Finitely many prime ideals lying over $\mathfrak{p}$

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...

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.