Finitely many prime ideals lying over $\mathfrak{p}$ 
Let $A$ be a commutative ring with identity and $B$ a finitely generated $A$-algebra that is integral over $A$.  If $\mathfrak{p}$ is a prime ideal of $A$, there are only finitely many prime ideals $P$ of $B$ such that $P\cap A=\mathfrak{p}$.

Let me say that I am aware of this answer, but I can't follow through the hint.  Also, I don't know how to extend to work for algebras rather than extension rings.

Let $f:A\to B$ be a finite ring homomorphism and $\mathfrak p$ a prime ideal of $A$. Consider $S=A-\mathfrak p$ and the ring homomorphism $S^{-1}f:S^{-1}A\to S^{-1}B$. This is also a finite ring homomorphism and all primes of $B$ lying over $\mathfrak p$ "survive" in $S^{-1}B$. This shows that one can assume $A$ local.
If $A\to B$ is a finite ring homomorphism and $(A,\mathfrak m)$ is a local ring, then all prime ideals of $B$ lying over $\mathfrak m$ contain $\mathfrak mB$. This allows us to replace $A\to B$ by $A/\mathfrak m\to B/\mathfrak mB$ (which is also a finite ring homomorphism), and therefore one can assume that $B$ is a finite algebra over a field. Now we have to prove that any finite algebra over a field has only finitely many prime (maximal) ideals. But such algebra is an artinian ring...
