Integral extension implies that the induced map on prime spectra is closed

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 $\mathfrak{P}$?

Yes, your candidate for $J$ is the right one. – Georges Elencwajg Apr 5 '13 at 17:48
Apply lying over to $R/I^c \hookrightarrow S/I$.
Ah, yes! Because passing to quotients preserves integrity (integrality?), and prime ideals $\mathfrak{p}$ containing $I^c$ correspond to prime ideals $\mathfrak{p}/I^c$in the quotient... Thus there exists a prime ideal $\mathfrak{P}/I$ such that $(\mathfrak{P}/I)^c = p/I^c$, and from there we can easily conclude that $\mathfrak{P}^c = \mathfrak{p}$. – Alex Provost Apr 5 '13 at 20:34