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

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!

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Hi, welcome to math.SE! People are more likely to read and answer your question if you provide details of what you've worked out so far. That way, they don't feel like they are slaving to do all your work for you. Good luck! –  rschwieb Jan 4 '13 at 18:07
If this is homework it would be better to tag it as such. –  A.P. Mar 17 '13 at 20:46
This is also the exercise 9.3 in Matsumura, Commutative Ring Theory. –  user26857 Mar 17 '13 at 20:53

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?
I was trying to prove that $B_p/pB_p$ is an Artinian Ring, so it has a finite number of maximal ideals. Hence $B_p$ has only finitely many maximal ideals containing $pB_p$. Is this the right way? –  Corra Jan 4 '13 at 19:06
Could you suggest me a way to prove that $B_p/pB_p$ is Artinian? –  Corra Jan 4 '13 at 19:08
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.