4
$\begingroup$

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?

  • 11
    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...2011-08-27
  • 2
    I assumed he was trying to prove that $M(P)$ is bounded as $P$ varies.2011-08-27
  • 1
    I'm sorry I made some mistakes. Yes, Dylan guessed what I mean. And thank you for your suggestions , dear Pierre.2011-08-27

1 Answers 1

1

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?