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

Question

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!

Answer 1

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 $\mathscr{O}_\infty$.

Answer 2

The case where $k[x] \subseteq S$ is easy. Let $m$ be maximal ideal of $S$. Thus $k[x] \cap m$ is a prime ideal in $k[x]$. Since $k[x]$ is a PID, $k[x] \cap m$ is either zero ideal or a maximal ideal. Its easy to dismiss the zero ideal case (check if $k[x] \cap m$ is zero ideal then we get $S$ is a field). So $k[x] \cap m$ is a maximal ideal. Since $k$ is algebraically closed, $k[x] \cap m = \langle x-a \rangle$ for some $a \in k$. Let $\frac{F}{G} \in S \setminus \{0\}$ where $F$ and $G$ have no common factors. Its easy to see $\frac{F}{G} \in \mathcal{O_{a} (\mathbb{A^{1}})}$ iff $G(a) \not = 0$ iff $x-a$ is not a factor of $G$. If $G$ had $x-a$ as a factor then $G \in k[x] \cap m \subseteq m$. So $F=\frac{F}{G}G \in m$ and since $F \in k[x]$, $F \in k[x] \cap m \subseteq m = \langle x-a \rangle$. So $F$ has $x-a$ as a factor too, which contradicts the assumption that $F$ and $G$ have no factors in common. So $\frac{F}{G} \in \mathcal{O_{a} (\mathbb{A^{1}})}$. Now we have to show maximal ideal of $\mathcal{O_{a} (\mathbb{A^{1}})}$ contains $m$. Now let $\frac{F}{G} \in m \setminus \{0\}$ (with $F$ and $G$ having no common factors). Since $G \in S$, $F=G \frac{F}{G} \in m$ and so $F \in k[x] \cap m = \langle x - a \rangle$. Thus $F$ has $x-a$ as a factor. Its straightforward to see $\frac{F}{G}$ is not a unit in $\mathcal{O_{a} (\mathbb{A^{1}})}$ (recall $F$ and $G$ has no common factors and $k[x]$ is a UFD). Then using the result mentioned in this post (its problem $2.26$ in Fulton's book), we have $S=\mathcal{O_{a} (\mathbb{A^{1}})}$.

Suppose $k[x] \not \subseteq S$. So for any $a \in k$, $x-a \not \in S$ (or else $k[x] \subseteq S$, recall $k \subseteq S$ and use polynomial long division to see this). So $\frac{1}{x-a} \in m$ for all $a \in k$. Note $\frac{x-b}{x-a} = 1 + \frac{a-b}{x-a}$ ($\frac{a-b}{x-a} \in m$ since $a-b \in k \subseteq S$) and since $S$ is a local ring, $\frac{x-b}{x-a}$ is a unit in $S$. Let $\nu$ be the discrete valuation on $k(x)$ associated with $S$. Then $\nu ( \frac{x-b}{x-a} ) = 0$ and so $\nu (x-a) = \nu (x-b) =: \theta$ for all $a,b \in k$ (note $\theta < 0$). So if $P$ is a nonzero polynomial, then $\nu(P) = deg(P) \theta$. Let $\frac{F}{G} \in S \setminus \{0\}$. Then $\nu(\frac{F}{G}) = (deg(F)-deg(G))\theta$. But $\nu(\frac{F}{G}) \ge 0$ and $\theta < 0$ and so we conclude $deg(F) \le deg(G)$ ie. $\frac{F}{G} \in \mathcal{O_{\infty}}$. Let $\frac{F}{G} \in m \setminus \{0\}$. So $(deg(F)-deg(G))\theta = \nu(\frac{F}{G}) > 0$ and we get $deg(F) <deg(G)$ (which implies $\frac{F}{G}$ is not a unit in $\mathcal{O_{\infty}}$).