If $J$ is a subset of an affine variety $X$ , then show $\bar J= V(I(J))$.

Question

If $J$ is a subset of an affine variety $X$ , then show $\bar J= V(I(J))$.

attempt: We need to show that the closure of $J$, is the smallest closed variety set of $X$ that contains $J$. Let $J \subseteq X$. Then by definition, we have $J \subseteq V(I(J))$. And since $V(I(J))$ is a variety , we have $\bar J \subseteq V(I(J)).$

Conversely, if $J \subseteq \bar J$, we have $V(I(J)) \subseteq V(I(\bar J))$. I am not sure, could someone please help me? Any feedback would really help. Thank you!

Answer 1

$V(I(J))$ is closed and contains $J$.This implies that $\bar J\subset V(I(J))$, since $\bar J$ is the intersection of the closed subsets which contain $J$.

On the other hand, you can write $\bar J=V(L)$ since it is closed, since $J\subset V(L)=\bar J, L\subset I(J)$. We deduce that $V(I(J))\subset V(L)=\bar J$.