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I am studying Zariski Topology. Here is a problem I am trying to work on:

Let $X\subset A^n$ be an arbitrary subset. Prove that $V(I(X))=\bar{X}$.

This is my work so far:

(1) Since $\bar{X}$ is the intersection of all varieties containing $X$, it is a variety. So $V(I(\bar{X}))=\bar{X}$.

(2) Now $X\subset\bar{X}$, so $V(I(X))\subset V(I(\bar{X}))$.

(3) It remains to show that $\bar{X} \subset V(I(X))$.

(4) For any $x\in \bar{X}$, it is in all varieties that contains $X$. Specifically, $V(I(X))$ is a variety that contains $X$, so we are done.

Is the proof correct? I am especially not sure about step (b). Please help. Thanks!

KittyL
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1 Answers1

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If $X$ is contained in $V(J)$ then any function in $J$ vanishes on $X$ so it is in $I(X)$. In fact we have the equivalence:

$$ X \subset V(J) \iff J \subset I(X)$$

Therefore amongst the $J$ so that $V(J) \supset X$ there exists a largest one, $I(X)$. This will give the smallest closed subset $V(I(X))$ containing $X$

orangeskid
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