I am trying to understand the proof on Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry.
The theorem I am trying to prove is:
Let $k$ be an algebraically closed field. If $I \subset k[x_{1},...x_{n}]$ is an ideal, then $I(Z(I))=\mathrm{rad}(I)$.
"The proof goes like follows: we know that points of $X \subset A^{r}(k)$ are in 1-1 correspondence to maximal ideals in $A(X)$, thus to maximal ideals in $k[x_{1},...,x_{n}]$ containing $I(X)$. Thus points in $Z(I)$ corresponds to the maximal ideals in $k[x_{1}, ...x_{n}]$ containing $I$. Thus $I(Z(I))$ is the intersection of all maximal ideals containing $I$."
I can't see "Thus points in Z(I) corresponds to the ideals in $k[x_{1}, ...x_{n}]$ containing I. Thus $I(Z(I))$ is the intersection of all maximal ideals containing I." Any help will be appreciated.