Exercise 5.1

Question

Let $f: A 	o B$ be an integral homomorphism of rings. Show that $f^* : \operatorname{Spec}(B) 	o \operatorname{Spec}(A)$ is a \emph{closed} mapping, i.e., that it maps closed sets to closed sets.

Amit Awasthi · Accepted Answer

\begin{proof}
  Let $\mathfrak{q} \in \operatorname{Spec}(B)$; we claim $f^*(\operatorname{\mathbf{V}}(\mathfrak{q})) = \operatorname{\mathbf{V}}(f^*(\mathfrak{q}))$. By AM Exercise $1.21iii)$, we see $f^*(\operatorname{\mathbf{V}}(\mathfrak{q})) \subseteq \overline{f^*(\operatorname{\mathbf{V}}(\mathfrak{q}))} = \operatorname{\mathbf{V}}(f^{-1}(\mathfrak{q})) = \operatorname{\mathbf{V}}(f^*(\mathfrak{q}))$. Conversely, if $\mathfrak{p} \in \operatorname{\mathbf{V}}(f^*(\mathfrak{q}))$, then $f^*(\mathfrak{q}) \subseteq \mathfrak{p}$, and so $f(f^*(\mathfrak{q})) \subseteq f(\mathfrak{p})$ is a chain of prime ideals in $f(A)$. Since $f(f^*(\mathfrak{q})) = f(f^{-1}(\mathfrak{q})) = \mathfrak{q} \cap f(A)$, we have that $\mathfrak{q}$ lies over $f(f^*(\mathfrak{q}))$, and so since $B$ is integral over $f(A)$, by Going Up there exists $\mathfrak{r} \in \operatorname{Spec}(B)$ such that $\mathfrak{r} \supseteq \mathfrak{q}$ and $\mathfrak{r} \cap f(A) = f(\mathfrak{p})$. Thus, $\mathfrak{p} = f^{-1}(f(\mathfrak{p})) = f^{-1}(\mathfrak{r} \cap f(A)) = f^{-1}(\mathfrak{r}) = f^*(\mathfrak{r})$, where $\mathfrak{r} \in \operatorname{\mathbf{V}}(\mathfrak{q})$. This implies $\mathfrak{p} \in f^*(\operatorname{\mathbf{V}}(\mathfrak{q}))$; thus, $f^*(\operatorname{\mathbf{V}}(\mathfrak{q})) = \operatorname{\mathbf{V}}(f^*(\mathfrak{q}))$, i.e., $f^*$ is closed.
\end{proof}

Exercise 5.1

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Exercise 5.1

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