From Matsumura p.34
Let $k$ be a field, $A$ a ring which is finitely generated over $k$, and $I$ a proper ideal of $A$; then the radical of $I$ is the intersection of all maximal ideals containing $I$.
I understand that the radical of $I$ is contained in this intersection but I'm having troublr understanding the other direction. The proof says this follows from part 2 of Hilbert's Nullstellensatz
Given a subset $\Phi$ of $k[X_1, ... ,X_n]$ and an element $f \in k[X_1, ... ,X_n]$ suppose that $f$ vanishes at every algebraic zero of $\Phi$. Then some power of $f$ belongs to the ideal generated by $\Phi$.
So in order for the theorem to follow from this...don't we need $k$ to be algebraically closed? Because otherwise the maximal ideals are not necessarily of the form $(X_1 - a_1, ... , X_n - a_n)$. So we can't say that f vanishes at $(a_1, ... , a_n)$.