Polynomial equations remember more than their zero sets at first glance: they also remember the algebra of functions that survive on those zero sets. If two polynomials agree at every point of a geometric object, the geometry has no way to distinguish them as functions on that object. The coordinate ring is the construction that makes this identification official, turning an affine algebraic set into a commutative ring whose elements are the polynomial functions living on it. The motivating tension is already visible in the plane. Let $k$ be a field, and write $k[x,y]$ for the [polynomial ring](/page/Polynomial%20Ring) in the two variables $x$ and $y$ with coefficients in $k$. The polynomials $y$ and $x^2$ are different elements of $k[x,y]$, but on the parabola cut out by the equation $y-x^2=0$ they give the same function. Over an infinite field, this is the only polynomial relation forced by the parabola; over a finite field, extra pointwise relations can appear because a nonzero polynomial may vanish at every $k$-point. The coordinate ring records the correct equality by taking a quotient of the polynomial ring by the full vanishing ideal, not merely by the ideal generated by the equations we happened to write down. [example: Polynomial Functions on a Parabola] Let $k$ be an infinite field and let \begin{align*} X=\{(a,b)\in k^2\mid b=a^2\}\subset k^2. \end{align*} For any point $(a,b)\in X$, the defining condition gives $b=a^2$, so evaluating the two ambient polynomials $y$ and $x^2$ at $(a,b)$ gives \begin{align*} y(a,b)=b=a^2=x^2(a,b). \end{align*} Thus $y$ and $x^2$ define the same function on $X$. For example, at a point $(a,a^2)\in X$, \begin{align*} (x^3+2y)(a,a^2)=a^3+2a^2. \end{align*} Also \begin{align*} (x^3+2x^2)(a,a^2)=a^3+2a^2. \end{align*} Therefore $x^3+2y$ and $x^3+2x^2$ have equal restrictions to $X$. Now let $f\in k[x,y]$ vanish on $X$. Since $y-x^2$ is monic as a polynomial in $y$, polynomial division in the variable $y$ gives \begin{align*} f(x,y)=q(x,y)(y-x^2)+r(x) \end{align*} for some $q\in k[x,y]$ and some $r\in k[x]$. Evaluating at $(a,a^2)\in X$ gives \begin{align*} 0=f(a,a^2)=q(a,a^2)(a^2-a^2)+r(a)=r(a). \end{align*} So $r(a)=0$ for every $a\in k$. Because $k$ is infinite, a one-variable polynomial over $k$ with infinitely many roots is the zero polynomial, hence $r=0$. Therefore \begin{align*} f(x,y)=q(x,y)(y-x^2). \end{align*} So $I(X)=(y-x^2)$, and the coordinate ring is \begin{align*} k[X]=k[x,y]/(y-x^2). \end{align*} In this quotient, the class of $y$ equals the class of $x^2$, so every polynomial function on the parabola can be represented using only the single coordinate $x$. [/example] The point of the construction is not to replace geometry by algebra for its own sake. It is to make geometric operations visible in a ring where ordinary algebraic tools apply. The coordinate ring sits at the meeting point of [Ring](/page/Ring), ideals, and affine algebraic sets, and it is one of the basic bridges into [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry) and [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). ## Definition The construction should identify two ambient polynomials precisely when they give the same value at every point of the algebraic set. That means the first object to define is the quotient itself: the ring obtained by killing all polynomial functions that vanish on the set. [definition: Coordinate Ring] Let $k$ be a field and let $X\subset k^n$ be an affine algebraic set. Let \begin{align*} I(X)=\{f\in k[x_1,\ldots,x_n]\mid f(a)=0 \text{ for every } a\in X\}. \end{align*} The coordinate ring of $X$ over $k$ is the [quotient ring](/page/Quotient%20Ring) \begin{align*} k[X]=k[x_1,\ldots,x_n]/I(X). \end{align*} [/definition] This quotient definition is compact, but it depends on two pieces of background that deserve their own names. The next section isolates the geometric input and the vanishing ideal that makes the quotient independent of the particular equations chosen. ## Algebraic Sets and Regular Functions Before a coordinate ring can be formed, we need to know which subsets of [affine space](/page/Affine%20Space) are legitimate geometric objects for this chapter. The guiding requirement is that the subset should be described by polynomial tests: a point belongs to the set exactly when it satisfies a chosen family of polynomial equations. ### Polynomial Equation Sets The basic geometric object should remember only polynomial conditions, not an arbitrary list of points. This definition packages the common solution set of any chosen family of polynomial equations, so later algebraic constructions can refer to the set $X$ itself rather than to one particular presentation of its equations. [definition: Affine Algebraic Set] Let $k$ be a field and let $n\in\mathbb N$. An affine algebraic set in $k^n$ is a subset $X\subset k^n$ for which there exists a set of polynomials $S\subset k[x_1,\ldots,x_n]$ such that \begin{align*} X=\{a\in k^n\mid f(a)=0 \text{ for every } f\in S\}. \end{align*} [/definition] The same set $X$ can be cut out by many different collections of equations. To build the correct ring of functions on $X$, we need the collection of every polynomial that vanishes on $X$, since those are precisely the polynomials that represent the zero function on $X$. [definition: Vanishing Ideal] Let $k$ be a field and let $X\subset k^n$. The vanishing ideal of $X$ is the subset \begin{align*} I(X)=\{f\in k[x_1,\ldots,x_n]\mid f(a)=0 \text{ for every } a\in X\}. \end{align*} [/definition] The notation $k[X]$ emphasizes that elements of the quotient are functions on $X$, not just formal polynomial classes. The affine algebraic set definition tells us where those functions are evaluated, and the vanishing ideal tells us which ambient polynomials become zero. This raises the next question: which functions on $X$ should count as algebraic functions coming from its ambient affine space? ### Regular Functions and Coordinate Classes Once $X$ is fixed, we want functions that are intrinsic to the algebraic geometry of $X$, not arbitrary set-theoretic functions on its points. For this introductory discussion, we first use the polynomial-restriction version: the algebraic functions on $X$ are obtained by evaluating ambient polynomials and then identifying polynomials that agree on $X$. [definition: Regular Function on an Affine Algebraic Set] Let $k$ be a field and let $X\subset k^n$ be an affine algebraic set. A regular function on $X$ is a function $\varphi:X\to k$ for which there exists a polynomial $f\in k[x_1,\ldots,x_n]$ satisfying \begin{align*} \varphi(a)=f(a) \end{align*} for every $a\in X$. [/definition] There is a possible mismatch between the two descriptions: an element of $k[X]$ is a polynomial class modulo $I(X)$, while a regular function is an actual function $X\to k$. For the polynomial-restriction definition just given, the comparison is direct: the class of $f$ gives the function $a\mapsto f(a)$ on $X$, and two polynomials give the same function exactly when their difference lies in $I(X)$. The quoted theorem below uses the standard algebraic-geometric formulation over an [algebraically closed field](/page/Algebraically%20Closed%20Field). In that setting $\mathbb A_k^n$ denotes affine $n$-space over $k$ and $\mathcal O(X)$ denotes the ring of regular functions on $X$. There, a regular function may be described locally: near each point of $X$, on some Zariski open neighbourhood, it can be written as a quotient $p/q$ of polynomials with $q$ nonzero on that neighbourhood. A $k$-algebra homomorphism is a ring homomorphism that fixes the scalars from $k$. The content is that, for affine algebraic sets over algebraically closed fields, this local quotient definition gives the same function ring as the coordinate-ring quotient. [quotetheorem:9542] Thus the theorem should be read as a strengthening of the polynomial-restriction comparison, not as a change in the meaning of the earlier definition. The earlier definition is enough for the examples in this note, where we explicitly compute polynomial classes in $k[X]$. The theorem explains why the same quotient also matches the more flexible local rational notion of regularity used in algebraic geometry, under the additional algebraically closed field hypothesis. Outside that setting, especially over finite or non-algebraically closed fields, pointwise polynomial functions and geometric regular functions require more care, so the quotient definition remains the safest object to compute with. ## Relations as Geometry The quotient by $I(X)$ turns equations into algebraic relations. This is the first practical way to use a coordinate ring: write down generators for the ideal, then read the resulting quotient as a ring with named relations. ### Smooth Plane Curves [example: The Coordinate Ring of a Line] Let $k$ be an infinite field and let \begin{align*} X=\{(a,b)\in k^2\mid b-3a=0\}. \end{align*} For every $(a,b)\in X$, one has $b=3a$, so \begin{align*} (y-3x)(a,b)=b-3a=0. \end{align*} Thus $y-3x\in I(X)$. We now show that no further polynomial relations vanish on $X$. Let $f\in k[x,y]$ vanish on $X$. Since $y-3x$ is monic as a polynomial in $y$, division in the variable $y$ gives \begin{align*} f(x,y)=q(x,y)(y-3x)+r(x) \end{align*} for some $q\in k[x,y]$ and $r\in k[x]$. Evaluating at the point $(a,3a)\in X$ gives \begin{align*} 0=f(a,3a)=q(a,3a)(3a-3a)+r(a) \end{align*} and hence \begin{align*} r(a)=0. \end{align*} This holds for every $a\in k$. Since $k$ is infinite, a one-variable polynomial over $k$ with infinitely many roots is the zero polynomial, so $r=0$. Therefore \begin{align*} f(x,y)=q(x,y)(y-3x), \end{align*} which proves $I(X)=(y-3x)$. By the definition of the coordinate ring, \begin{align*} k[X]=k[x,y]/I(X)=k[x,y]/(y-3x). \end{align*} In this quotient, \begin{align*} \overline{y}-3\overline{x}=\overline{y-3x}=0, \end{align*} so \begin{align*} \overline{y}=3\overline{x}. \end{align*} Thus every polynomial expression in $\overline{x}$ and $\overline{y}$ can be rewritten using only $\overline{x}$; geometrically, one coordinate parameter $x$ already determines every regular function on the line. [/example] The line example shows the quotient removing redundant coordinates. A curve in the plane may be parametrized by one variable, but the coordinate ring still remembers how it is embedded in affine space through the relations among the coordinate functions. [example: The Coordinate Ring of the Parabola] Let $k$ be an infinite field and let \begin{align*} X=\{(a,b)\in k^2\mid b=a^2\}. \end{align*} The polynomial $y-x^2$ vanishes on $X$, since for every $(a,a^2)\in X$ one has \begin{align*} (y-x^2)(a,a^2)=a^2-a^2=0. \end{align*} Thus $(y-x^2)\subset I(X)$. Conversely, let $f\in I(X)$. Since $y-x^2$ is monic as a polynomial in $y$, division in the variable $y$ gives \begin{align*} f(x,y)=q(x,y)(y-x^2)+r(x) \end{align*} for some $q\in k[x,y]$ and $r\in k[x]$. Evaluating at $(a,a^2)\in X$ gives \begin{align*} 0=f(a,a^2)=q(a,a^2)(a^2-a^2)+r(a) \end{align*} and therefore \begin{align*} r(a)=0. \end{align*} This holds for every $a\in k$. Since $k$ is infinite, the one-variable polynomial $r$ has infinitely many roots, so $r=0$. Hence \begin{align*} f(x,y)=q(x,y)(y-x^2), \end{align*} so $f\in (y-x^2)$. Therefore \begin{align*} I(X)=(y-x^2). \end{align*} By the definition of the coordinate ring, \begin{align*} k[X]=k[x,y]/I(X)=k[x,y]/(y-x^2). \end{align*} In this quotient, \begin{align*} \overline{y}-\overline{x}^2=\overline{y-x^2}=0, \end{align*} so \begin{align*} \overline{y}=\overline{x}^2. \end{align*} Define a $k$-algebra homomorphism \begin{align*} \varphi:k[t]\to k[X] \end{align*} by $\varphi(t)=\overline{x}$. To construct its inverse, first define \begin{align*} \psi:k[x,y]\to k[t] \end{align*} by $\psi(x)=t$ and $\psi(y)=t^2$. Then \begin{align*} \psi(y-x^2)=t^2-t^2=0, \end{align*} so $(y-x^2)\subset\ker(\psi)$. Therefore $\psi$ descends to a homomorphism \begin{align*} \overline{\psi}:k[x,y]/(y-x^2)\to k[t] \end{align*} satisfying \begin{align*} \overline{\psi}(\overline{x})=t \end{align*} and \begin{align*} \overline{\psi}(\overline{y})=t^2. \end{align*} Now \begin{align*} (\overline{\psi}\circ\varphi)(t)=\overline{\psi}(\overline{x})=t, \end{align*} so $\overline{\psi}\circ\varphi$ is the identity on $k[t]$. Also \begin{align*} (\varphi\circ\overline{\psi})(\overline{x})=\varphi(t)=\overline{x} \end{align*} and \begin{align*} (\varphi\circ\overline{\psi})(\overline{y})=\varphi(t^2)=\overline{x}^2=\overline{y}. \end{align*} Since $\overline{x}$ and $\overline{y}$ generate $k[X]$ as a $k$-algebra, $\varphi\circ\overline{\psi}$ is the identity on $k[X]$. Thus \begin{align*} k[X]\cong k[t]. \end{align*} The coordinate ring of the parabola is therefore a one-variable polynomial ring, with the ambient coordinate $y$ remembered only through the relation $\overline{y}=\overline{x}^2$. [/example] Not every quotient has such a simple one-variable form. Singularities and reducible geometry leave visible algebraic traces in the coordinate ring. ### Reducible Geometry [example: Two Axes Crossing] Let $k$ be an infinite field, and let $X\subset k^2$ be the union of the coordinate axes: \begin{align*} X=\{(a,0)\mid a\in k\}\cup\{(0,b)\mid b\in k\}. \end{align*} The polynomial $xy$ vanishes on $X$, because \begin{align*} (xy)(a,0)=a\cdot 0=0 \end{align*} and \begin{align*} (xy)(0,b)=0\cdot b=0. \end{align*} Thus $(xy)\subset I(X)$. Conversely, let $f\in I(X)$, and write \begin{align*} f(x,y)=\sum_{i,j} c_{ij}x^i y^j \end{align*} with only finitely many nonzero coefficients $c_{ij}\in k$. Evaluating on the $x$-axis gives \begin{align*} 0=f(a,0)=\sum_i c_{i0}a^i \end{align*} for every $a\in k$. Since $k$ is infinite, the one-variable polynomial $\sum_i c_{i0}x^i$ is zero, so $c_{i0}=0$ for every $i$. Evaluating on the $y$-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 $j$. Hence every remaining nonzero term of $f$ has both $i\geq 1$ and $j\geq 1$, and therefore \begin{align*} f(x,y)=xy\sum_{i\geq 1,\ j\geq 1} c_{ij}x^{i-1}y^{j-1}. \end{align*} Thus $f\in (xy)$, so $I(X)=(xy)$. 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, \begin{align*} \overline{x}\,\overline{y}=\overline{xy}=0. \end{align*} However, $\overline{x}\neq 0$ because $x\notin (xy)$, and $\overline{y}\neq 0$ because $y\notin (xy)$. Thus $k[X]$ has two nonzero coordinate functions whose product is zero, reflecting that $X$ is made from two components, the two coordinate axes, meeting at the origin. [/example] The axes example shows one mechanism for zero divisors: a function can vanish on one component while another vanishes on the other, so their product vanishes everywhere. To use coordinate rings as an algebraic test for whether a space is a single geometric piece, we need a criterion that works without already knowing a decomposition into components. The subtle point is that reducibility is stated geometrically, while being an [integral domain](/page/Integral%20Domain) is stated purely inside the ring $k[X]$. A bridge between them must explain how a decomposition $X=X_1\cup X_2$ creates nonzero functions whose product vanishes, and also why every zero divisor in $k[X]$ witnesses such a decomposition when $k$ is algebraically closed. This is the point at which the hypothesis that $k$ is algebraically closed becomes structurally important: it lets algebraic equations in the coordinate ring detect closed subsets of $X$ without losing geometric information. The criterion below is therefore not just a convenient translation of vocabulary. It is the test that allows later arguments to replace the geometric phrase "one irreducible piece" with the ring-theoretic condition that no product of two nonzero functions is zero. The precise question is whether reducibility always produces zero divisors in the coordinate ring, and conversely whether the absence of zero divisors forces the algebraic set to be irreducible when the ground field is algebraically closed. This is not automatic from the definitions: a geometric union has to be converted into vanishing functions, while a ring-theoretic factorization has to be converted back into proper closed subsets. The theorem records exactly when these two translations are faithful, making the coordinate ring a usable detector of irreducible geometry rather than only a container for polynomial functions. [quotetheorem:9418] This result explains why algebraic geometers often study integral domains when they want a single piece of geometry. Reducible sets force products to vanish for geometric reasons, and those products appear inside the coordinate ring. ## Ideals and Closed Subsets Once $k[X]$ is available, closed subsets of $X$ can be described by ideals of the coordinate ring. The ambient polynomial ring is no longer the natural place to work; after quotienting, the functions already know they live on $X$. [definition: Closed Subset Defined in a Coordinate Ring] Let $k$ be a field, let $X\subset k^n$ be an affine algebraic set, and let $J\subset k[X]$ be an ideal. The closed subset of $X$ defined by $J$ is \begin{align*} V_X(J)=\{a\in X\mid g(a)=0 \text{ for every } g\in J\}. \end{align*} [/definition] The subscript in $V_X(J)$ matters because the same equations are being evaluated only on points of $X$. In the coordinate ring, adding equations corresponds to cutting $X$ down further. [example: A Point on the Parabola] Let $k$ be an infinite field, let $X\subset k^2$ be the parabola $y=x^2$, and let $p=(1,1)\in X$. From the computation of the parabola's vanishing ideal, \begin{align*} k[X]\cong k[x,y]/(y-x^2). \end{align*} Write $\overline{x}$ and $\overline{y}$ for the classes of $x$ and $y$ in $k[X]$, and set \begin{align*} J=(\overline{x}-1)\subset k[X]. \end{align*} A point of $X$ has the form $(a,a^2)$. It lies in $V_X(J)$ exactly when the generator $\overline{x}-1$ vanishes there, which means \begin{align*} (\overline{x}-1)(a,a^2)=a-1. \end{align*} Thus $(a,a^2)\in V_X(J)$ if and only if $a-1=0$, equivalently $a=1$. Substituting $a=1$ into the equation of $X$ gives \begin{align*} (a,a^2)=(1,1^2)=(1,1). \end{align*} Therefore \begin{align*} V_X(J)=\{(1,1)\}. \end{align*} Inside $k[X]$, the defining relation gives \begin{align*} \overline{y}=\overline{x}^2. \end{align*} Hence \begin{align*} \overline{y}-1=\overline{x}^2-1=(\overline{x}-1)(\overline{x}+1). \end{align*} So $\overline{y}-1\in(\overline{x}-1)=J$. The single condition $\overline{x}=1$ already forces $\overline{y}=1$ because points of $X$ satisfy the relation $\overline{y}=\overline{x}^2$. [/example] The example shows that different generating sets can impose the same conditions on points of $X$. More seriously, an ideal and its radical define the same zero set, because requiring $g^m$ to vanish at a point already forces $g$ to vanish there. The ambient form of this principle is [Hilbert's Nullstellensatz](/theorems/2124): in affine space over an algebraically closed field, closed algebraic subsets correspond to radical ideals in the polynomial ring. The obstruction is that equations do not remember the original ideal uniquely: enlarging an ideal by functions that already vanish on the same points, or replacing an ideal by its radical, does not change the zero set. To classify closed algebraic sets by equations, one must identify exactly which ideals are recovered from their zero sets over an algebraically closed field. [quotetheorem:9414] The adjective radical is the algebraic memory of the fact that zero sets ignore nilpotent powers. For a fixed affine algebraic set $X\subset k^n$, the relative statement for closed subsets of $X$ is obtained by applying the theorem to subsets $Y\subset X$ and then passing from ideals of $k[x_1,\ldots,x_n]$ containing $I(X)$ to ideals of the quotient $k[X]=k[x_1,\ldots,x_n]/I(X)$. Thus, over an algebraically closed field, radical ideals of $k[X]$ are the ideals that faithfully encode closed subsets of $X$. ## Morphisms Through Rings The coordinate ring also turns maps between affine algebraic sets around. A geometric map $X\to Y$ pulls functions on $Y$ back to functions on $X$, so it produces a homomorphism $k[Y]\to k[X]$ in the opposite direction. [definition: Morphism of Affine Algebraic Sets] Let $k$ be a field, let $X\subset k^n$ and $Y\subset k^m$ be affine algebraic sets. A morphism $\Phi:X\to Y$ is a function for which there exist polynomials $f_1,\ldots,f_m\in k[x_1,\ldots,x_n]$ such that \begin{align*} \Phi(a)=(f_1(a),\ldots,f_m(a)) \end{align*} for every $a\in X$. [/definition] ### Pullbacks of Regular Functions The definition says that each coordinate of the target point is a regular function on the source. To compare morphisms algebraically, we therefore need a named construction that sends target functions to source functions. [definition: Pullback Homomorphism] Let $\Phi:X\to Y$ be a morphism of affine algebraic sets over $k$. The pullback homomorphism associated to $\Phi$ is the ring homomorphism \begin{align*} \Phi^\ast:k[Y]\to k[X]. \end{align*} It is defined by $\Phi^\ast(g)=g\circ\Phi$ for each $g\in k[Y]$. [/definition] The pullback reverses direction because a function on the target is tested only after a point has been sent there. This gives one direction of comparison between geometry and algebra: every morphism produces a homomorphism of coordinate rings. ### Morphisms Versus Ring Homomorphisms For coordinate rings to be a complete algebraic language for affine maps, the reverse direction must also hold. The obstruction is not merely choosing images for functions in $k[Y]$: those images must respect all relations among the coordinate functions on $Y$. Otherwise the proposed coordinate functions on $X$ may define a point of affine space that does not actually lie in $Y$. The issue is whether every homomorphism $k[Y]\to k[X]$ comes from evaluating the coordinate functions of $Y$ on some polynomial map $X\to Y$, and whether the defining relations of $Y$ are exactly what makes those chosen functions land back in $Y$. In the quoted theorem, "[affine variety](/page/Affine%20Variety)" is used in the broad classical convention of an affine algebraic set over an algebraically closed field; no irreducibility assumption is being added to the discussion. [quotetheorem:9420] This theorem is the operational heart of coordinate rings. To describe a map into $Y$, specify where the coordinate functions of $Y$ go, subject to the relations defining $Y$. [example: Maps Into the Parabola] Let $Y\subset k^2$ be the parabola with coordinates $u,v$ and equation $v=u^2$, so its coordinate ring is \begin{align*} k[Y]\cong k[u,v]/(v-u^2). \end{align*} Write $\overline{u}$ and $\overline{v}$ for the classes of $u$ and $v$ in $k[Y]$. Since $v-u^2$ is zero in the quotient, we have \begin{align*} \overline{v}-\overline{u}^2=\overline{v-u^2}=0. \end{align*} Thus \begin{align*} \overline{v}=\overline{u}^2. \end{align*} If $\Phi:X\to Y$ is a morphism, then its pullback $\Phi^\ast:k[Y]\to k[X]$ is a $k$-algebra homomorphism. Set \begin{align*} h=\Phi^\ast(\overline{u}). \end{align*} Applying $\Phi^\ast$ to the relation $\overline{v}=\overline{u}^2$ gives \begin{align*} \Phi^\ast(\overline{v})=\Phi^\ast(\overline{u}^2). \end{align*} Because $\Phi^\ast$ is a ring homomorphism, \begin{align*} \Phi^\ast(\overline{u}^2)=\Phi^\ast(\overline{u})^2=h^2. \end{align*} Therefore every morphism into the parabola determines an element $h\in k[X]$ with \begin{align*} \Phi^\ast(\overline{u})=h,\qquad \Phi^\ast(\overline{v})=h^2. \end{align*} Conversely, given $h\in k[X]$, define a $k$-algebra homomorphism \begin{align*} \theta:k[u,v]\to k[X] \end{align*} by $\theta(u)=h$ and $\theta(v)=h^2$. Then \begin{align*} \theta(v-u^2)=\theta(v)-\theta(u)^2=h^2-h^2=0. \end{align*} Hence $(v-u^2)\subset\ker(\theta)$, so $\theta$ descends to a homomorphism \begin{align*} \overline{\theta}:k[u,v]/(v-u^2)\to k[X]. \end{align*} Under the identification $k[Y]\cong k[u,v]/(v-u^2)$, this is exactly a pullback homomorphism with \begin{align*} \overline{\theta}(\overline{u})=h \end{align*} and \begin{align*} \overline{\theta}(\overline{v})=h^2. \end{align*} Thus a map into the parabola is controlled by one regular function $h$ on $X$: the second coordinate is forced to be the square of the first. [/example] This way of thinking replaces coordinate formulas by universal constraints. The target equations become the relations that images of generators must satisfy in the source ring. ## Points, Maximal Ideals, and Fields Points of an affine algebraic set can also be read inside the coordinate ring. Evaluating all regular functions at one point gives a homomorphism to the ground field. [definition: Evaluation Homomorphism] Let $k$ be a field, let $X\subset k^n$ be an affine algebraic set, and let $p\in X$. The evaluation homomorphism at $p$ is \begin{align*} \operatorname{ev}_p:k[X]\to k. \end{align*} It is defined by $\operatorname{ev}_p(g)=g(p)$ for each $g\in k[X]$. [/definition] The kernel of evaluation consists of the functions vanishing at the chosen point. To use this kernel as an algebraic name for the point, we isolate it as a special ideal. [definition: Maximal Ideal of a Point] Let $k$ be a field, let $X\subset k^n$ be an affine algebraic set, and let $p\in X$. The maximal ideal of $p$ in $k[X]$ is \begin{align*} \mathfrak m_p=\ker(\operatorname{ev}_p). \end{align*} [/definition] The construction sends each point to a maximal ideal. The converse is the point-classification question for the coordinate ring: if an ideal of $k[X]$ is maximal, must it be the kernel of evaluation at some point of $X$? Over an algebraically closed field, [Hilbert's Nullstellensatz](/theorems/2940) answers yes. [remark: Coordinate-Ring Form of the Nullstellensatz] Let $k$ be algebraically closed, let $X\subset k^n$ be an affine algebraic set, and let $k[X]=k[x_1,\dots,x_n]/I(X)$. The maximal ideals of $k[X]$ are exactly the ideals \begin{align*} \mathfrak m_p=\{f\in k[X]:f(p)=0\} \end{align*} for points $p\in X$. Equivalently, if $\operatorname{mspec}(A)$ denotes the set of maximal ideals of a ring $A$, then the assignment \begin{align*} p\longmapsto \mathfrak m_p \end{align*} defines a bijection from $X$ to $\operatorname{mspec}(k[X])$. [/remark] This formulation is the version needed here because the page works with regular functions as classes in the coordinate ring, not with arbitrary Zariski-local rational functions. The algebraically closed hypothesis is essential: over a non-algebraically-closed field, maximal ideals can encode closed points whose residue fields are finite algebraic extensions of $k$, rather than ordinary $k$-valued points. For affine algebraic sets over an algebraically closed field, however, no such extra closed points appear, so the maximal ideals of $k[X]$ recover exactly the original points of $X$. This also explains why the coordinate ring is more than a bookkeeping device. Its maximal ideals remember the points of the algebraic set, while its prime ideals record larger irreducible geometric pieces. ## Nilpotents and Reduced Structure The definition above uses $I(X)$, the full ideal of polynomials vanishing on the set of points. This choice removes nilpotent behavior, because any polynomial whose power vanishes on $X$ already vanishes on $X$. [definition: Reduced Ring] A commutative ring $A$ is reduced if the only element $a\in A$ satisfying \begin{align*} a^m=0 \end{align*} for some $m\in\mathbb N$ is $a=0$. [/definition] Reducedness is a way of saying that the ring has no hidden infinitesimal thickness. To connect this ring-theoretic condition back to coordinate rings, we need to check that quotienting by the full vanishing ideal $I(X)$ really prevents nilpotent residue classes from surviving. The possible obstruction is a residue class $\overline{f}\in k[X]$ whose power is zero even though $\overline{f}$ itself is not. If such an element existed, the quotient would remember more than pointwise values on $X$: it would contain infinitesimal structure invisible to the underlying set. The formal statement below rules this out for ordinary affine algebraic sets, showing that their coordinate rings contain pointwise function information but no infinitesimal thickening. [quotetheorem:9543] This distinction becomes important when comparing algebraic sets with affine schemes. A quotient such as $k[x]/(x^2)$ has a nilpotent class $\overline{x}$, so it carries infinitesimal information that no ordinary subset of affine space can see. [example: A Double Point Is Not an Ordinary Coordinate Ring] In the quotient $k[x]/(x^2)$, write $\overline{x}$ for the class of $x$. Its square is \begin{align*} \overline{x}^2=\overline{x^2}=0, \end{align*} because $x^2\in (x^2)$. The element $\overline{x}$ is not zero: if $\overline{x}=0$, then $x\in (x^2)$, so there would be some $h(x)\in k[x]$ with \begin{align*} x=x^2h(x). \end{align*} If $h=0$, the right-hand side is $0$, not $x$. If $h\neq 0$, then $\deg(x^2h)=2+\deg h\geq 2$, while $\deg x=1$, again impossible. Hence $\overline{x}\neq 0$, but $\overline{x}^2=0$. Now compare this with the ordinary algebraic set cut out by $x^2$. Since $k$ is a field, $a^2=0$ implies $a=0$, so the zero set of $x^2$ in $k$ is \begin{align*} X=\{0\}. \end{align*} Its vanishing ideal is \begin{align*} I(X)=\{f\in k[x]\mid f(0)=0\}. \end{align*} Writing $f(x)=c_0+c_1x+\cdots+c_nx^n$, the condition $f(0)=0$ says $c_0=0$, so \begin{align*} f(x)=x(c_1+c_2x+\cdots+c_nx^{n-1}). \end{align*} Thus $I(X)=(x)$, and the coordinate ring of the underlying point is \begin{align*} k[X]=k[x]/(x). \end{align*} The quotient by $(x^2)$ therefore keeps an extra nilpotent element $\overline{x}$ that is invisible to the point set $\{0\}$; it describes a thickened point rather than the coordinate ring of the ordinary algebraic set. [/example] The example marks the boundary of this chapter. Classical affine algebraic sets use reduced coordinate rings, while scheme theory allows nonreduced coordinate rings as geometric objects in their own right. ## Beyond and Connected Topics Coordinate rings are the entry point to the affine dictionary between geometry and algebra. The next step is the Zariski topology, where closed sets are defined by ideals and the topology is intentionally coarser than metric intuition suggests. The commutative algebra behind coordinate rings is developed through ideals, prime ideals, localization, integral extensions, and Noetherian rings. [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra) is the natural companion for the algebraic side of the story. The geometric continuation is affine algebraic geometry and morphisms, where irreducibility, dimension, singularity, and regular maps are studied through the associated rings. [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry) develops this dictionary in a systematic way. The parent algebraic structure remains the [Ring](/page/Ring): coordinate rings are commutative $k$-algebras, and many of their geometric features are expressed using ring-theoretic language introduced in [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Scheme theory extends the idea by allowing every commutative ring to define an affine scheme. In that setting, nilpotents, local rings, and prime spectra are not defects; they are part of the geometry. ## References Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Ring](/page/Ring). Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry). Atiyah and Macdonald, *Introduction to Commutative Algebra* (1969). Hartshorne, *Algebraic Geometry* (1977). Shafarevich, *Basic Algebraic Geometry 1* (1977).