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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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    I'm sorry I made some mistakes. Yes, Dylan guessed what I mean. And thank you for your suggestions , dear Pierre.2011-08-27

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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?