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.

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    $\begingroup$ 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 $I$. $\endgroup$ – user26857 May 10 '13 at 12:59
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    $\begingroup$ 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 $\endgroup$ – Martin Brandenburg May 10 '13 at 13:06
  • $\begingroup$ Related: math.stackexchange.com/questions/1283839/… $\endgroup$ – user26857 May 17 '15 at 13:37

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