a general hyperplane, rather than collapsing points of $\mathbb P^1$ or producing a lower-degree image. Pulling a hyperplane back to $\mathbb P^1$ turns the question into the zero count of a degree $n$ binary form, which is the calculation needed to identify the embedded degree. [quotetheorem:9441] [citeproof:9441] This theorem depends on choosing a general hyperplane; if the pulled-back binary form has repeated roots, the set-theoretic intersection may have fewer than $n$ points even though the multiplicity count remains $n$. The injectivity of $\nu_n$ on closed points is what lets zeros on $\mathbb P^1$ be counted as intersection points on $C$ without further identifications. The theorem does not say that every degree $n$ rational curve is presented in this form, but it provides the standard reference embedding against which other rational curves are compared. This example links degree with parametrization: although $C$ is abstractly a copy of $\mathbb P^1$, its embedding in $\mathbb P^n$ has degree $n$. [example: Twisted Cubic] For $n=3$, the rational normal curve is the image \begin{align*} \nu_3:\mathbb P^1\to \mathbb P^3,\quad [s:t]\mapsto [s^3:s^2t:st^2:t^3]. \end{align*} Let $H=V(a_0X_0+a_1X_1+a_2X_2+a_3X_3)$ be a plane in $\mathbb P^3$. A point $\nu_3([s:t])$ lies in $H$ exactly when \begin{align*} a_0s^3+a_1s^2t+a_2st^2+a_3t^3=0. \end{align*} Thus $H\cap \nu_3(\mathbb P^1)$ is counted by the zeros on $\mathbb P^1$ of the pulled-back homogeneous cubic \begin{align*} G(s,t)=a_0s^3+a_1s^2t+a_2st^2+a_3t^3. \end{align*} Since $H$ is a plane, not all $a_i$ are zero, and the four monomials $s^3,s^2t,st^2,t^3$ are linearly independent, so $G$ is not the zero polynomial. Over the algebraically closed field $k$, the nonzero binary cubic $G$ factors as \begin{align*} G(s,t)=\lambda(\beta_1s-\alpha_1t)(\beta_2s-\alpha_2t)(\beta_3s-\alpha_3t) \end{align*} with $\lambda\ne 0$, where repeated linear factors are counted repeatedly. Each factor $\beta_js-\alpha_jt$ vanishes at the point $[\alpha_j:\beta_j]\in \mathbb P^1$, so $G$ has exactly three zeros on $\mathbb P^1$ counted with multiplicity. The map $\nu_3$ identifies no two points of $\mathbb P^1$: if $s\ne 0$, then $t/s=X_1/X_0$ on the image, while if $t\ne 0$, then $s/t=X_2/X_3$. Therefore the three zeros of $G$ give three intersection points of the plane with the twisted cubic, counted with multiplicity. Hence a general plane meets the twisted cubic in $3$ points counted with multiplicity, so the twisted cubic has degree $3$. [/example] ## Dimension and Degree as Working Invariants The chapter's two invariants play different roles in later intersection theory. Dimension tells us how many equations are needed before a finite answer can be expected. Degree tells us what that finite answer is for general linear equations in projective space. [remark: Intrinsic and Extrinsic Data] Dimension is intrinsic to a variety: isomorphic varieties have the same dimension. Degree is extrinsic: it depends on the chosen projective embedding. For instance, $\mathbb P^1$ embedded as a line in $\mathbb P^1$ has degree $1$, while the same abstract variety embedded as the rational normal curve in $\mathbb P^n$ has degree $n$. [/remark] This distinction will matter when divisors and intersections are introduced. Dimension controls which intersections are expected to be finite, while degree gives the stable count for linear slices in projective space. After projective constructions, the natural invariants are numerical: dimension records how many parameters a variety has, and degree measures how it meets linear spaces. These quantities summarize much of the geometry encoded in the coordinate ring and in the projective embedding. The next chapter refines that local picture by studying tangent spaces and singularities, where dimension behaves differently from the naive linear approximation. # 9. Tangent Spaces and Singularities This chapter turns the local intuition of a curve having a direction at a point into an algebraic construction. It builds on affine and projective varieties, coordinate rings, maximal ideals, Krull dimension, and the local rings introduced in Chapter 4. The main question is how the coordinate ring records first-order motion near a point, and how this detects when the variety locally looks as smooth as its dimension predicts. We then use the [Jacobian matrix](/page/Jacobian%20Matrix) to compute tangent spaces and isolate singular points in explicit affine and projective examples. ## First-Order Functions at a Point A point $p \in X$ determines a maximal ideal of functions vanishing at $p$. To retain only first-order information, we should ignore products of functions that both vanish at $p$, since such products vanish to at least second order. [definition: Maximal Ideal at a Point] Let $X \subset \mathbb A_k^n$ be an affine variety over an algebraically closed field $k$, with coordinate ring $A(X)=k[x_1,\dots,x_n]/I(X)$. For $p \in X$, the maximal ideal at $p$ is \begin{align*} \mathfrak m_p := \{f \in A(X): f(p)=0\}. \end{align*} [/definition] The maximal ideal records every function that vanishes at $p$, not just its first-order part. The obstruction is that two functions that both vanish at $p$ can have a product that vanishes too strongly to affect first-order motion. Quotienting by $\mathfrak m_p^2$ discards this second-order noise, and a tangent vector can then be read as a linear measurement of the remaining first-order function germs. [definition: Zariski Tangent Space] Let $X$ be an affine variety over $k$ and let $p \in X$. The Zariski tangent space of $X$ at $p$ is the $k$-vector space \begin{align*} T_pX := \operatorname{Hom}_k(\mathfrak m_p/\mathfrak m_p^2,k). \end{align*} [/definition] This definition does not mention an embedding of $X$ into affine space, so it is suited to intrinsic arguments. We still need to connect it with the familiar operation of differentiating functions along a direction, which motivates the derivation formulation. [definition: Derivation at a Point] Let $A$ be a finitely generated $k$-algebra and let $p$ be a $k$-point with evaluation homomorphism $\operatorname{ev}_p:A\to k$. A derivation at $p$ is a $k$-[linear map](/page/Linear%20Map) $D:A\to k$ satisfying \begin{align*} D(fg)=f(p)D(g)+g(p)D(f) \end{align*} for all $f,g\in A$. [/definition] A derivation kills constants and kills products of two functions vanishing at $p$, so it factors through first-order functions. This gives the bridge between the quotient definition and directional differentiation, and it is the form we use when deriving the Jacobian formula. [quotetheorem:9442] [citeproof:9442] This equivalence gives a coordinate-free definition while keeping the computational feel of directional derivatives. The affine and $k$-point hypotheses are part of why the formula is so clean: the evaluation map lands in $k$, so subtracting the scalar $f(p)$ keeps us inside the same algebra and makes $\mathfrak m_p/\mathfrak m_p^2$ a $k$-vector space. For a non-$k$-rational closed point the residue field $\kappa(p)$ replaces $k$, and on a non-affine variety the construction must be made in the local ring $\mathcal O_{X,p}$ rather than in one global coordinate ring. In affine space it recovers the expected vector space and fixes the interpretation of the coordinate derivations. [example: Tangent Space of Affine Space] Let $X=\mathbb A_k^n$, so $A(X)=k[x_1,\dots,x_n]$, and fix $p=(a_1,\dots,a_n)$. Put $u_i=x_i-a_i$. Evaluation at $p$ sends $u_i$ to $0$, so \begin{align*} \mathfrak m_p=(u_1,\dots,u_n)=(x_1-a_1,\dots,x_n-a_n). \end{align*} Every polynomial $f\in\mathfrak m_p$ can be written uniquely in the shifted variables as a sum of monomials in $u_1,\dots,u_n$ with no constant term. Its class modulo $\mathfrak m_p^2$ keeps only the degree-one part, because $\mathfrak m_p^2$ is generated by products $u_i u_j$ and therefore contains exactly the terms of degree at least $2$ in the shifted expansion. Thus \begin{align*} f \equiv c_1u_1+\cdots+c_nu_n \pmod{\mathfrak m_p^2} \end{align*} for unique scalars $c_1,\dots,c_n\in k$. Hence the classes of $x_1-a_1,\dots,x_n-a_n$ form a basis of $\mathfrak m_p/\mathfrak m_p^2$. Therefore \begin{align*} T_p\mathbb A_k^n=\operatorname{Hom}_k(\mathfrak m_p/\mathfrak m_p^2,k)\cong k^n \end{align*} by sending a functional $\lambda$ to the tuple $(\lambda(\overline{x_1-a_1}),\dots,\lambda(\overline{x_n-a_n}))$. Under the derivation interpretation, the vector $v=(v_1,\dots,v_n)$ corresponds to the derivation satisfying $D_v(x_i-a_i)=v_i$; since $D_v(a_i)=0$ for the constant $a_i$, this gives $D_v(x_i)=v_i$. Thus affine $n$-space has the expected $n$ independent first-order coordinate directions at every point. [/example] ## Equations and the Jacobian Matrix The intrinsic definition is compact, but computations start from equations. If $X\subset\mathbb A_k^n$ is cut out by polynomials, the tangent vectors in $\mathbb A_k^n$ that remain tangent to $X$ should satisfy the linearized equations. There is a subtle point: the equations must generate the ideal defining the variety, not merely have the same zero set. For instance, $V(x^2)=V(x)\subset\mathbb A_k^1$ as sets, but differentiating $x^2$ at $0$ gives no linear condition, while the reduced variety $V(x)$ has zero tangent space. [definition: Jacobian Matrix of an Affine System] Let $f_1,\dots,f_r\in k[x_1,\dots,x_n]$ and let $p\in\mathbb A_k^n$. The Jacobian matrix at $p$ is \begin{align*} J(f_1,\dots,f_r)_p = \left(\frac{\partial f_i}{\partial x_j}(p)\right)_{1\le i\le r,\,1\le j\le n} \in k^{r\times n}. \end{align*} [/definition] The Jacobian matrix packages the first derivatives of all defining equations at once. The issue is whether solving these linearized equations merely gives a heuristic tangent plane or actually recovers the intrinsic space $\operatorname{Hom}_k(\mathfrak m_p/\mathfrak m_p^2,k)$. To use Jacobians as a reliable computational substitute for the intrinsic tangent-space definition, the equations must be the actual defining equations in the coordinate ring, not just equations with the same set of solutions. With that convention, first-order geometry is computed by differentiating the defining functions and solving the resulting linear equations. Without it, replacing $x$ by $x^2$ at the origin in $\mathbb A_k^1$ would incorrectly produce the whole ambient line as the tangent space. This is the computational bridge we need: the tangent space is read from the kernel of the Jacobian matrix attached to the defining ideal. The dimension of that kernel is controlled by the rank of the Jacobian, which makes tangent lines and tangent planes quick to compute. [example: Tangent Line to a Plane Curve] Let $X=V(f)\subset\mathbb A_k^2$ and let $p=(a,b)\in X$. A tangent vector at $p$ has coordinate form $v=(v_1,v_2)$, where $v_1$ is the first-order change in the $x$-coordinate and $v_2$ is the first-order change in the $y$-coordinate. By the linearized-equation rule for a hypersurface, the tangent vectors are exactly the solutions of the single linear equation \begin{align*} \frac{\partial f}{\partial x}(p)v_1+\frac{\partial f}{\partial y}(p)v_2=0. \end{align*} If we write a point of the affine tangent line as $(x,y)=(a+v_1,b+v_2)$, then $v_1=x-a$ and $v_2=y-b$, so the same equation becomes \begin{align*} \frac{\partial f}{\partial x}(p)(x-a)+\frac{\partial f}{\partial y}(p)(y-b)=0. \end{align*} Thus, when at least one of the two partial derivatives is nonzero, \begin{align*} T_pX=V\left(\frac{\partial f}{\partial x}(p)(x-a)+\frac{\partial f}{\partial y}(p)(y-b)\right)\subset\mathbb A_k^2. \end{align*} For $f=x^2+y^2-1$ over a field with $\operatorname{char}k\ne 2$, the partial derivatives are \begin{align*} \frac{\partial f}{\partial x}=2x \end{align*} and \begin{align*} \frac{\partial f}{\partial y}=2y. \end{align*} At $p=(1,0)$ these evaluate to \begin{align*} \frac{\partial f}{\partial x}(1,0)=2 \end{align*} and \begin{align*} \frac{\partial f}{\partial y}(1,0)=0. \end{align*} Substituting $a=1$ and $b=0$ into the tangent-line equation gives \begin{align*} 2(x-1)+0(y-0)=0. \end{align*} Since $2\ne 0$ in $k$, this is equivalent to $x-1=0$, so the tangent line at $(1,0)$ is $x=1$. [/example] The example shows that one independent equation usually cuts the tangent dimension down by one. To state the higher-codimension version, we first isolate the situation where the number of equations matches the local codimension. [definition: Affine Complete Intersection at a Point] Let $X\subset\mathbb A_k^n$ be an affine variety with ideal $I(X)$, and let $p\in X$. We say that $X$ is a local complete intersection at $p$ of codimension $c$ with local equations $f_1,\dots,f_c\in I(X)$ if the images of $f_1,\dots,f_c$ generate $I(X)$ in the local ring $k[x_1,\dots,x_n]_{\mathfrak m_p}$ and form a regular sequence there. [/definition] For a local complete intersection, the expected tangent dimension is $n-c$. The regular sequence hypothesis rules out the possibility that the chosen equations impose the right set-theoretic dimension while hiding algebraic dependence in the local ring. The only remaining question is whether the $c$ linearized equations are independent at $p$, and the Jacobian rank measures exactly that independence. [quotetheorem:9443] [citeproof:9443] This criterion is often the shortest route from equations to geometry, but it is not a rank test for arbitrary lists of equations. If the local equations fail to generate the local ideal or fail to form a regular sequence, the number $c$ need not be the local codimension, so full row rank may test the wrong expectation. The hypersurface case avoids this extra complication because a nonzero reduced hypersurface is locally cut out by one equation, and there is only one row in the Jacobian matrix. [example: Smooth Points of a Hypersurface] Let $X=V(f)\subset\mathbb A_k^n$ be a hypersurface and let $p\in X$. By *Tangent Space by Linearized Equations*, a vector $v=(v_1,\dots,v_n)\in k^n$ lies in $T_pX$ exactly when the one-row Jacobian equation is satisfied: \begin{align*} \left(\frac{\partial f}{\partial x_1}(p),\dots,\frac{\partial f}{\partial x_n}(p)\right) \begin{pmatrix}v_1&\cdots&v_n\end{pmatrix}^\top=0. \end{align*} Multiplying the row vector by the column vector gives the linear condition \begin{align*} \frac{\partial f}{\partial x_1}(p)v_1+\cdots+\frac{\partial f}{\partial x_n}(p)v_n=0. \end{align*} Thus $T_pX$ is the kernel of the linear map \begin{align*} L_p:k^n\to k,\qquad L_p(v)=\sum_{j=1}^n \frac{\partial f}{\partial x_j}(p)v_j. \end{align*} If at least one [partial derivative](/page/Partial%20Derivative) $\frac{\partial f}{\partial x_j}(p)$ is nonzero, then $L_p$ is a nonzero linear map from $k^n$ to $k$. Its image is a nonzero subspace of the one-dimensional vector space $k$, so $\dim_k\operatorname{im}L_p=1$. By rank-nullity, \begin{align*} \dim_kT_pX=\dim_k\ker L_p=n-\dim_k\operatorname{im}L_p=n-1. \end{align*} This matches the expected dimension of a hypersurface. If instead \begin{align*} \frac{\partial f}{\partial x_1}(p)=\cdots=\frac{\partial f}{\partial x_n}(p)=0, \end{align*} then $L_p(v)=0$ for every $v\in k^n$, so \begin{align*} T_pX=\ker L_p=k^n. \end{align*} Thus a hypersurface has the expected tangent dimension at $p$ exactly when its gradient does not vanish there. [/example] ## Smooth Points and the Jacobian Criterion The next problem is to separate a defect of a chosen equation from a genuine geometric singularity. The Jacobian criterion gives an intrinsic test over the fields used in this course. [definition: Smooth Point] Let $X$ be an affine variety and let $p\in X$. The point $p$ is smooth if \begin{align*} \dim_k T_pX = \dim_p X, \end{align*} where $\dim_pX$ is the maximum dimension of an irreducible component of $X$ containing $p$. [/definition] A smooth point is one where the first-order approximation has the expected size. When the tangent space is too large, the excess first-order directions signal that the local geometry cannot be modeled by the expected affine space at that point. The complementary class needs its own name because later tests will locate exactly the points where the expected-size condition fails. This separates the good local models from the exceptional locus that tangent-space computations are meant to detect. [definition: Singular Point] Let $X$ be an affine variety and let $p\in X$. The point $p$ is singular if $p$ is not smooth. [/definition] The definitions are intrinsic, but checking them directly from $\mathfrak m_p/\mathfrak m_p^2$ can be slow. This motivates the Jacobian criterion, which converts the comparison with local dimension into a rank computation on any chosen set of generators. The field hypothesis matters: over an imperfect field, a polynomial such as $x^p-a$ in characteristic $p$ can have all formal derivatives equal to zero for inseparability reasons rather than because the geometric tangent space is visibly too large. [quotetheorem:9444] [citeproof:9444] The theorem is local: it can be checked at one point using one finite generating set for the ideal. It does not say that the rank condition is independent of the local dimension appearing in the statement; at points where several components of different dimensions meet, the relevant dimension must be identified before the rank can be interpreted. The field hypothesis rules out inseparable derivative pathologies in this course's setting. It also implies that the set of bad points is cut out by minors of a matrix, which leads to a structural result about where smooth points occur. [quotetheorem:9445] [citeproof:9445] The irreducibility hypothesis is used to turn nonempty open into dense; this is a topological statement about the whole variety, not just a pointwise rank computation. If one studies a reducible variety component by component, the smooth locus on each component may behave well while the intersections of components create singular points for the union. The complement of the smooth locus is therefore a proper closed subset when $X$ is irreducible. We now define this subset as the singular locus, so that its equations can be studied uniformly rather than point by point. [definition: Singular Locus] Let $X\subset\mathbb A_k^n$ be an affine variety. The singular locus $\operatorname{Sing}(X)$ is the set of singular points of $X$. [/definition] For a general affine variety the singular locus is obtained by rank conditions on the Jacobian matrix. This motivates the hypersurface formula below, where those rank conditions reduce to the vanishing of one gradient vector. The hypersurface assumption is essential for this simple formula: for several generators, singularity is controlled by the failure of sufficiently large minors to have the required rank, not by the vanishing of every partial derivative of every generator. [quotetheorem:9446] [citeproof:9446] For a hypersurface, singularity means exactly that the first-order part of the defining equation disappears at the point. The formula does not extend by replacing one gradient with all gradients in a multi-equation presentation: the curve $X=V(y^2-x^3,z)\subset\mathbb A_k^3$ is singular at the origin, but the generator $z$ has nonzero differential there. A test requiring every partial derivative of every generator to vanish would miss this singular point, while the Jacobian rank condition detects it because the matrix has rank $1$ instead of the expected rank $2$. This illustrates why complete intersections and higher-codimension varieties require minors of the Jacobian rather than a single gradient-vanishing equation. The hypersurface formula supplies the singular set, while the next section studies the finer local shape once a singular point has been found. ## Plane Curve Singularities Plane curves provide the first useful dictionary between algebraic rank defects and visible geometry. The lowest-degree nonzero homogeneous part of the defining equation controls the tangent cone and separates cusps from nodes. [definition: Multiplicity of a Plane Curve at a Point] Let $C=V(f)\subset\mathbb A_k^2$ be a plane curve, and let $p=(a,b)\in C$. Write the Taylor expansion of $f(a+u,b+v)$ as a sum of homogeneous terms \begin{align*} f(a+u,b+v)=F_m(u,v)+F_{m+1}(u,v)+\cdots, \end{align*} with $F_m\ne 0$. The multiplicity of $C$ at $p$ is $m$. [/definition] Multiplicity tells us the order at which the curve first appears near the point, but it does not by itself distinguish repeated and distinct directions. This motivates the tangent cone, which keeps the first nonzero homogeneous equation so that those directions can be read from its factorization. [definition: Tangent Cone of a Plane Curve] Let $C=V(f)\subset\mathbb A_k^2$ and let $p\in C$ have multiplicity $m$, with first nonzero homogeneous part $F_m(u,v)$. The tangent cone of $C$ at $p$ is the plane curve in the tangent coordinate plane defined by \begin{align*} F_m(u,v)=0. \end{align*} [/definition] This construction refines the Zariski tangent space. The Zariski tangent space can be too large at a singular point, while the tangent cone remembers the leading directions of branches. [example: Cusp] Let $C=V(f)\subset\mathbb A_k^2$ with $f=y^2-x^3$, where $\operatorname{char}k\ne 2,3$, and let $p=(0,0)$. The partial derivatives are \begin{align*} \frac{\partial f}{\partial x}=-3x^2 \end{align*} and \begin{align*} \frac{\partial f}{\partial y}=2y. \end{align*} Evaluating at $p$ gives \begin{align*} \frac{\partial f}{\partial x}(0,0)=-3\cdot 0^2=0 \end{align*} and \begin{align*} \frac{\partial f}{\partial y}(0,0)=2\cdot 0=0. \end{align*} Since $f(0,0)=0^2-0^3=0$, the point $p$ lies on $C$, and by *Singular Locus of a Hypersurface* it is singular. To compute the tangent cone, introduce tangent coordinates $u=x-0$ and $v=y-0$. Then \begin{align*} f(0+u,0+v)=v^2-u^3. \end{align*} The term $v^2$ is homogeneous of degree $2$, and the term $-u^3$ is homogeneous of degree $3$, so the first nonzero homogeneous part is \begin{align*} F_2(u,v)=v^2. \end{align*} Thus the tangent cone is defined by \begin{align*} v^2=0. \end{align*} Because $v^2=0$ has the same underlying line $v=0$ but with multiplicity two, the tangent cone is the double line $y=0$. This records one tangent direction at the cusp, counted twice. [/example] The cusp has one repeated tangent direction. A node instead has two distinct tangent directions, corresponding to two local branches crossing. [example: Node] Let \begin{align*} f=y^2-x^2(x+1)=y^2-x^3-x^2. \end{align*} At $p=(0,0)$ we have \begin{align*} f(0,0)=0^2-0^2(0+1)=0, \end{align*} so $p\in C$. The partial derivatives are \begin{align*} \frac{\partial f}{\partial x}=-\frac{\partial}{\partial x}(x^3+x^2)=-3x^2-2x \end{align*} and \begin{align*} \frac{\partial f}{\partial y}=2y. \end{align*} Evaluating at the origin gives \begin{align*} \frac{\partial f}{\partial x}(0,0)=-3\cdot 0^2-2\cdot 0=0 \end{align*} and \begin{align*} \frac{\partial f}{\partial y}(0,0)=2\cdot 0=0. \end{align*} By *Singular Locus of a Hypersurface*, the origin is singular on $C$. To compute the tangent cone, use tangent coordinates $u=x-0$ and $v=y-0$. Then \begin{align*} f(0+u,0+v)=v^2-u^2(u+1)=v^2-u^3-u^2. \end{align*} Grouping by total degree gives \begin{align*} v^2-u^3-u^2=(v^2-u^2)-u^3. \end{align*} The degree-$2$ part is therefore \begin{align*} F_2(u,v)=v^2-u^2. \end{align*} Factoring the difference of squares, \begin{align*} F_2(u,v)=v^2-u^2=(v-u)(v+u). \end{align*} Since $\operatorname{char}k\ne 2$, the linear forms $v-u$ and $v+u$ are distinct. Hence the tangent cone is the union of the two distinct lines $v=u$ and $v=-u$, equivalently $y=x$ and $y=-x$. The singularity has multiplicity $2$ and two distinct tangent directions, so the origin is an ordinary double point. [/example] The same derivative test works in higher-dimensional hypersurfaces. Singularities then measure the failure of a hypersurface to have a well-defined tangent hyperplane. [example: Quadric Cone] Let $X=V(f)\subset\mathbb A_k^3$ with \begin{align*} f=x^2+y^2-z^2. \end{align*} Assume $\operatorname{char}k\ne 2$. The first partial derivatives are \begin{align*} \frac{\partial f}{\partial x}=2x. \end{align*} \begin{align*} \frac{\partial f}{\partial y}=2y. \end{align*} \begin{align*} \frac{\partial f}{\partial z}=-2z. \end{align*} By *Singular Locus of a Hypersurface*, a point $(x,y,z)\in X$ is singular exactly when \begin{align*} 2x=0,\qquad 2y=0,\qquad -2z=0. \end{align*} Since $2\ne 0$ in $k$, these equations are equivalent to \begin{align*} x=0,\qquad y=0,\qquad z=0. \end{align*} The origin lies on $X$ because \begin{align*} f(0,0,0)=0^2+0^2-0^2=0. \end{align*} Therefore the origin is the unique singular point of the quadric cone. Now let $q=(a,b,c)\in X$ with $q\ne(0,0,0)$. Since $q\in X$, \begin{align*} a^2+b^2-c^2=0. \end{align*} The gradient at $q$ is \begin{align*} \nabla f(q)=(2a,2b,-2c). \end{align*} Because $q\ne(0,0,0)$ and $2\ne 0$, this vector is nonzero, so $q$ is smooth. By *Tangent Space by Linearized Equations*, a tangent vector $v=(v_1,v_2,v_3)$ at $q$ satisfies \begin{align*} 2av_1+2bv_2-2cv_3=0. \end{align*} Dividing by $2$ gives \begin{align*} av_1+bv_2-cv_3=0. \end{align*} Writing a point of the affine tangent plane as $(x,y,z)=(a+v_1,b+v_2,c+v_3)$, we have $v_1=x-a$, $v_2=y-b$, and $v_3=z-c$. Substitution gives \begin{align*} a(x-a)+b(y-b)-c(z-c)=0. \end{align*} Expanding each term gives \begin{align*} ax-a^2+by-b^2-cz+c^2=0. \end{align*} Moving the constant terms to the right gives the tangent plane \begin{align*} ax+by-cz=a^2+b^2-c^2. \end{align*} Since $a^2+b^2-c^2=0$ on $X$, this same plane may also be written as $ax+by-cz=0$. Thus the cone has a well-defined tangent plane at every point away from its vertex, while the vertex is the unique point where all first-order terms vanish. [/example] ## Projective Tangent Spaces Projective varieties need a version of tangent space compatible with homogeneous coordinates. Ordinary affine tangent vectors to the cone include the radial direction coming from rescaling a homogeneous lift, and this direction should not count as motion in projective space. The affine cone is the most efficient bridge: first compute the tangent space upstairs, then quotient out the radial direction through the point. [definition: Projective Tangent Space] Let $X\subset\mathbb P_k^n$ be a projective variety and let $p=[a_0:\cdots:a_n]\in X$. Let $\widehat X\subset\mathbb A_k^{n+1}$ be the affine cone over $X$ and choose a nonzero lift $a=(a_0,\dots,a_n)\in\widehat X$. The projective tangent space is \begin{align*} T_pX := \mathbb P\left(T_a\widehat X / ka\right), \end{align*} viewed as a projective linear subspace of $\mathbb P(k^{n+1}/ka)$. [/definition] The quotient by the radial line removes the ambiguity caused by scaling homogeneous coordinates. This motivates the tangent hyperplane formula for a projective hypersurface, where Euler's relation for homogeneous polynomials explains why the affine cone construction descends to projective space. [quotetheorem:9447] [citeproof:9447] This is the projective analogue of the tangent line formula for affine plane curves. The smoothness hypothesis is needed because if all partial derivatives of $F$ vanish at a lift of $p$, the displayed equation becomes $0=0$ and no hyperplane is determined. The formula also depends on homogeneity: Euler's relation is what makes the affine tangent space contain the radial line and hence descend to projective space. Conics give the basic model calculation. [example: Tangent Lines to Projective Conics] Let $C=V(F)\subset\mathbb P_k^2$ with \begin{align*} F=X_0X_2-X_1^2 \end{align*} and assume $\operatorname{char}k\ne 2$. For $p=[s^2:st:t^2]$ with $(s,t)\ne(0,0)$, the point lies on $C$ because \begin{align*} F(s^2,st,t^2)=s^2t^2-(st)^2=s^2t^2-s^2t^2=0. \end{align*} The first partial derivatives are \begin{align*} \frac{\partial F}{\partial X_0}=X_2 \end{align*} \begin{align*} \frac{\partial F}{\partial X_1}=-2X_1 \end{align*} \begin{align*} \frac{\partial F}{\partial X_2}=X_0. \end{align*} Evaluating at the homogeneous lift $(s^2,st,t^2)$ gives \begin{align*} \frac{\partial F}{\partial X_0}(s^2,st,t^2)=t^2 \end{align*} \begin{align*} \frac{\partial F}{\partial X_1}(s^2,st,t^2)=-2st \end{align*} \begin{align*} \frac{\partial F}{\partial X_2}(s^2,st,t^2)=s^2. \end{align*} Since $(s,t)\ne(0,0)$, at least one of $s^2$ and $t^2$ is nonzero, so this gradient is nonzero and $p$ is a smooth point of the conic. By *[Tangent Hyperplane to a Projective Hypersurface](/theorems/9447)*, the projective tangent line is \begin{align*} \frac{\partial F}{\partial X_0}(p)X_0+\frac{\partial F}{\partial X_1}(p)X_1+\frac{\partial F}{\partial X_2}(p)X_2=0. \end{align*} Substituting the three evaluated partial derivatives gives \begin{align*} t^2X_0+(-2st)X_1+s^2X_2=0, \end{align*} that is, \begin{align*} t^2X_0-2stX_1+s^2X_2=0. \end{align*} For $p=[1:0:0]$, take $s=1$ and $t=0$. The tangent-line equation becomes \begin{align*} 0^2X_0-2\cdot 1\cdot 0\,X_1+1^2X_2=0, \end{align*} so it reduces to \begin{align*} X_2=0. \end{align*} Thus the parametrized conic has tangent line $t^2X_0-2stX_1+s^2X_2=0$ at $[s^2:st:t^2]$, and at the endpoint $[1:0:0]$ this is the line at infinity direction $X_2=0$. [/example] The chapter's conclusion is that singularities are not extra structure added to a variety; they are already encoded by first-order algebra in the coordinate ring. The quotient $\mathfrak m_p/\mathfrak m_p^2$ gives the intrinsic object, the Jacobian matrix computes it from equations, and examples such as cusps, nodes, cones, and conics show how the rank of derivatives becomes visible geometry. Tangent spaces make the local geometry visible through first-order algebra, and singularities mark the points where that linear approximation fails. The Jacobian criterion turns equations into a practical test, while examples such as cusps, nodes, and cones show how the failure of smoothness is detected algebraically. Once local behavior is understood, the next step is to record zeros of functions globally through divisors. # 10. Divisors and Hypersurface Sections in the Classical Setting This chapter turns the local study of varieties into a first global bookkeeping device: divisors. Earlier chapters described functions, morphisms, tangent spaces, projective embeddings, and degree; here we record where a function or a homogeneous equation vanishes, and with what multiplicity. The guiding case is a nonsingular curve, where zeros and poles are points with integer weights, but the same language also records codimension-one subvarieties on higher-dimensional nonsingular varieties. The chapter stays in the classical setting. We use local rings and orders of vanishing only at the elementary level needed for curves and hypersurface sections, avoiding the scheme-theoretic divisor theory that comes later. The main point is that projective geometry turns equations into divisors, and rational functions compare such divisors through linear equivalence. ## Codimension-One Cycles and Weil Divisors How should a hypersurface inside a variety be counted when it appears with multiplicity? A single equation may vanish to order $2$ along one component and order $1$ along another, so the correct object is not only a subset but a formal integer combination of irreducible codimension-one pieces. [definition: Prime Divisor] Let $X$ be an irreducible variety over an algebraically closed field $k$. A prime divisor on $X$ is an irreducible closed subvariety $Y \subset X$ of codimension $1$. [/definition] Prime divisors identify the pieces that can occur as hypersurface components. A single subset is not enough data, because different equations can have the same support but vanish to different orders along the same component. The bookkeeping object must therefore remember two pieces of data at once: which prime divisors occur and what integer coefficient each one carries. This is why the next formal object is a finite formal sum rather than a single subvariety. [definition: Weil Divisor] Let $X$ be an irreducible variety. A Weil divisor on $X$ is a finite formal sum \begin{align*} D = \sum_i n_i Y_i, \end{align*} where each $Y_i$ is a prime divisor on $X$ and each $n_i \in \mathbb Z$. The group of Weil divisors on $X$ is denoted $\operatorname{Div}(X)$. [/definition] Weil divisors form an abelian group under addition of coefficients, so they allow both hypersurfaces and formal differences of hypersurfaces. This flexibility creates a distinction: some divisors have negative coefficients and should be interpreted as formal subtraction rather than as an actual positive cycle. When a divisor is meant to represent an actual hypersurface or zero locus, its coefficients should record multiplicities that are present, not components that have been subtracted away. This motivates isolating the nonnegative divisors as a special class. [definition: Effective Divisor] A Weil divisor $D=\sum_i n_iY_i$ is effective if $n_i\ge 0$ for every $i$. In this case we write $D\ge 0$. [/definition] Effectivity is the divisor-theoretic version of being cut out by equations rather than by quotients. The following example prepares for the later definition of order of vanishing by showing why a component must carry multiplicity data. [example: Plane Curve Divisor] Let $X=\mathbb A^2=\operatorname{Spec} k[x,y]$ and let $C=V(f)$, where \begin{align*} f=x^2(y-1)\in k[x,y]. \end{align*} The quotient rings are \begin{align*} k[x,y]/(x)\cong k[y] \end{align*} and \begin{align*} k[x,y]/(y-1)\cong k[x]. \end{align*} Both $k[y]$ and $k[x]$ are integral domains, so $(x)$ and $(y-1)$ are prime ideals. Therefore $V(x)$ and $V(y-1)$ are irreducible closed subvarieties of codimension $1$ in $\mathbb A^2$, hence prime divisors. For a point $(a,b)\in \mathbb A^2$, \begin{align*} f(a,b)=a^2(b-1). \end{align*} Since $k$ is a field, $a^2(b-1)=0$ exactly when $a^2=0$ or $b-1=0$. Also $a^2=0$ exactly when $a=0$. Hence \begin{align*} f(a,b)=0 \end{align*} exactly when \begin{align*} a=0 \end{align*} or \begin{align*} b=1. \end{align*} Thus \begin{align*} V(f)=V\bigl(x^2(y-1)\bigr)=V(x)\cup V(y-1). \end{align*} To compute the coefficient of $V(x)$, work at the generic point of the line $V(x)$. In the quotient by $(x)$, the factor $y-1$ maps to \begin{align*} y-1\in k[x,y]/(x)\cong k[y]. \end{align*} This element is nonzero in $k[y]$, so $y-1$ does not vanish identically along the component $V(x)$. Therefore it is a unit for the order computation at the generic point of $V(x)$. The factorization \begin{align*} f=x^2\cdot (y-1) \end{align*} has exactly two factors of the local defining equation $x$ and one unit factor $y-1$ at that generic point. Hence the coefficient of $V(x)$ is $2$. To compute the coefficient of $V(y-1)$, work at the generic point of the line $V(y-1)$. In the quotient by $(y-1)$, the factor $x^2$ maps to \begin{align*} x^2\in k[x,y]/(y-1)\cong k[x]. \end{align*} This element is nonzero in $k[x]$, so $x^2$ does not vanish identically along $V(y-1)$ and is a unit for the order computation at its generic point. Rewriting the same polynomial as \begin{align*} f=(y-1)\cdot x^2 \end{align*} shows exactly one factor of the local defining equation $y-1$ and one unit factor $x^2$ at that generic point. Hence the coefficient of $V(y-1)$ is $1$. No other prime divisor occurs, because the complete factorization \begin{align*} f=x^2(y-1) \end{align*} has only the irreducible factors $x$ and $y-1$. Therefore the divisor cut out by $f$ on the nonsingular surface $\mathbb A^2$ is \begin{align*} 2V(x)+V(y-1). \end{align*} The support remembers only the two lines $V(x)$ and $V(y-1)$, while the divisor records the additional fact that $V(x)$ occurs with multiplicity $2$. [/example] The affine calculation above used visible powers in a polynomial factorisation. On a general nonsingular variety we still need a way to measure the multiplicity of a rational function along a prime divisor, and this motivates the order function. [definition: Order Along a Prime Divisor] Let $X$ be a nonsingular irreducible variety and let $Y\subset X$ be a prime divisor. The order along $Y$ is the map \begin{align*} \operatorname{ord}_Y:k(X)^{\times}\to \mathbb Z \end{align*} which sends a nonzero rational function $f\in k(X)^{\times}$ to its order of vanishing along $Y$, with negative value for a pole along $Y$. [/definition] The order along a prime divisor records only one component at a time. A rational function usually has zeros on some prime divisors and poles on others, so looking at one order loses the global pattern of cancellation. To compare divisors using rational functions, we need a single divisor that records every zero and pole of the function at once. The construction below packages all componentwise orders into the formal sum produced by that rational function. [definition: Principal Divisor] Let $X$ be a nonsingular irreducible variety. The principal divisor map is \begin{align*} \operatorname{div}:k(X)^{\times}\to \operatorname{Div}(X), \qquad f\mapsto \sum_Y \operatorname{ord}_Y(f)Y, \end{align*} where the sum runs over all prime divisors $Y\subset X$ and only finitely many terms have nonzero coefficient. [/definition] Principal divisors are the divisor-theoretic shadow of rational functions, and they should behave well under multiplication of functions. The obstruction is that this construction would be useless for equivalence unless multiplication of functions became addition of divisor coefficients. In the nonsingular algebraic setting used here, the order along each prime divisor behaves like a valuation: for $f,g\in k(X)^\times$, one has $\operatorname{ord}_Y(fg)=\operatorname{ord}_Y(f)+\operatorname{ord}_Y(g)$ for every prime divisor $Y$. Therefore \begin{align*} \operatorname{div}(fg)=\operatorname{div}(f)+\operatorname{div}(g), \end{align*} so the [principal divisors form a subgroup](/theorems/7043) of $\operatorname{Div}(X)$. The nonsingularity hypothesis is what makes this valuation picture reliable along prime divisors. On singular varieties it can fail in the form used here; for instance, at the node of the affine curve $V(y^2-x^2(x+1))$, a codimension-one point no longer has the same regular local behaviour as a nonsingular point. Even in the nonsingular case, this does not say that every divisor is principal: on a smooth projective plane cubic, a single point $P$ cannot be principal because every principal divisor has degree $0$, while $\deg[P]=1$. Thus principal divisors provide the allowed changes of equation, not all divisors. The next definition turns that subgroup into a comparison relation: two divisors should be considered the same for rational-function purposes precisely when their difference is one of these principal divisors. [definition: Linear Equivalence] Let $X$ be a nonsingular irreducible variety. Two Weil divisors $D,E\in \operatorname{Div}(X)$ are linearly equivalent, written $D\sim E$, if there exists $f\in k(X)^{\times}$ such that \begin{align*} D-E=\operatorname{div}(f). \end{align*} [/definition] Linear equivalence keeps track of divisors up to rational change of equation. On a curve, every prime divisor is a point, so a divisor is a finite weighted list of points; the most basic question is the net total weight. This number forgets where the points lie but remembers how many zeros remain after poles are subtracted, making it the first numerical test for whether a divisor could be principal. [definition: Degree of a Divisor on a Curve] Let $C$ be a nonsingular irreducible curve over an algebraically closed field $k$. For a divisor \begin{align*} D=\sum_i n_iP_i \end{align*} on $C$, where $P_i\in C$ are closed points, the degree map is \begin{align*} \deg:\operatorname{Div}(C)\to \mathbb Z, \end{align*} defined by \begin{align*} \deg D=\sum_i n_i. \end{align*} [/definition] Degree counts total zeros minus total poles when a divisor is principal. A rational function on a projective curve should not be able to create more zeros than poles globally, because its values define a map to $\mathbb P^1$ whose fibers over $0$ and $\infty$ must balance with multiplicity. This balancing is the first global obstruction to principality: any divisor with nonzero degree cannot come from a rational function. [quotetheorem:2177] [citeproof:2177] This theorem is used here as a global balancing law for rational functions on projective curves. Geometrically, zeros and poles of such a function behave like the fibres over $0$ and $\infty$ of the associated map to $\mathbb P^1$, counted with multiplicity. Projectivity is part of the balancing mechanism, not a cosmetic assumption: on the affine line $\mathbb A^1$, the regular function $t-a$ has divisor $[a]$ if we record only affine points, so its degree is $1$. Passing to $\mathbb P^1$ restores the missing pole at infinity and turns the divisor into $[a]-[\infty]$. [example: Zeros and Poles on Projective Line] Use the affine coordinate $t=X_1/X_0$ on $\mathbb P^1$, and let \begin{align*} f(t)=\frac{(t-a)^2}{t-b} \end{align*} with $a\ne b$. We compute the order of $f$ at the points where one of the factors $t-a$, $t-b$, or the coordinate $t$ has a zero or pole. Near $[a]=[1:a]$, use the local coordinate $z=t-a$, so $t=a+z$. Then \begin{align*} t-b=(a+z)-b=(a-b)+z. \end{align*} Since $a-b\ne 0$, the value of $(a-b)+z$ at $z=0$ is $a-b$, so this factor is a unit in the local ring at $[a]$. Also \begin{align*} (t-a)^2=z^2. \end{align*} Therefore near $[a]$, \begin{align*} f=\frac{z^2}{(a-b)+z}. \end{align*} The denominator is a unit and the numerator is exactly $z^2$, so $f$ has two factors of the local parameter $z$ and no local-parameter factor in the denominator. Hence \begin{align*} \operatorname{ord}_{[a]}(f)=2. \end{align*} Near $[b]=[1:b]$, use the local coordinate $w=t-b$, so $t=b+w$. Then \begin{align*} t-a=(b+w)-a=(b-a)+w. \end{align*} Since $b-a\ne 0$, the value of $(b-a)+w$ at $w=0$ is nonzero, so $(b-a)+w$ is a unit in the local ring at $[b]$. Also \begin{align*} t-b=w. \end{align*} Thus \begin{align*} f=\frac{((b-a)+w)^2}{w}. \end{align*} The numerator is a unit and the denominator contributes exactly one factor of the local parameter $w$, so \begin{align*} \operatorname{ord}_{[b]}(f)=-1. \end{align*} It remains to compute the order at $\infty=[0:1]$. On the chart $X_1\ne 0$, use the local coordinate \begin{align*} u=X_0/X_1. \end{align*} Then $u=0$ at $\infty$, and on the overlap with the affine chart $X_0\ne 0$, \begin{align*} t=X_1/X_0=1/u. \end{align*} Substituting $t=1/u$ into the two factors gives \begin{align*} t-a=1/u-a=(1-au)/u. \end{align*} Similarly, \begin{align*} t-b=1/u-b=(1-bu)/u. \end{align*} Therefore \begin{align*} f=\frac{((1-au)/u)^2}{(1-bu)/u}. \end{align*} Squaring the numerator gives \begin{align*} f=\frac{(1-au)^2/u^2}{(1-bu)/u}. \end{align*} Dividing by $(1-bu)/u$ means multiplying by $u/(1-bu)$, so \begin{align*} f=\frac{(1-au)^2}{u^2}\cdot\frac{u}{1-bu}. \end{align*} Multiplying the fractions gives \begin{align*} f=\frac{u(1-au)^2}{u^2(1-bu)}. \end{align*} Canceling the common factor $u$ gives \begin{align*} f=\frac{(1-au)^2}{u(1-bu)}. \end{align*} At $u=0$, the factors $1-au$ and $1-bu$ both have value $1$, so both are units in the local ring at $\infty$. Thus the numerator is a unit and the denominator contributes exactly one factor of the local parameter $u$, giving \begin{align*} \operatorname{ord}_{\infty}(f)=-1. \end{align*} At any affine point $[c]=[1:c]$ with $c\ne a,b$, use the local coordinate $r=t-c$. Then $t=c+r$, so \begin{align*} t-a=(c-a)+r. \end{align*} Also \begin{align*} t-b=(c-b)+r. \end{align*} Because $c-a\ne 0$ and $c-b\ne 0$, both factors are units at $r=0$, so $f$ has order $0$ at $[c]$. Hence the only nonzero orders occur at $[a]$, $[b]$, and $\infty$, and the principal divisor is \begin{align*} \operatorname{div}(f)=2[a]-[b]-[\infty]. \end{align*} By the definition of degree on a divisor on a curve, \begin{align*} \deg\operatorname{div}(f)=2+(-1)+(-1)=0. \end{align*} The pole at infinity is the term that balances the double affine zero at $[a]$ against the affine pole at $[b]$. [/example] This example also shows why projectivity matters: a polynomial on an affine line can have zeros without visible poles, but the missing pole appears at infinity after compactification. ## Hyperplane Sections and the Hyperplane Class Projective varieties come with a preferred supply of hypersurfaces: intersections with hyperplanes. The question is how to record such an intersection when the hyperplane meets the variety with multiplicity or contains special tangent behaviour. [definition: Hyperplane Section Divisor] Let $X\subset \mathbb P^n$ be a positive-dimensional nonsingular irreducible projective variety not contained in the hyperplane $H=V(L)$, where $L\in k[X_0,\dots,X_n]$ is a nonzero homogeneous linear form. The hyperplane section divisor cut out by $H$ is the effective divisor on $X$ denoted \begin{align*} H|_X. \end{align*} Its prime components are the codimension-one irreducible components of $X\cap H$, counted with their intersection multiplicities. [/definition] The condition that $X$ is not contained in $H$ prevents the equation from vanishing identically on $X$. The remaining issue is whether the local orders contributed by the linear form can ever produce negative coefficients, which would make the object a formal difference rather than an actual section. For hyperplane intersections, the coefficients should come only from vanishing, not from poles. [quotetheorem:9448] [citeproof:9448] The non-containment hypothesis is essential: if $X\subset H$, then the linear form vanishes identically on $X$ and no codimension-one divisor is being selected. Irreducibility keeps the hyperplane section attached to one ambient function field; for a reducible union of two components, a hyperplane could vanish on one component while cutting the other, so a single divisor on one function field would not capture the situation. Nonsingularity is what makes the local orders used in the divisor construction part of the present classical setup; on a singular curve such as the nodal cubic, the local ring at the node is not the regular one-dimensional local ring used by this order-of-vanishing definition. The theorem only proves nonnegative coefficients; it does not compare this hyperplane section with any other one. That comparison is the next question, because a projective embedding should determine one common divisor class rather than unrelated classes for each linear equation. [quotetheorem:9449] [citeproof:9449] The non-containment assumptions ensure that $L_0/L_1$ is a genuine nonzero rational function on $X$ rather than an expression with an identically zero numerator or denominator. If one hyperplane contained $X$, its would-be section would not be a divisor in this sense, so the displayed principal divisor formula would have no valid left-hand comparison. The positive-dimensional hypothesis keeps codimension-one divisor language active; a zero-dimensional projective variety has no prime divisors of this kind. Irreducibility puts both sections in the same function field $k(X)$, while nonsingularity is the classical condition under which the orders along prime divisors used in the proof behave as valuations. Projectivity supplies the hyperplanes and the homogeneous coordinate ratios whose classes will be compared across the embedding. The theorem also does not assert that $H_0|_X$ and $H_1|_X$ are equal as divisors; for a smooth conic in $\mathbb P^2$, one line may meet it in two distinct points while a tangent line gives twice one point. Their supports and multiplicities can therefore be different, but their difference is principal. This is exactly the level of sameness needed for intersection computations and for the later Picard-group viewpoint. Since every allowable hyperplane produces the same linear equivalence class, the embedding supplies a distinguished divisor class. The previous theorem makes this class independent of the particular hyperplane, so it is now useful to name the common class once and reuse it. Later statements compare arbitrary hypersurface sections to repeated hyperplane sections, and those comparisons should be phrased in terms of this named class rather than in terms of a newly chosen hyperplane each time. [definition: Hyperplane Class] Let $X\subsetneq \mathbb P^n$ be a positive-dimensional nonsingular irreducible projective variety. The hyperplane class of $X$ is the linear equivalence class of $H|_X$ for any hyperplane $H\subset \mathbb P^n$ not containing $X$. [/definition] The hyperplane class depends on the chosen projective embedding, not only on the abstract variety. This dependence is productive: it remembers how the variety sits in projective space. [example: Divisor of a Line on a Smooth Plane Conic] Let $C=V(q)\subset \mathbb P^2$ be a smooth conic and let $\ell=V(L)$ be a line not containing $C$. The divisor $\ell|_C$ is obtained by restricting $L$ to $C$ and recording the order of vanishing at each point of $C\cap \ell$. First suppose that $\ell$ meets $C$ at two distinct points $P$ and $Q$. Choose projective coordinates with $\ell=V(Z)$, $P=[1:0:0]$, and $Q=[0:1:0]$. The restriction $q(X,Y,0)$ is a homogeneous quadratic in $X,Y$, so it has the form \begin{align*} q(X,Y,0)=\alpha X^2+\beta XY+\gamma Y^2. \end{align*} Since $P\in C\cap \ell$, \begin{align*} 0=q(1,0,0)=\alpha. \end{align*} Since $Q\in C\cap \ell$, \begin{align*} 0=q(0,1,0)=\gamma. \end{align*} Therefore \begin{align*} q(X,Y,0)=\beta XY. \end{align*} Write $a=\beta$. If $a=0$, then $q(X,Y,0)=0$ for every point $[X:Y:0]$ of the line $Z=0$, so $\ell\subset C$, contradicting the assumption that $\ell$ is not contained in $C$. Hence $a\ne 0$. Near $P$, use the affine chart $X\ne 0$ and set \begin{align*} u=Y/X,\quad v=Z/X. \end{align*} Because $q$ is homogeneous of degree $2$, the local equation $q(1,u,v)$ is a polynomial of degree at most $2$ in $u,v$. Its value at $v=0$ is \begin{align*} q(1,u,0)=au. \end{align*} Therefore every remaining term in $q(1,u,v)-au$ is divisible by $v$, so \begin{align*} q(1,u,v)=au+v h(u,v) \end{align*} for some linear polynomial $h(u,v)$. On $C$, \begin{align*} au+v h(u,v)=0. \end{align*} The partial derivative with respect to $u$ is \begin{align*} \frac{\partial}{\partial u}\bigl(au+v h(u,v)\bigr)=a+v\frac{\partial h}{\partial u}(u,v). \end{align*} At $(u,v)=(0,0)$ this equals $a\ne 0$, so $v=Z/X$ is a local parameter on the smooth curve $C$ at $P$. The line equation is $L=Z$, and on the chart $X\ne 0$ we have \begin{align*} Z=Xv. \end{align*} Since $X$ is nonzero on this chart, $X$ is a unit in the local ring at $P$. Hence $L|_C$ has the same order as $v$, and because $v$ is a local parameter, \begin{align*} \operatorname{ord}_P(L|_C)=\operatorname{ord}_P(v)=1. \end{align*} Near $Q$, use the affine chart $Y\ne 0$ and set \begin{align*} r=X/Y,\quad s=Z/Y. \end{align*} The restriction to $s=0$ is \begin{align*} q(r,1,0)=ar. \end{align*} Thus the local equation has the form \begin{align*} q(r,1,s)=ar+s h'(r,s) \end{align*} for some linear polynomial $h'(r,s)$. The partial derivative with respect to $r$ is \begin{align*} \frac{\partial}{\partial r}\bigl(ar+s h'(r,s)\bigr)=a+s\frac{\partial h'}{\partial r}(r,s), \end{align*} which equals $a\ne 0$ at $(0,0)$. Therefore $s=Z/Y$ is a local parameter on $C$ at $Q$. Since \begin{align*} Z=Ys \end{align*} and $Y$ is a unit on the chart $Y\ne 0$, the restricted line equation has order \begin{align*} \operatorname{ord}_Q(L|_C)=\operatorname{ord}_Q(s)=1. \end{align*} There are no other intersection points with $\ell$, because on the projective line $Z=0$ the equation is \begin{align*} q(X,Y,0)=aXY, \end{align*} and with $a\ne 0$ its zeros are exactly $[1:0:0]$ and $[0:1:0]$. Hence \begin{align*} \ell|_C=P+Q. \end{align*} Now suppose that $\ell$ is tangent to $C$ at $P$. Choose coordinates with $\ell=V(Z)$ and $P=[1:0:0]$. On the line $Z=0$, tangency at $P$ means that the binary quadratic $q(X,Y,0)$ has a double zero at $[1:0]$, so \begin{align*} q(X,Y,0)=aY^2 \end{align*} with $a\ne 0$. On the chart $X\ne 0$, set \begin{align*} u=Y/X,\quad v=Z/X. \end{align*} Then \begin{align*} q(1,u,0)=au^2. \end{align*} Every remaining term in $q(1,u,v)-au^2$ is divisible by $v$, so \begin{align*} q(1,u,v)=au^2+v h(u,v) \end{align*} for some linear polynomial $h(u,v)$. Since $C$ is smooth at $P$, the gradient of $q(1,u,v)$ at $(0,0)$ is nonzero. The partial derivative with respect to $u$ is \begin{align*} 2au+v\frac{\partial h}{\partial u}(u,v), \end{align*} which is $0$ at $(0,0)$. Therefore the partial derivative with respect to $v$ must be nonzero. That derivative is \begin{align*} h(u,v)+v\frac{\partial h}{\partial v}(u,v), \end{align*} so at $(0,0)$ it equals $h(0,0)$. Hence \begin{align*} h(0,0)\ne 0. \end{align*} Because $h(0,0)\ne 0$, the function $h(u,v)$ is a unit in the local ring at $P$. The equation of $C$ gives \begin{align*} v h(u,v)=-au^2. \end{align*} Multiplying by the unit $h(u,v)^{-1}$ gives \begin{align*} v=-a u^2 h(u,v)^{-1}. \end{align*} The coordinate $u$ is a local parameter on $C$ at $P$, since the partial derivative with respect to $v$ is nonzero. Thus $u$ has order $1$, $u^2$ has order $2$, and the factor $-a h(u,v)^{-1}$ is a unit. Therefore \begin{align*} \operatorname{ord}_P(v)=2. \end{align*} As before, $L=Z=Xv$ on the chart $X\ne 0$, and $X$ is a unit at $P$, so \begin{align*} \operatorname{ord}_P(L|_C)=2. \end{align*} On the line $\ell$, the restricted conic equation is \begin{align*} q(X,Y,0)=aY^2, \end{align*} whose only zero on $Z=0$ is $[1:0:0]$, with multiplicity $2$. Therefore \begin{align*} \ell|_C=2P. \end{align*} Line sections of the same smooth conic can therefore have different support, but in both cases the coefficients add to $2$. [/example] For curves in projective space, the degree of the hyperplane class recovers the usual degree of the curve. The conic has degree $2$ because a general line meets it in two points counted with multiplicity. [example: Hyperplane Sections of the Twisted Cubic] Let $C\subset \mathbb P^3$ be the twisted cubic parametrised by \begin{align*} \phi:\mathbb P^1\to C,\qquad [s:t]\mapsto [s^3:s^2t:st^2:t^3]. \end{align*} For the hyperplane $H=V(L)$ with \begin{align*} L=a_0X_0+a_1X_1+a_2X_2+a_3X_3, \end{align*} pulling back $L$ along $\phi$ gives \begin{align*} (L\circ\phi)([s:t])=a_0s^3+a_1s^2t+a_2st^2+a_3t^3. \end{align*} The homogeneous cubic forms $s^3$, $s^2t$, $st^2$, and $t^3$ form a basis for the vector space of degree-$3$ binary forms, so this pullback is the zero form exactly when \begin{align*} a_0=a_1=a_2=a_3=0. \end{align*} Thus a nonzero hyperplane equation whose hyperplane does not contain $C$ pulls back to a nonzero binary cubic. Since $k$ is algebraically closed, this nonzero binary cubic factors into homogeneous linear forms, counted with multiplicity: \begin{align*} a_0s^3+a_1s^2t+a_2st^2+a_3t^3=c\ell_1(s,t)\ell_2(s,t)\ell_3(s,t) \end{align*} with $c\in k^\times$ and each $\ell_i$ a nonzero linear form. Let $R=[\alpha:\beta]\in\mathbb P^1$, and suppose exactly $m$ of the factors $\ell_i$ vanish at $R$. If $\alpha\ne 0$, write $R=[1:\lambda]$ and use the local coordinate \begin{align*} z=t/s-\lambda. \end{align*} On the chart $s\ne 0$, the coordinate $s$ is a unit, so for a linear form $\ell(s,t)=As+Bt$ the order of $\ell$ is the order of \begin{align*} \ell(s,t)/s=A+B(t/s). \end{align*} Since $t/s=\lambda+z$, this becomes \begin{align*} A+B(t/s)=A+B(\lambda+z). \end{align*} Expanding the right side gives \begin{align*} A+B(\lambda+z)=(A+B\lambda)+Bz. \end{align*} If $\ell(R)\ne 0$, then \begin{align*} A+B\lambda\ne 0, \end{align*} so $(A+B\lambda)+Bz$ has nonzero value at $z=0$ and is a unit in the local ring at $R$. If $\ell(R)=0$, then \begin{align*} A+B\lambda=0. \end{align*} In that case $A=-B\lambda$. If also $B=0$, then $A=0$, contradicting that $\ell$ is a nonzero linear form. Hence $B\ne 0$, and the expression is \begin{align*} (A+B\lambda)+Bz=Bz. \end{align*} Thus a vanishing linear factor contributes exactly one factor of the local parameter $z$, while a nonvanishing linear factor contributes a unit. If $R=[0:1]$, use the local coordinate \begin{align*} u=s/t. \end{align*} On the chart $t\ne 0$, the coordinate $t$ is a unit, so for $\ell(s,t)=As+Bt$ the order of $\ell$ is the order of \begin{align*} \ell(s,t)/t=A(s/t)+B. \end{align*} Since $s/t=u$, this is \begin{align*} Au+B. \end{align*} If $\ell(R)\ne 0$, then $B\ne 0$, so $Au+B$ has nonzero value at $u=0$ and is a unit. If $\ell(R)=0$, then $B=0$; because $\ell$ is nonzero, $A\ne 0$, and the expression is \begin{align*} Au. \end{align*} Thus at infinity too, a vanishing linear factor contributes one factor of the local parameter and a nonvanishing linear factor contributes a unit. Therefore, in a local coordinate $w$ at any point $R\in\mathbb P^1$, each of the $m$ linear factors vanishing at $R$ has the form $w$ times a unit, and each of the remaining $3-m$ factors is a unit. Multiplying the factors gives \begin{align*} L\circ\phi=cw^mU \end{align*} where $U$ is a unit at $R$. Since $c\in k^\times$ is also a unit, \begin{align*} \operatorname{ord}_R(L\circ\phi)=m. \end{align*} The parametrisation identifies $\mathbb P^1$ with $C$: away from $[0:0:0:1]$ the inverse is read from $[X_0:X_1]$, and at $[0:0:0:1]$ it is $[0:1]$. Hence the order of $L\circ\phi$ at $R$ is the coefficient of $\phi(R)$ in the hyperplane section divisor. Thus \begin{align*} H|_C=\sum_{R\in \mathbb P^1,\ L(\phi(R))=0}\operatorname{ord}_R(L\circ\phi)\,\phi(R). \end{align*} The three linear factors $\ell_1,\ell_2,\ell_3$ contribute total multiplicity \begin{align*} 1+1+1=3. \end{align*} Therefore every hyperplane section of the twisted cubic is an effective divisor of degree $3$: a general hyperplane gives three distinct points of $C$, while a special hyperplane gives repeated points whose coefficients still add to $3$. [/example] The twisted cubic example is the model for reading projective degree from hyperplane sections: pull the linear equation back to a homogeneous form on the parameter space and count its zeros with multiplicity. ## Divisors Cut Out by Homogeneous Forms Hyperplanes are only the first case. A homogeneous form of degree $d$ also cuts out a hypersurface in projective space, and its restriction to a projective variety gives a divisor comparable with $d$ times the hyperplane class. [definition: Divisor Cut Out by a Homogeneous Form] Let $X\subset \mathbb P^n$ be a positive-dimensional nonsingular irreducible projective variety and let $F\in k[X_0,\dots,X_n]$ be a nonzero homogeneous form of degree $d$ whose restriction to $X$ is not identically zero. The divisor cut out by $F$ on $X$ is the effective divisor denoted \begin{align*} (F)_0|_X, \end{align*} whose prime components are the codimension-one components of $X\cap V(F)$ counted with their vanishing orders. [/definition] The notation emphasises that this divisor records zeros. Comparing two zero divisors requires a rational function whose zeros come from one equation and whose poles come from the other. Homogeneous quotients only define rational functions on projective space when the two forms have the same degree, so equal degree is the condition that makes such a comparison meaningful. [quotetheorem:9450] [citeproof:9450] The equal-degree hypothesis is essential because only then is $F/G$ invariant under rescaling homogeneous coordinates, so only then does it define a rational function on projective space and on $X$. If the degrees differ, the quotient is not projectively well-defined until it is corrected by another homogeneous factor, and the missing factor is exactly what the hyperplane class records. The assumption that neither form restricts to zero on $X$ is equally important: an identically zero restriction would cut out all of $X$ rather than a codimension-one divisor, and it would also make the quotient zero or undefined as an element of $k(X)^\times$. The positive-dimensional, irreducible, nonsingular hypotheses keep the argument inside the classical divisor setting: there are prime divisors to sum over, there is one function field $k(X)$, and orders along prime divisors are the valuation-like orders used in the proof. The theorem is therefore not a statement about arbitrary quotients of equations; it is a statement about rational functions obtained after projectivizing with the correct degree. With that limitation in mind, the hyperplane class gives a standard reference point for every degree-$d$ equation. [quotetheorem:9451] [citeproof:9451] The condition that $F$ not vanish identically on $X$ is necessary because an identically zero restriction would cut out all of $X$, not a codimension-one divisor. The chosen hyperplane must also avoid containing $X$, since $L^d$ is used as the comparison form and must define the divisor $d(H|_X)$. Positive dimension ensures that these intersections are being recorded by prime divisors rather than by an empty codimension-one theory, irreducibility keeps the ratio $F/L^d$ inside a single function field, and nonsingularity is the standing classical hypothesis for reading vanishing orders as divisor coefficients. Projectivity is what makes homogeneous degree meaningful and supplies the hyperplane class used as the reference. The conclusion is linear equivalence, not equality: the hypersurface $V(F)$ and the chosen multiple hyperplane may meet $X$ in quite different subsets, but their difference is principal. This statement explains why a plane curve of degree $m$ meets a line in $m$ points counted with multiplicity, and it motivates the next example, where a degree-$2$ equation on a degree-$2$ curve gives total degree $4$. [example: Quadratic Form on a Plane Conic] Let $C\subset \mathbb P^2$ be a smooth conic and let $F$ be a quadratic form whose restriction to $C$ is not identically zero. We first compute the total degree of the zero divisor $(F)_0|_C$. By *Degree d Hypersurfaces Represent d Times the Hyperplane Class* with $d=2$, \begin{align*} (F)_0|_C\sim 2(H|_C). \end{align*} By the definition of linear equivalence, this means that \begin{align*} (F)_0|_C-2(H|_C) \end{align*} is a principal divisor. Since principal divisors on nonsingular projective curves have degree $0$ by *Degree of a Principal Divisor on a Projective Curve*, \begin{align*} \deg\bigl((F)_0|_C-2(H|_C)\bigr)=0. \end{align*} Degree is additive on divisors, so \begin{align*} \deg\bigl((F)_0|_C\bigr)-\deg\bigl(2(H|_C)\bigr)=0. \end{align*} Therefore \begin{align*} \deg\bigl((F)_0|_C\bigr)=\deg\bigl(2(H|_C)\bigr). \end{align*} For a line $H$ meeting the smooth conic $C$, the line section has degree $2$: if $H|_C=P+Q$, then \begin{align*} \deg(H|_C)=\deg(P+Q)=1+1=2, \end{align*} while if $H|_C=2P$, then \begin{align*} \deg(H|_C)=\deg(2P)=2. \end{align*} Using $\deg(nD)=n\deg(D)$ for divisors on a curve, \begin{align*} \deg\bigl((F)_0|_C\bigr)=\deg\bigl(2(H|_C)\bigr)=2\deg(H|_C)=2\cdot 2=4. \end{align*} Suppose first that $F=L_1L_2$, where the two lines $\ell_i=V(L_i)$ meet $C$ transversely and their line sections have no common point. Write \begin{align*} \ell_1|_C=P_1+P_2 \end{align*} and \begin{align*} \ell_2|_C=Q_1+Q_2. \end{align*} At each $P_j$, the section $L_1|_C$ has order $1$ because $P_j$ occurs with coefficient $1$ in $\ell_1|_C$. The section $L_2|_C$ does not vanish at $P_j$, since the two line sections share no point, so \begin{align*} \operatorname{ord}_{P_j}(L_2|_C)=0. \end{align*} By additivity of order under multiplication, \begin{align*} \operatorname{ord}_{P_j}(F|_C)=\operatorname{ord}_{P_j}(L_1|_C)+\operatorname{ord}_{P_j}(L_2|_C)=1+0=1. \end{align*} Similarly, at each $Q_j$, the section $L_2|_C$ has order $1$, while $L_1|_C$ does not vanish there, so \begin{align*} \operatorname{ord}_{Q_j}(F|_C)=\operatorname{ord}_{Q_j}(L_1|_C)+\operatorname{ord}_{Q_j}(L_2|_C)=0+1=1. \end{align*} If $R\in C$ is not one of $P_1,P_2,Q_1,Q_2$, then $L_1|_C$ and $L_2|_C$ are both units in the local ring at $R$, so their product $F|_C$ is also a unit there and \begin{align*} \operatorname{ord}_R(F|_C)=0. \end{align*} Thus the zero divisor is \begin{align*} (F)_0|_C=P_1+P_2+Q_1+Q_2. \end{align*} Its degree is \begin{align*} \deg(P_1+P_2+Q_1+Q_2)=1+1+1+1=4. \end{align*} Now suppose that one factor is tangent to $C$ at $P$, say \begin{align*} \ell_1|_C=2P, \end{align*} and the other line meets $C$ transversely at two points $Q_1$ and $Q_2$ away from $P$: \begin{align*} \ell_2|_C=Q_1+Q_2. \end{align*} At $P$, the section $L_1|_C$ has order $2$, while $L_2|_C$ is a unit because $P$ is not in $\ell_2|_C$. Hence \begin{align*} \operatorname{ord}_P(F|_C)=\operatorname{ord}_P(L_1|_C)+\operatorname{ord}_P(L_2|_C)=2+0=2. \end{align*} At each $Q_j$, the section $L_2|_C$ has order $1$, while $L_1|_C$ is a unit there, so \begin{align*} \operatorname{ord}_{Q_j}(F|_C)=\operatorname{ord}_{Q_j}(L_1|_C)+\operatorname{ord}_{Q_j}(L_2|_C)=0+1=1. \end{align*} At any other point $R\in C$, both $L_1|_C$ and $L_2|_C$ are units, so $F|_C$ has order $0$ at $R$. Therefore \begin{align*} (F)_0|_C=2P+Q_1+Q_2. \end{align*} Its degree is \begin{align*} \deg(2P+Q_1+Q_2)=2+1+1=4. \end{align*} In both cases the coefficients add to $4$, so tangency changes which points carry the multiplicity, not the total degree of the divisor. [/example] The affine picture of zeros and poles is recovered by choosing a chart and comparing the homogeneous equation with a power of the coordinate that defines the hyperplane at infinity. [example: Affine Polynomial and the Pole at Infinity] On $\mathbb P^1$ with homogeneous coordinates $[X_0:X_1]$, use the affine coordinate $t=X_1/X_0$ on the chart $X_0\ne 0$. After collecting equal roots, write \begin{align*} p(t)=\prod_{i=1}^r(t-a_i)^{m_i}, \end{align*} where the $a_i$ are distinct and each $m_i>0$. Since each factor $(t-a_i)^{m_i}$ has degree $m_i$ and the leading coefficient of the product is $1$, the degree of $p$ is \begin{align*} d=\sum_{i=1}^r m_i. \end{align*} The degree-$d$ homogenisation is obtained by substituting $t=X_1/X_0$ and multiplying by $X_0^d$: \begin{align*} P(X_0,X_1)=X_0^d p(X_1/X_0). \end{align*} Using the factorisation of $p$ gives \begin{align*} P(X_0,X_1)=X_0^d\prod_{i=1}^r\left(\frac{X_1}{X_0}-a_i\right)^{m_i}. \end{align*} For each $i$, \begin{align*} \frac{X_1}{X_0}-a_i=\frac{X_1}{X_0}-\frac{a_iX_0}{X_0}=\frac{X_1-a_iX_0}{X_0}. \end{align*} Therefore \begin{align*} P(X_0,X_1)=X_0^d\prod_{i=1}^r\left(\frac{X_1-a_iX_0}{X_0}\right)^{m_i}. \end{align*} Raising each fraction to the exponent gives \begin{align*} P(X_0,X_1)=X_0^d\prod_{i=1}^r\frac{(X_1-a_iX_0)^{m_i}}{X_0^{m_i}}. \end{align*} Multiplying the denominators gives the power $X_0^{\sum_{i=1}^r m_i}=X_0^d$, so \begin{align*} P(X_0,X_1)=X_0^d\cdot\frac{\prod_{i=1}^r(X_1-a_iX_0)^{m_i}}{X_0^d}. \end{align*} Cancelling the common factor $X_0^d$ gives \begin{align*} P(X_0,X_1)=\prod_{i=1}^r(X_1-a_iX_0)^{m_i}. \end{align*} Thus the corresponding rational function on $\mathbb P^1$ is \begin{align*} \frac{P(X_0,X_1)}{X_0^d}=\prod_{i=1}^r\left(\frac{X_1-a_iX_0}{X_0}\right)^{m_i}. \end{align*} On the affine chart $X_0\ne 0$, \begin{align*} \frac{X_1-a_iX_0}{X_0}=\frac{X_1}{X_0}-a_i=t-a_i. \end{align*} Hence on this chart, \begin{align*} \frac{P(X_0,X_1)}{X_0^d}=\prod_{i=1}^r(t-a_i)^{m_i}=p(t). \end{align*} At the affine point $[a_i]=[1:a_i]$, use the local coordinate $z=t-a_i$, so $t=a_i+z$. For $j\ne i$, \begin{align*} t-a_j=(a_i+z)-a_j. \end{align*} Rearranging terms gives \begin{align*} t-a_j=(a_i-a_j)+z. \end{align*} Since the roots are distinct, $a_i-a_j\ne 0$, so $(a_i-a_j)+z$ has nonzero value at $z=0$ and is a unit in the local ring at $[a_i]$. The $i$th factor is \begin{align*} t-a_i=z. \end{align*} Therefore near $[a_i]$, \begin{align*} p(t)=z^{m_i}\prod_{j\ne i}\bigl((a_i-a_j)+z\bigr)^{m_j}. \end{align*} Each factor in the product over $j\ne i$ is a unit, and a product of units is a unit. Thus $p(t)$ has exactly $m_i$ factors of the local parameter $z$, so \begin{align*} \operatorname{ord}_{[a_i]}(p)=m_i. \end{align*} It remains to compute the order at $\infty=[0:1]$. On the chart $X_1\ne 0$, use the local coordinate \begin{align*} u=X_0/X_1. \end{align*} Then $u=0$ at $\infty$, and on the overlap with the affine chart $X_0\ne 0$, \begin{align*} t=X_1/X_0=1/u. \end{align*} For each $i$, \begin{align*} t-a_i=\frac{1}{u}-a_i. \end{align*} Putting the two terms over the common denominator $u$ gives \begin{align*} t-a_i=\frac{1-a_iu}{u}. \end{align*} Substituting this expression into the factorisation of $p$ yields \begin{align*} p(t)=\prod_{i=1}^r\left(\frac{1-a_iu}{u}\right)^{m_i}. \end{align*} Separating numerator and denominator in each factor gives \begin{align*} p(t)=\prod_{i=1}^r\frac{(1-a_iu)^{m_i}}{u^{m_i}}. \end{align*} Multiplying the numerators together and the denominators together gives \begin{align*} p(t)=\frac{\prod_{i=1}^r(1-a_iu)^{m_i}}{u^{\sum_{i=1}^r m_i}}. \end{align*} Since $d=\sum_{i=1}^r m_i$, this becomes \begin{align*} p(t)=\frac{\prod_{i=1}^r(1-a_iu)^{m_i}}{u^d}. \end{align*} At $u=0$, each factor $1-a_iu$ has value $1$, so each factor is a unit in the local ring at $\infty$, and their product is also a unit. The denominator contributes exactly $d$ powers of the local parameter $u$, so \begin{align*} \operatorname{ord}_{\infty}(p)=-d. \end{align*} If $[c]=[1:c]$ is an affine point with $c\ne a_i$ for every $i$, use the local coordinate $r=t-c$. Then $t=c+r$, and for each $i$, \begin{align*} t-a_i=(c-a_i)+r. \end{align*} Since $c-a_i\ne 0$, the factor $(c-a_i)+r$ is a unit at $r=0$. Hence every factor of $p(t)$ is a unit at $[c]$, so \begin{align*} \operatorname{ord}_{[c]}(p)=0. \end{align*} Thus the only affine zeros are the points $[a_i]$, with orders $m_i$, and the only pole is at $\infty$, with order $d$. Therefore \begin{align*} \operatorname{div}(p)=\sum_{i=1}^r m_i[a_i]-d[\infty]. \end{align*} The affine zeros have total multiplicity $d$, and the single pole of order $d$ at infinity balances them on $\mathbb P^1$. [/example] This comparison is the practical rule for moving between affine and projective calculations. Homogenisation remembers the missing behaviour at infinity, while ratios of homogeneous forms keep track of poles as well as zeros. [remark: Classical Scope] In this chapter, multiplicities are used in the standard nonsingular and hypersurface-section cases. A later scheme-theoretic treatment replaces these classical orders by local algebra at codimension-one points and explains how Cartier and Weil divisors compare on more general varieties. [/remark] Divisors turn local vanishing data into a global bookkeeping device, measuring where hypersurfaces meet a variety and with what multiplicity. This language depends on the local analysis of Chapter 9 and on the projective framework developed earlier, since multiplicity becomes meaningful only when intersections are counted correctly. With divisors in place, the course can move to explicit intersection computations in projective space. # 11. First Intersection Theory Examples This chapter turns the geometric dictionary into a counting tool. Using projective closure from Chapter 6, tangent spaces from Chapter 9, and divisors from Chapter 10, we now count intersections of plane curves with multiplicity. The guiding problem is that the visible set-theoretic intersection often gives the wrong number: tangent points, singular points, and points at infinity must contribute extra weight. ## Local Intersection Multiplicity for Plane Curves If two affine plane curves meet at a point, the first question is how much that point should count. A transverse crossing should count once, while a tangency should count more, and the algebra must detect this distinction without depending on a parametrisation or a chosen picture. [definition: Local Ring of the Affine Plane at a Point] Let $k$ be an algebraically closed field and let $p=(a,b)\in \mathbb A^2$. The local ring of $\mathbb A^2$ at $p$ is \begin{align*} \mathcal O_{\mathbb A^2,p}=k[x,y]_{\mathfrak m_p}, \qquad \mathfrak m_p=(x-a,y-b). \end{align*} [/definition] This ring keeps track only of functions near $p$: a polynomial that does not vanish at $p$ becomes invertible. The next definition uses this local ring to convert the pair of curve equations into a finite vector-space length, which is the local number we want to attach to the point. [definition: Local Intersection Multiplicity] Let $F,G\in k[x,y]$ define plane affine curves with no common component through $p\in \mathbb A^2$. The local intersection multiplicity of $F$ and $G$ at $p$ is \begin{align*} I_p(F,G)=\dim_k \mathcal O_{\mathbb A^2,p}/(F,G). \end{align*} [/definition] The condition excluding a common component through $p$ ensures that the quotient is a finite-dimensional $k$-vector space. If the curves share a component, the intersection near $p$ is not isolated, so a finite local number is not the right invariant. [example: Transverse Coordinate Axes] At $p=(0,0)$, the local ring is $\mathcal O_{\mathbb A^2,p}=k[x,y]_{(x,y)}$. For $F=x$ and $G=y$, the quotient by the two curve equations is \begin{align*} \mathcal O_{\mathbb A^2,p}/(x,y)=k[x,y]_{(x,y)}/(x,y). \end{align*} The evaluation homomorphism \begin{align*} k[x,y]_{(x,y)}\longrightarrow k,\qquad \frac{h(x,y)}{u(x,y)}\longmapsto \frac{h(0,0)}{u(0,0)} \end{align*} is well-defined because every denominator $u\notin (x,y)$ satisfies $u(0,0)\ne 0$. Its kernel is exactly the ideal generated by $x$ and $y$, since $\frac{h}{u}$ maps to $0$ precisely when $h(0,0)=0$, equivalently $h\in (x,y)$. Therefore \begin{align*} k[x,y]_{(x,y)}/(x,y)\cong k. \end{align*} Thus $I_p(x,y)=\dim_k k=1$. The two coordinate axes contribute one intersection at the origin because the two equations remove both local coordinates and leave only constant functions. [/example] A tangency should force the quotient to remember first-order contact along one direction. The next computation shows how nilpotents in the quotient record that extra contact. [example: Tangent Line to a Parabola] Let $F=y-x^2$ and $G=y$ at $p=(0,0)$, so $\mathcal O_{\mathbb A^2,p}=k[x,y]_{(x,y)}$. We compute the local quotient \begin{align*} \mathcal O_{\mathbb A^2,p}/(F,G)=k[x,y]_{(x,y)}/(y-x^2,y). \end{align*} Since $y$ lies in the ideal, the relation $y=0$ holds in the quotient. Since $y-x^2$ also lies in the ideal, substituting $y=0$ into $y-x^2=0$ gives $-x^2=0$, hence $x^2=0$. Define \begin{align*} \phi:k[x,y]_{(x,y)}\to k[x]_{(x)},\qquad \phi\!\left(\frac{h(x,y)}{u(x,y)}\right)=\frac{h(x,0)}{u(x,0)}. \end{align*} This is well-defined because if $u\notin (x,y)$, then $u(0,0)\ne 0$, so $u(x,0)\notin (x)$ and is invertible in $k[x]_{(x)}$. The ideal $(y)$ is the kernel of $\phi$, so \begin{align*} k[x,y]_{(x,y)}/(y)\cong k[x]_{(x)}. \end{align*} Under this isomorphism, the remaining relation $y-x^2=0$ becomes $-x^2=0$, so \begin{align*} k[x,y]_{(x,y)}/(y-x^2,y)\cong k[x]_{(x)}/(x^2). \end{align*} Every class in $k[x]_{(x)}/(x^2)$ has a unique representative $a+bx$: higher powers vanish because $x^2=0$, and any denominator with nonzero constant term is a unit whose inverse reduces to a linear expression modulo $x^2$. Thus the quotient has $k$-basis $1,x$, and \begin{align*} I_p(F,G)=\dim_k k[x]_{(x)}/(x^2)=2. \end{align*} The tangent line $y=0$ therefore contributes two intersections with the parabola at the origin: one for the point itself and one for the first-order tangency remembered by the nilpotent class of $x$. [/example] The algebra also separates higher contact from accidental simplification. If a line cuts a branch to order $r$, the local quotient often becomes a truncated polynomial algebra of length $r$. [example: Cubic Flex and Its Tangent] Let $F=y-x^3$ and $G=y$ at $p=(0,0)$, so $\mathcal O_{\mathbb A^2,p}=k[x,y]_{(x,y)}$. We compute \begin{align*} \mathcal O_{\mathbb A^2,p}/(F,G)=k[x,y]_{(x,y)}/(y-x^3,y). \end{align*} Because $y$ lies in the ideal, every class satisfies $y=0$. Because $y-x^3$ also lies in the ideal, the same quotient also satisfies $y-x^3=0$, and substituting $y=0$ gives $-x^3=0$, hence $x^3=0$. The substitution map \begin{align*} \phi:k[x,y]_{(x,y)}\to k[x]_{(x)},\qquad \phi\!\left(\frac{h(x,y)}{u(x,y)}\right)=\frac{h(x,0)}{u(x,0)} \end{align*} is well-defined because $u\notin (x,y)$ implies $u(0,0)\ne 0$, so $u(x,0)\notin (x)$ and is invertible in $k[x]_{(x)}$. Its kernel is $(y)$, since $\phi(h/u)=0$ exactly when $h(x,0)=0$, equivalently $h(x,y)$ is divisible by $y$. Therefore \begin{align*} k[x,y]_{(x,y)}/(y)\cong k[x]_{(x)}. \end{align*} Under this isomorphism, the remaining generator $y-x^3$ maps to $-x^3$, so \begin{align*} k[x,y]_{(x,y)}/(y-x^3,y)\cong k[x]_{(x)}/(x^3). \end{align*} Every element of $k[x]_{(x)}/(x^3)$ has a representative $a+bx+cx^2$: powers $x^n$ with $n\ge 3$ vanish because $x^3=0$, and if $u(x)=u_0+u_1x+u_2x^2+\cdots$ with $u_0\ne 0$, then its inverse modulo $x^3$ has the form $v_0+v_1x+v_2x^2$, determined by matching coefficients in $u(x)(v_0+v_1x+v_2x^2)\equiv 1 \pmod{x^3}$. The classes $1,x,x^2$ are linearly independent because $a+bx+cx^2\in (x^3)$ forces $a=b=c=0$. Hence this quotient has $k$-basis $1,x,x^2$, so \begin{align*} I_p(F,G)=\dim_k k[x]_{(x)}/(x^3)=3. \end{align*} The tangent line $y=0$ therefore contributes three intersections with the cubic at the flex: the quotient keeps the point, the first-order class $x$, and the second-order class $x^2$. [/example] The definition is local, but it still agrees with the first-order picture when the tangent directions are distinct. This gives the multiplicity-one criterion used in computations. [quotetheorem:9452] [citeproof:9452] This theorem is often the fastest way to count ordinary intersection points, but each hypothesis is doing real work. If the tangent lines are not distinct, multiplicity can increase even for smooth curves, as the tangent line to a parabola already showed. If nonsingularity is dropped, the tangent-line picture can be misleading: at a node there are several branch directions, and a line may meet the singularity with total multiplicity coming from more than one branch. The theorem therefore identifies the clean multiplicity-one situation; outside it, the quotient-length definition is the reliable tool. [example: Two Branches at a Node] Let $F=y^2-x^2-x^3$ and $G=y$ at $p=(0,0)$, so $\mathcal O_{\mathbb A^2,p}=k[x,y]_{(x,y)}$. The lowest-degree homogeneous part of $F$ is \begin{align*} F_2=y^2-x^2=(y-x)(y+x), \end{align*} so the tangent cone of $F=0$ has the two directions $y=x$ and $y=-x$, while the line $G=0$ has tangent direction $y=0$. We compute the local quotient \begin{align*} \mathcal O_{\mathbb A^2,p}/(F,G)=k[x,y]_{(x,y)}/(y^2-x^2-x^3,y). \end{align*} Because $y$ lies in the ideal, the quotient satisfies $y=0$. Substituting this relation into the other generator gives \begin{align*} y^2-x^2-x^3=0^2-x^2-x^3=-x^2(1+x). \end{align*} Thus the same quotient is \begin{align*} k[x,y]_{(x,y)}/(y^2-x^2-x^3,y)\cong k[x]_{(x)}/(x^2(1+x)). \end{align*} In $k[x]_{(x)}$, the element $1+x$ is a unit because its value at $x=0$ is $1\ne 0$. Multiplication by this unit does not change the generated ideal, so \begin{align*} (x^2(1+x))=(x^2). \end{align*} Therefore \begin{align*} k[x]_{(x)}/(x^2(1+x))\cong k[x]_{(x)}/(x^2). \end{align*} Every class in $k[x]_{(x)}/(x^2)$ has a unique representative $a+bx$: powers $x^n$ with $n\ge 2$ vanish, and any denominator with nonzero constant term has an inverse modulo $x^2$. Hence the quotient has $k$-basis $1,x$, so \begin{align*} I_p(F,G)=\dim_k k[x]_{(x)}/(x^2)=2. \end{align*} The line $y=0$ is not tangent to either branch direction $y=x$ or $y=-x$, but it passes through both local branches at the node, so the total local contribution is $2$. [/example] ## Bezout's Theorem for Projective Plane Curves Local multiplicities solve the problem at a fixed point, but a global counting theorem must also decide where the points live. The affine plane misses intersections at infinity, while projective space supplies the compact setting in which the total number is stable. [definition: Projective Plane Curve] A projective plane curve over $k$ is the zero locus $V_+(F)\subset\mathbb P^2$ of a nonzero homogeneous polynomial $F\in k[X,Y,Z]$. Its degree is the degree of $F$. [/definition] For a projective point $p\in\mathbb P^2$, local intersection multiplicity is computed by passing to an affine chart containing $p$ and dehomogenising the two equations. The result is independent of the chart because changing affine coordinates induces an isomorphism of local rings at $p$. [definition: Projective Local Intersection Multiplicity] Let $C=V_+(F)$ and $D=V_+(G)$ be projective plane curves with no common component, and let $p\in C\cap D$. Choose an affine chart $U\cong\mathbb A^2$ containing $p$, and let $f,g$ be the corresponding dehomogenisations of $F,G$ on $U$. The projective local intersection multiplicity is \begin{align*} I_p(C,D)=I_p(f,g). \end{align*} [/definition] This definition extends quotient-length counting to all projective points, including points on the line at infinity. The natural global question is whether the sum of these local contributions depends on the equations themselves or only on the degrees of the two curves. [quotetheorem:2131] [citeproof:2131] Bezout is not just a counting statement; it is a warning against counting only visible affine points. The hypotheses also explain the main ways the formula can fail if misused. If $C$ and $D$ share a component, then $C\cap D$ contains infinitely many points along that component, so a finite sum of isolated local multiplicities is not the right object. If $k$ is not algebraically closed, some intersection points may only appear after extending scalars; for instance, two real conics can have non-real complex intersection points that still contribute to the degree count over $\mathbb C$. With the hypotheses in place, the missing contributions are exactly tangencies, singular local lengths, and intersections on the line at infinity. [example: Two Conics Meeting in Four Points] Let $C=V_+(X^2+Y^2-Z^2)$ and $D=V_+(X^2-YZ)$ over an algebraically closed field $k$ with $\operatorname{char}(k)\ne 2,5$. First check that there are no intersections on the line at infinity. If $Z=0$, the equation $X^2-YZ=0$ gives $X^2=0$, hence $X=0$, and then $X^2+Y^2-Z^2=0$ gives $Y^2=0$, hence $Y=0$. This would be the invalid projective point $[0:0:0]$, so every intersection lies in the affine chart $Z=1$. In this chart the equations are \begin{align*} f=x^2+y^2-1,\qquad g=x^2-y. \end{align*} The equation $g=0$ gives $y=x^2$. Substituting this into $f=0$ gives \begin{align*} x^2+(x^2)^2-1=x^4+x^2-1=0. \end{align*} Put $t=x^2$. Then $t$ must satisfy \begin{align*} q(t)=t^2+t-1=0. \end{align*} The discriminant of $q$ is \begin{align*} 1^2-4(1)(-1)=5, \end{align*} which is nonzero because $\operatorname{char}(k)\ne 5$. Since $k$ is algebraically closed, $q$ has two distinct roots $\alpha$ and $\beta$. Neither root is $0$, because $q(0)=-1\ne 0$. For each nonzero root $t$, the equation $x^2=t$ has two distinct solutions because $\operatorname{char}(k)\ne 2$. Thus the four affine intersection points are \begin{align*} (x,y)=(x,\alpha)\text{ with }x^2=\alpha,\qquad (x,y)=(x,\beta)\text{ with }x^2=\beta, \end{align*} two points over $\alpha$ and two points over $\beta$. It remains to check that each of these four points is transverse. The gradients are \begin{align*} \nabla f=(2x,2y),\qquad \nabla g=(2x,-1). \end{align*} Their determinant is \begin{align*} (2x)(-1)-(2y)(2x)=-2x-4xy=-2x(1+2y). \end{align*} At an intersection point, $y=t$ where $t^2+t-1=0$, and $x^2=t\ne 0$, so $x\ne 0$. If $1+2t=0$, then $t=-1/2$, and substituting into $q$ gives \begin{align*} q(-1/2)=(-1/2)^2+(-1/2)-1=1/4-1/2-1=-5/4\ne 0, \end{align*} because $\operatorname{char}(k)\ne 2,5$. Hence $1+2y\ne 0$, so the determinant is nonzero at every intersection point. Therefore the tangent lines are distinct at all four points, and by *Transverse Intersections Have Multiplicity One*, each point has local intersection multiplicity $1$. The total contribution is \begin{align*} 1+1+1+1=4, \end{align*} matching the Bezout number $\deg(C)\deg(D)=2\cdot 2=4$. [/example] When roots merge, Bezout does not fail; multiplicities increase. The tangent line to a conic is the basic example of a degree-one curve meeting a degree-two curve with all intersection concentrated at one point. [example: Tangent Line to a Smooth Conic] Let $C=V_+(XZ-Y^2)$ and $L=V_+(X)$. A point of $C\cap L$ satisfies $X=0$, and then the conic equation gives \begin{align*} XZ-Y^2=0\cdot Z-Y^2=-Y^2=0. \end{align*} Since $k$ is a field, $Y^2=0$ implies $Y=0$. The point cannot be $[0:0:0]$, so $Z\ne 0$, and the unique intersection point is $p=[0:0:1]$. Work in the affine chart $Z=1$, with coordinates $x=X/Z$ and $y=Y/Z$. Then $p=(0,0)$), and the two equations become \begin{align*} f=x-y^2,\qquad g=x. \end{align*} Thus the local quotient is \begin{align*} \mathcal O_{\mathbb A^2,p}/(f,g)=k[x,y]_{(x,y)}/(x-y^2,x). \end{align*} Because $x$ lies in the ideal, every class satisfies $x=0$. The other generator then gives \begin{align*} x-y^2=0-y^2=-y^2, \end{align*} so the quotient satisfies $y^2=0$. The substitution map \begin{align*} k[x,y]_{(x,y)}\to k[y]_{(y)},\qquad h(x,y)/u(x,y)\mapsto h(0,y)/u(0,y) \end{align*} is well-defined because $u(0,0)\ne 0$ implies $u(0,y)\notin (y)$. Its kernel is $(x)$, so \begin{align*} k[x,y]_{(x,y)}/(x)\cong k[y]_{(y)}. \end{align*} Under this isomorphism, $x-y^2$ maps to $-y^2$, hence \begin{align*} k[x,y]_{(x,y)}/(x-y^2,x)\cong k[y]_{(y)}/(y^2). \end{align*} Every class in $k[y]_{(y)}/(y^2)$ has a unique representative $a+by$, so the quotient has $k$-basis $1,y$. Therefore \begin{align*} I_p(C,L)=\dim_k k[y]_{(y)}/(y^2)=2. \end{align*} The tangent line $X=0$ contributes the full Bezout number $\deg C\deg L=2\cdot 1=2$ at the single point $[0:0:1]$. [/example] ## Tangencies, Singularities, and Points at Infinity The final issue is diagnostic: given a count that seems too small, where did the missing intersection number go? There are three standard answers in this first theory: tangent contact, singular local structure, and projective closure. [definition: Transverse Intersection of Plane Curves] Let $C=V(F)$ and $D=V(G)$ be affine plane curves meeting at $p$. They intersect transversely at $p$ if $C$ and $D$ are nonsingular at $p$ and their tangent lines in $T_p\mathbb A^2$ are distinct. [/definition] Transversality is the local condition under which set-theoretic and multiplicity-theoretic counting should agree at a point. The obstruction to quick counting is that intersection multiplicity is defined by a local algebra calculation, and in general that calculation can be larger than the visible number of common points. Distinct tangent directions are exactly the situation where no hidden higher contact should remain. [quotetheorem:9453] [citeproof:9453] The theorem is a detection result for multiplicity one, not a complete classification of higher multiplicity. If the curves are smooth but tangent, the local quotient can have length greater than one, as for a tangent line to a conic. If one curve is singular at $p$, a line with a distinct-looking direction may still collect contributions from several branches, as in the nodal example above. The same method gives a practical workflow: compute tangent lines first, count the transverse points immediately, and reserve quotient-length calculations for tangent, singular, or projective-boundary points. [example: Affine Intersections Lost at Infinity] In $\mathbb A^2$, the equations $y=0$ and $xy=1$ have no common solution, because substituting $y=0$ into $xy=1$ gives $0=1$. Their projective closures are \begin{align*} Y=0 \end{align*} and \begin{align*} XY-Z^2=0. \end{align*} On the line at infinity $Z=0$, a common point satisfies $Y=0$ and $XY-Z^2=X\cdot 0-0^2=0$. Since $[0:0:0]$ is not a projective point, $X\ne 0$, so the unique point at infinity is $p=[1:0:0]$. Work in the affine chart $X=1$, with coordinates $y=Y/X$ and $z=Z/X$. Then $p=(0,0)$), and the two local equations become \begin{align*} g_1=y \end{align*} and \begin{align*} g_2=y-z^2. \end{align*} Thus the local quotient is \begin{align*} \mathcal O_{\mathbb A^2,p}/(g_1,g_2)=k[y,z]_{(y,z)}/(y,y-z^2). \end{align*} Because $y$ lies in the ideal, the quotient satisfies $y=0$. The second generator then gives \begin{align*} y-z^2=0-z^2=-z^2, \end{align*} so the quotient also satisfies $z^2=0$. The substitution map \begin{align*} k[y,z]_{(y,z)}\to k[z]_{(z)},\qquad h(y,z)/u(y,z)\mapsto h(0,z)/u(0,z) \end{align*} is well-defined because $u(0,0)\ne 0$ implies $u(0,z)\notin (z)$. Its kernel is $(y)$, so \begin{align*} k[y,z]_{(y,z)}/(y)\cong k[z]_{(z)}. \end{align*} Under this isomorphism, $y-z^2$ maps to $-z^2$, hence \begin{align*} k[y,z]_{(y,z)}/(y,y-z^2)\cong k[z]_{(z)}/(z^2). \end{align*} Every class in $k[z]_{(z)}/(z^2)$ has a unique representative $a+bz$: powers $z^n$ with $n\ge 2$ vanish, and every denominator with nonzero constant term has an inverse modulo $z^2$. Therefore the quotient has $k$-basis $1,z$, and \begin{align*} I_p= \dim_k k[z]_{(z)}/(z^2)=2. \end{align*} Thus the affine chart contains no visible intersections, while the unique point at infinity carries local multiplicity $2$, accounting for the degree product $1\cdot 2$ predicted by *[Bezout Theorem](/theorems/728) for Plane Curves*. [/example] This example shows why projective closure is not a cosmetic step: affine intersections may disappear from the visible chart while the projective local lengths remain conserved. The next theorem makes this conservation precise by separating the affine contributions from the contributions on the line at infinity. [quotetheorem:9454] [citeproof:9454] The theorem is the first form of a general principle: intersection numbers are stable when the correct compactification and local multiplicities are used. The condition that the affine curves have no common component is essential, because a shared component remains a positive-dimensional intersection after projective closure and cannot be measured by a finite Bezout sum of isolated points. The use of projective closure is also essential: without the line at infinity, affine counts can drop when roots escape the chosen chart. The degree assumptions enter through the homogenised equations, so the theorem does not by itself locate the missing points or compute their individual lengths; it only says that the affine contributions and the boundary contributions together add to $mn$. Later theories replace this quotient-length calculation by intersection products in Chow groups or by scheme-theoretic lengths, but the plane-curve examples already contain the essential idea. [example: Cubic and Line Through a Flex] Let $C=V_+(YZ^2-X^3)$ and $L=V_+(Y)$. A point of $C\cap L$ satisfies $Y=0$, and then the cubic equation becomes \begin{align*} YZ^2-X^3=0\cdot Z^2-X^3=-X^3=0. \end{align*} Since $k$ is a field, $X^3=0$ implies $X=0$. The point cannot be $[0:0:0]$, so $Z\ne 0$, and the unique intersection point is $p=[0:0:1]$. Work in the affine chart $Z=1$, with coordinates $x=X/Z$ and $y=Y/Z$. Then $p=(0,0)$, and the equations become \begin{align*} f=y-x^3 \end{align*} and \begin{align*} g=y. \end{align*} Thus the local quotient is \begin{align*} k[x,y]_{(x,y)}/(y-x^3,y). \end{align*} Because $y$ lies in the ideal, every class satisfies $y=0$. The other generator then gives \begin{align*} y-x^3=0-x^3=-x^3, \end{align*} so the quotient also satisfies $x^3=0$. The substitution map \begin{align*} k[x,y]_{(x,y)}\to k[x]_{(x)},\qquad h(x,y)/u(x,y)\mapsto h(x,0)/u(x,0) \end{align*} is well-defined because $u(0,0)\ne 0$ implies $u(x,0)\notin (x)$. Its kernel is $(y)$, so \begin{align*} k[x,y]_{(x,y)}/(y)\cong k[x]_{(x)}. \end{align*} Under this isomorphism, $y-x^3$ maps to $-x^3$, hence \begin{align*} k[x,y]_{(x,y)}/(y-x^3,y)\cong k[x]_{(x)}/(x^3). \end{align*} Every class in $k[x]_{(x)}/(x^3)$ has a unique representative $a+bx+cx^2$: powers $x^n$ with $n\ge 3$ vanish, and every denominator with nonzero constant term has an inverse modulo $x^3$. Therefore the quotient has $k$-basis $1,x,x^2$, and \begin{align*} I_p(C,L)=\dim_k k[x]_{(x)}/(x^3)=3. \end{align*} The line through the flex contributes all three intersections with the cubic at the single point $[0:0:1]$, matching the Bezout number $3\cdot 1$. [/example] Intersection theory makes the dictionary computational: once projective closure, tangent spaces, and divisors are available, intersection multiplicities can be counted rather than guessed. The flex-line example shows why scheme-theoretic multiplicity is needed, since a single point can account for all expected intersections. The final chapter then gathers these constructions into a practical summary of how algebra reads geometry and geometry reads algebra. # 12. Synthesis: The Classical Dictionary in Practice ## Reading Geometry from Algebra When a variety is presented by equations, the first task is not merely to draw its zero set. We want a systematic translation from the defining ideal to geometric information: functions on the variety, irreducible components, local behaviour, dimension, and tangent directions. [definition: Affine Algebraic Dictionary] Let $k$ be an algebraically closed field, let $X \subset \mathbb A^n$ be an affine variety with ideal $I(X) \trianglelefteq k[x_1, \dots, x_n]$, and let $p \in X$. The coordinate ring of $X$ is \begin{align*} k[X] := k[x_1, \dots, x_n]/I(X). \end{align*} The local ring of $X$ at $p$ is \begin{align*} \mathcal O_{X,p} := k[X]_{\mathfrak m_p}, \end{align*} where $\mathfrak m_p = \{f \in k[X] : f(p)=0\}$. The Zariski tangent space is \begin{align*} T_pX := \{v \in k^n : \sum_{i=1}^n (\partial_i f)(p)v_i = 0 \text{ for all } f \in I(X)\}. \end{align*} [/definition] The coordinate ring records global regular functions, while the local ring isolates what happens near a chosen point. The tangent space is the first-order part of the same local information. To use these constructions reliably, we need the theorem that identifies which algebraic features of $k[X]$ correspond to points, subvarieties, functions, and dimension. [quotetheorem:9455] [citeproof:9455] This theorem is the working version of the affine half of the course, but each hypothesis is doing real work. Algebraic closedness is what turns maximal ideals into actual points of affine space; over a non-algebraically closed field, maximal ideals may correspond instead to Galois orbits or closed points with residue field extensions. The theorem also speaks about reduced varieties through $I(X)$, not about every quotient $k[x_1,\dots,x_n]/I$ with the same zero set: nonreduced quotients retain nilpotent and multiplicity information that the underlying classical variety discards. It does not say that every algebraic feature of an arbitrary presentation is geometric in the reduced sense; before applying the dictionary to the visible set, replace the ideal by its radical. The next example isolates this distinction, since it has the same point set as a line but a quotient ring with extra nilpotent structure. [example: Double Axis Ideal] Let $I=(x^2,xy)\trianglelefteq k[x,y]$. A point $(a,b)\in \mathbb A^2$ lies in $V(I)$ exactly when \begin{align*} a^2=0 \quad \text{and} \quad ab=0. \end{align*} Since $k$ is a field, $a^2=0$ implies $a=0$, and then $ab=0$ imposes no condition on $b$. Hence \begin{align*} V(I)=\{(0,b):b\in k\}=V(x). \end{align*} The ideal of this zero set is $(x)$: if $f\in k[x,y]$ vanishes on every $(0,b)$, then $f(0,y)=0$ as a polynomial in $k[y]$, so every term of $f$ is divisible by $x$. Thus $I(V(I))=(x)$. Also $\sqrt I=(x)$, because $x^2\in I$ gives $x\in \sqrt I$, so $(x)\subseteq \sqrt I$; conversely $I\subseteq (x)$, and $(x)$ is prime since $k[x,y]/(x)\cong k[y]$ is a domain, so $\sqrt I\subseteq (x)$. Therefore the coordinate ring of the reduced variety is \begin{align*} k[x,y]/(x)\cong k[y]. \end{align*} In the nonreduced quotient $A=k[x,y]/(x^2,xy)$, the class $\bar x$ is nonzero but satisfies \begin{align*} \bar x^2=0 \quad \text{and} \quad \bar y\bar x=0. \end{align*} Indeed, $x\notin (x^2,xy)$: if $x=x^2a+xyb$, then $x=x(xa+yb)$, so $1=xa+yb$ in the domain $k[x,y]$, which becomes $1=0$ after setting $x=0$ and $y=0$. Thus radicalisation replaces the nonreduced algebra $A$ by the ordinary coordinate ring of the visible $y$-axis and discards the nilpotent direction represented by $\bar x$. [/example] The example shows how radicalisation discards nilpotent information. A different algebraic operation, primary or prime decomposition, detects whether the visible set itself breaks into smaller pieces. [example: Coordinate Axes] For $(a,b)\in \mathbb A^2$, the equation $xy=0$ becomes $ab=0$. Since $k$ is a field, $ab=0$ implies $a=0$ or $b=0$, so \begin{align*} V(xy)=\{(0,b):b\in k\}\cup \{(a,0):a\in k\}=V(x)\cup V(y). \end{align*} In $k[x,y]$ we have $(xy)\subseteq (x)\cap (y)$ because $xy$ is divisible by both $x$ and $y$. Conversely, if $f\in (x)\cap (y)$, then $f=xg$ for some $g\in k[x,y]$, and also $y\mid f$. Since $y\nmid x$ and $k[x,y]$ is a [unique factorisation domain](/page/Unique%20Factorisation%20Domain), $y\mid g$, so $g=yh$ for some $h\in k[x,y]$. Hence $f=xyh\in (xy)$, and therefore \begin{align*} (xy)=(x)\cap (y). \end{align*} Thus the coordinate ring is \begin{align*} k[V(xy)]=k[x,y]/(xy). \end{align*} The ideals $(x)$ and $(y)$ are prime because $k[x,y]/(x)\cong k[y]$ and $k[x,y]/(y)\cong k[x]$ are domains. If $P$ is any prime ideal containing $(xy)$, then $xy\in P$, so $x\in P$ or $y\in P$; hence $P$ contains $(x)$ or $(y)$. Therefore the two minimal primes of $k[x,y]/(xy)$ are $(x)/(xy)$ and $(y)/(xy)$, corresponding to the two axes. At the origin, the defining polynomial is $f=xy$, with \begin{align*} \partial_x f(0,0)=0 \quad \text{and} \quad \partial_y f(0,0)=0. \end{align*} For $v=(v_1,v_2)\in k^2$, the tangent condition is \begin{align*} \partial_x f(0,0)v_1+\partial_y f(0,0)v_2=0v_1+0v_2=0. \end{align*} Every $v\in k^2$ satisfies this equation, so $T_{(0,0)}V(xy)=k^2$. The tangent space remembers both crossing directions at once, even though each irreducible component is only a line. [/example] The coordinate axes show that tangent spaces can detect more than set-theoretic membership: at the crossing point, the first-order directions remember both branches at once. We therefore need a computable criterion that turns the abstract definition of $T_pX$ into a Jacobian calculation and compares it with dimension. [quotetheorem:9456] [citeproof:9456] The criterion is especially useful because it separates features visible in the set from features visible only infinitesimally, but it must be applied to the ideal of the reduced variety. A presentation such as $V(x^2)$ has the same underlying point as $V(x)$, yet differentiating only the nonreduced equation $x^2$ gives a larger first-order solution space than the tangent space of the reduced point. The comparison is also local: tangent dimension has to be compared with the dimension of the component through $p$, or with the local dimension $\dim_p X$, rather than with an unrelated ambient or global number when several components are present. The Jacobian test therefore detects singularity for classical affine varieties once the correct ideal and local dimension are in hand; it does not by itself describe embedded nilpotents, multiplicities, or the full analytic shape of a singularity. [example: Cusp Tangent Space] Let $f=y^2-x^3$, so $X=V(f)\subset \mathbb A^2$. The partial derivatives are \begin{align*} \partial_x f=-3x^2,\qquad \partial_y f=2y. \end{align*} Thus at $p=(a,b)\in X$, the Jacobian row is \begin{align*} J_p=(-3a^2,2b), \end{align*} and the tangent condition from *Jacobian Tangent Criterion for Affine Varieties* is \begin{align*} -3a^2v_1+2bv_2=0 \end{align*} for $v=(v_1,v_2)\in k^2$. Assume $p\ne (0,0)$. Since $p\in X$, we have $b^2=a^3$. If $b=0$, then $a^3=0$, hence $a=0$ because $k$ is a field, contradicting $p\ne (0,0)$. Therefore $b\ne 0$, and, in the usual characteristic assumptions for this computation, $2b\ne 0$. The tangent equation can then be solved as \begin{align*} v_2=\frac{3a^2}{2b}v_1. \end{align*} So $v_1$ is free and $v_2$ is determined by it, giving $\dim_k T_pX=1$ away from the origin. At the origin, \begin{align*} J_{(0,0)}=(-3\cdot 0^2,2\cdot 0)=(0,0), \end{align*} so the tangent condition is \begin{align*} 0v_1+0v_2=0. \end{align*} Every vector $v=(v_1,v_2)\in k^2$ satisfies this equation, hence \begin{align*} T_{(0,0)}X=k^2. \end{align*} Since $X$ is a curve, its dimension is $1$, and the tangent space at the origin has dimension $2>1$; the origin is singular. The parametrisation $\gamma(t)=(t^2,t^3)$ reflects the same infinitesimal failure, because \begin{align*} \gamma'(t)=(2t,3t^2) \end{align*} and therefore \begin{align*} \gamma'(0)=(0,0). \end{align*} The cusp has one visible branch through the origin, but its defining equation imposes no first-order linear condition there. [/example] ## Building Geometry from Algebra The reverse problem begins with an algebraic operation and asks what geometric construction it represents. Quotients cut out subvarieties, homomorphisms reverse arrows between varieties, homogeneous ideals define projective sets, and elimination describes images and closures. [definition: Regular Morphism via Coordinate Rings] Let $X\subset \mathbb A^n$ and $Y\subset \mathbb A^m$ be affine varieties over $k$. A regular morphism $\varphi:X\to Y$ is a map whose coordinate functions are regular functions on $X$. The pullback homomorphism is the $k$-algebra map \begin{align*} \varphi^*:k[Y]\to k[X],\qquad \varphi^*(g)=g\circ \varphi. \end{align*} [/definition] This definition makes morphisms contravariant: maps of spaces become maps of rings in the opposite direction. The question is whether every compatible homomorphism of coordinate rings actually comes from a geometric map, or whether pullback only produces some special examples. For affine varieties, the coordinate ring has enough functions to recover the map itself. [quotetheorem:2127] [citeproof:2127] Affineness is essential in this theorem: the coordinate ring of an affine variety contains enough global functions to reconstruct the variety and its morphisms, while projective varieties do not have such a direct ring-of-regular-functions description. The theorem is also contravariant, so geometric operations often reverse the expected algebraic direction; closed inclusions become quotient maps of coordinate rings. A further limitation is that a bijective regular map need not be an isomorphism, since the inverse map may fail to be regular and the pullback on coordinate rings may fail to be onto. The cusp is the standard example where a parametrisation captures the set of points but not the local ring at the singular point, so it tests the difference between set-theoretic bijection and isomorphism of varieties. [example: Cuspidal Parametrisation] Let $\varphi:\mathbb A^1\to \mathbb A^2$ be given by $\varphi(t)=(t^2,t^3)$, and let $C=V(y^2-x^3)$. For every $t\in k$, \begin{align*} (t^3)^2-(t^2)^3=t^6-t^6=0. \end{align*} Thus $\varphi(t)\in C$, so $\operatorname{im}\varphi\subseteq C$. Conversely, let $(x,y)\in C$, so $y^2=x^3$. If $x=0$, then $y^2=0$, hence $y=0$ because $k$ is a field, and $(x,y)=(0,0)=\varphi(0)$. If $x\ne 0$, set $t=y/x$. Then \begin{align*} t^2=\left(\frac{y}{x}\right)^2=\frac{y^2}{x^2}=\frac{x^3}{x^2}=x. \end{align*} Using this equality, \begin{align*} t^3=t\cdot t^2=\frac{y}{x}\cdot x=y. \end{align*} Hence $(x,y)=(t^2,t^3)$, and therefore $\operatorname{im}\varphi=C$. The pullback on coordinate rings is \begin{align*} \varphi^*:k[x,y]/(y^2-x^3)\to k[t],\qquad \overline{x}\mapsto t^2,\quad \overline{y}\mapsto t^3. \end{align*} It is well-defined because the defining relation maps to zero: \begin{align*} \varphi^*(\overline{y}^2-\overline{x}^3)=(t^3)^2-(t^2)^3=t^6-t^6=0. \end{align*} The image of $\varphi^*$ is exactly the subring $k[t^2,t^3]\subset k[t]$, since every polynomial expression in $\overline{x}$ and $\overline{y}$ maps to the same expression in $t^2$ and $t^3$, and the elements $t^2,t^3$ themselves occur as images of $\overline{x},\overline{y}$. This subring is not all of $k[t]$. Indeed, every element of $k[t^2,t^3]$ has the form $F(t^2,t^3)$ for some $F(u,v)\in k[u,v]$. Each monomial $u^av^b$ maps to \begin{align*} (t^2)^a(t^3)^b=t^{2a+3b}. \end{align*} If $a=b=0$, this gives a constant term; otherwise $2a+3b\ge 2$. Thus no element of $k[t^2,t^3]$ has a nonzero $t^1$ term, while $t\in k[t]$ does. Hence $t\notin k[t^2,t^3]$, so $\varphi^*$ is not surjective. Therefore the parametrisation reaches every point of the cusp, but it is not an isomorphism of affine varieties: the coordinate ring of the cusp maps into $k[t]$ as $k[t^2,t^3]$, and the missing function $t$ records the singular algebra hidden at the cusp. [/example] The cusp computation is affine, but many geometric questions require adding points at infinity before the algebra becomes stable under projection and intersection. This motivates the projective version of the dictionary, where homogeneous equations replace ordinary equations. [definition: Projective Algebraic Dictionary] Let $S=k[x_0,\dots,x_n]$ be graded by total degree, and let $X\subset \mathbb P^n$ be a projective variety with homogeneous ideal $I(X)\trianglelefteq S$. The homogeneous coordinate ring of $X$ is \begin{align*} S(X):=S/I(X). \end{align*} For $D_+(x_i)=\{[x_0:\cdots:x_n]\in \mathbb P^n:x_i\ne 0\}$, the affine chart $X\cap D_+(x_i)$ has coordinate ring obtained by dehomogenising with $x_i=1$. [/definition] The homogeneous coordinate ring is not the ring of all regular functions on a projective variety; instead, its grading stores how equations behave under scaling. Once this graded object is in place, the next question is how an affine variety sits inside projective space and how its missing limiting points are recovered algebraically. The projective-closure theorem [Projective Closure of an Affine Variety via Homogenisation of the Ideal](/theorems/2141) says that if $Y=V(I)\subset\mathbb A^n$, then its closure in the standard projective chart is cut out in $\mathbb P^n$ by the homogenised ideal $I^h$. This result explains why affine computations can miss intersections, but it also warns against homogenising a short generating list without checking the ideal it generates. The theorem uses all polynomials in $I(Y)$; homogenising only chosen generators can leave extra components on the hyperplane at infinity unless the resulting homogeneous ideal is saturated with respect to $x_0$ and then radicalised for the set-theoretic closure. The construction also describes the closure of the reduced affine variety, not a canonical projective scheme structure for an arbitrary nonreduced quotient. Its main geometric value is that it adds precisely the missing points at infinity, where projective intersection statements such as [Bezout's theorem](/theorems/2131) become stable. [example: Affine and Projective Intersections] In $\mathbb A^2$, the two affine lines are \begin{align*} L_0=V(y) \end{align*} and \begin{align*} L_1=V(y-1). \end{align*} A point $(a,b)\in \mathbb A^2$ lies in $L_0\cap L_1$ exactly when $b=0$ and $b-1=0$. These equations would imply $0=1$, which is impossible in a field, so $L_0\cap L_1=\varnothing$. Embed $\mathbb A^2$ into $\mathbb P^2$ by $(x,y)\mapsto [1:x:y]$, using homogeneous coordinates $[Z:X:Y]$. The equation $y=0$ homogenises to \begin{align*} Y=0, \end{align*} and the equation $y-1=0$ homogenises to \begin{align*} Y-Z=0. \end{align*} Thus the projective closures are the projective lines $\overline{L_0}=V(Y)$ and $\overline{L_1}=V(Y-Z)$. Their intersection consists of projective points $[Z:X:Y]$ satisfying \begin{align*} Y=0 \end{align*} and \begin{align*} Y-Z=0. \end{align*} Substituting $Y=0$ into $Y-Z=0$ gives $-Z=0$, hence $Z=0$. Therefore any point of the intersection has the form $[0:X:0]$. Since projective coordinates cannot all be zero, $X\ne 0$, and rescaling by $X^{-1}$ gives \begin{align*} [0:X:0]=[0:1:0]. \end{align*} So the affine lines are disjoint in the chart $Z=1$, but their projective closures meet at the unique point $[0:1:0]$ on the line at infinity $Z=0$. This point records the fact that both affine lines have the same horizontal direction. [/example] ## Worked Projective Varieties A complete calculation should move back and forth between equations, parametrisations, coordinate rings, affine charts, tangent spaces, and intersections. The following examples show the dictionary operating as a single method rather than as a list of separate facts. [example: Twisted Cubic] The twisted cubic $C\subset \mathbb P^3$ is the image of \begin{align*} \nu:\mathbb P^1\to \mathbb P^3,\qquad [s:t]\mapsto [s^3:s^2t:st^2:t^3]. \end{align*} Writing coordinates as $[X_0:X_1:X_2:X_3]$, set \begin{align*} q_1=X_0X_2-X_1^2,\qquad q_2=X_0X_3-X_1X_2,\qquad q_3=X_1X_3-X_2^2. \end{align*} These quadrics vanish on the parametrisation because \begin{align*} s^3\cdot st^2-(s^2t)^2=s^4t^2-s^4t^2=0, \end{align*} \begin{align*} s^3\cdot t^3-(s^2t)(st^2)=s^3t^3-s^3t^3=0, \end{align*} and \begin{align*} (s^2t)t^3-(st^2)^2=s^2t^4-s^2t^4=0. \end{align*} They are the $2\times 2$ minors of the catalecticant matrix whose rows are $(X_0,X_1,X_2)$ and $(X_1,X_2,X_3)$, and they generate the homogeneous ideal of $C$. On the chart $X_0\ne 0$, put $X_0=1$ and write $u=X_1$. The equation $q_1=0$ gives \begin{align*} X_2=X_1^2=u^2. \end{align*} Then $q_2=0$ gives \begin{align*} X_3=X_1X_2=u\cdot u^2=u^3. \end{align*} The third equation is then satisfied because \begin{align*} X_1X_3-X_2^2=u\cdot u^3-(u^2)^2=u^4-u^4=0. \end{align*} Thus this affine part is parametrised by \begin{align*} u\mapsto (u,u^2,u^3). \end{align*} If $X_0=0$, then $q_1=0$ gives $-X_1^2=0$, hence $X_1=0$, and $q_3=0$ gives $-X_2^2=0$, hence $X_2=0$. Since projective coordinates are not all zero, $X_3\ne 0$, so the only point over $X_0=0$ is \begin{align*} [0:0:0:1]=\nu([0:1]). \end{align*} The Jacobian matrix has rows \begin{align*} (X_2,-2X_1,X_0,0), \end{align*} \begin{align*} (X_3,-X_2,-X_1,X_0), \end{align*} and \begin{align*} (0,X_3,-2X_2,X_1). \end{align*} At a point of the curve with $s\ne 0$ and $t\ne 0$, the $2\times 2$ minor using the first two rows and first two columns is \begin{align*} (st^2)(-st^2)-(-2s^2t)(t^3)=-s^2t^4+2s^2t^4=s^2t^4\ne 0. \end{align*} So the rank is at least $2$ there. At $[1:0:0:0]$, the first two rows are \begin{align*} (0,0,1,0) \end{align*} and \begin{align*} (0,0,0,1), \end{align*} so the rank is at least $2$. At $[0:0:0:1]$, the second and third rows are \begin{align*} (1,0,0,0) \end{align*} and \begin{align*} (0,1,0,0), \end{align*} so the rank is again at least $2$. The rank is not larger than $2$ along the cone over $C$: the radial vector $(X_0,X_1,X_2,X_3)$ lies in the kernel because each $q_i$ is homogeneous of degree $2$ and vanishes at the point, and the derivative of the parametrised curve gives another tangent vector in the kernel. Hence the affine cone tangent space has dimension $4-2=2$, and quotienting by the radial direction leaves projective tangent dimension $1$ at every point. Therefore the twisted cubic is nonsingular. Projections of $C$ can still create singular plane images: secant directions may be identified, and projection from a tangent-related centre can collapse first-order information into a singular plane cubic. [/example] The twisted cubic is the clean rational model: its parametrisation is an isomorphism from $\mathbb P^1$ onto its image. Plane cubics with a node are also rational, but the parametrisation is birational rather than an isomorphism because the singular point encodes two branches. [example: Nodal Plane Cubic] Let $C\subset \mathbb P^2$ be the cubic \begin{align*} Y^2Z=X^2(X+Z). \end{align*} On the affine chart $Z=1$, write $x=X/Z$ and $y=Y/Z$. The equation becomes \begin{align*} y^2=x^2(x+1). \end{align*} Equivalently, \begin{align*} f(x,y)=y^2-x^2(x+1)=y^2-x^3-x^2. \end{align*} At the origin, \begin{align*} f(0,0)=0^2-0^3-0^2=0. \end{align*} The partial derivatives are \begin{align*} \partial_x f=-3x^2-2x,\qquad \partial_y f=2y. \end{align*} Thus \begin{align*} \partial_x f(0,0)=-3\cdot 0^2-2\cdot 0=0,\qquad \partial_y f(0,0)=2\cdot 0=0. \end{align*} So the origin is singular. The degree-$2$ part of $f$ is \begin{align*} y^2-x^2=(y-x)(y+x), \end{align*} which gives two distinct tangent directions $y=x$ and $y=-x$ when $\operatorname{char} k\ne 2$. Hence the singularity is a node. A line through the node with finite slope $t$ has equation $y=tx$. Substituting this into the affine cubic gives \begin{align*} (tx)^2=x^2(x+1). \end{align*} The left side is $t^2x^2$, so the equation is \begin{align*} t^2x^2=x^3+x^2. \end{align*} Moving all terms to one side gives \begin{align*} 0=x^3+x^2-t^2x^2=x^2(x+1-t^2). \end{align*} Thus the intersection with the line consists of the node $x=0$, counted by the factor $x^2$, and the residual point determined by \begin{align*} x+1-t^2=0. \end{align*} Therefore \begin{align*} x=t^2-1. \end{align*} Using $y=tx$, the corresponding $y$-coordinate is \begin{align*} y=t(t^2-1). \end{align*} This gives the affine parametrisation \begin{align*} t\mapsto (t^2-1,\;t(t^2-1)). \end{align*} It lands on the cubic because \begin{align*} \bigl(t(t^2-1)\bigr)^2=t^2(t^2-1)^2 \end{align*} and \begin{align*} (t^2-1)^2\bigl((t^2-1)+1\bigr)=(t^2-1)^2t^2=t^2(t^2-1)^2. \end{align*} For $t=1$, the parametrised point is \begin{align*} (1^2-1,\;1(1^2-1))=(0,0). \end{align*} For $t=-1$, it is \begin{align*} ((-1)^2-1,\;(-1)((-1)^2-1))=(0,0). \end{align*} Thus the two tangent directions $y=x$ and $y=-x$ correspond to the two parameter values $t=1$ and $t=-1$, both mapping to the same node. The parametrisation recovers the cubic away from the node, while the node records the two branches that have been identified. [/example] This line-through-the-singular-point construction is the prototype for recognising rational curves. It raises the next problem: when does a parametrisation method produce a birational equivalence rather than only a set-theoretic parametrisation? The following theorem supplies the criterion needed to justify the nodal cubic computation and similar rational examples. [quotetheorem:9457] [citeproof:9457] The theorem gives a practical test for the examples in this chapter, with several built-in limits. Irreducibility prevents the residual intersection construction from jumping between components, and the singularity assumption is what makes a line through $p$ contribute multiplicity at least $2$ before leaving one further point on a cubic. The degree $3$ hypothesis is also essential: for higher-degree curves, a line through a singular point may leave several residual points, so projection need not produce a birational inverse. Exceptional lines, including tangent directions and lines meeting other singular behaviour, are removed from the dense open sets on which the rational maps are inverse. Finally, rational parametrisation is weaker than isomorphism of varieties: a nodal or cuspidal cubic is birational to $\mathbb P^1$, but its singular local ring prevents it from being isomorphic to $\mathbb P^1$. [example: Comparing Chart and Homogeneous Tangents] For the projective conic $Q=V(XZ-Y^2)\subset \mathbb P^2$, work first on the affine chart $Z=1$ with coordinates \begin{align*} x=X/Z,\qquad y=Y/Z. \end{align*} Substituting $Z=1$, the homogeneous equation becomes \begin{align*} XZ-Y^2=0\quad \Longleftrightarrow \quad x-y^2=0. \end{align*} The point $[0:0:1]$ corresponds to $(x,y)=(0,0)$. For $g(x,y)=x-y^2$, the partial derivatives are \begin{align*} \partial_x g=1,\qquad \partial_y g=-2y. \end{align*} At $(0,0)$ these become \begin{align*} \partial_x g(0,0)=1,\qquad \partial_y g(0,0)=0. \end{align*} Thus a tangent vector $(u,v)\in k^2$ satisfies \begin{align*} 1\cdot u+0\cdot v=0. \end{align*} Hence $u=0$, so the affine tangent line in the chart is $x=0$. Now compute the same condition homogeneously. Let \begin{align*} F(X,Y,Z)=XZ-Y^2. \end{align*} The partial derivatives are \begin{align*} \partial_XF=Z,\qquad \partial_YF=-2Y,\qquad \partial_ZF=X. \end{align*} At $[0:0:1]$ they give \begin{align*} \partial_XF(0,0,1)=1,\qquad \partial_YF(0,0,1)=0,\qquad \partial_ZF(0,0,1)=0. \end{align*} Therefore a homogeneous first-order displacement $(U,V,W)$ satisfies \begin{align*} 1\cdot U+0\cdot V+0\cdot W=0. \end{align*} So $U=0$, and the corresponding projective tangent line is \begin{align*} X=0. \end{align*} On the chart $Z=1$, the equation $X=0$ becomes $x=X/Z=0$, exactly the affine tangent line found above. Thus the chart calculation and the homogeneous calculation describe the same tangent direction. [/example] ## The Dictionary as a Workflow The classical dictionary is most useful as a sequence of checks. Start with the ideal, replace by its radical when studying the underlying variety, keep the original quotient when nilpotent or multiplicity data matters, pass to local rings for local questions, and use tangent spaces to test smoothness. [explanation: Practical Computation Template] For an affine variety $X=V(I)\subset \mathbb A^n$, compute $\sqrt I$ to identify the underlying closed set and its components. Form $k[X]=k[x_1,\dots,x_n]/I(X)$ to study regular functions, and localise at $\mathfrak m_p$ when the question concerns a point $p$. Use the Jacobian matrix to compute $T_pX$, comparing its dimension to the dimension of the component through $p$. For a projective variety, first check that the ideal is homogeneous. Work chart by chart by setting a nonzero coordinate equal to $1$, but return to homogeneous equations to control points at infinity and intersection counts. When a parametrisation is present, compute the induced map on coordinate rings or function fields to decide whether it is an isomorphism, a closed embedding, or only birational. [/explanation] The course has built these tools in stages because each tool answers a different question. The synthesis is that no single computation is privileged: geometry is recovered by matching the ring, its primes, its localisations, its graded pieces, and its first-order quotients with the corresponding visible features of the variety. [remark: What Survives into Scheme Theory] The classical dictionary developed across the previous chapters already points beyond classical varieties. Nilpotents lost by passing from $I$ to $I(V(I))$, local rings at points, and homogeneous coordinate rings all become central rather than auxiliary in scheme theory. The next conceptual step is to stop discarding this algebraic information and to treat prime ideals themselves as points of a geometric object. [/remark] ## Beyond and Connections The affine dictionary developed here is the entry point for several later Androma topics. The Zariski topology leads naturally to irreducible spaces, generic points, and the language of schemes, where prime ideals are treated as points rather than only as tools for decomposing varieties. The Nullstellensatz explains why algebraically closed fields make the classical point-set picture work, while its radical form prepares the transition from varieties to coordinate rings and from visible solution sets to nilpotent structure. Projective methods connect this affine story to compactness-like phenomena in algebraic geometry: homogenization, projective closure, and divisors all record what happens at infinity. The degree-zero property of principal divisors is the first shadow of a broader intersection-theoretic principle, where local multiplicities assemble into global invariants. From here, natural continuations include [Affine Space](/page/Affine%20Space), [Coordinate Ring of Affine Space](/theorems/9339), [Affine Pieces of Projective Varieties Are Affine Varieties](/theorems/2138), [Projective Closure of an Affine Variety via Homogenisation of the Ideal](/theorems/2141), [Birational Equivalence via Function Fields](/theorems/2144), and [Principal Divisors Have Degree Zero](/theorems/2177). ## References - [Affine Space](/page/Affine%20Space) - [Coordinate Ring of Affine Space](/theorems/9339) - [Affine Pieces of Projective Varieties Are Affine Varieties](/theorems/2138) - [Projective Closure of an Affine Variety via Homogenisation of the Ideal](/theorems/2141) - [Rational Maps from Smooth Curves Are Morphisms](/theorems/2172) - [Birational Equivalence via Function Fields](/theorems/2144) - [Principal Divisors Have Degree Zero](/theorems/2177) - Robin Hartshorne, *Algebraic Geometry*, Chapter I. - David Mumford, *The Red Book of Varieties and Schemes*, Lecture I.