Let $W$ be a subset of $C^{n}$. If $\overline{W}$ is the closure of $W$ in the $k$-Zariski topology on $C^{n}$, show that $\overline{W} = Z(I(W))$.
I know that $Z(I(W))$ is closed, so $\overline{W} \subseteq Z(I(W))$, since $W \subset C^{n}$. How to do the converse?
$Z$ is zero; $I$ is an ideal of $W$.
If necessary, I can write the book definitions.
On
You can see this more clearly in an abstract way:https://math.liveadvances.com/c-ding/if-sigma-is-a-reverse-mapping-such-that-sigma2geq-operatornameid-then-sigma3sigma/ Note that Z,I satisfy the conditions of Proposition.
For any closed set $Z(J)\supset W$,$W\subset Z(I(W))\subset Z(I(Z(J)))=Z(J)$. This means precisely that $Z(I(W))$ is the smallest closed set that contains $W$.
If $J$ is an ideal, then $Z(I(Z(J))) = Z(J)$. It's better to carefully write out what this is saying and convince yourself that it is true, than to bother reading a proof of this.
Since $\overline{W}$ is closed, we can write $\overline{W} = Z(J)$ for some ideal $J$. Since $W \subseteq Z(J)$, we have $I(Z(J)) \subseteq I(W)$, hence $Z(I(W)) \subseteq Z(I(Z(J))) = Z(J) = \overline{W}$.