Intersection of all maximal ideals containing a given ideal

The inclusion $\displaystyle \sqrt{I} \subseteq \bigcap_{m \supseteq I \text{ maximal}} m$ always holds. Suppose $\displaystyle f \in \bigcap_{m \supseteq I \text{ maximal}} m$. Since $k$ is algebraically closed, the maximal ideals containing $I$ are precisely the ideals of the form $(x_1 - a_1, \ldots, x_n - a_n)$, where $(a_1, \ldots, a_n)$ are the points of $V(I) \subseteq k^n$. This says that $f$ vanishes at every point of $V(I)$, so $f \in I(V(I)) = \sqrt{I}$, by the Nullstellensatz.

This in fact still holds even if $k$ is not algebraically closed, and says that $k[x_1, \ldots, x_n]$ (for any field $k$) is a Jacobson ring.