Exercise 5.10

Question

A ring homomorphism $f : A 	o B$ is said to have the \emph{going-up property} \emph{(}resp.~the \emph{going-down property)} if the conclusion of the going-up theorem $(5.11)$ \emph{(}resp.~the \emph{going-down theorem (5.16))} holds for $B$ and its subring $f(A)$.
  \par Let $f^* : \operatorname{Spec}(B) 	o \operatorname{Spec}(A)$ be the mapping associated with $f$.
  \par (i) Consider the following three statements:
  \begin{enumerate}[label=(\alph*)]
    \item $f^*$ is a closed mapping.
    \item $f$ has the going-up property.
    \item Let $\mathfrak{q}$ be any prime ideal of $B$ and let $\mathfrak{p} = \mathfrak{q}^c$. Then $f^* : \operatorname{Spec}(B/\mathfrak{q}) 	o \operatorname{Spec}(A/\mathfrak{p})$ is surjective.
  \end{enumerate}
  Prove that $(a) \Rightarrow (b) \Leftrightarrow (c)$.
ewline

(ii) In the same situation as in Problem $3.1$, show that $f$ has the going-down property $\iff$ for any prime ideal $\mathfrak{q}$ of $B$, if $\mathfrak{p} = \mathfrak{q}^c$, then $f^*:\operatorname{Spec}(B_{\mathfrak{q}}) 	o \operatorname{Spec}(A_{\mathfrak{p}})$ is surjective.

Amit Awasthi · Accepted Answer

\begin{proof}
  $(a) \Rightarrow (b)$. Suppose $\mathfrak{p}_1 \subseteq \mathfrak{p}_2$ a chain of prime ideals in $f(A)$ and $\mathfrak{q}_1 \subseteq B$ lies over $\mathfrak{p}_1$. Since $f^*(\mathfrak{q}_1) = f^{-1}(\mathfrak{p}_1) \subseteq f^{-1}(\mathfrak{p}_2)$, we have $f^{-1}(\mathfrak{p}_2) \in \operatorname{\mathbf{V}}(f^*(\mathfrak{q}_1))$. Note $\operatorname{\mathbf{V}}(f^*(\mathfrak{q}_1)) = \operatorname{\mathbf{V}}(f^{-1}(\mathfrak{q}_1)) = \overline{f^*(\operatorname{\mathbf{V}}(\mathfrak{q}_1))} = f^*(\operatorname{\mathbf{V}}(\mathfrak{q}_1))$ by the AM Exercise $1.21iii)$ and the fact that $f^*$ is closed. Thus, $f^{-1}(\mathfrak{p}_2) \in f^*(\operatorname{\mathbf{V}}(\mathfrak{q}_1))$, and so there exists $\mathfrak{q}_2 \in \operatorname{\mathbf{V}}(\mathfrak{q}_1)$ such that $f^*(\mathfrak{q}_2) = f^{-1}(\mathfrak{p}_2)$. Since $f^*(\mathfrak{q}_2) = f^{-1}(\mathfrak{q}_2 \cap f(A))$, we then see that $\mathfrak{p}_2 = \mathfrak{q}_2 \cap f(A)$, and so $\mathfrak{q}_2$ lies over $\mathfrak{p}_2$. By induction as in the original proof of Going Up, we are done.

\par $(b) \Rightarrow (c)$. Suppose $\mathfrak{q} \in \operatorname{Spec}(B)$ and let $\mathfrak{p} = f^{-1}(\mathfrak{q}) = f^{-1}(\mathfrak{q} \cap f(A))$. Consider $\mathfrak{p}^{\prime} \in \operatorname{\mathbf{V}}(\mathfrak{p})$; note that $\operatorname{\mathbf{V}}(\mathfrak{p}) \xleftrightarrow{1-1:} \operatorname{Spec}(A/\mathfrak{p})$, and so showing that $\mathfrak{p}^{\prime} \in \operatorname{im} f^*$ suffices. Then, $\mathfrak{p} \subseteq \mathfrak{p}^{\prime}$, and so $f(\mathfrak{p}) \subseteq f(\mathfrak{p}^{\prime})$ is a chain of prime ideals in $f(A)$ and $\mathfrak{q} \cap f(A) = f(\mathfrak{p})$. By $(b)$, the Going Up property, there exists $\mathfrak{q}^{\prime} \subseteq B$ such that $\mathfrak{q}^{\prime} \cap f(A) = f(\mathfrak{p}^{\prime})$. Finally, $f^{-1}(\mathfrak{q}^{\prime} \cap f(A)) = f^*(\mathfrak{q}^{\prime}) = f^{-1}(f(\mathfrak{p}^{\prime})) = \mathfrak{p}^{\prime}$, and so $f^*$ is surjective.

\par $(c) \Rightarrow (b)$. Suppose $\mathfrak{p}_1 \subseteq \mathfrak{p}_2$ is a chain of prime ideals in $f(A)$ and $\mathfrak{q}_1 \subseteq B$ lies over $\mathfrak{p}_1$. Note that $f^{-1}(\mathfrak{p}_1)$ is prime since it is a contraction of a prime ideal, and also $f^{-1}(\mathfrak{p}_1) = f^{-1}(\mathfrak{q}_1 \cap f(A)) = (\mathfrak{q}_1)^c$. By $(c)$, $f^* : \operatorname{\mathbf{V}}(\mathfrak{q}_1) 	o \operatorname{\mathbf{V}}(f^{-1}(\mathfrak{p}_1))$ is surjective, and so $f^{-1}(\mathfrak{p}_2) \in \operatorname{\mathbf{V}}(f^{-1}(\mathfrak{p}_1))$ has preimage $\mathfrak{q}_2 = (f^*)^{-1}(f^{-1}(\mathfrak{p}_2)) \in \operatorname{\mathbf{V}}(\mathfrak{q}_1)$. $\mathfrak{q}_2 \in \operatorname{\mathbf{V}}(\mathfrak{q}_1) \implies \mathfrak{q}_1 \subseteq \mathfrak{q}_2$, and $f^{-1}(\mathfrak{p}_2) = f^*(\mathfrak{q}_2) = f^{-1}(\mathfrak{q}_2 \cap f(A)) \implies \mathfrak{p}_2 = \mathfrak{q}_2 \cap f(A)$, i.e., $\mathfrak{q}_2$ lies over $\mathfrak{p}_2$. By induction as in the original proof of Going Up, we are done.
\end{proof}

\begin{proof}[Proof of (ii)]
  $(b) \Rightarrow (c)$. Suppose $\mathfrak{q} \in \operatorname{Spec}(B)$ and let $\mathfrak{p} = f^{-1}(\mathfrak{q}) = f^{-1}(\mathfrak{q} \cap f(A))$. Consider $\mathfrak{p}^{\prime} \in \operatorname{Spec}(A) \setminus \operatorname{\mathbf{V}}(\mathfrak{p})$; note that $\operatorname{Spec}(A) \setminus \operatorname{\mathbf{V}}(\mathfrak{p}) \xleftrightarrow{1-1\:} \operatorname{Spec}(A_{\mathfrak{p}})$, and so showing that $\mathfrak{p}^{\prime} \in \operatorname{im} f^*$ suffices. Then, $\mathfrak{p}^{\prime} \subseteq \mathfrak{p}$, and so we have $f(\mathfrak{p}^{\prime}) \subseteq f(\mathfrak{p})$ a chain of prime ideals in $f(A)$ and $\mathfrak{q} \cap f(A) = f(\mathfrak{p})$. By $(b)$, the Going Down property, there exists $\mathfrak{q}^{\prime} \subseteq B$ such that $\mathfrak{q}^{\prime} \cap f(A) = f(\mathfrak{p}^{\prime})$. Finally, $f^{-1}(\mathfrak{q}^{\prime} \cap f(A)) = f^*(\mathfrak{q}^{\prime}) = f^{-1}(f(\mathfrak{p}^{\prime})) = \mathfrak{p}^{\prime}$, and so $f^*$ is surjective.
  \par $(c) \Rightarrow (b)$. Suppose $\mathfrak{p}_1 \subseteq \mathfrak{p}_2$ is a chain of prime ideals in $f(A)$ and $\mathfrak{q}_2 \subseteq B$ lies over $\mathfrak{p}_2$. Note that $f^{-1}(\mathfrak{p}_2)$ is prime since it is a contraction of a prime ideal, and also $f^{-1}(\mathfrak{p}_2) = f^{-1}(\mathfrak{q}_2 \cap f(A)) = (\mathfrak{q}_2)^c$. By $(c)$, $f^* : \operatorname{Spec}(B_{\mathfrak{q}_2}) 	o \operatorname{Spec}(A_{f^{-1}(\mathfrak{p}_2)})$ is surjective, and so $f^{-1}(\mathfrak{p}_1) \in \operatorname{Spec}(A_{f^{-1}(\mathfrak{p}_2)})$ has preimage $\mathfrak{q}_1 = (f^*)^{-1}(f^{-1}(\mathfrak{p}_1)) \in \operatorname{Spec}(B_{\mathfrak{q}_2})$. $\mathfrak{q}_1 \in \operatorname{Spec}(B_{\mathfrak{q}_2}) \implies \mathfrak{q}_1 \subseteq \mathfrak{q}_2$, and $f^{-1}(\mathfrak{p}_1) = f^*(\mathfrak{q}_1) = f^{-1}(\mathfrak{q}_1 \cap f(A)) \implies \mathfrak{p}_1 = \mathfrak{q}_1 \cap f(A)$, i.e., $\mathfrak{q}_1$ lies over $\mathfrak{p}_1$. By induction as in the original proof of Going Down, we are done.
\end{proof}

Exercise 5.10

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Exercise 5.10

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