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Let $V = \mathbb A^1(k)$ ($k$ is an algebraically closed field), $\Gamma(V) = k[X]$ and let $K = k(V) = k(X)$. Prove that for each $a \in k = V$, $\mathcal{O}_a(V) := \{f\in K(V): f$ is defined at $a\}$ is a DVR with uniformizing parameter $t = X - a$.

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I did my best for the LaTeX format. Please define Oa(V). – user18119 Nov 6 '12 at 13:42
Oa(V):={f∈K(V); f is defined at a} – Miguel Nov 6 '12 at 13:46
Welcome to Math.Se! Is it homework? What did you try to do? Where did you get stuck? – Giovanni De Gaetano Nov 6 '12 at 16:00
Use the fact that $F(x)\in\Gamma(V)$ has a root at $a$ iff $x-a~\vert~F(x).$ – Andrew Nov 6 '12 at 16:59

The ring $\mathcal O_a(V)$ coincides with the localization $k[X]_{(X-a)}$. This is a PID because it is a localization of a PID. We must show it has only one nonzero prime ideal. To each prime $Q$ in $\mathcal O_a(V)$ there corresponds a (unique) prime $P\subset k[X]$ contained in $(X-a)$. But $P$, being nonzero, is also maximal so $P=(X-a)$ and the unique nonzero prime is $Q=(X-a)k[X]_{(X-a)}$. This is indeed generated by $X-a$, as claimed.

[However, as an aside, the local rings of a nonsingular curve are always DVRs.]

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Thank you so much! – Miguel Nov 8 '12 at 15:06
@Miguel: Can you please "accept" the answer, if it is ok for you? – Brenin Nov 8 '12 at 16:58

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