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.

One very useful tool is: 

For a ring map $A\to B$, and $\mathfrak{p}$ a prime ideal of $A$, the prime ideals which contract to $\mathfrak{p}$ are in $1:1$ correspondence to the prime ideals of $\kappa(\mathfrak{p})\otimes_AB$, where $\kappa(p)=Q(A/\mathfrak{p})$ is the quotient field of the domain $A/\mathfrak{p}$.

Now we come back to the question. Since $\kappa(\mathfrak{p})\otimes_AB$ is integral over $\kappa(\mathfrak{p})$, all prime ideals of $\kappa(\mathfrak{p})\otimes_AB$ must be maximal (in fancy words, the Krull dimension is zero). And note that $\kappa(\mathfrak{p})\otimes_AB$ is finitely generated over $\kappa(\mathfrak{p})$, then $\kappa(\mathfrak{p})\otimes_AB$ is noetherian. A noetherian ring whose prime ideals are all maximal is an Artinian ring. An Artinian ring has only finitely many prime ideals.
