# Let $A$ be an integrally closed noetherian domain, and $(R, \mathfrak{m})$ local with $A \subseteq R \subseteq K(A)$. Is $R$ a localization of $A$?

Let $$K(A)$$ denote the fraction field of $$A$$. For context, I'm trying to prove $$A = \bigcap_{\text{ht}(\mathfrak{p}) = 1} A_{\mathfrak{p}}$$ for an integrally closed domain $$A$$, from Atiyah-Macdonald's corollary 5.22 which states that $$A$$ is the intersection of all valuation rings containing it. This would follow from the following conjecture: $$R$$ is a valuation ring of $$K(A)$$ iff $$R = A_{\frak{p}}$$ for some $$\mathfrak{p} \subseteq A$$ of height 1.

Before that, though, I want to prove an intermediate result that if $$(R, \frak{m})$$ is a local ring in $$K(A)$$ containing $$A$$, then $$R = A_{\frak{p}}$$ for some prime $$\frak{p}$$. The choice for $$\mathfrak{p}$$ is very natural, we take $$\mathfrak{p} = A \cap \mathfrak{m}$$. Then, if $$x \in A \setminus \mathfrak{p}$$, we would have that $$x$$ is a unit in $$R$$, since $$x \notin \frak{m}$$. Hence, we have an induced homomorphism $$A_{\mathfrak{p}} \to R$$. I think this is injective by definition, but I am not sure why it would be surjective.

Alternatively, I thought that $$R$$ may just satisfy the universal property of $$A_{\mathfrak{p}}$$ but this is still unclear. Is this just not true given the hypotheses or is there something I am missing?

Thanks!

This is not true. For instance, let $$k$$ be a field and $$A=k[x,y]$$, and let $$B=A[y/x]$$. Note that $$B=k[x,y,y/x]$$ can be identified with a polynomial ring $$k[x,z]$$ by mapping $$y$$ to $$xz$$ and $$y/x$$ to $$z$$. In particular, $$(x,y/x)$$ is a maximal ideal of $$B$$; let $$R$$ be the localization of $$B$$ at this maximal ideal. I claim $$R$$ is not a localization of $$A$$. To see this, note that if $$\mathfrak{m}$$ is the maximal ideal of $$R$$, then $$\mathfrak{m}\cap A=(\mathfrak{m}\cap B)\cap A=(x,y/x)\cap A$$ is the ideal $$(x,y)$$ of $$A$$. So if $$R$$ is a localization of $$A$$, it must be the localization at $$(x,y)$$. But $$y$$ is not divisible by $$x$$ in that localization, whereas $$y$$ is divisible by $$x$$ in $$B$$.