Exercise 5.29

Question

Let $A$ be a valuation ring of a field $K$. Show that every subring of $K$ which contains $A$ is a local ring of $A$.

Amit Awasthi · Accepted Answer

\begin{lemma*}[AM Exercise 5.27]
  Let $A,B$ be two local rings. $B$ is said to \emph{dominate} $A$ if $A$ is a subring of $B$ and the maximal ideal $\mathfrak{m}$ of $A$ is contained in the maximal ideal $\mathfrak{n}$ of $B$ (or, equivalently, $\mathfrak{m} = \mathfrak{n} \cap A$). Let $K$ be a field and let $\Sigma$ be the set of all local subrings of $K$. If $\Sigma$ is ordered by the relation of domination, show that $\Sigma$ has maximal elements and that $A \in \Sigma$ is maximal if and only if $A$ is a valuation ring of $K$.
\end{lemma*}
\begin{proof}[Proof of Lemma]
  Let $\{A_{\alpha}\}$ be a chain in $\Sigma$; let $A = \bigcup A_{\alpha}, \mathfrak{m} = \bigcup \mathfrak{m}_\alpha$. We claim $A$ is a local ring with maximal ideal $\mathfrak{m}$. $x \in A \setminus \mathfrak{m}$ implies $x \in A_{\alpha} \setminus \mathfrak{m}_\alpha$ for some $\alpha$. By MR Corollary $1.9$, we then see $x \in A_{\alpha}^	imes$, and so $x \in A^	imes$. By AM Proposition $1.6i$, then, $(A,\mathfrak{m})$ is local. By Zorn's lemma, $\Sigma$, therefore, has a maximal element.
  \par $\Rightarrow$. Letting $f_\alpha : A_{\alpha} \hookrightarrow \overline{K}$ be the canonical inclusions, $\Sigma^{\prime} = \{(A_{\alpha},f_\alpha)\}$ has the same maximal elements as $\Sigma$, and moreover satisfies the construction on p.~$65$ in AM. Thus, by AM Theorem $5.21$, if $A \in \Sigma$ is maximal, it is a valuation ring of $K$.
  \par $\Leftarrow$. Suppose $(A,\mathfrak{m})$ is a valuation ring properly dominated by $(B,\mathfrak{n})$. Choose $x \in B \setminus A$ such that $x^{-1} \in A$, which exists since $A$ is a valuation ring. Then, $x^{-1} \in \mathfrak{m}$ since $x 
otin A \implies x^{-1} 
otin A^	imes$, and then by MR Corollary $1.9$. But $x \in B$ implies $x^{-1} 
otin \mathfrak{n}$ by MR Corollary $1.9$, and so $\mathfrak{m} 
ot\subseteq \mathfrak{n}$, a contradiction.
\end{proof}
\begin{proof}
  Let $A \subseteq B \subseteq K$; we want to show $B = A_{\mathfrak{p}}$ for some $\mathfrak{p} \subseteq A$. By AM Proposition $5.18i,ii$, we see $B$ is a valuation ring of $K$, and is moreover local. Let $\mathfrak{M}$ be the maximal ideal of $B$. Let $\mathfrak{p} = A \cap \mathfrak{M}$, which is prime since contractions of prime ideals of prime.
  \par We claim $B = A_{\mathfrak{p}}$. First, consider $x \in A \setminus \mathfrak{p}$. Then, $x 
otin \mathfrak{M}$ by definition of $\mathfrak{p}$, and so $x \in B^	imes$ by MR Corollary $1.9$. Thus, $a \in A \implies a/x \in B$, and so $A_{\mathfrak{p}} \subseteq B$. Now since $A_{\mathfrak{p}} \subseteq B$, we also have $\mathfrak{p}A_{\mathfrak{p}} \subseteq \mathfrak{M}$, where both ideals are maximal, i.e., $B$ dominates $A_{\mathfrak{p}}$. Since $A_{\mathfrak{p}}$ is a valuation ring by AM Exercise 5.28, we see that $A_{\mathfrak{p}}$ is maximal with respect to domination by AM Exercise $5.27$, i.e., $B = A_{\mathfrak{p}}$.
\end{proof}

Exercise 5.29

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Exercise 5.29

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