Let $k$ be a field. Affine algebraic geometry begins with a deceptively simple ambient space: $k^n$. At first it looks like ordinary coordinate space, but its real role is not metric or Euclidean. It is the place where formal polynomial expressions can be evaluated at points, where equations such as $x_1x_2-1=0$ cut out geometric objects, and where algebra can read geometry through the ring $k[x_1,\ldots,x_n]$. [example: A Curve as a Polynomial Equation] Let $k$ be a field and consider $f=x_1x_2-1 \in k[x_1,x_2]$. For a point $a=(a_1,a_2)\in k^2$, the coordinate functions satisfy $x_1(a)=a_1$ and $x_2(a)=a_2$, so evaluating $f$ at $a$ gives \begin{align*} f(a)=x_1(a)x_2(a)-1=a_1a_2-1. \end{align*} Therefore the zero set of $f$ in $\mathbb{A}^2_k=k^2$ is \begin{align*} V(f)=\{(a_1,a_2)\in k^2 : f(a_1,a_2)=0\}=\{(a_1,a_2)\in k^2 : a_1a_2-1=0\}. \end{align*} Equivalently, a point $(a_1,a_2)$ lies on this set exactly when $a_1a_2=1$. This is not being introduced as a graph in the calculus sense; it is the solution set of one polynomial equation inside affine space. The ambient field is part of the data: the same formal expression $x_1x_2-1$ may have different sets of solutions when evaluated on $\mathbb{R}^2$, $\mathbb{C}^2$, or $k^2$ for a finite field $k$. [/example] The same word "affine" also appears in affine geometry, where the emphasis is on points, translations, lines, and the absence of a distinguished origin. That viewpoint is useful, and this page returns to it later. The graph-context object for algebraic geometry, however, is the algebraic affine space: the set $k^n$ together with the formal coordinate algebra whose elements are evaluated on points. ## Definition The first definition packages the ambient object for polynomial equations. The set of points alone is not enough for algebraic geometry: over finite fields, two different formal polynomials may give the same set-theoretic function on $k^n$. To avoid losing algebraic information, affine space remembers the coordinate functions and the formal polynomial algebra that supplies expressions to be evaluated. [definition: Affine Space] Let $k$ be a field and let $n \in \mathbb{N}$. The $n$-dimensional affine space over $k$ is the set of points \begin{align*} \mathbb{A}^n_k=k^n \end{align*} equipped with its coordinate functions $x_1,\ldots,x_n:\mathbb{A}^n_k \to k$ and the formal coordinate algebra $k[x_1,\ldots,x_n]$ whose elements are evaluated on $\mathbb{A}^n_k$. [/definition] This definition turns a point $a=(a_1,\ldots,a_n)$ into something polynomials can evaluate on. The notation $\mathbb{A}^n_k$ reminds us that the field is part of the data: the same equations may have different solution sets over $\mathbb{R}$, $\mathbb{C}$, or a finite field. The next task is to make the variables $x_i$ into actual operations on points. Without this step, the symbol $x_i$ would only be a formal indeterminate in a ring; the coordinate function is what lets an equation read the $i$-th entry of a point in $\mathbb{A}^n_k$. [definition: Coordinate Function on Affine Space] For $\mathbb{A}^n_k=k^n$, the $i$-th coordinate function is the map \begin{align*} x_i:\mathbb{A}^n_k \to k. \end{align*} It is defined by \begin{align*} x_i(a_1,\ldots,a_n)=a_i \end{align*} for $1 \le i \le n$. [/definition] Once the coordinate functions are available, the next question is which scalar-valued functions count as algebraic. Algebraic geometry does not permit arbitrary maps $\mathbb{A}^n_k \to k$; it permits the functions obtained by evaluating formal expressions built from the coordinates by addition and multiplication. [definition: Polynomial Function on Affine Space] Let $\mathbb{A}^n_k$ be affine space over $k$. A polynomial function on $\mathbb{A}^n_k$ is a function $F:\mathbb{A}^n_k \to k$ obtained by evaluating a polynomial $f \in k[x_1,\ldots,x_n]$: \begin{align*} F(a_1,\ldots,a_n)=f(a_1,\ldots,a_n). \end{align*} [/definition] This definition is pointwise: it describes the induced function on $k$-points. The formal polynomial that produces the function remains part of the algebraic background, because pointwise equality can collapse distinct expressions over small fields. [example: Different Polynomials, Same Function over a Finite Field] Over $k=\mathbb{F}_p$, every element of $k$ has the form $n\cdot 1$ for some $n\in\{0,\ldots,p-1\}$. Since $k$ has characteristic $p$, the binomial coefficients $\binom{p}{j}$ vanish in $k$ for $1\le j\le p-1$, so \begin{align*} (u+v)^p=u^p+v^p \end{align*} for all $u,v\in k$. Starting from $0^p=0$, induction gives \begin{align*} ((n+1)\cdot 1)^p=(n\cdot 1+1)^p=(n\cdot 1)^p+1^p=n\cdot 1+1=(n+1)\cdot 1. \end{align*} Thus $a^p=a$ for every $a\in\mathbb{F}_p$, and the two polynomial functions $\mathbb{A}^1_k\to k$ induced by $x^p$ and $x$ agree pointwise: \begin{align*} x^p(a)=a^p=a=x(a). \end{align*} As formal polynomials in $k[x]$, however, they are not equal, because \begin{align*} x^p-x\ne 0 \end{align*} has nonzero coefficient $1$ on the monomial $x^p$ and nonzero coefficient $-1$ on the monomial $x$. The point is that equality as functions on $k$-points can identify distinct formal polynomial expressions, so algebraic geometry keeps track of coordinate rings and not only set-theoretic functions. [/example] The finite-field example shows why the formal algebra cannot be replaced by raw functions too soon. The next thing the theory needs is a disciplined way to pass from equations to geometry: given a family of formal polynomials, we want the set of points on which all of them vanish simultaneously. This construction is the basic bridge from algebra to subsets of affine space, and it is flexible enough to handle one equation, many equations, and even an infinite family of equations at once. [definition: Affine Algebraic Set] Let $S \subset k[x_1,\ldots,x_n]$. The affine algebraic set cut out by $S$ is \begin{align*} V(S)=\{a \in \mathbb{A}^n_k : f(a)=0 \text{ for every } f \in S\}. \end{align*} [/definition] Algebraic sets turn systems of polynomial equations into geometry. The notation $V(S)$ records the common vanishing locus of the equations in $S$, but the word "variety" is less uniform across the literature. We need a local convention before using it, because some authors reserve "affine variety" for irreducible algebraic sets while others use it for any affine algebraic set over an [algebraically closed field](/page/Algebraically%20Closed%20Field). [definition: Affine Variety] In this page, an affine variety over $k$ is an affine algebraic set $X\subset \mathbb{A}^n_k$. [/definition] This convention keeps the focus on the ambient affine-space construction. Irreducibility, reducedness, and scheme structure refine the notion later; the next example shows why that broad convention includes geometrically meaningful reducible sets rather than excluding them at the first pass. [example: Coordinate Axes as an Algebraic Set] For $a=(a_1,a_2)\in \mathbb{A}^2_k=k^2$, evaluating the polynomial $x_1x_2$ gives \begin{align*} (x_1x_2)(a)=x_1(a)x_2(a)=a_1a_2. \end{align*} Thus \begin{align*} V(x_1x_2)=\{(a_1,a_2)\in k^2:a_1a_2=0\}. \end{align*} Because $k$ is a field, $a_1a_2=0$ is equivalent to $a_1=0$ or $a_2=0$: if $a_1=0$ or $a_2=0$, then $a_1a_2=0$; conversely, if $a_1a_2=0$ and $a_1\ne 0$, then $a_1^{-1}$ exists and \begin{align*} a_2=1a_2=(a_1^{-1}a_1)a_2=a_1^{-1}(a_1a_2)=a_1^{-1}0=0. \end{align*} Therefore \begin{align*} V(x_1x_2)=\{(a_1,a_2)\in k^2:a_1=0\text{ or }a_2=0\}. \end{align*} The condition $a_2=0$ is the $x_1$-axis, and the condition $a_1=0$ is the $x_2$-axis, so one factored equation cuts out the union of the two coordinate axes. This is why algebraic sets naturally include reducible objects: the factorization $x_1x_2$ records two components in the geometry. [/example] The coordinate-axes example shows that geometry can remember algebraic features such as factorization and components. To go back in the other direction, from a subset of affine space to algebra, we need to ask which polynomial expressions cannot be detected on that subset because they vanish at every one of its points. Collecting exactly those expressions produces an ideal, and that ideal is the algebraic shadow of the point set. [definition: Vanishing Ideal] Let $X\subset \mathbb{A}^n_k$. The vanishing ideal of $X$ is \begin{align*} I(X)=\{f \in k[x_1,\ldots,x_n] : f(a)=0 \text{ for every } a \in X\}. \end{align*} [/definition] The vanishing ideal records a point-set convention: it consists of formal polynomials that vanish on the chosen $k$-points of $X$. The next construction uses that convention to build a ring of polynomial functions on those points, while keeping in view that scheme-theoretic algebraic geometry may instead start from a specified defining ideal before taking its radical or its $k$-point vanishing ideal. [definition: Coordinate Ring of an Affine Algebraic Set] Let $X\subset \mathbb{A}^n_k$ be an affine algebraic set, regarded as a set of $k$-points. The point-set coordinate ring of $X$ is \begin{align*} k[X]=k[x_1,\ldots,x_n]/I(X). \end{align*} [/definition] The coordinate ring records formal polynomial expressions up to equality on $X$. Over algebraically closed fields, [Hilbert's Nullstellensatz](/theorems/2124) later explains how radical ideals and algebraic sets determine each other; over finite or non-algebraically closed fields, using $I(X)$ is a deliberate point-set choice and may differ from the quotient by a chosen list of defining equations. For the whole affine space, the quotient depends on whether nonzero formal polynomials can vanish on every $k$-point. This question matters because the coordinate ring is supposed to be the algebra of polynomial functions available on the ambient space itself. Over an infinite field, there is no hidden collapse: vanishing on all of $k^n$ forces the formal polynomial to be zero, so the point-set coordinate ring recovers the [polynomial ring](/page/Polynomial%20Ring) rather than a smaller quotient. [quotetheorem:9339] This theorem is a statement about the point-set coordinate-ring convention used above. It explains why affine space is the first arena of algebraic geometry over infinite fields: its coordinate ring is the polynomial ring itself, and its closed subsets are described by ideals in that ring. The next question is how maps between affine spaces should interact with this algebra. A plain set map can ignore equations completely; the algebraic maps are those whose coordinate components come from formal polynomials, so that substitution makes sense inside coordinate rings. [definition: Polynomial Map Between Affine Spaces] A polynomial map $F:\mathbb{A}^n_k \to \mathbb{A}^m_k$ is a map of the form \begin{align*} F(a)=(f_1(a),\ldots,f_m(a)) \end{align*} for polynomials $f_1,\ldots,f_m \in k[x_1,\ldots,x_n]$. [/definition] Polynomial maps are the morphisms of affine space. They pull coordinate functions on the target back to polynomial functions on the source, which is the first glimpse of the contravariant link between geometry and algebra. For algebraic subsets, the same idea needs one extra condition: the map must land in the target set, so that equations valid on the target become equations valid on the source after substitution. [definition: Morphism of Affine Algebraic Sets] Let $X\subset \mathbb{A}^n_k$ and $Y\subset \mathbb{A}^m_k$ be affine algebraic sets. A morphism $F:X\to Y$ is a map for which there exist polynomials $f_1,\ldots,f_m \in k[x_1,\ldots,x_n]$ such that \begin{align*} F(a)=(f_1(a),\ldots,f_m(a)) \end{align*} for every $a\in X$. [/definition] Thus morphisms between affine algebraic sets are restrictions of polynomial maps whose images satisfy the defining equations of the target. The reason this definition is useful is that every target polynomial can now be composed with the map; composition turns functions on the target into functions on the source and is the algebraic mechanism behind changing coordinates. [quotetheorem:9340] This pullback is the basic reason morphisms are studied through coordinate rings. Instead of checking all functions on $X$ directly, one can transport algebra from $Y$ to $X$ by substitution, and equations on the target become equations on the source. The construction is contravariant: a geometric map $X\to Y$ produces an algebra homomorphism in the opposite direction. Later, this is what allows properties of algebraic sets and maps to be translated into statements about polynomial functions modulo the equations defining the sets. The rest of the page separates two structures that coexist on $\mathbb{A}^n_k$: its algebraic structure coming from the formal coordinate algebra and its affine-geometric structure as a space of points with translations and lines. ## Underlying Affine Geometry The algebraic affine space $\mathbb{A}^n_k$ has an underlying geometric structure: points can be translated by vectors in $k^n$, differences of points are vectors, and lines are parametrized by affine formulas. This structure does not replace the polynomial-function viewpoint; it supplies the language of lines, spans, and affine linear maps inside the ambient space. [definition: Affine Space Modeled on a Vector Space] Let $k$ be a field and let $V$ be a [vector space](/page/Vector%20Space) over $k$. An affine space modeled on $V$ is a nonempty set $A$ together with a map $A \times V \to A$, written $(p,v)\mapsto p+v$, such that \begin{align*} p+0=p \end{align*} \begin{align*} (p+v)+w=p+(v+w), \end{align*} for all $p \in A$ and $v,w \in V$, and for every pair $p,q \in A$ there is a unique vector $v \in V$ such that $q=p+v$. [/definition] This torsor-style definition is the affine-geometric abstraction of the translation structure on $k^n$. In this page it should be read as a companion viewpoint, while $\mathbb{A}^n_k$ with its coordinate algebra remains the primary algebraic object. After this definition, the next question is how to name the unique vector that carries one point to another. That vector is the basic measurable change in an affine space: it is not a new point, it does not depend on coordinates, and it is the object that later defines directions, spans, and equations of subspaces. [definition: Displacement Vector] Let $A$ be an affine space modeled on $V$. For $p,q \in A$, the displacement vector from $p$ to $q$ is the unique vector $q-p \in V$ such that \begin{align*} q=p+(q-p). \end{align*} [/definition] The notation $q-p$ is useful because it records the only subtraction that affine geometry permits. Subtracting points gives a vector; adding that vector back to a point gives a point. [example: The Standard Affine Space $k^n$] Let $A=k^n$ be regarded as a set, and let $V=k^n$ be the usual vector space. Define the action of $V$ on $A$ by coordinatewise addition: for \begin{align*} p=(p_1,\ldots,p_n)\in A \quad \text{and} \quad v=(v_1,\ldots,v_n)\in V, \end{align*} set \begin{align*} p+v=(p_1+v_1,\ldots,p_n+v_n). \end{align*} This satisfies the affine-space axioms. If $0=(0,\ldots,0)$, then \begin{align*} p+0=(p_1+0,\ldots,p_n+0)=(p_1,\ldots,p_n)=p. \end{align*} If $w=(w_1,\ldots,w_n)$, then \begin{align*} (p+v)+w=((p_1+v_1)+w_1,\ldots,(p_n+v_n)+w_n). \end{align*} By associativity of addition in $k$ in each coordinate, this is \begin{align*} (p_1+(v_1+w_1),\ldots,p_n+(v_n+w_n))=p+(v+w). \end{align*} Now take \begin{align*} p=(p_1,\ldots,p_n) \quad \text{and} \quad q=(q_1,\ldots,q_n). \end{align*} The vector \begin{align*} v=(q_1-p_1,\ldots,q_n-p_n) \end{align*} satisfies \begin{align*} p+v=(p_1+(q_1-p_1),\ldots,p_n+(q_n-p_n))=(q_1,\ldots,q_n)=q. \end{align*} If $u=(u_1,\ldots,u_n)$ also satisfies $p+u=q$, then for each $i$ we have $p_i+u_i=q_i$, so adding $-p_i$ to both sides gives $u_i=q_i-p_i$. Hence $u=v$, and the displacement vector from $p$ to $q$ is \begin{align*} q-p=(q_1-p_1,\ldots,q_n-p_n). \end{align*} The same underlying set $k^n$ can therefore be read in two ways: as the algebraic affine space $\mathbb{A}^n_k$ when the coordinate algebra is emphasized, and as a torsor under $k^n$ when translations and affine combinations are emphasized. [/example] A point can be promoted to an origin, but that promotion is a choice rather than part of the affine-geometric structure. We need a separate name for this operation because many coordinate computations begin by making such a choice, and the reader must know exactly where the hidden origin has entered the argument. [definition: Origin Choice] Let $A$ be an affine space modeled on $V$. An origin choice is a point $o \in A$. [/definition] The chosen origin determines the map $\phi_o:A \to V$ given by $\phi_o(p)=p-o$. The same affine space has many coordinate vector-space descriptions, one for each origin. The next theorem is needed to justify the common practice of doing affine calculations in vectors after choosing $o$: it says that no points are lost and that vector addition in $V$ exactly records translation in $A$. [quotetheorem:9341] This theorem explains why affine spaces often look like vector spaces in coordinates. The warning is that the bijection depends on $o$, so equations that use position vectors may hide a chosen origin. ## Lines and Affine Combinations ### Affine Formulas Without an Origin The first constructions in affine geometry are the ones that can be made from points using coefficients whose sum is $1$. This condition is what makes the expression independent of an origin. [definition: Affine Combination] Let $A$ be an affine space modeled on $V$. An affine combination is the value of the map from \begin{align*} A^m\times\{(\lambda_1,\ldots,\lambda_m)\in k^m:\lambda_1+\cdots+\lambda_m=1\} \end{align*} to $A$ that sends $((p_1,\ldots,p_m),(\lambda_1,\ldots,\lambda_m))$ to the point denoted \begin{align*} \lambda_1 p_1+\cdots+\lambda_m p_m \end{align*} and defined by choosing any point $o \in A$ and setting \begin{align*} \lambda_1 p_1+\cdots+\lambda_m p_m =o+\sum_{i=1}^m \lambda_i(p_i-o). \end{align*} [/definition] The condition on the coefficients cancels the effect of moving the origin. Still, the formula defining the point visibly mentions a temporary point $o$, so there is a real ambiguity to rule out: choosing a different temporary origin must not change the point produced by the same weighted list. Otherwise affine combinations would be coordinate artifacts rather than geometric points. This origin-independence is the basic well-definedness result that makes affine combinations usable in later constructions. Before using them to define lines, spans, and coordinates, we need the theorem that the weighted point depends only on the points and coefficients, not on the auxiliary origin chosen to evaluate the formula. [quotetheorem:9342] The theorem turns the displayed formula from a coordinate recipe into genuine affine notation. It says that the expression $\lambda_1p_1+\cdots+\lambda_mp_m$ may be used without naming an origin, provided the coefficients sum to $1$. The restriction on the sum is essential: ordinary linear combinations of points still depend on where the origin is placed, while affine combinations are exactly the combinations whose translation terms cancel. This is why barycentric coordinates, midpoints, and parametrized lines can be discussed inside an affine space rather than only after choosing coordinates. ### Lines and Direction Affine combinations give a coordinate-free definition of the line through two points. This is the first geometric object built from the barycentric rule, and the parameter is allowed to range over the whole field so that the construction gives the full affine line rather than just the segment between the points. [definition: Affine Line] Let $A$ be an affine space over $k$, and let $p,q \in A$ with $p \ne q$. The affine line through $p$ and $q$ is the subset \begin{align*} L(p,q)=\{(1-t)p+tq : t \in k\}. \end{align*} [/definition] Once the line has been defined as a set of points, we still need to record the vector information that controls motion along it. This direction is what survives translation of the line, and it is the invariant used to compare two different lines. [definition: Direction of an Affine Line] Let $L=L(p,q)$ be an affine line in an affine space modeled on $V$. The direction of $L$ is the one-dimensional subspace \begin{align*} \operatorname{dir}(L)=k(q-p)\subset V. \end{align*} [/definition] Parallelism can now be stated without measuring angles or lengths. The definition is needed because affine geometry does not have a metric by default, so two lines must be compared through their displacement directions rather than through angle measurements. [definition: Parallel Affine Lines] Two affine lines $L$ and $M$ in an affine space modeled on $V$ are parallel if \begin{align*} \operatorname{dir}(L)=\operatorname{dir}(M). \end{align*} [/definition] [example: Midpoints Are Affine but Not Purely Incidence-Theoretic] Assume $\operatorname{char}(k)\ne 2$, so $2=1+1$ is nonzero in $k$ and therefore has an inverse $\frac{1}{2}\in k$. For two points $p,q\in A$, define \begin{align*} m=\frac{1}{2}p+\frac{1}{2}q. \end{align*} Since $\frac{1}{2}+\frac{1}{2}=1$, this is an affine combination. Taking $p$ as the temporary origin in the definition of affine combination gives \begin{align*} m=p+\frac{1}{2}(p-p)+\frac{1}{2}(q-p). \end{align*} The displacement $p-p$ is $0$, because $p=p+0$ and the displacement vector is unique, so \begin{align*} m=p+\frac{1}{2}(q-p). \end{align*} By uniqueness of displacement from $p$ to $m$, this proves \begin{align*} m-p=\frac{1}{2}(q-p). \end{align*} For the other displacement, use $q=p+(q-p)$ and the affine action law: \begin{align*} m+\frac{1}{2}(q-p)=\left(p+\frac{1}{2}(q-p)\right)+\frac{1}{2}(q-p). \end{align*} Thus \begin{align*} m+\frac{1}{2}(q-p)=p+\left(\frac{1}{2}(q-p)+\frac{1}{2}(q-p)\right). \end{align*} By distributivity in the model vector space, \begin{align*} \frac{1}{2}(q-p)+\frac{1}{2}(q-p)=\left(\frac{1}{2}+\frac{1}{2}\right)(q-p)=q-p. \end{align*} Therefore \begin{align*} m+\frac{1}{2}(q-p)=p+(q-p)=q. \end{align*} By uniqueness of displacement from $m$ to $q$, \begin{align*} q-m=\frac{1}{2}(q-p). \end{align*} Thus the midpoint has equal displacement to $p$ and to $q$ along the same affine line, and the construction depends on the field admitting the scalar $\frac{1}{2}$. [/example] ## Affine Subspaces ### Translated Linear Subspaces Linear subspaces pass through the origin. Affine subspaces are their translated versions, so they are the right objects for solution sets of inhomogeneous linear equations. [definition: Affine Subspace] Let $A$ be an affine space modeled on $V$. A subset $B \subset A$ is an affine subspace if there is a point $p \in A$ and a linear subspace $W \subset V$ such that \begin{align*} B=p+W=\{p+w : w \in W\}. \end{align*} [/definition] The displayed presentation $B=p+W$ uses a point $p$, and a different point of $B$ would give another presentation. We therefore need a base-point-free direction space, because later statements about dimension and parallelism should depend on the subset $B$ itself and not on a chosen description. [definition: Direction Space of an Affine Subspace] Let $B \subset A$ be an affine subspace. Its direction space is \begin{align*} \operatorname{dir}(B)=\{q-p : p,q \in B\}\subset V. \end{align*} [/definition] The definition using pairwise displacement should recover the same $W$ from any translated presentation. The possible obstruction is that $B=p+W$ names both a point and a subspace, while $\operatorname{dir}(B)$ only remembers differences of points in $B$. We need these two descriptions to determine the same direction space before dimension and parallelism can be treated as properties of $B$ itself. [quotetheorem:9343] The most familiar affine subspaces are solution sets of linear systems with a nonzero right-hand side. They are translated kernels. [example: Inhomogeneous Linear Equations] Let $T:V\to U$ be a [linear map](/page/Linear%20Map) between vector spaces over $k$, fix $b\in U$, and suppose $x_0\in V$ satisfies $T(x_0)=b$. We show that the solutions of $T(x)=b$ are exactly the translate of the kernel by $x_0$: \begin{align*} \{x\in V:T(x)=b\}=x_0+\ker T. \end{align*} First let $x\in V$ satisfy $T(x)=b$. Since $T(x_0)=b$, linearity gives \begin{align*} T(x-x_0)=T(x)-T(x_0)=b-b=0. \end{align*} Thus $x-x_0\in\ker T$, so $x=x_0+(x-x_0)$ lies in $x_0+\ker T$. Conversely, let $x\in x_0+\ker T$. Then $x=x_0+w$ for some $w\in\ker T$, so $T(w)=0$. By linearity, \begin{align*} T(x)=T(x_0+w)=T(x_0)+T(w)=b+0=b. \end{align*} Thus $x$ is a solution of $T(x)=b$, proving the equality of the two sets. Therefore the solution set is an affine subspace of the affine space underlying $V$: it is the translate of the linear subspace $\ker T$ by the point $x_0$. Its direction space is $\ker T$, and the homogeneous equation $T(x)=0$ is the special case $b=0$, where one may take $x_0=0$ and the affine subspace passes through the chosen origin. [/example] ### Hyperplanes and Linear Equations Affine subspaces also explain why coordinates for a plane in three-dimensional space usually contain constants. The constants place the plane, while the homogeneous part gives the direction. Hyperplanes deserve a separate definition because they are the codimension-one affine subspaces that appear as single linear equations and as separating walls in geometry. [definition: Affine Hyperplane] Let $A$ be an affine space modeled on a finite-dimensional vector space $V$. An affine hyperplane is an affine subspace $H \subset A$ whose direction space has codimension $1$ in $V$. [/definition] [example: A Hyperplane in $k^3$] Choose a scalar $c\in k$, and define \begin{align*} H=\{(x_1,x_2,x_3)\in k^3:x_1+2x_2-x_3=c\}. \end{align*} Let $p=(c,0,0)$ and let \begin{align*} W=\{(u_1,u_2,u_3)\in k^3:u_1+2u_2-u_3=0\}. \end{align*} We show that $H=p+W$. If $x=(x_1,x_2,x_3)\in H$, then $x_1+2x_2-x_3=c$, and \begin{align*} x-p=(x_1-c,x_2,x_3). \end{align*} This displacement lies in $W$, since \begin{align*} (x_1-c)+2x_2-x_3=(x_1+2x_2-x_3)-c=c-c=0. \end{align*} Hence $x=p+(x-p)\in p+W$. Conversely, if $x\in p+W$, then $x=p+u$ for some $u=(u_1,u_2,u_3)\in W$. Thus \begin{align*} x=(c+u_1,u_2,u_3). \end{align*} Since $u\in W$, we have $u_1+2u_2-u_3=0$, so \begin{align*} (c+u_1)+2u_2-u_3=c+(u_1+2u_2-u_3)=c+0=c. \end{align*} Therefore $x\in H$, proving \begin{align*} H=p+W. \end{align*} The direction set $W$ is a linear subspace: if $u,v\in W$ and $\lambda\in k$, then \begin{align*} (u_1+v_1)+2(u_2+v_2)-(u_3+v_3)=(u_1+2u_2-u_3)+(v_1+2v_2-v_3)=0+0=0 \end{align*} and \begin{align*} \lambda u_1+2\lambda u_2-\lambda u_3=\lambda(u_1+2u_2-u_3)=\lambda 0=0. \end{align*} Moreover, \begin{align*} W=\{(s,t,s+2t):s,t\in k\}=k(1,0,1)+k(0,1,2). \end{align*} The two displayed spanning vectors are linearly independent because \begin{align*} \alpha(1,0,1)+\beta(0,1,2)=(0,0,0) \end{align*} forces $\alpha=0$ from the first coordinate and $\beta=0$ from the second coordinate. Thus $W$ has dimension $2$, so it has codimension $1$ in $k^3$. Hence $H$ is an affine hyperplane: the point $p=(c,0,0)$ places the hyperplane, while $W$ is its direction space. [/example] ## Affine Maps Maps between affine spaces should preserve the constructions that do not use an origin. This leads to maps that carry affine combinations to affine combinations. [definition: Affine Map] Let $A$ and $B$ be affine spaces modeled on vector spaces $V$ and $W$ over the same field $k$. A map $f:A \to B$ is affine if there exists a linear map $L:V \to W$ such that \begin{align*} f(p+v)=f(p)+L(v) \end{align*} for every $p \in A$ and every $v \in V$. [/definition] To compare two affine maps, compose affine maps, or decide whether an affine map is invertible, the translation data and the displacement data must be separated. The next definition names the displacement data, because this linear component is the part that acts on direction spaces, kernels, and coordinate changes. [definition: Linear Part of an Affine Map] Let $f:A \to B$ be an affine map between affine spaces modeled on $V$ and $W$. The linear part of $f$ is the linear map $L:V \to W$ satisfying \begin{align*} f(p+v)=f(p)+L(v). \end{align*} [/definition] The definition of an affine map is written using one point $p$ and one displacement vector $v$, but affine geometry often describes points by barycentric formulas involving several points at once. The issue is whether the one-displacement rule is strong enough to control those multi-point formulas. We need this compatibility before using affine maps to transport lines, parallelism, and ratios along lines. [quotetheorem:9344] [example: Matrix Plus Translation] Let $A=k^n$ and $B=k^m$ be standard affine spaces. Fix $M=(M_{ij})\in k^{m\times n}$ and $b=(b_1,\ldots,b_m)\in k^m$, and define $f:k^n\to k^m$ by \begin{align*} f(x)=Mx+b. \end{align*} We show that $f$ is affine with linear part $L:k^n\to k^m$ given by $L(v)=Mv$. First $L$ is linear. For $u,v\in k^n$ and $\lambda\in k$, the $i$-th coordinate of $M(u+v)$ is \begin{align*} \sum_{j=1}^n M_{ij}(u_j+v_j)=\sum_{j=1}^n M_{ij}u_j+\sum_{j=1}^n M_{ij}v_j. \end{align*} Thus \begin{align*} M(u+v)=Mu+Mv. \end{align*} Similarly, the $i$-th coordinate of $M(\lambda u)$ is \begin{align*} \sum_{j=1}^n M_{ij}(\lambda u_j)=\lambda\sum_{j=1}^n M_{ij}u_j, \end{align*} so \begin{align*} M(\lambda u)=\lambda Mu. \end{align*} Therefore $L(v)=Mv$ is a linear map. Now take $p,v\in k^n$. Using additivity of matrix multiplication, \begin{align*} f(p+v)=M(p+v)+b=(Mp+Mv)+b. \end{align*} By associativity and commutativity of vector addition in $k^m$, \begin{align*} (Mp+Mv)+b=(Mp+b)+Mv. \end{align*} Since $f(p)=Mp+b$ and $L(v)=Mv$, this gives \begin{align*} f(p+v)=f(p)+L(v). \end{align*} Hence $f$ is affine. The vector $b=f(0)$ records where the chosen origin is sent, while the matrix $M$ determines every displacement, since \begin{align*} f(p+v)-f(p)=Mv=L(v). \end{align*} [/example] Not every function that sends points to points is affine. Curvature, nonlinear stretching, and projective effects all break affine combinations. [example: Squaring Is Not Affine] Over a field $k$ with $\operatorname{char}(k)\ne 2$, the element $2=1+1$ is nonzero, so $\frac{1}{2}\in k$ exists. Consider $f:k\to k$ given by $f(x)=x^2$. The midpoint of $0$ and $2$ is the affine combination \begin{align*} \frac{1}{2}\cdot 0+\frac{1}{2}\cdot 2=0+\frac{2}{2}=1. \end{align*} If $f$ were affine, it would preserve this affine combination, so it would satisfy \begin{align*} f(1)=\frac{1}{2}f(0)+\frac{1}{2}f(2). \end{align*} The left-hand side is \begin{align*} f(1)=1^2=1. \end{align*} The right-hand side is \begin{align*} \frac{1}{2}f(0)+\frac{1}{2}f(2)=\frac{1}{2}\cdot 0^2+\frac{1}{2}\cdot 2^2. \end{align*} Since $0^2=0$ and $2^2=(1+1)^2=1+1+1+1=4$, this becomes \begin{align*} \frac{1}{2}\cdot 0+\frac{1}{2}\cdot 4=0+2=2. \end{align*} Thus affine preservation of the midpoint would force $1=2$, hence $0=1$ after subtracting $1$ from both sides, which is impossible in a field. Therefore $x\mapsto x^2$ is not an affine map over any field of characteristic not equal to $2$. [/example] ## Coordinates, Frames, and Barycentric Data ### Frames and Point Coordinates Coordinates in an affine space require more than a basis of the model vector space. They require a point where the coordinate grid is anchored. [definition: Affine Frame] Let $A$ be an affine space modeled on an $n$-dimensional vector space $V$. An affine frame is a tuple \begin{align*} (p;e_1,\ldots,e_n) \end{align*} where $p \in A$ and $(e_1,\ldots,e_n)$ is a basis of $V$. [/definition] An affine frame gives an anchored grid, but a point $q$ is still an element of the affine space rather than a tuple of scalars. To compute with $q$, we must translate the displacement $q-p$ into the chosen vector-space basis. The following definition names the unique scalars that perform that translation. [definition: Affine Coordinates] Let $(p;e_1,\ldots,e_n)$ be an affine frame for $A$. The affine coordinates of a point $q \in A$ are the scalars $x_1,\ldots,x_n \in k$ satisfying \begin{align*} q=p+x_1e_1+\cdots+x_ne_n. \end{align*} [/definition] ### Barycentric Coordinates Barycentric coordinates provide another coordinate system, one tied to a simplex rather than to an origin and basis. Before defining those coordinates, we need to know when the chosen points form a non-degenerate affine coordinate system; this is the role of affine independence. [definition: Affinely Independent Points] Points $p_0,\ldots,p_m$ in an affine space modeled on $V$ are affinely independent if the vectors \begin{align*} p_1-p_0,\ldots,p_m-p_0 \end{align*} are linearly independent in $V$. [/definition] Barycentric coordinates then express a point using weights on affinely independent vertices. The definition is needed because many geometric constructions in triangles, tetrahedra, and simplices are most stable when written as weights whose total is $1$ rather than as coordinates from a chosen origin. [definition: Barycentric Coordinates] Let $p_0,\ldots,p_m$ be affinely independent points in an affine space $A$. Barycentric coordinates of a point $q$ in their affine span are scalars $\lambda_0,\ldots,\lambda_m \in k$ such that \begin{align*} \lambda_0+\cdots+\lambda_m=1 \end{align*} and \begin{align*} q=\lambda_0p_0+\cdots+\lambda_mp_m. \end{align*} [/definition] Affinely independent points play the role of a coordinate basis for a flat simplex. A point in their affine span might a priori admit several weighted descriptions with coefficients summing to $1$. To treat barycentric coordinates as coordinates, not just as one possible expression, that ambiguity has to be eliminated. The next point is therefore a uniqueness principle for this coordinate system. It is what justifies speaking of the barycentric coordinates of a point and prepares the examples below, where computations in triangles and simplices rely on the coefficients being determined by the point. [quotetheorem:9345] [example: Barycentric Coordinates in a Triangle] Let $p_0,p_1,p_2$ be non-collinear points in a real affine plane. Non-collinearity means that the displacement vectors $p_1-p_0$ and $p_2-p_0$ are linearly independent, so barycentric coordinates relative to these three points are unique. Indeed, if \begin{align*} q=\lambda_0p_0+\lambda_1p_1+\lambda_2p_2 \end{align*} with \begin{align*} \lambda_0+\lambda_1+\lambda_2=1, \end{align*} then taking $p_0$ as the temporary origin in the affine-combination formula gives \begin{align*} q=p_0+\lambda_0(p_0-p_0)+\lambda_1(p_1-p_0)+\lambda_2(p_2-p_0). \end{align*} Since $p_0-p_0=0$, this is equivalent to \begin{align*} q-p_0=\lambda_1(p_1-p_0)+\lambda_2(p_2-p_0). \end{align*} If the same point also has barycentric coordinates $\mu_0,\mu_1,\mu_2$, then \begin{align*} q-p_0=\mu_1(p_1-p_0)+\mu_2(p_2-p_0). \end{align*} Subtracting the two vector equations gives \begin{align*} 0=(\lambda_1-\mu_1)(p_1-p_0)+(\lambda_2-\mu_2)(p_2-p_0). \end{align*} Because $p_1-p_0$ and $p_2-p_0$ are linearly independent, $\lambda_1-\mu_1=0$ and $\lambda_2-\mu_2=0$, hence $\lambda_1=\mu_1$ and $\lambda_2=\mu_2$. Using the two sum conditions, \begin{align*} \lambda_0=1-\lambda_1-\lambda_2=1-\mu_1-\mu_2=\mu_0. \end{align*} Thus the three barycentric coordinates are determined by $q$. The centroid is the point obtained by taking equal weights \begin{align*} \lambda_0=\lambda_1=\lambda_2=\frac{1}{3}. \end{align*} These are valid barycentric coordinates because \begin{align*} \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1. \end{align*} Therefore \begin{align*} c=\frac{1}{3}p_0+\frac{1}{3}p_1+\frac{1}{3}p_2 \end{align*} is the centroid, and relative to the temporary origin $p_0$ it satisfies \begin{align*} c-p_0=\frac{1}{3}(p_1-p_0)+\frac{1}{3}(p_2-p_0). \end{align*} Over $\mathbb{R}$, the closed triangular region with vertices $p_0,p_1,p_2$ is exactly the set of convex affine combinations of the vertices. Thus a point lies inside or on the boundary of the triangle exactly when its barycentric coordinates satisfy \begin{align*} \lambda_0\ge 0,\quad \lambda_1\ge 0,\quad \lambda_2\ge 0, \end{align*} together with \begin{align*} \lambda_0+\lambda_1+\lambda_2=1. \end{align*} [/example] ## Affine Span and Dimension ### Generated Affine Subspaces A set of points generates the smallest affine subspace containing it. This construction is the affine analogue of linear span, but it uses affine combinations rather than linear combinations. [definition: Affine Span] Let $S$ be a nonempty subset of an affine space $A$. The affine span of $S$, denoted $\operatorname{Aff}(S)$, is the intersection of all affine subspaces of $A$ containing $S$. [/definition] The definition by intersection is intrinsic, but it is not computational: it describes $\operatorname{Aff}(S)$ by all affine subspaces containing $S$, rather than by points one can build from $S$. To work with the span, we need an internal description. The obstruction is the usual affine one: after choosing a base point of $S$, the resulting formula must recover exactly the smallest affine subspace and not depend on hidden extra choices. [quotetheorem:9346] This formula replaces the intersection definition with a usable calculation. To find the affine span of finitely many points, choose one point as a base point, form the displacement vectors from it to the other points, take their ordinary linear span in the model vector space, and translate that vector subspace back to the base point. The intersection definition remains the intrinsic characterization, but the displacement-vector description is what one actually uses to test membership, compute equations, and identify the directions available in the generated affine subspace. ### Dimension Dimension is inherited from the direction space. This keeps the dimension independent of the chosen point used to present the affine space, and it explains why an affine line, plane, or hyperplane has the same dimension as its vector direction. [definition: Dimension of an Affine Space] The dimension of an affine space $A$ modeled on a vector space $V$ is \begin{align*} \dim A=\dim V. \end{align*} [/definition] For subsets that are already affine subspaces, dimension should measure the number of independent directions available inside the subset rather than the number of ambient coordinates. This separate definition is needed because an affine plane inside $k^5$ should still be two-dimensional. [definition: Dimension of an Affine Subspace] The dimension of an affine subspace $B \subset A$ is \begin{align*} \dim B=\dim \operatorname{dir}(B). \end{align*} [/definition] [example: Three Points in the Plane] Let $A$ be a real affine plane modeled on a two-dimensional real vector space $V$, and let $p_0,p_1,p_2\in A$. For two distinct points $p_0\ne p_1$, the displacement $v_1=p_1-p_0$ is nonzero, because $v_1=0$ would give $p_1=p_0+0=p_0$. By *[Affine Span from Displacement Vectors](/theorems/9346)*, their affine span is \begin{align*} \operatorname{Aff}\{p_0,p_1\}=p_0+\operatorname{span}\{p_1-p_0\}. \end{align*} Since $p_1-p_0\ne 0$, the subspace $\operatorname{span}\{p_1-p_0\}$ is one-dimensional, so this affine span is the line through $p_0$ and $p_1$. Now write \begin{align*} v_1=p_1-p_0 \end{align*} and \begin{align*} v_2=p_2-p_0. \end{align*} Again by *Affine Span from Displacement Vectors*, \begin{align*} \operatorname{Aff}\{p_0,p_1,p_2\}=p_0+\operatorname{span}\{v_1,v_2\}. \end{align*} If $p_0,p_1,p_2$ are collinear, then $p_2$ lies on the line through $p_0$ and $p_1$, so there is some $t\in\mathbb{R}$ with \begin{align*} p_2=p_0+t(p_1-p_0). \end{align*} Subtracting $p_0$ in the affine sense gives \begin{align*} p_2-p_0=t(p_1-p_0). \end{align*} Thus $v_2=tv_1$, so $v_1$ and $v_2$ are linearly dependent, and \begin{align*} \operatorname{span}\{v_1,v_2\}=\operatorname{span}\{v_1\}. \end{align*} Therefore the affine span is a line. If the three points are not collinear, then $v_1$ and $v_2$ cannot be linearly dependent. Indeed, if $v_2=tv_1$ for some $t\in\mathbb{R}$, then \begin{align*} p_2=p_0+v_2=p_0+t(p_1-p_0), \end{align*} so $p_2$ would lie on the line through $p_0$ and $p_1$. Hence $v_1$ and $v_2$ are linearly independent. Since $V$ has dimension $2$, two linearly independent vectors span $V$, so \begin{align*} \operatorname{span}\{v_1,v_2\}=V. \end{align*} Consequently \begin{align*} \operatorname{Aff}\{p_0,p_1,p_2\}=p_0+V=A. \end{align*} Thus three collinear points generate a line, while three non-collinear points generate the whole affine plane. [/example] ## What Affine Geometry Preserves Affine geometry does not preserve distances or angles. It preserves incidence, parallelism, affine dimension, ratios along a fixed line, and convex combinations over ordered fields such as $\mathbb{R}$. [quotetheorem:9347] This invariance explains why affine geometry is the natural language for changing coordinate frames in Euclidean space before a metric is introduced. The metric adds lengths and angles; the affine structure already knows which subsets are lines, planes, and their higher-dimensional analogues. [example: A Shear Preserves Lines but Not Angles] Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be defined by \begin{align*} f(x,y)=(x+y,y). \end{align*} Define $L:\mathbb{R}^2\to\mathbb{R}^2$ by \begin{align*} L(v_1,v_2)=(v_1+v_2,v_2). \end{align*} This is the linear map represented by the matrix entries \begin{align*} M_{11}=1,\qquad M_{12}=1,\qquad M_{21}=0,\qquad M_{22}=1. \end{align*} For $p=(p_1,p_2)$ and $v=(v_1,v_2)$, we have \begin{align*} f(p+v)=f(p_1+v_1,p_2+v_2)=(p_1+v_1+p_2+v_2,p_2+v_2). \end{align*} Also, \begin{align*} f(p)+L(v)=(p_1+p_2,p_2)+(v_1+v_2,v_2)=(p_1+v_1+p_2+v_2,p_2+v_2). \end{align*} Thus $f(p+v)=f(p)+L(v)$, so $f$ is affine with linear part $L$. Now let \begin{align*} \ell=\{p+td:t\in\mathbb{R}\} \end{align*} be a line with nonzero direction vector $d=(d_1,d_2)$. For every $t\in\mathbb{R}$, \begin{align*} f(p+td)=f(p)+L(td). \end{align*} Since $L$ is linear, \begin{align*} L(td)=tL(d). \end{align*} Therefore \begin{align*} f(\ell)=\{f(p)+tL(d):t\in\mathbb{R}\}. \end{align*} Here \begin{align*} L(d)=(d_1+d_2,d_2). \end{align*} If $L(d)=(0,0)$, then $d_2=0$ from the second coordinate, and then $d_1+d_2=0$ gives $d_1=0$. This contradicts $d\ne 0$, so $L(d)\ne 0$. Hence the image of $\ell$ is again a line. If two lines are parallel, their direction spaces are the same one-dimensional subspace $\mathbb{R}d$; their images have direction space $\mathbb{R}L(d)$, so the images remain parallel. The coordinate axes have direction vectors \begin{align*} e_1=(1,0) \end{align*} and \begin{align*} e_2=(0,1). \end{align*} They are perpendicular in the Euclidean sense because \begin{align*} e_1\cdot e_2=1\cdot 0+0\cdot 1=0. \end{align*} Under the shear, \begin{align*} L(e_1)=L(1,0)=(1,0) \end{align*} and \begin{align*} L(e_2)=L(0,1)=(1,1). \end{align*} Their Euclidean dot product is \begin{align*} (1,0)\cdot(1,1)=1\cdot 1+0\cdot 1=1. \end{align*} Since this dot product is nonzero, the image directions are not perpendicular. Thus the shear preserves the affine data of lines and parallelism, but it does not preserve Euclidean angles. [/example] Affine ideas also sit between linear and projective geometry. Linear geometry fixes an origin. Affine geometry forgets the origin but keeps parallelism. Projective geometry goes further and adds points at infinity so that parallel lines meet. [remark: Affine Space versus Vector Space] Every vector space gives an affine space by forgetting which point is the zero vector. The reverse direction requires choosing an origin. Statements that depend on $0$ belong to linear geometry; statements invariant under moving $0$ belong to affine geometry. [/remark] ## Beyond and Connected Topics Affine space is the starting point for the affine schemes and affine varieties studied in [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). There the set $k^n$ is studied through its coordinate algebra, and algebraic subsets are cut out by polynomial equations. The affine structure supplies the ambient coordinates, while the polynomial equations supply the geometry. In differential geometry, an affine space is the local flat model that coordinate charts imitate. The pages [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry) and [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry) move beyond a single affine model by allowing coordinate patches to vary from point to point. Tangent spaces then record infinitesimal displacement vectors attached to individual points. In complex geometry, affine coordinate patches are used to describe local holomorphic data before global curvature or projective compactification enters. The notes [Several Complex Variables IV: Complex Geometry and Curvature](/page/Several%20Complex%20Variables%20IV%3A%20Complex%20Geometry%20and%20Curvature) are a later setting where local coordinate models interact with analytic and metric structure. Affine geometry also underlies convexity. Over $\mathbb{R}$, convex combinations are affine combinations with nonnegative coefficients, and convex sets are exactly subsets closed under those combinations. This connection is why barycentric coordinates, affine spans, and hyperplanes appear throughout optimization and geometric analysis. ## References Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). Androma, [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry). Androma, [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry). Androma, [Several Complex Variables IV: Complex Geometry and Curvature](/page/Several%20Complex%20Variables%20IV%3A%20Complex%20Geometry%20and%20Curvature). Marcel Berger, *Geometry I* (1987). Robin Hartshorne, *Algebraic Geometry* (1977). Miles Reid, *Undergraduate Algebraic Geometry* (1988).