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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!

yandz
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  • Since $F$ is a field and $B$ is a finitely generated $F$-algebra then so is $B/M$, thus it is an algebraic extension of $F$. You obtain $F \subset A/(M \cap A)\subset B/M$, therefore $ A/(M \cap A)$ is also an algebraic extension of $F$. – reuns Oct 18 '17 at 22:05
  • @reuns No Zariski Lemma! – user26857 Oct 19 '17 at 07:31
  • If it is true. take $A=k$, then $B$ is integral over $k$. Equivalently,$B $ is finite over $k$,that is , Any finitely generated k-algebra is a finite k-module. Of course, it’s not true, a counterexample is $k[x]$ where $x$ is an indeterminate element.So I think it’s supposed to add some extra conditions on $A$. – Jiabin Du Dec 03 '17 at 13:33

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