Question about localizations of discrete valuation rings.

Let $R$ be a discrete valuation ring. Then $R$ has only two prime ideals: $0$ and the maximal ideal $\mathfrak{m}$. It is said in Hartshorne, page 74, Example 2.3.2 that the localization of $R$ at $0$ is $K$, where $K$ is the field of fractions of $R$ and the localization of $R$ at $\mathfrak{m}$ is $R$. In the book, $0$ is denoted by $t_2$ ($t_2=0$ is open) and $\mathfrak{m}$ is denoted by $t_1$ ($t_1=\mathfrak{m}$ is closed).

But according to Wikipedia it seems that $R_{\mathfrak{m}}$ is also $K$ since $R-\mathfrak{m}$ does not contain $0$.

I am confused. Could you explain this? Thank you very much. • A discrete valuation ring $R$ is local and so localizing at the maximal ideal $\mathfrak{m}$ just gives you back your ring again: you don't get the fraction field. I don't see from the wikipedia article that they claim that $R_{\mathfrak{m}}$ is a field. Sure $R - \mathfrak{m}$ does not contain $0$ but why does this mean that $R_{\mathfrak{m}}$ is a field? For example that $R =\Bbb{Z}$ and $\mathfrak{m} = (2)$. Then $R_{(2)}$ is all fractions whose denominator is not a multiple of $2$. Now my question is: If this is a field then what is the inverse of $2$? – user38268 Aug 27 '13 at 14:09
• Any multiplicative set that you invert should not contain zero, or any zerodivisors, otherwise the localization is trivial. In this case, $R_{(0)}$ means invert everything except zero, while $R_{\mathfrak m}$ means invert everything except elements of $\mathfrak m$, which means not every element gets an inverse. – Andrew Aug 27 '13 at 14:20
• I looked rather quickly through the Wikipedia article, and, like user38268, I saw no claim that $R_{\mathfrak m}$ is $K$. The closest thing I found was that if you invert all of the non-zero-divisors (which would be all of $R-\{0\}$ in this case), then you get $K$. – Andreas Blass Aug 27 '13 at 20:29
• @user38268, thank you very much. – LJR Aug 29 '13 at 4:01
• @Andrew, thank you very much. – LJR Aug 29 '13 at 4:01