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This question already has an answer here:

I've got two noetherian rings $A\subset B$ such that $B$ is a finite $A$-module. Now, if I consider the associated map between spectra that given $q \in \operatorname{Spec} B$ consider $q \cap A \in \operatorname{Spec} A$ and I should demonstrate it has finite fibers.

I really don't know how to even start. I tried to use the fact that the extension is integral etc but i didn't manage to do anything. Ty for the help

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marked as duplicate by user26857 commutative-algebra Nov 2 '17 at 17:11

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Fibers over $\mathfrak{p}$ for any morphism $A \to B$ correspond to elements of $\mathrm{frac}(A/\mathfrak{p}) \otimes_A B$.

But then $\mathrm{frac}(A/\mathfrak{p}) \otimes_A B$ is a finite $\mathrm{frac}(A/\mathfrak{p})$-algebra, so there are only finitely many prime ideals.

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  • $\begingroup$ Sorry about the horrendous notation, and kind of low quality answer. I don't have time to fill in details, but this is the idea. $\endgroup$ – Andres Mejia Nov 2 '17 at 15:45

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