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I think this is a basic property of $\mathcal{I}(-)$, but I'm having trouble seeing it.

I denote by $\mathbb{A}^n$ the affine $n$-space over an algebraically closed field $k$, where if $A\subseteq\mathbb{A}^n$, then $$\mathcal{I}(A)=\{f\in k[x_1,\dots,x_n]:f(a_i)=0,\forall\ (a_i)\in A\}.$$

If $A,B\subseteq\mathbb{A}^n$, why does $A\subseteq\overline{B}\iff\mathcal{I}(B)\subseteq\mathcal{I}(A)$?

For one way, supposing $A\subseteq\overline{B}$, if $f\in I(B)$, then $f$ vanishes on all points in $B$, so I suspect $f$ vanishes on all points of $\overline{B}$, hence on all of $A$, so $f\in I(A)$.

I know $I(-)$ is inclusion reversing, so if $A\subseteq\overline{B}$, then $I(\overline{B})\subseteq I(A)$, but $I(\overline{B})\subseteq I(B)$ so I'm at a loss. What's the best way to verify this? Thanks.

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I think what you are missing is that $VI(B)$ is the closure of $B$, and $IVI(B) = I(B)$ (by nullstellensatz).

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  • $\begingroup$ The Nullstellensatz has nothing to do with your (correct) equalities $VI(B)=\bar B$ and $ IVI(B) = I(B)$ . These equalities are immediate consequences of the definition of the Zariski topology and are valid over non-algebraically closed fields. $\endgroup$ – Georges Elencwajg Oct 17 '14 at 9:42
  • $\begingroup$ You're right, thanks for pointing that out. I was using the corollary that $I$ and $V$ induce a correspondence between radical ideals and algebraic sets, but it is even easier as Jared has stated. $\endgroup$ – Ben Oct 18 '14 at 3:31
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Any function $f$ vanishing on $B$ must also vanish on $\overline{B}$. If not, then the intersection of $\overline{B}$ with the zero set of $f$ would be a closed set containing $B$ properly contained in $\overline{B}$. This is a contradiction, as the closure can be defined to be the smallest closed set containing $B$.

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