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

Question

I'm working on Problem 5.29 of Atiyah & Macdonald's Commutative Algebra text. Problem 5.29 is the following:

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

I'll note a few items before presenting my (incomplete) proof:

• To clarify, to say that the subring of $K$ is a "local ring of $A$" just means that the subring is a localization of $A$ at a prime ideal of $A$.
• Let $B$ be an integral domain, $K$ its field of fractions. $B$ is called a valuation ring of $K$ if, for each $x \neq 0$, either $x \in B$ or $x^{-1} \in B$.
• Proposition 5.18 in the text states that if $B$ is a valuation ring of $K$, then (i) $B$ is a local ring, ii) If $B'$ is a ring such that $B \subseteq B' \subseteq K$, then $B'$ is a valuation ring of $K$, and iii) $B$ is integrally closed (in $K$).

Here is my proof so far:

Let $B$ be a subring of $K$ which contains $A$. We need to show that $B$ is the localization of $A$ at a prime ideal $p$ of $A$. By Proposition 5.18ii), $B$ is a valuation ring of $K$. By Proposition 5.18i), $B$ is a local ring with unique maximal ideal $m$. Let $p = m \cap A$. If $s \in A\setminus p$, then $s \in B\setminus m$. Since every non-unit of a nonzero ring $R$ is contained in some maximal ideal of $R$, this means that $s$ is a unit of $B$. By the universal property of localization, there exists a unique ring homomorphism $h:A_p \rightarrow B$ such that $g = h \circ f$, where $g:A \rightarrow B$ is the inclusion map and $f:A \rightarrow A_p$ is the canonical localization homomorphism.

In order to show that $h$ is an isomorphism, it's left for me to show that $h$ is both surjective and injective. How can I do this?

Thanks!

Answer 1

I hope it's okay by you if I just write $A_p$ as a subring of $K$, the precise statement I suppose is that the natural map $A_p\to K$ is injective, which is easy to check using that $A$ is an integral domain.

Now, as you've noted if we have $a/s\in A_p$ where $s\notin p$, then $s\in B^\times$ so $a/s\in B$ showing $A_p\subseteq B$. We want to know the reverse inequality holds as well.

If we had an element $b\in B\smallsetminus A_p$, then because $A_p$ is a valuation ring of $K$ (this is from one of the properties you listed) we deduce $b^{-1}\in A_p$, and in fact $b^{-1}\in pA_p$ because otherwise one would deduce $b\in A_p$ using the fact that $pA_p$ is the unique maximal ideal of $A_p$. Notice also $pA_p\subseteq m_B\cap A_p\subseteq m_B$, so as a result one has $b^{-1}\in m_B$, and then $1=b\,b^{-1}\in m_B$, which is a contradiction. Going back we see it is impossible to have an element $b\in B\smallsetminus A_p$, so $A_p=B$.