# Integral homomorphism induces a closed map on spectra

I'm trying to prove the following:

Let $f:A\rightarrow B$ be an integral homomorphism (e.g. $B/f(A)$ is a integral extension). Consider $f^{*}: \operatorname{Spec}B \rightarrow \operatorname{Spec}A$ given by $f^{*}(Q)=f^{-1}(Q)$. Show that $f^{*}$ is a closed map.

My problem here is that a I don't know how to describe the closed sets in the Zariski topology, since I have to show that given $Q$ closed in $\operatorname{Spec}B$ then $f^{*}(Q)$ is closed in $\operatorname{Spec}A$.

Thank you for any help.

• As far as I know the closed sets in Zariski topology are of the form $V(I)$, the set of prime ideals containing a given ideal $I$. – user26857 May 10 '13 at 12:59
• Why do you want to prove this without knowing the definition of the topological spaces involved?! The claim follows from Going-Up, see also the almost identical question math.stackexchange.com/questions/383080 – Martin Brandenburg May 10 '13 at 13:06
• – user26857 May 17 '15 at 13:37