Lying-over theorem without Axiom of Choice

Solution 1:

First we show

Lemma If $A\to B$ is a finite homomorphism of rings with $A$ local and $B\ne 0$, then the maximal ideal $P$ of $A$ is the pre-image of a maximal ideal $Q$ of $B$.

Proof. By Nakayama's lemma, $PB\ne B$. The quotient $B/PB$ is a finite $k$-algebra (where $k$ is the field $A/P$) and is no zero. The set of proper ideals of $B/PB$ is non-empty and has an element of maximal $k$-vector space dimension. The latter is then a maximal ideal, hence equal to $Q/PB$ for some maximal ideal $Q$ of $B$ containing $P$. The pre-image $P'$ of $Q$ is maximal because $A/P'$ is contained in $B/Q$ and the later is finite over $A/P'$. So $P'=P$.

Now we prove your proposition. As $B$ is a finitely generated $A$-algebra, $B$ integral over $A$ implies that $B$ is finite over $A$. Hence $A_P\to A_P\otimes_A B$ is finite with $A_P\otimes_A B\ne 0$. By the above lemma, $PA_P$ is the pre-image of a maximal ideal of $A_P\otimes_A B$. The existence of $Q$ as desired follows from standard arguments on localizations.

This been said, I agree with the second part of Martin Brandenburg's comment.