# How to show there are only two discrete valuation rings with quotient field $k(x)$?

I want to show that only discrete valuation rings with quotient field as $$k(x)$$ containing $$k$$ are: $$\mathcal{O_{a} (\mathbb{A^{1}})}$$ for each $$a \in k$$ and $$\mathcal{O_{\infty}}$$; the former is the set of rational functions on $$\mathbb{A^{1}}$$ (affine 1-space, that is field $$k$$ here) that are defined at $$a \in k$$, it is a discrete valuation ring with uniformizing parameter $$x-a$$ and the latter is the ring $$\left\{\frac{F}{G} \in k(x) \mid \deg(G) \geq \deg(F) \right\}$$ with $$\frac{1}{x}$$ as its uniformizing parameter.

My idea was to first observe that if $$S$$ is any DVR, then it cannot be clearly field of quotients $$k(x)$$, since in the book (Fulton, Algebraic Curves) we have not defined them as fields. So, $$S\subset k(x)$$.

It will contain the ring $$k[x]$$. Now I will use a previous exercise that says that "If $$R$$ is a DVR with quotient field $$K$$ and $$m$$ as its maximal ideal then for $$z\in K, z \notin R$$, we must have $$z^{-1} \in m$$." and another that says that

"Further if $$R\subset S\subset K$$ and $$S$$ is also a DVR, and the maximal ideal of $$S$$ contains $$m$$ then $$S =R$$."

But I don't know how I can start.

Any hint would be appreciated, thanks!

• Is $k$ algebraically closed? Otherwise, I don't think the statement is true. (You'd have to restrict to discrete valuations that are trivial on $k$.) – André 3000 Feb 11 at 7:55
• It is algebraically closed. Also, thanks for editing I was not able to correct that editing error so left it as it is. – clear Feb 11 at 8:03

Let me start by pointing out the a DVR $$R\subset k(x)$$ may not contain $$k[x]$$, and in fact you $$\mathscr{O}_\infty$$ doesn't. Here's an outline of how to prove the result: (1) using the second exercise you mention (about maximality of DVRs), prove that a DVR R that does contain $$k[x]$$ must be the localization of $$k[x]$$ at some maximal ideal, and so isomorphic to some $$\mathscr{O}_a (\mathbb{A}^1)$$ (hint: if $$\mathfrak{m}$$ is the maximal ideal of $$R$$, what is $$\mathfrak{m}\cap k[x]$$?) (2) by the first exercise you cite, if $$R$$ does not contain $$k[x]$$, it must contain $$x^{-1}$$; use maximality again to prove that such a DVR must be $$\mathscr{O}_\infty$$.