[proofplan] We prove the object-level dictionary in both directions. Starting from an affine algebraic set, its coordinate ring is a quotient of a [polynomial ring](/page/Polynomial%20Ring) by the vanishing ideal, hence is finitely generated, and reducedness follows directly from the definition of a vanishing ideal. Conversely, a reduced finitely generated $k$-algebra admits a presentation by a polynomial ring; reducedness forces the kernel ideal to be radical, and the affine [closed set](/page/Closed%20Set) radical ideal correspondence identifies that ideal with the vanishing ideal of its zero set. Finally, the regular-map correspondence shows that isomorphisms of coordinate rings are exactly the pullbacks of regular isomorphisms, giving the claimed identification up to isomorphism and the contravariant map dictionary. [/proofplan] [step:Show that affine algebraic sets have reduced finitely generated coordinate rings] Let $X\subseteq \mathbb A_k^n$ be an affine algebraic set. Let $I(X)\trianglelefteq k[x_1,\dots,x_n]$ denote the vanishing ideal of $X$, namely the ideal of all polynomials in $k[x_1,\dots,x_n]$ that vanish at every point of $X$. By definition of the coordinate ring, there is a $k$-algebra isomorphism \begin{align*} k[X]\cong k[x_1,\dots,x_n]/I(X). \end{align*} Since $k[x_1,\dots,x_n]$ is generated as a $k$-algebra by $x_1,\dots,x_n$, its quotient $k[X]$ is generated as a $k$-algebra by the residue classes $\overline{x_1},\dots,\overline{x_n}$. Thus $k[X]$ is finitely generated over $k$. It remains to verify reducedness. Let $\overline f\in k[X]$ be the residue class of $f\in k[x_1,\dots,x_n]$, and suppose that $\overline f^m=0$ for some $m\in\mathbb N$. Then $f^m\in I(X)$, so for every point $a\in X$, \begin{align*} f(a)^m=0. \end{align*} Because $k$ is a field, this implies $f(a)=0$ for every $a\in X$. Hence $f\in I(X)$, so $\overline f=0$ in $k[X]$. Therefore $k[X]$ has no nonzero nilpotent elements, and so $k[X]$ is reduced. This proves the first direction. [/step] [step:Prove that a reduced presentation has radical kernel] Let $A$ be a reduced finitely generated $k$-algebra, and suppose that a $k$-algebra isomorphism \begin{align*} \psi:k[x_1,\dots,x_n]/I\to A \end{align*} is given. Let \begin{align*} q:k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]/I \end{align*} be the quotient homomorphism. Define the $k$-algebra homomorphism \begin{align*} \varphi:k[x_1,\dots,x_n]\to A \end{align*} by $\varphi=\psi\circ q$. Then $\ker \varphi=I$. We prove that $I$ is radical. Let $f\in k[x_1,\dots,x_n]$ and let $m\in\mathbb N$ satisfy $f^m\in I$. Then \begin{align*} \varphi(f)^m=\varphi(f^m)=0 \end{align*} in $A$. Since $A$ is reduced, $\varphi(f)=0$. Hence $f\in \ker\varphi=I$. Therefore $\sqrt I\subseteq I$. The reverse containment $I\subseteq \sqrt I$ holds for every ideal, so $I=\sqrt I$. [guided] The point of this step is to translate reducedness of the quotient algebra into radicality of the ideal defining the quotient. We have a quotient presentation of $A$, so choose the quotient map \begin{align*} q:k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]/I \end{align*} and compose it with the given isomorphism \begin{align*} \psi:k[x_1,\dots,x_n]/I\to A. \end{align*} This gives a $k$-algebra homomorphism \begin{align*} \varphi:k[x_1,\dots,x_n]\to A \end{align*} defined by $\varphi=\psi\circ q$. Its kernel is exactly $I$, because $q(f)=0$ precisely when $f\in I$, and $\psi$ is injective. Now take an element $f\in k[x_1,\dots,x_n]$ whose power lies in $I$. Thus there exists $m\in\mathbb N$ such that $f^m\in I$. Applying $\varphi$ gives \begin{align*} \varphi(f)^m=\varphi(f^m)=0. \end{align*} This says that $\varphi(f)$ is nilpotent in $A$. Since $A$ is reduced, the only nilpotent element of $A$ is $0$, so $\varphi(f)=0$. Therefore $f\in\ker\varphi=I$. We have shown that every element of $\sqrt I$ already belongs to $I$, so $\sqrt I\subseteq I$. Since every ideal is contained in its radical, $I\subseteq \sqrt I$. Hence $I=\sqrt I$, which is exactly the assertion that $I$ is radical. [/guided] [/step] [step:Apply the Nullstellensatz to recover the presentation from its zero set] Since $k$ is algebraically closed and $I\trianglelefteq k[x_1,\dots,x_n]$ is a radical ideal in the polynomial ring over $k$, the affine closed set radical ideal correspondence, equivalently the strong Nullstellensatz, gives \begin{align*} I(V(I))=I. \end{align*} This is precisely the radical-ideal direction of [citetheorem:9414]. By definition of the coordinate ring of $V(I)$, \begin{align*} k[V(I)]=k[x_1,\dots,x_n]/I(V(I)). \end{align*} Substituting $I(V(I))=I$ yields \begin{align*} k[V(I)]\cong k[x_1,\dots,x_n]/I. \end{align*} Composing this isomorphism with the given isomorphism $k[x_1,\dots,x_n]/I\cong A$ gives a $k$-algebra isomorphism \begin{align*} k[V(I)]\cong A. \end{align*} Equivalently, \begin{align*} A\cong k[V(I)]. \end{align*} [/step] [step:Use the regular-map correspondence to identify isomorphism classes and reverse arrows] Let $X\subseteq \mathbb A_k^n$ and $Y\subseteq \mathbb A_k^m$ be affine algebraic sets over the [algebraically closed field](/page/Algebraically%20Closed%20Field) $k$. We use [citetheorem:9420], the regular-map coordinate-ring correspondence, which applies because $X$ and $Y$ are affine algebraic sets over $k$: the assignment \begin{align*} F\longmapsto F^* \end{align*} gives a bijection from regular maps $F:X\to Y$ to $k$-algebra homomorphisms $F^*:k[Y]\to k[X]$. Suppose first that $\alpha:k[Y]\to k[X]$ is a $k$-algebra isomorphism. By [citetheorem:9420], there is a unique regular map \begin{align*} F:X\to Y \end{align*} such that $F^*=\alpha$. Applying the same theorem to the inverse isomorphism $\alpha^{-1}:k[X]\to k[Y]$, there is a unique regular map \begin{align*} G:Y\to X \end{align*} such that $G^*=\alpha^{-1}$. The pullback of the composite $G\circ F:X\to X$ is \begin{align*} (G\circ F)^*=F^*\circ G^*=\alpha\circ\alpha^{-1}=\operatorname{id}_{k[X]}. \end{align*} The identity map $\operatorname{id}_X:X\to X$ also has pullback $\operatorname{id}_{k[X]}$. By the uniqueness part of [citetheorem:9420], $G\circ F=\operatorname{id}_X$. Similarly, \begin{align*} (F\circ G)^*=G^*\circ F^*=\alpha^{-1}\circ\alpha=\operatorname{id}_{k[Y]}, \end{align*} and uniqueness gives $F\circ G=\operatorname{id}_Y$. Hence $F$ is a regular isomorphism with inverse $G$. Conversely, if $F:X\to Y$ is a regular isomorphism with inverse $G:Y\to X$, then $F^*:k[Y]\to k[X]$ and $G^*:k[X]\to k[Y]$ are inverse $k$-algebra homomorphisms, because \begin{align*} F^*\circ G^*=(G\circ F)^*=\operatorname{id}_{k[X]} \end{align*} and \begin{align*} G^*\circ F^*=(F\circ G)^*=\operatorname{id}_{k[Y]}. \end{align*} Thus $F^*$ is a $k$-algebra isomorphism. The preceding paragraphs prove that affine algebraic sets have isomorphic coordinate rings exactly when they are regularly isomorphic. Combining this with the realization of every reduced finitely generated $k$-algebra as $k[V(I)]$ proves the claimed identification of isomorphism classes. The same cited bijection, without imposing invertibility, gives the contravariant correspondence between regular maps $X\to Y$ and $k$-algebra homomorphisms $k[Y]\to k[X]$. [guided] The remaining point is not only that coordinate rings exist, but that they remember the affine algebraic set up to the correct notion of isomorphism. The precise external result we need is [citetheorem:9420]. Its hypotheses require affine algebraic sets over the algebraically closed field $k$, and these are exactly the objects under discussion: $X\subseteq \mathbb A_k^n$ and $Y\subseteq \mathbb A_k^m$. The theorem gives a bijection \begin{align*} \{\text{regular maps }X\to Y\}\longleftrightarrow \operatorname{Hom}_{k\text{-alg}}(k[Y],k[X]) \end{align*} by sending a regular map $F:X\to Y$ to its pullback $F^*:k[Y]\to k[X]$. Now suppose $\alpha:k[Y]\to k[X]$ is a $k$-algebra isomorphism. Since $\alpha$ is a homomorphism in the direction $k[Y]\to k[X]$, the contravariance in [citetheorem:9420] produces a unique regular map \begin{align*} F:X\to Y \end{align*} with $F^*=\alpha$. The inverse algebra isomorphism $\alpha^{-1}:k[X]\to k[Y]$ produces a unique regular map \begin{align*} G:Y\to X \end{align*} with $G^*=\alpha^{-1}$. We must check that $F$ and $G$ are inverse maps, not merely maps obtained from inverse algebra homomorphisms. Pullback reverses composition: for regular maps $F:X\to Y$ and $G:Y\to X$, the pullback of $G\circ F:X\to X$ is \begin{align*} (G\circ F)^*=F^*\circ G^*. \end{align*} Substituting $F^*=\alpha$ and $G^*=\alpha^{-1}$ gives \begin{align*} (G\circ F)^*=F^*\circ G^*=\alpha\circ\alpha^{-1}=\operatorname{id}_{k[X]}. \end{align*} The identity map $\operatorname{id}_X:X\to X$ also has pullback $\operatorname{id}_{k[X]}$. Because [citetheorem:9420] is a bijection, two regular maps $X\to X$ with the same pullback must be equal. Therefore $G\circ F=\operatorname{id}_X$. The other composite is the same calculation with the roles of $X$ and $Y$ reversed. We have \begin{align*} (F\circ G)^*=G^*\circ F^*=\alpha^{-1}\circ\alpha=\operatorname{id}_{k[Y]}. \end{align*} Since $\operatorname{id}_Y:Y\to Y$ has pullback $\operatorname{id}_{k[Y]}$, uniqueness in [citetheorem:9420] gives $F\circ G=\operatorname{id}_Y$. Thus $F$ is a regular isomorphism and $G$ is its inverse. Conversely, let $F:X\to Y$ be a regular isomorphism, and let $G:Y\to X$ be its inverse regular map. Pullback gives homomorphisms \begin{align*} F^*:k[Y]\to k[X] \end{align*} and \begin{align*} G^*:k[X]\to k[Y]. \end{align*} Since $G\circ F=\operatorname{id}_X$ and $F\circ G=\operatorname{id}_Y$, contravariance gives \begin{align*} F^*\circ G^*=(G\circ F)^*=\operatorname{id}_{k[X]} \end{align*} and \begin{align*} G^*\circ F^*=(F\circ G)^*=\operatorname{id}_{k[Y]}. \end{align*} Hence $F^*$ and $G^*$ are inverse $k$-algebra isomorphisms. This proves the missing object-level assertion: two affine algebraic sets are regularly isomorphic exactly when their coordinate rings are isomorphic as $k$-algebras. It also proves the full map-level assertion, because [citetheorem:9420] gives the bijection between every regular map $X\to Y$ and every $k$-algebra homomorphism $k[Y]\to k[X]$, with the direction of arrows reversed. [/guided] [/step]