In a finite ring extension there are only finitely many prime ideals lying over a given prime ideal [duplicate]

I'm trying to solve the exercise 6.7 of Miles Reid's Undergraduate Commutative Algebra (pag 93).

How can I prove that if $B$ is a finite ring extension of $A$, there are only finitely many prime ideals of $B$ whose intersection with $A$ is a given prime ideal of $A$?

Thank you!

Solution 1:

Hint: Let $\mathfrak{p} \subset A$ be a prime ideal. Localize at $\mathfrak{p}$, so that $A_{\mathfrak{p}}$ is a local ring. The primes $\mathfrak{q}$ in $B$ for which $\mathfrak{q} \cap A = \mathfrak{p}$ are called primes lying over $\mathfrak{p}$. The primes lying over $\mathfrak{p}$ remain in the ring $B_{\mathfrak{p}},$ and are distinct. All these primes lie over the maximal ideal $\mathfrak{p}A_{\mathfrak{p}}$ in $A_{\mathfrak{p}}$, so they are all maximal. Now, how many maximal ideals containing $\mathfrak{p}B_{\mathfrak{p}}$ can $B_{\mathfrak{p}}$ have?

Solution 2:

Hint: First show that it's enough to prove this for maximal $\mathfrak p$ (localize). Then show that it's enough to prove this when the maximal ideal is $(0)$ (factor). Then you have a finite $k$-algebra and must show that it has finitely many prime ideals.