Consider the integral homomorphism $f: A \rightarrow B$. How do I prove that the map $f^*:Spec(B) \rightarrow Spec(A)$ given by $f^*(Q) = f^{-1}(Q)$ is a closed application. Note: 1) I believe we need to show that $f^*(V(I)) = V(f^{-1}(I))$. I have already shown that $f^*(V(I))$ is contained in $V(f^{-1}(I))$ but I can not show the other inclusion. 2) $f^{-1}$ is the inverse image of $f$. 3) $I$ is any ideal of the ring $B$.
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Is it supposed to be $f^$ or $f_$? – kahen May 16 '13 at 03:02
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Presumably $f^*$ since it's contravariant. – Sammy Black May 16 '13 at 03:15
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http://math.stackexchange.com/questions/387544/ – Martin Brandenburg May 16 '13 at 08:46
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The massive hint that you need to solve your problem is the following result
Theorem 5.10 Atiyah - Macdonald: Let $f : A \to B$ be an integral morphism. For every prime ideal $\mathfrak{p} \subseteq A$ there is a prime ideal $\mathfrak{q} \subseteq B$ such that $A \cap \mathfrak{q} = \mathfrak{p}$.
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@Cesar: You accidentally created a duplicate account; with fewer than 50 reputation points, an account can only comment on its own questions and answers. Follow the instructions here to merge your accounts. – Zev Chonoles May 16 '13 at 14:40