I am trying to prove that given $B$, a finitely generated $F$-algebra (where $F$ is a field), and $A$ an $F$-subalgebra (which I assume is a subring of $B$), for any maximal ideal $M$ of $B$, $A \cap M$ is a maximal ideal of $A$.
I'm aware that there's an answer here: In an extension of finitely generated $k$-algebras the contraction of a maximal ideal is also maximal
but since my sketch of the proof is a little different and we have not learned Zariski's Lemma, I'd like to avoid using it. Essentially, I use the fact that if $B$ is integral in $A$, then $B/M$ is integral in $A/(A \cap M)$, so $B/M$ is a field iff $A/(A \cap M)$ is a field. The only issue is proving that $B$ is integral in $A$. I tried going in several directions, first trying to prove that $\forall b \in B$, $A[b]$ is finitely generated over $A$, and using the Noether Normalization Lemma to see if that leads me anywhere. It didn't.
Any suggestions are appreciated. Thanks in advance!