Sign up ×  Say we have an integral extension $f:R
\hookrightarrow S$ of rings. I want to show that the induced map
$f^*:Spec(S) \twoheadrightarrow Spec(R)$ is closed. In other
words, let $V(I) = \{\mathfrak{P} \in Spec(S) | \mathfrak{P}
\supset I\}$, for some ideal $I \in S$, be a closed subset in
$Spec(S)$. I want to find an ideal $J \in R$ such that
$f^*(V(I)) = V(J)$.  I thought about using $J = I^{c} = I \cap
R$ (contraction), but one direction is unclear to me. To be more
precise, let $\mathfrak{p}$ be a prime ideal of $R$ containing
$I^c$. By the lying over property of the extension, we know that
there exists a $\mathfrak{P} \in Spec(S)$ such that
$\mathfrak{p} = \mathfrak{P}^c$. But is it even true then that
$I$ is contained in $\mathfrak{P}$?  Thanks for your time.
share|cite|improve this question Yes, your candidate for $J$ is
the right one. –  Georges Elencwajg Apr 5 '13 at 17:48  1 Answer
1  up vote 2 down vote accepted  Apply lying over to $R/I^c
\hookrightarrow S/I$.  share|cite|improve this ans