Why does $V(I(S))=\overline{S}$? Let $S\subset\operatorname{Spec}A$, where $A$ is a commutative ring with $1$. I am having trouble seeing why $V(I(S))=\overline{S}$, where $\overline{S}$ is the Zariski closure of $S$.
My attempt is as follows. It is not hard to see that $S\subset V(I(S))$: $V(I(S))$ is the set of all prime ideals of $A$ containing $I(S)$, and $I(S)$ is the intersection of all elements (prime ideals) of $S$. Since every element of $S$ contains $I(S)$, it follows that $S\subset V(I(S))$. This implies that $\overline{S}\subset V(I(S))$.
The issue I'm having is proving the reverse inclusion. Suppose $V(J)$ is any closed set containing $S$. Then $I(V(J))\subset I(S)$ since $I(\cdot)$ is inclusion reversing. I'm not sure where to go from here or even if this is the correct line of thought to have. What am I missing?
 A: The idea of hitting $V(J)$ with $I$ is a good idea and can probably be made to work, but I think it's more straightforward to reason directly with $V(J)$ and $J$. It's a matter of unwinding the definitions completely:
Suppose $V(J)\supset S$ as you said. What is $V(J)$? It is the set of all primes containing $J$, where $J$ is some ideal in $A$. What does $S\subset V(J)$ tell us about the relationship between $J$ and the prime ideals that are members of $S$? Well, $\mathfrak{p}\in V(J)$ means $\mathfrak{p}\supset J$, by definition. So $S\subset V(J)$ means that for all $\mathfrak{p}\in S$, we have $J\subset \mathfrak{p}$. Thus, intersecting over all these primes, we get
$$J\subset \bigcap_{\mathfrak{p}\in S} \mathfrak{p}.$$
Recalling that $I(S)$ is nothing but the right side of this, we have
$$J\subset I(S).$$
But then this means that any prime ideal containing $I(S)$ also contains $J$! In other words, any member of $V(I(S))$ is also a member of $V(J)$! This is the assertion that
$$V(I(S))\subset V(J).$$
So any closed set containing $S$ actually contains $V(I(S))$, and it follows that $V(I(S))\subset \overline S$, as desired.
(To me it feels like saying the same sentence over and over again in slightly different language until it becomes the desired statement.)
A: Recall that $I(V(J)) = \sqrt{J}$. Since $I(V(J))\subseteq I(S)$, applying $V$, then
$$
V(J) = V(\sqrt{J}) = V(I(V(J))) \supseteq V(I(S)).
$$
Thus $V(I(S))$ is contained in any closed subset containing $S$, so $V(I(S)) = \overline{S}$.
