TITLE: If $B$ is finitely generated as a $k$-algebra, and $\phi:A\to B$ is a $k$-algebra map, is $\phi^{-1}(M)$ maximal for any maximal $M\subset B$?
QUESTION [9 upvotes]: Suppose that $A$ and $B$ are commutative rings containing a field $k$, and $B$ is finitely generated $k$-algebra. Let $\phi: A\rightarrow B$ be a ring homomorphism with $\phi|_k =\mathrm{Id}$. I am trying to prove that if $M\subset B$ is a maximal ideal, then $\phi^{-1}(M)$ is a maximal ideal of $A$. 

The case when $A \subset B$ is an integral extension of rings is well-known. I think I can also prove the result when $\phi$ is surjective. 
Inverse Image of Maximal Ideals discusses the case when $B$ is a finitely generated $\mathbb{Z}$-algebra but I am not sure how to generalize this.

REPLY [4 votes]: We have $k\subset A/\phi^{-1}(M)\subset B/M$. The extension $k\subset B/M$ is finite (see here, Corollary 8.3.9), so the extension $k\subset A/\phi^{-1}(M)$ is finite, too. This shows that $A/\phi^{-1}(M)$ is a field, so $\phi^{-1}(M)$ is maximal.