Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $t = X - a$.

share|improve this question
    
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
1  
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

1 Answer 1

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.]

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.