[proofplan] Represent a rational map by a regular map on a nonempty open subset and record its ambient affine coordinate functions. The fact that the image lies in $Y$ forces every polynomial equation in $I(Y)$ to vanish after substitution, and the irreducibility of $X$ lets us interpret this vanishing in the function field $k(X)$. Conversely, a tuple of rational functions gives a regular map to [affine space](/page/Affine%20Space) on a common domain of regularity; the equations defining $Y$ force this map to land in $Y$. Finally, equality of rational functions on a smaller nonempty open subset proves that these two constructions are independent of representatives and inverse to each other. [/proofplan] [step:Send a rational map to its tuple of pulled back affine coordinates] Let $\Phi:X\dashrightarrow Y$ be a rational map, and choose a representative regular map \begin{align*} \Phi_U:U\to Y \end{align*} defined on a nonempty open subset $U\subset X$. Let \begin{align*} \iota:Y\hookrightarrow \mathbb A_k^m \end{align*} denote the inclusion. For each $i\in\{1,\dots,m\}$, let $y_i\in k[\mathbb A_k^m]=k[y_1,\dots,y_m]$ denote the $i$-th ambient affine coordinate regular function, viewed as the map $\mathbb A_k^m\to k$ sending $(a_1,\dots,a_m)$ to $a_i$; its restriction $y_i|_Y$ is a regular function on $Y$. Define \begin{align*} \varphi_i:U\to k \end{align*} by \begin{align*} \varphi_i(p)=y_i(\iota(\Phi_U(p))) \end{align*} for $p\in U$. Since $\Phi_U$ is regular and $y_i|_Y$ is regular on $Y$, each $\varphi_i$ is regular on $U$. By the definition of the function field of the irreducible variety $X$, a regular function on a nonempty open subset of $X$ determines an element of $k(X)$. Thus the representative $\Phi_U$ determines a tuple \begin{align*} (\varphi_1,\dots,\varphi_m)\in k(X)^m. \end{align*} [/step] [step:Verify that pulled back coordinates satisfy the equations of $Y$] Let $F\in I(Y)$. For every $p\in U$, the point $\Phi_U(p)$ lies in $Y$, so $F$ vanishes at the ambient affine coordinates of $\Phi_U(p)$. Hence \begin{align*} F(\varphi_1,\dots,\varphi_m)(p)=0 \end{align*} for every $p\in U$. The expression $F(\varphi_1,\dots,\varphi_m)$ is a rational function on $X$ represented by the regular function \begin{align*} U\to k,\qquad p\mapsto F(\varphi_1(p),\dots,\varphi_m(p)). \end{align*} Since this representative is identically zero on the nonempty open subset $U$, it is the zero element of $k(X)$. Equivalently, the rational-function part of [citetheorem:9427] says that on an irreducible affine variety, a rational function represented by zero on a nonempty open subset is zero in the function field. Applying this to $X$ gives \begin{align*} F(\varphi_1,\dots,\varphi_m)=0 \end{align*} in $k(X)$ for every $F\in I(Y)$. [guided] Fix a polynomial $F\in I(Y)$. The condition $F\in I(Y)$ means precisely that $F$ vanishes on every point of the affine algebraic set $Y$. Since the representative map \begin{align*} \Phi_U:U\to Y \end{align*} has image contained in $Y$, every point $\Phi_U(p)$ satisfies the equation $F=0$. Written in coordinates, this says \begin{align*} F(y_1(\Phi_U(p)),\dots,y_m(\Phi_U(p)))=0 \end{align*} for every $p\in U$. By the definition of $\varphi_i$ as the pullback of $y_i$, this is exactly \begin{align*} F(\varphi_1,\dots,\varphi_m)(p)=0 \end{align*} for every $p\in U$. Now we must translate pointwise vanishing on $U$ into an equality inside the function field $k(X)$. The rational function $F(\varphi_1,\dots,\varphi_m)$ is represented on $U$ by the regular function \begin{align*} U\to k,\qquad p\mapsto F(\varphi_1(p),\dots,\varphi_m(p)). \end{align*} This regular representative is identically zero on the nonempty open subset $U$. Since $X$ is irreducible, the rational-function [identity principle](/theorems/3357) in [citetheorem:9427] applies: two rational functions on $X$ are equal if they agree on a nonempty open subset where both are defined. Applying it to $F(\varphi_1,\dots,\varphi_m)$ and the zero rational function gives \begin{align*} F(\varphi_1,\dots,\varphi_m)=0 \end{align*} in $k(X)$. [/guided] [/step] [step:Construct a rational map from a tuple satisfying the equations of $Y$] Conversely, let \begin{align*} (\varphi_1,\dots,\varphi_m)\in k(X)^m \end{align*} satisfy \begin{align*} F(\varphi_1,\dots,\varphi_m)=0 \end{align*} in $k(X)$ for every $F\in I(Y)$. By the definition of $k(X)$, for each $i\in\{1,\dots,m\}$ there exists a nonempty open subset $U_i\subset X$ on which $\varphi_i$ is represented by a regular function. Because $X$ is irreducible, the finite intersection \begin{align*} U:=U_1\cap\cdots\cap U_m \end{align*} is again a nonempty open subset of $X$. On $U$, define \begin{align*} \psi_U:U\to \mathbb A_k^m \end{align*} by \begin{align*} \psi_U(p)=(\varphi_1(p),\dots,\varphi_m(p)). \end{align*} By the coordinate characterization of regular maps into affine space, since each coordinate function $\varphi_i|_U:U\to k$ is regular, the map $\psi_U$ is regular. We claim that $\psi_U(U)\subset Y$. Let $p\in U$ and let $F\in I(Y)$. On $U$, the rational function $F(\varphi_1,\dots,\varphi_m)$ is represented by the regular function $g_F:U\to k$ defined by $g_F(q)=F(\psi_U(q))$. Since $F(\varphi_1,\dots,\varphi_m)=0$ in $k(X)$, the rational-function identity principle in [citetheorem:9427] gives a nonempty open subset $O\subset U$ on which $g_F$ agrees with the zero function. We cannot apply the affine regular-function identity principle directly to $U$, because a nonempty open subset of an affine variety need not itself be affine. Instead, fix the point $p\in U$. Since [distinguished opens form a basis](/theorems/9424) for the Zariski topology by [citetheorem:9424], choose a distinguished open subset $W\subset X$ such that $p\in W\subset U$. The subset $W$ is affine by the regular-functions-on-distinguished-opens description in [citetheorem:9425], and $W$ is irreducible because it is a nonempty open subset of the irreducible variety $X$. The intersection $O\cap W$ is nonempty because $O$ and $W$ are nonempty open subsets of the irreducible space $U$. The regular function $g_F|_W:W\to k$ vanishes on $O\cap W$, so the regular-function part of [citetheorem:9427] applied to the irreducible affine variety $W$ gives $g_F|_W=0$. Since $p\in W$, this implies \begin{align*} F(\psi_U(p))=g_F(p)=0. \end{align*} This holds for every $F\in I(Y)$, so $\psi_U(p)\in V(I(Y))$, where $V(I(Y))$ denotes the common zero set of the ideal $I(Y)$ in $\mathbb A_k^m$. Since $Y$ is an affine algebraic subset and $I(Y)$ is its vanishing ideal over the [algebraically closed field](/page/Algebraically%20Closed%20Field) $k$, we have $V(I(Y))=Y$. Thus $\psi_U(U)\subset Y$. Therefore $\psi_U$ factors as a regular map \begin{align*} \Phi_U:U\to Y, \end{align*} and this regular map represents a rational map $X\dashrightarrow Y$. [/step] [step:Check independence of the chosen common domain] Suppose $U,V\subset X$ are two nonempty open subsets on which all $\varphi_i$ are regular. Since $X$ is irreducible, $U\cap V$ is nonempty. On $U\cap V$, both maps \begin{align*} p\mapsto (\varphi_1(p),\dots,\varphi_m(p)) \end{align*} are defined using the same regular representatives of the same rational functions. Hence their coordinate functions agree on $U\cap V$, and therefore the two regular maps to $\mathbb A_k^m$, and hence to $Y$, agree on $U\cap V$. By the [equivalence relation](/page/Equivalence%20Relation) defining rational maps, the resulting rational map $X\dashrightarrow Y$ is independent of the chosen common domain of regularity. [/step] [step:Prove that the two constructions are inverse] Starting with a rational map represented by \begin{align*} \Phi_U:U\to Y, \end{align*} we form the tuple \begin{align*} (y_1\circ \Phi_U,\dots,y_m\circ \Phi_U). \end{align*} The construction from a tuple then gives, on the same open subset $U$, the map \begin{align*} p\mapsto (y_1(\Phi_U(p)),\dots,y_m(\Phi_U(p))). \end{align*} This is exactly the ambient coordinate description of $\Phi_U(p)\in Y\subset \mathbb A_k^m$, so it represents the same rational map. Conversely, starting with a tuple \begin{align*} (\varphi_1,\dots,\varphi_m)\in k(X)^m \end{align*} satisfying the equations of $Y$, the constructed rational map is represented on a common nonempty open subset $U\subset X$ by \begin{align*} p\mapsto (\varphi_1(p),\dots,\varphi_m(p)). \end{align*} Pulling back the affine coordinate function $y_i$ along this representative gives $\varphi_i|_U$ for every $i\in\{1,\dots,m\}$. Hence the recovered element of $k(X)^m$ is the original tuple. The two assignments are therefore mutually inverse, giving the desired bijection. [/step]