# Finite extension of Noetherian rings [duplicate]

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

## marked as duplicate by user26857 commutative-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 2 '17 at 17:11

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.