[proofplan] The proof is the strong Hilbert Nullstellensatz applied in both directions. First, if an affine algebraic set is written as $X=V(J)$, then the Nullstellensatz identifies its vanishing ideal with $\sqrt J$, so $I(X)$ is radical and applying $V$ returns $X$. Conversely, if $I$ is already radical, the same identity gives $I(V(I))=I$. The inclusion-reversing property is then checked directly from the definitions of $I(X)$ and $V(I)$. [/proofplan] [step:Fix the ambient notation and the Nullstellensatz identity] Let \begin{align*} R:=k[x_1,\dots,x_n]. \end{align*} For an ideal $J\trianglelefteq R$, let $\sqrt J$ denote its radical: \begin{align*} \sqrt J:=\{f\in R : f^m\in J \text{ for some } m\in \mathbb N\}. \end{align*} We use the strong Hilbert Nullstellensatz in the following form: because $k$ is algebraically closed, every ideal $J\trianglelefteq R$ satisfies \begin{align*} I(V(J))=\sqrt J. \end{align*} This is the essential algebraic input of the proof. The external result used here is the strong Hilbert Nullstellensatz: for an [algebraically closed field](/page/Algebraically%20Closed%20Field) $k$ and every ideal $J\trianglelefteq k[x_1,\dots,x_n]$, the vanishing ideal of the zero locus of $J$ is the radical $\sqrt J$. Its hypotheses match the present setting because $k$ is algebraically closed and $J$ is an ideal of $R=k[x_1,\dots,x_n]$. [guided] We first isolate the only non-formal ingredient. The ring in the theorem is \begin{align*} R:=k[x_1,\dots,x_n], \end{align*} and for any ideal $J\trianglelefteq R$ its radical is \begin{align*} \sqrt J:=\{f\in R : f^m\in J \text{ for some } m\in \mathbb N\}. \end{align*} The strong Hilbert Nullstellensatz says precisely that, when $k$ is algebraically closed, the polynomials vanishing on the common zero set of $J$ are exactly the radical of $J$: \begin{align*} I(V(J))=\sqrt J. \end{align*} This is where the hypothesis that $k$ is algebraically closed is used. Without that hypothesis, this equality can fail in this form, and the correspondence with ordinary radical ideals need not hold. The external result used here is the strong Hilbert Nullstellensatz: for an algebraically closed field $k$ and every ideal $J\trianglelefteq k[x_1,\dots,x_n]$, the vanishing ideal of the zero locus of $J$ is the radical $\sqrt J$. The hypotheses match the present theorem exactly: $k$ is algebraically closed by assumption, and $J$ is an ideal of the [polynomial ring](/page/Polynomial%20Ring) $R=k[x_1,\dots,x_n]$. [/guided] [/step] [step:Show that an affine algebraic set is recovered from its vanishing ideal] Let $X\subseteq \mathbb A_k^n$ be an affine algebraic set. By definition, there is an ideal $J\trianglelefteq R$ such that \begin{align*} X=V(J). \end{align*} By the strong Hilbert Nullstellensatz, \begin{align*} I(X)=I(V(J))=\sqrt J. \end{align*} Thus $I(X)$ is radical. It remains to check that $V(I(X))=X$. Since $I(X)=\sqrt J$, it is enough to prove $V(\sqrt J)=V(J)$. If $a\in V(J)$ and $f\in \sqrt J$, choose $m\in \mathbb N$ such that $f^m\in J$. Then \begin{align*} f(a)^m=(f^m)(a)=0. \end{align*} Because $k$ is a field, this implies $f(a)=0$, so $a\in V(\sqrt J)$. Hence $V(J)\subseteq V(\sqrt J)$. Conversely, since $J\subseteq \sqrt J$, every point vanishing on all elements of $\sqrt J$ also vanishes on all elements of $J$, so \begin{align*} V(\sqrt J)\subseteq V(J). \end{align*} Therefore \begin{align*} V(I(X))=V(\sqrt J)=V(J)=X. \end{align*} [guided] We now prove that an affine algebraic set is recovered after applying first $I$ and then $V$. Let $X\subseteq \mathbb A_k^n$ be an affine algebraic set. By the definition of affine algebraic set, there exists an ideal $J\trianglelefteq R$ such that \begin{align*} X=V(J). \end{align*} The strong Hilbert Nullstellensatz applies because $k$ is algebraically closed and $J$ is an ideal of $R=k[x_1,\dots,x_n]$. It gives \begin{align*} I(X)=I(V(J))=\sqrt J. \end{align*} Since radicals are radical ideals, this proves that $I(X)$ is radical. It remains to show that applying $V$ returns the original set $X$. Because $I(X)=\sqrt J$, it is enough to prove \begin{align*} V(\sqrt J)=V(J). \end{align*} First take $a\in V(J)$. To prove $a\in V(\sqrt J)$, let $f\in \sqrt J$. By the definition of radical, there exists $m\in \mathbb N$ such that $f^m\in J$. Since $a\in V(J)$, every element of $J$ vanishes at $a$, so \begin{align*} 0=(f^m)(a)=f(a)^m. \end{align*} Because $k$ is a field, the only element whose positive power is $0$ is $0$ itself. Hence $f(a)=0$. Since this holds for every $f\in \sqrt J$, we have $a\in V(\sqrt J)$, and therefore \begin{align*} V(J)\subseteq V(\sqrt J). \end{align*} Conversely, $J\subseteq \sqrt J$ by the definition of radical. Hence any point that vanishes on every polynomial in $\sqrt J$ also vanishes on every polynomial in $J$, so \begin{align*} V(\sqrt J)\subseteq V(J). \end{align*} The two inclusions give $V(\sqrt J)=V(J)$, and therefore \begin{align*} V(I(X))=V(\sqrt J)=V(J)=X. \end{align*} [/guided] [/step] [step:Show that a radical ideal is recovered from its zero locus] Let $I\trianglelefteq R$ be a radical ideal. Applying the strong Hilbert Nullstellensatz to $I$ gives \begin{align*} I(V(I))=\sqrt I. \end{align*} Since $I$ is radical, $\sqrt I=I$. Hence \begin{align*} I(V(I))=I. \end{align*} Thus every radical ideal is recovered after applying $V$ and then $I$. [guided] We now prove the reverse recovery statement for ideals. Let $I\trianglelefteq R$ be a radical ideal. The strong Hilbert Nullstellensatz applies because $k$ is algebraically closed and $I$ is an ideal of $R=k[x_1,\dots,x_n]$. It gives \begin{align*} I(V(I))=\sqrt I. \end{align*} The hypothesis that $I$ is radical means precisely that $I=\sqrt I$. Substituting this into the Nullstellensatz identity gives \begin{align*} I(V(I))=I. \end{align*} Thus a radical ideal loses no information when we pass to its common zero locus and then take all polynomials vanishing on that zero locus. [/guided] [/step] [step:Verify that both assignments reverse inclusions] Let $I,J\trianglelefteq R$ be ideals with $I\subset J$. If $a\in V(J)$, then every polynomial in $J$ vanishes at $a$. Since every polynomial in $I$ belongs to $J$, every polynomial in $I$ vanishes at $a$, so $a\in V(I)$. Hence \begin{align*} V(J)\subset V(I). \end{align*} Let $X,Y\subseteq \mathbb A_k^n$ satisfy $X\subset Y$. If $f\in I(Y)$, then $f$ vanishes at every point of $Y$. Since every point of $X$ is a point of $Y$, the polynomial $f$ vanishes at every point of $X$, so $f\in I(X)$. Hence \begin{align*} I(Y)\subset I(X). \end{align*} [guided] We verify directly from the definitions that both operations reverse inclusions. First let $I,J\trianglelefteq R$ be ideals with $I\subset J$. A point $a\in V(J)$ is, by definition, a point of $\mathbb A_k^n$ at which every polynomial in $J$ vanishes. Since $I\subset J$, every polynomial in $I$ is also a polynomial in $J$. Therefore every polynomial in $I$ vanishes at $a$, so $a\in V(I)$. This proves \begin{align*} V(J)\subset V(I). \end{align*} Now let $X,Y\subseteq \mathbb A_k^n$ satisfy $X\subset Y$. A polynomial $f\in I(Y)$ is, by definition, a polynomial in $R$ that vanishes at every point of $Y$. Since every point of $X$ is also a point of $Y$, the same polynomial $f$ vanishes at every point of $X$. Hence $f\in I(X)$, and therefore \begin{align*} I(Y)\subset I(X). \end{align*} This is exactly the inclusion-reversing property recorded in [citetheorem:9405], proved here from the definitions in the present notation. [/guided] [/step] [step:Conclude the mutually inverse inclusion-reversing bijection] The second step shows that for every affine algebraic set $X\subseteq \mathbb A_k^n$, \begin{align*} V(I(X))=X. \end{align*} The third step shows that for every radical ideal $I\trianglelefteq R$, \begin{align*} I(V(I))=I. \end{align*} Therefore the two assignments are mutually inverse between affine algebraic subsets of $\mathbb A_k^n$ and radical ideals of $R$. The fourth step shows that each assignment reverses inclusions, completing the proof. [guided] We assemble the two recovery statements. For every affine algebraic set $X\subseteq \mathbb A_k^n$, the second step proved \begin{align*} V(I(X))=X. \end{align*} Thus the operation $X\mapsto I(X)$ followed by $I\mapsto V(I)$ is the identity on affine algebraic subsets. For every radical ideal $I\trianglelefteq R$, the third step proved \begin{align*} I(V(I))=I. \end{align*} Thus the operation $I\mapsto V(I)$ followed by $X\mapsto I(X)$ is the identity on radical ideals. These two identities show that the assignments are mutually inverse bijections between affine algebraic subsets of $\mathbb A_k^n$ and radical ideals of $R$. The preceding inclusion check shows that both assignments reverse inclusions, so the bijection is inclusion-reversing in both directions. [/guided] [/step]