[proofplan] We first construct the evaluation map from ambient polynomials to functions on $X$ and identify its kernel with the vanishing ideal $I(X)$. This gives an injective $k$-algebra homomorphism from the quotient $k[x_1,\dots,x_n]/I(X)$ into the algebra of functions on $X$. The only substantive point is surjectivity: a regular function, locally written as a quotient of polynomial functions, must be represented globally by a single coordinate-ring element. We prove this by using a finite distinguished-[open cover](/page/Open%20Cover) and the affine Nullstellensatz dictionary to patch the local polynomial representatives. [/proofplan] [step:Construct the evaluation homomorphism and compute its kernel] Let \begin{align*} R := k[x_1,\dots,x_n]. \end{align*} Let $\operatorname{Fun}(X,k)$ denote the $k$-algebra of all set-theoretic functions from $X$ to $k$, with pointwise addition and multiplication. Define the evaluation map \begin{align*} \operatorname{ev}: R &\longrightarrow \operatorname{Fun}(X,k) \end{align*} \begin{align*} F &\longmapsto \left(p \mapsto F(p)\right). \end{align*} For $F,G \in R$, $\lambda \in k$, and $p \in X$, evaluation gives \begin{align*} \operatorname{ev}(F+G)(p)=F(p)+G(p), \end{align*} \begin{align*} \operatorname{ev}(FG)(p)=F(p)G(p), \end{align*} and \begin{align*} \operatorname{ev}(\lambda)(p)=\lambda. \end{align*} Thus $\operatorname{ev}$ is a $k$-algebra homomorphism. By definition of the vanishing ideal, \begin{align*} I(X)=\{F\in R:F(p)=0 \text{ for every } p\in X\}. \end{align*} Therefore \begin{align*} \ker(\operatorname{ev})=I(X). \end{align*} By the [universal property of quotient rings](/theorems/8255), $\operatorname{ev}$ induces a unique injective $k$-algebra homomorphism \begin{align*} \Phi: R/I(X) &\longrightarrow \operatorname{Fun}(X,k) \end{align*} \begin{align*} \overline{F} &\longmapsto \left(p \mapsto F(p)\right). \end{align*} Its image consists of polynomial functions on $X$. [guided] The first issue is well-definedness: two ambient polynomials may define the same function on $X$. We encode this by starting before quotienting. Let \begin{align*} R := k[x_1,\dots,x_n]. \end{align*} The evaluation map is the function \begin{align*} \operatorname{ev}: R &\longrightarrow \operatorname{Fun}(X,k) \end{align*} \begin{align*} F &\longmapsto \left(p \mapsto F(p)\right), \end{align*} where $\operatorname{Fun}(X,k)$ is the $k$-algebra of all functions $X\to k$. This map respects the algebra operations because polynomial evaluation respects addition, multiplication, and scalar multiplication pointwise. For every $F,G\in R$, every $\lambda\in k$, and every $p\in X$, we have \begin{align*} \operatorname{ev}(F+G)(p)=F(p)+G(p), \end{align*} \begin{align*} \operatorname{ev}(FG)(p)=F(p)G(p), \end{align*} and \begin{align*} \operatorname{ev}(\lambda)(p)=\lambda. \end{align*} So $\operatorname{ev}$ is a $k$-algebra homomorphism. Now we compute the kernel. A polynomial $F\in R$ lies in $\ker(\operatorname{ev})$ precisely when the function $p\mapsto F(p)$ is identically zero on $X$. By the definition of the vanishing ideal, this condition is exactly \begin{align*} F\in I(X)=\{G\in R:G(p)=0 \text{ for every } p\in X\}. \end{align*} Hence \begin{align*} \ker(\operatorname{ev})=I(X). \end{align*} Because the kernel is $I(X)$, the quotient construction removes exactly the ambiguity of choosing different ambient polynomial representatives. The universal property of quotient rings gives a unique injective $k$-algebra homomorphism \begin{align*} \Phi: R/I(X) &\longrightarrow \operatorname{Fun}(X,k) \end{align*} \begin{align*} \overline{F} &\longmapsto \left(p \mapsto F(p)\right). \end{align*} The injectivity follows because the kernel has already been divided out. [/guided] [/step] [step:Reduce surjectivity to patching a locally regular function] It remains to show that every regular function belongs to the image of $\Phi$. Let \begin{align*} f: X \longrightarrow k \end{align*} be a regular function. By the local quotient definition of regularity, for every $p\in X$ there exist an open neighbourhood $U_p\subseteq X$ of $p$ and polynomials $A_p,B_p\in R$ such that $B_p(q)\ne 0$ for every $q\in U_p$ and \begin{align*} f(q)=\frac{A_p(q)}{B_p(q)} \end{align*} for every $q\in U_p$. Since $X$ is a closed subspace of the Noetherian space $\mathbb A_k^n$, [citetheorem:9411] and [citetheorem:9412] imply that $X$ is Noetherian. In particular, every open cover of $X$ has a [finite subcover](/page/Finite%20Subcover). By [citetheorem:9424], [distinguished opens form a basis](/theorems/9424) for the Zariski topology on $X$, so each $U_p$ contains a distinguished open neighbourhood on which the same quotient formula remains valid. Thus we obtain polynomials $g_1,\dots,g_r,A_1,\dots,A_r,B_1,\dots,B_r\in R$ such that \begin{align*} X=D_X(g_1)\cup\cdots\cup D_X(g_r), \end{align*} where \begin{align*} D_X(g_i):=\{p\in X:g_i(p)\ne 0\}, \end{align*} and for any polynomials $h_1,\dots,h_s\in R$ we write \begin{align*} V_X(h_1,\dots,h_s):=\{p\in X:h_1(p)=\cdots=h_s(p)=0\}. \end{align*} For every $i\in\{1,\dots,r\}$ and every $q\in D_X(g_i)$, \begin{align*} B_i(q)\ne 0 \end{align*} and \begin{align*} f(q)=\frac{A_i(q)}{B_i(q)}. \end{align*} [/step] [step:Replace each local quotient by a polynomial on a distinguished open] Fix $i\in\{1,\dots,r\}$. Since $B_i$ is nonzero on $D_X(g_i)$, the closed subset \begin{align*} D_X(g_i)\cap V_X(B_i) \end{align*} is empty. Equivalently, \begin{align*} V_X(g_iB_i)=V_X(g_i). \end{align*} By the affine radical ideal correspondence [citetheorem:9414], the equality of these closed sets implies that the radicals of their defining ideals in $k[X]=R/I(X)$ are equal. Hence the element $\overline{g_i}$ lies in the radical of the principal ideal generated by $\overline{g_iB_i}$ in $k[X]$. Therefore there exists an integer $N_i\ge 1$ and an element $\overline{C_i}\in k[X]$ such that \begin{align*} \overline{g_i}^{N_i}=\overline{C_i}\,\overline{g_iB_i}. \end{align*} Choose a polynomial representative $C_i\in R$ of $\overline{C_i}$ and define \begin{align*} H_i:=g_iC_iA_i\in R. \end{align*} On $D_X(g_i)$, division by $g_i(q)$ is valid, so the preceding identity gives \begin{align*} g_i(q)^{N_i-1}=C_i(q)B_i(q). \end{align*} Multiplying the local expression for $f$ by $g_i(q)$ times this identity gives \begin{align*} g_i(q)^{N_i}f(q)=H_i(q) \end{align*} for every $q\in D_X(g_i)$. If $q\in X\setminus D_X(g_i)$, then $g_i(q)=0$, so both $g_i(q)^{N_i}f(q)$ and $H_i(q)=g_i(q)C_i(q)A_i(q)$ are zero. Hence \begin{align*} g_i(q)^{N_i}f(q)=H_i(q) \end{align*} for every $q\in X$. [guided] The denominator $B_i$ is the obstruction to writing $f$ by a polynomial on $D_X(g_i)$. We remove that denominator using the fact that $B_i$ has no zeros on the [open set](/page/Open%20Set) where the formula is used. Fix $i\in\{1,\dots,r\}$. On $D_X(g_i)$, the polynomial $B_i$ never vanishes. This says \begin{align*} D_X(g_i)\cap V_X(B_i)=\varnothing. \end{align*} Equivalently, a point of $X$ satisfies $g_i(p)\ne 0$ only if $B_i(p)\ne 0$. Thus the zero sets of $g_i$ and $g_iB_i$ on $X$ are the same: \begin{align*} V_X(g_iB_i)=V_X(g_i). \end{align*} We now translate equality of zero sets into an algebra statement in the coordinate ring. Let $k[X]=R/I(X)$, and write $\overline{G}$ for the class of a polynomial $G\in R$ in $k[X]$. By the affine radical ideal correspondence [citetheorem:9414], equality of the closed sets cut out by $\overline{g_iB_i}$ and $\overline{g_i}$ implies equality of the corresponding radical ideals. In particular, \begin{align*} \overline{g_i}\in \sqrt{(\overline{g_iB_i})}. \end{align*} By the definition of radical membership, there exist an integer $N_i\ge 1$ and an element $\overline{C_i}\in k[X]$ such that \begin{align*} \overline{g_i}^{N_i}=\overline{C_i}\,\overline{g_iB_i}. \end{align*} Choose a polynomial representative $C_i\in R$ of $\overline{C_i}$. Evaluating this equality at any $q\in X$ gives \begin{align*} g_i(q)^{N_i}=C_i(q)g_i(q)B_i(q). \end{align*} If $q\in D_X(g_i)$, then $g_i(q)\ne 0$, so we may divide by $g_i(q)$ and obtain \begin{align*} g_i(q)^{N_i-1}=C_i(q)B_i(q). \end{align*} Now use the local formula \begin{align*} f(q)=\frac{A_i(q)}{B_i(q)} \end{align*} on $D_X(g_i)$. Multiplying by the identity above gives \begin{align*} g_i(q)^{N_i-1}f(q)=C_i(q)A_i(q). \end{align*} Multiplying once more by $g_i(q)$ gives \begin{align*} g_i(q)^{N_i}f(q)=g_i(q)C_i(q)A_i(q). \end{align*} The right-hand side is the evaluation of the polynomial \begin{align*} H_i:=g_iC_iA_i\in R. \end{align*} Thus \begin{align*} g_i(q)^{N_i}f(q)=H_i(q) \end{align*} for every $q\in D_X(g_i)$. This identity also holds at every $q\in X\setminus D_X(g_i)$, because then $g_i(q)=0$ and both sides are zero. Therefore we have arranged the global identity \begin{align*} g_i(q)^{N_i}f(q)=H_i(q) \end{align*} for every $q\in X$. [/guided] [/step] [step:Patch the local polynomial data into one global polynomial] Because the distinguished opens $D_X(g_1),\dots,D_X(g_r)$ cover $X$, the common zero set of the powers $g_1^{N_1},\dots,g_r^{N_r}$ on $X$ is empty: \begin{align*} V_X(g_1^{N_1},\dots,g_r^{N_r})=\varnothing. \end{align*} Applying [citetheorem:9414] to the empty [closed set](/page/Closed%20Set) gives \begin{align*} 1\in \sqrt{(\overline{g_1}^{N_1},\dots,\overline{g_r}^{N_r})}\subseteq k[X]. \end{align*} By the definition of radical membership, this means that $1^m=1$ lies in $(\overline{g_1}^{N_1},\dots,\overline{g_r}^{N_r})$ for some integer $m\ge 1$. Hence \begin{align*} 1\in (\overline{g_1}^{N_1},\dots,\overline{g_r}^{N_r}). \end{align*} Thus there exist polynomials $E_1,\dots,E_r\in R$ such that, in $k[X]$, \begin{align*} 1=\sum_{i=1}^r \overline{E_i}\,\overline{g_i}^{N_i}. \end{align*} Define \begin{align*} F:=\sum_{i=1}^r E_iH_i\in R. \end{align*} For every $q\in X$, evaluating the preceding identity and using the global identities $g_i(q)^{N_i}f(q)=H_i(q)$ gives \begin{align*} f(q)=\sum_{i=1}^r E_i(q)g_i(q)^{N_i}f(q)=\sum_{i=1}^r E_i(q)H_i(q)=F(q). \end{align*} Therefore $f$ is represented by the ambient polynomial $F$ on all of $X$. [guided] The point of the preceding step was to obtain identities that hold at every point of $X$, not just on one member of the cover. For each $i\in\{1,\dots,r\}$, we have a polynomial $H_i\in R$ and an integer $N_i\ge 1$ such that \begin{align*} g_i(q)^{N_i}f(q)=H_i(q) \end{align*} for every $q\in X$. Now we need coefficients that add the functions $g_i^{N_i}$ to $1$ on $X$. Since $D_X(g_1),\dots,D_X(g_r)$ cover $X$, there is no point of $X$ at which all $g_i$ vanish. Equivalently, there is no point of $X$ at which all powers $g_i^{N_i}$ vanish, so \begin{align*} V_X(g_1^{N_1},\dots,g_r^{N_r})=\varnothing. \end{align*} By [citetheorem:9414], the ideal of functions vanishing on this empty closed set has radical equal to the whole coordinate ring. Therefore \begin{align*} 1\in \sqrt{(\overline{g_1}^{N_1},\dots,\overline{g_r}^{N_r})}\subseteq k[X]. \end{align*} Unpacking radical membership, there is an integer $m\ge 1$ such that $1^m$ belongs to the ideal $(\overline{g_1}^{N_1},\dots,\overline{g_r}^{N_r})$. Since $1^m=1$, we get \begin{align*} 1\in (\overline{g_1}^{N_1},\dots,\overline{g_r}^{N_r}). \end{align*} Thus there exist polynomials $E_1,\dots,E_r\in R$ such that, in $k[X]$, \begin{align*} 1=\sum_{i=1}^r \overline{E_i}\,\overline{g_i}^{N_i}. \end{align*} Define \begin{align*} F:=\sum_{i=1}^r E_iH_i\in R. \end{align*} Evaluating the coordinate-ring identity at any point $q\in X$ gives \begin{align*} 1=\sum_{i=1}^r E_i(q)g_i(q)^{N_i}. \end{align*} Multiplying by $f(q)$ and then substituting the global identities $g_i(q)^{N_i}f(q)=H_i(q)$ gives \begin{align*} f(q)=\sum_{i=1}^r E_i(q)g_i(q)^{N_i}f(q)=\sum_{i=1}^r E_i(q)H_i(q)=F(q). \end{align*} Since $q\in X$ was arbitrary, the regular function $f$ is the restriction to $X$ of the ambient polynomial $F$. [/guided] [/step] [step:Conclude that the evaluation map is an isomorphism of $k$-algebras] The previous step shows that every regular function $f\in \mathcal O(X)$ is equal to $\Phi(\overline{F})$ for some $F\in R$. Hence the map \begin{align*} \Phi: R/I(X)\longrightarrow \mathcal O(X) \end{align*} is surjective. The first step showed that $\Phi$ is injective and that it preserves addition, multiplication, and scalar multiplication. Therefore $\Phi$ is an isomorphism of $k$-algebras. [/step]