# 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.

• Just so you know, you don't have to delete and reask in order to edit your post. The edit button at the bottom works just fine. Jun 5 at 18:14
• @KReiser thanks for the comment, mistake on my part Jun 5 at 18:16
• 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. Jun 5 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. Jun 5 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. Jun 5 at 18:31