# 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.

• en.wikipedia.org/wiki/Galois_connection – Hoot Jun 28 '15 at 3:58
• No such an $X$ does not exist. This question should be reopened: the accepted answer assumes that the base field is algebraically closed, which is not necessary, and uses the Nullstellensatz which is overkill and actually obscures the trivial and purely formal reason why $I(V(I(X))=I(X)$ for any subset $X\subset \mathbb A^n$. – Georges Elencwajg Jun 28 '15 at 9:36
• @GeorgesElencwajg I get more or less your point. However there is absolutely no effort shown from OP and so I don't see any good reason to reopen the question. – Surb Jun 28 '15 at 9:51
• I evoked this question on meta – Georges Elencwajg Jun 28 '15 at 10:09
• @Voyska I'm definitely not a "question closing fan" and I think often that some questions are closed for no good reason. But I think that this question is very badly asked: what is $\mathbb{A}^n$? what are $I$ and $V$? I mean even if I would like to, just with the question in its actual state, I'd have no way to understand it. Googling A,I or V will not bring me anywhere... – Surb Jun 28 '15 at 10:57

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 $V\left(\sqrt{S}\right)= 8\mathbb{Z}/8\mathbb{Z} \subsetneq V(S)$.

• I asked this for a friend: He was having trouble with his internet connection and sent me the question via SMS. As you may have noticed, this is far beyond the level of my habitual questions. He said that your answer is satisfactory and told me to accept it. – Billy Rubina Jun 28 '15 at 4:22
• This answer is not satisfactory for reasons I outlined in a comment above: the field $K$ should not be assumed algebraically closed and the Nullstellensatz should not be invoked. (I didn't downvote though). – Georges Elencwajg Jun 28 '15 at 9:39
• Well, Voyska is the OP, and he accepted the answer. So, I guess that he just forgot to mention that $K$ is algebraically closed. Anyway, I still think that the claim holds even if $K$ is not algebraically closed, or even if $K$ is a commutative integral ring. I just haven't found a good proof yet. (Frankly, I don't care if you would downvote my comment.) – Batominovski Jun 28 '15 at 10:01