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Put $A = k[x]$, where $k$ is an algebraically closed field and $x$ is an indeterminate. Let $B$ be a ring and $f: A \rightarrow B$ be finite integral morphism.

How can one show that the number of prime ideals of $B$ which lie over a given prime ideal of $A$ is finite and bounded?

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Is there a question here? –  Arturo Magidin Aug 27 '11 at 3:14
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Dear Chan: You seem to be a new user. Welcome to Mathematics Stack Exchange! I suggest that you read the faq, in particular the part about how to ask questions. I was wondering if you couldn't state your question in a clearer way, and give it a more suggestive title. When I saw your question, it had 2 downvotes. I voted for it. - To the others: Is it really good idea to downvote questions asked by users with almost no reputation points? Aren't there other ways to interact with them? I imagine myself asking a question on a Chinese forum... –  Pierre-Yves Gaillard Aug 27 '11 at 4:25
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I assumed he was trying to prove that $M(P)$ is bounded as $P$ varies. –  Dylan Moreland Aug 27 '11 at 5:54
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Dear Chan, you are most welcome! Another trick: when you write a comment for me, put an "@Pierre" in it, so I get a notification. (This works for any user of course.) Since the present comment is a comment to your question, you will be notified even if I don't put an "@Chan". –  Pierre-Yves Gaillard Aug 27 '11 at 8:09
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I am amazed at the gentleness and usefulness of the comments by @Pierre. His scenario of us asking questions on a Chinese site is thought-provoking. Bravo, Pierre-Yves! –  Georges Elencwajg Aug 27 '11 at 8:40

1 Answer 1

up vote 2 down vote accepted

You needn't that $B$ is the ring of polynomials over an algebraically closed field. It suffices that $A \to B$ is a finite extension: if $B$ can be generated by a set of $n$ elements as $A$-module, then for every prime ideal $P$ of $A$ the number of primes of $B$ lying over $P$ is $\leq n$.

For a proof see my answer Is the number of prime ideals of a zero-dimensional ring stable under base change?

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