A $X \subseteq \mathbb{A}^n$ such that $I(X) \neq I(V(I(X)))$? Let $\mathbb{A}^n$ be the affine $n$-space over a field $K$. Denote by $V(S)$ the zero locus of a $S \subseteq K[x_1, \dots, x_n]$ and let $I(X)$ be the ideal of a $X \subseteq \mathbb{A}^n$. Is there any $X \subseteq \mathbb{A}^n$ such that $I(X) \neq I(V(I(X)))$? If yes, give an example, please.
 A: EDIT (with Georges Elencwajg's Suggestion):  We assume that $K$ is a commutative integral ring (not necessarily unital).  Let $S$ be an ideal of $K\left[x_1,x_2,\ldots,x_n\right]$ and $X\subseteq K\mathbb{A}^n$.  We claim that $\sqrt{I(X)}=I(X)=I\Big(V\big(I(X)\big)\Big)$ and that $V\big(\sqrt{S}\big)=V(S)=V\Big(I\big(V(S)\big)\Big)$.  We invoke five properties: 
(1) $Y\subseteq V\big(I(Y)\big)$ for every $Y\subseteq K\mathbb{A}^n$,
(2) $T\subseteq I\big(V(T)\big)$ for every ideal $T$ of $K\left[x_1,x_2,\ldots,x_n\right]$,
(3) If $Y_1,Y_2\subseteq K\mathbb{A}^n$ are such that $Y_1\subseteq Y_2$, then $I\left(Y_1\right)\supseteq I\left(Y_2\right)$,
(4) If $T_1,T_2$ are ideals of $K\left[x_1,x_2,\ldots,x_n\right]$ such that $T_1\subseteq T_2$, then $V\left(T_1\right)\supseteq V\left(T_2\right)$, and
(5) For any ideal $T$ of $K\left[x_1,x_2,\ldots,x_n\right]$, $T\subseteq \sqrt{T}$.
From (2), as $I(X)$ is an ideal of $K\left[x_1,x_2,\ldots,x_n\right]$, we have $I(X)\subseteq I\Big(V\big(I(X)\big)\Big)$.  From (1), we have $X\subseteq V\big(I(X)\big)$, which means $I(X)\supseteq I\Big(V\big(I(X)\big)\Big)$, due to (3).
Similarly, by (1), as $V(S)\subseteq K\mathbb{A}^n$, we get $V(S)\subseteq V\Big(I\big(V(S)\big)\Big)$.  From (2), $S\subseteq I\big(V(S)\big)$, which leads to $V(S)\supseteq V\Big(I\big(V(S)\big)\Big)$, where (4) is applied.
Now, to show $I(X)=\sqrt{I(X)}$, we have $I(X)\subseteq \sqrt{I(X)}$ from (5).  Suppose $f\in\sqrt{I(X)}$.  Then, $f^k \in I(X)$ for some $k \in \mathbb{N}$.  That is, for all $p \in X$, $\big(f(p)\big)^k=0_K$, but, as $K$ is an integral ring, $f(p)=0_K$.  Therefore, $f\in I(X)$, or $\sqrt{I(X)}\subseteq I(X)$, as required.
To show that $V(S)=V\left(\sqrt{S}\right)$, we note from (5) that $S\subseteq \sqrt{S}$, which leads to $V\left(\sqrt{S}\right)\subseteq V(S)$, by (4).  Suppose that $p \in V\left(S\right)$.  Then, for $f\in \sqrt{S}$, we have $f^k\in S$ for some $k\in\mathbb{N}$.  Consequently, $\big(f(p)\big)^k=0_K$, or $f(p)=0_K$, since $K$ is an integral ring.  Ergo, $p \in V\left(\sqrt{S}\right)$.  Thus, $V(S)\subseteq V\left(\sqrt{S}\right)$, as desired.
If $K$ is a non-integral commutative ring, we only have $\sqrt{I(X)}\supseteq I(X)=I\Big(V\big(I(X)\big)\Big)$ and that $V\big(\sqrt{S}\big)\subseteq V(S)=V\Big(I\big(V(S)\big)\Big)$.  For example, if $K:=\mathbb{Z}/8\mathbb{Z}$, we can take $n:=1$, $X:=2\mathbb{Z}/8\mathbb{Z}$, and $S:=\left(4x_1\right)$.  In this case, $I(X)=\left(4x_1,2x_1+x_1^2,2x_1^2,x_1^3\right)$, whereas $\sqrt{I(X)}=\left(2,x_1\right) \supsetneq I(X)$.  Also, we have $\sqrt{S}=\left(2\right)$, so $V(S)=2\mathbb{Z}/8\mathbb{Z}$, but $V\left(\sqrt{S}\right)= 8\mathbb{Z}/8\mathbb{Z} \subsetneq V(S)$.
