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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?

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Hint:

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.

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