# Is the localization at a prime ideal of any polynomial ring always a valuation ring?

I observed that all examples of localizations of polynomial rings at prime ideals I've encountered have been, so far, valuation rings, and so I started wondering whether this is true in all cases. I think the following might be an argument.

We have the field of fractions of the localized polynomial ring: $$\text{Frac}(R[x_1,\dots,x_n]_{\mathfrak p}) = \{ \frac{f_1s_2}{f_2s_1}: f_1,f_2,s_1,s_2 \in R[x_1,\dots,x_n]; s_1, s_2 \notin \mathfrak p \}$$ Consider an arbitrary element of $$R(x_1, \dots, x_n)$$. Then the numerator is either in $$\mathfrak p$$ or not. If it is, then $$\text{num} = f \cdot 1$$. If it isn't, then $$\text{num} = 1 \cdot s$$. The same argument can be done for denominator, and therefore we have $$\text{Frac}(R[x_1,\dots,x_n]_{\mathfrak p}) = R(x_1 \dots, x_n)$$, from which it clearly follows $$f \in R(x_1, \dots, x_n)$$ implies $$f^{-1}$$ or $$f \in R[x_1,\dots,x_n]_{\mathfrak p}$$, since either the denominator is in $$\mathfrak p$$ or it is not.

• Do we know the localization is an integral domain, in general? If not, we can't define the field of fractions. Indeed, it is unclear that $R(x_1,\dots,x_n)$ is defined except when $R$ is an integral domain. Commented Jun 5, 2023 at 18:40
• @ThomasAndrews I'm trying to find the extent to which it could or not be true, so I hadn't specifically thought about having the integral domain condition, so thanks for that note! It does seem like we need it to be an integral domain. Commented Jun 5, 2023 at 18:50

For a field $$F$$, even $$R=F[x,y]$$ does not have the property that all its localizations at primes are valuation rings.
There are a lot of ways to see it, but try this one: $$R$$ has Krull dimension $$2$$, and so will its localization $$S$$ at $$(x,y)$$. $$S$$ is already a Noetherian domain, and if it were a valuation ring it would have to be a PID, hence have Krull dimension less or equal to $$1$$. So $$(x,y)$$ is a prime (maximal, even) ideal of $$R$$, for which the localization is not a valuation ring.
• You can be pretty hands on here, too: an element of the fraction field of $R$ like $x/y$ doesn't belong to $R$, and neither does its inverse. Commented Jun 5, 2023 at 18:31