# 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'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!

• Isn't a valuation ring already a local ring? Jun 30, 2021 at 11:54
• @Bernard Yes! But, it's different here to ask that the subring is a local ring of $A$. Jun 30, 2021 at 12:16
• What's a local ring of $A$? Jun 30, 2021 at 12:35
• But a localisation cannot be a subring. On the contrary, it can be considered as an overring. Jun 30, 2021 at 12:46
• It seems I misread as ‘every subring of $A$…’ Sorry for this. Jun 30, 2021 at 13:12

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 $$A_p=B$$.
• Thanks! But, what is $m_B$? Jul 2, 2021 at 6:28