# Integral ring homomorphism implies induced map on spectra is closed

Let $$\varphi: A \rightarrow B$$ be an integral ring homomorphism. Show that the induced morphism $$\tilde{\varphi}:\mathrm{Spec}B \rightarrow \mathrm{Spec}A$$ is closed.

My idea:

Let $$I$$ be an ideal of $$B$$. We have $$\tilde{\varphi}(V(I)) \subset V(\varphi^{-1}(I))$$ since $$\tilde{\varphi}(\mathfrak{q}) = \varphi^{-1}(\mathfrak{q}) \in \mathrm{Spec}A$$ and $$\varphi^{-1}(\mathfrak{q}) \supset \varphi^{-1}(I)$$ for any prime $$\mathfrak{q} \in \mathrm{Spec} B$$. Now suppose $$\mathfrak{p} \in V(\varphi^{-1}(I))$$: we have to prove there exists $$\mathfrak{q} \in V(I)$$ such that $$\mathfrak{q} \cap A=\mathfrak{p}$$. Since $$\mathfrak{p} \supset \varphi^{-1}(I) \supset \varphi^{-1}(0) = \mathrm{Ker \ } \varphi$$ by Lying Over Theorem there is a prime ideal $$\mathfrak{q} \subset B$$ such that $$\mathfrak{q} \cap A=\mathfrak{p}$$, i.e. $$\tilde{\varphi}(\mathfrak{q})=\mathfrak{p}$$ and it follows that $$\tilde{\varphi}(V(I)) = V(\varphi^{-1}(I))$$, so $$\tilde{\varphi}$$ is a closed mapping.

The detail I'm worried about is why $$\mathfrak{q} \in V(I)$$. Is my reasoning correct?

Your reasoning is correct, except there's no argument proving that $$\mathfrak q \in V(I)$$. You can show it considering the following commutative diagram: $$\begin{matrix} A\hskip-2em&\xrightarrow{\quad\varphi\quad}&B \\[-0.5ex] \qquad\downarrow&&\downarrow \\[-1ex] A/\varphi^{-1}(I)&\xrightarrow{\quad\overline\varphi\quad}&B /I \end{matrix}$$ in which the vertical maps are the canonical maps. As the bottom map is is an injective integral homomorphism, you can apply the lying over theorem. The rest is simple diagram-chasing.