6
$\begingroup$

To clarify, for a field $K$, a valuation $v$ on $K$ is a map $v:K^{\times}\to G$ for $G$ an ordered group (written additively) such that for any $a,b\in K^{\times}$:

1) $v(ab)=v(a)+v(b)$;

2) $v(a+b)\geq\min\{v(a),v(b)\}$ if $a+b\neq 0$.

We extend $v$ to $K$ by setting $v(0)=\infty$. A valuation is a discrete valuation if $v(K^{\times})$ is $\mathbb{Z}$.

There is a hint which says we could start by determining all the valuation rings. But I do not know how to use this. So I started from the basic by determining $v(c)$ for $c\in\mathbb{C}^{\times}$ and $v(x)$--which then determines $v$.

If we assume $v(c)=0$ for all $c\in\mathbb{C}^{\times}$, then:

For $v(x)$, we consider the cases $v(x)<0$ and $v(x)\geq 0$--that is, $v(x)\in\mathcal{O}_v$ or $v(x)\notin\mathcal{O}_v$.

Case $v(x)<0$: Then $v(x^{-1})>0$ and for any $f\in\mathbb{C}[x]$ of degree $d$, $v(f)=dv(x)=-dv(-x)$. Then for $q=f/g\in\mathbb{C}(x)$, $v(q)=(-deg(f)+deg(g))\cdot v(-x)$. So it seems to me that in this case, if we require $v$ to be a discrete valuation, $v(\mathbb{C}(x))$ is infinitely cyclic with $v(x)$ be its generator. Namely, we get a discrete valuation which sends $q=f/g$ to $v(q)=-deg(f)+deg(g)$.

Case $v(x)\geq 0$: Then $\mathbb{C}[x]\subset \mathcal{O}_v$. Consider $I=\mathbb{C}[x]\cap\mathcal{m}_v$. Notice that $I\neq(0)$. Since $m_v$ is the maximal ideal of $\mathcal{O}_v$, $I$ is a prime ideal of $\mathbb{C}[x]$, which is a PID. Then I is maximal. Then $I=(x-\alpha)$ for some $\alpha\in\mathbb{C}$. For any $r\in\mathbb{C}(x)$, write $r=(x-\alpha)^m\cdot\frac{f}{g}$ such that $f,g\in\mathbb{C}[x]\setminus(p)$. Then $v(r)=v((x-\alpha)^m)$ since $v(f)=v(g)=0$. Then $v(r)=mv(x-\alpha)$. If we require $v$ to be discrete, then $v(x-\alpha)=1$.

We are done with the case for discrete valuations that are trivial on $\mathbb{C}^{\times}$.

What happens when the $v$ is not necessarily trivial on $\mathbb{C}^{\times}$?

Is it possible to find all valuation ring of $\mathbb{C}(x)$ first and then pick the discrete ones so that we can recover the valuation?

$\endgroup$
3
$\begingroup$

Here's a guide to finding the discrete valuations on $\mathbb{C}(x)$. I've put my solutions in spoiler boxes so you can try before you peek. Also, since you didn't mention it, $A$ is a valuation ring of $\mathbb{C}(x)$ iff it is a subring and $z\in A$ or $z^{-1}\in A$ for all non-zero $z\in\mathbb{C}(x)$. In particular, valuations rings are local.

The main idea is that for a discrete valuation ring $A$ with maximal ideal $\mathfrak{m}$, the valuation $v(x)$ is the greatest integer $n$ such that $x\in\mathfrak{m}^n$.

Suppose that $A$ is a valuation ring of $\mathbb{C}(x)$.

First, show that $\mathbb{C}\subseteq A$, if $A\cap\mathbb{C}$ is assumed to be a closed subset of $\mathbb{C}$. Without this assumption you may run into some set theoretic issues or non-discrete valuations.

For every $\zeta\in\mathbb{C}$ on the unit circle, $\zeta\in A$ or $\zeta^{-1}\in A$. Thus $A$ contains some $\zeta$ with argument an irrational multiple of $\pi$. Since $A\cap\mathbb{C}$ is closed and the orbit of $\zeta$ is dense, $A$ contains $S^1$. Since $A$ is a ring, $A$ contains $\mathbb{Z}$. The set $S^1\cdot(\mathbb{Z}+S^1)$ (i.e. integer translates of the circle followed by arbitrary rotations) yields all of $\mathbb{C}$. So $\mathbb{C}\subseteq A$.

Note that $\mathbb{C}(x)$ is a valuation ring of itself. What is the valuation in this case?

$G=\mathbb{C}(x)^{\times}/\mathbb{C}(x)^{\times}$ is an ordered (trivial) group. The quotient map $v:\mathbb{C}(x)^{\times}\rightarrow\mathbb{C}(x)^{\times}/\mathbb{C}(x)^{\times}=G$ has the properties that $v(ab)=v(a)+v(b)$ and $v(a-b)\geq\min(v(a),v(b))$, if $a\neq b$ (where we are writing $G$ additively). So we have obtained the trivial valuation.

Now assume that $A$ is a proper subring. Since $A$ is a valuation ring of $\mathbb{C}(x)$, $x\in A$ or $x^{-1}\in A$.

Find an explicit description of $A$ assuming that $x\in A$.

Since $x\in A$, $\mathbb{C}[x]$ is a subring of $A$. Since $A$ is a valuation ring, it is local. Let $\mathfrak{m}$ be the unique maximal ideal of $A$. The preimage of $\mathfrak{m}$ under the inclusion $j:\mathbb{C}[x]\rightarrow A$ is a prime ideal of $\mathbb{C}[x]$. Since $\mathbb{C}$ is algebraically closed, $j^{-1}(\mathfrak{m})=(x-a)$ for some $a\in\mathbb{C}$. Thus every element of $\mathbb{C}[x]$ not in $(x-a)\mathbb{C}[x]$ is invertible in $A$. By the universal property of localization, $j$ lifts to some map $\overline{j}:\mathbb{C}[x]_{(x-a)}\rightarrow A$. Furthermore, $\overline{j}$ is injective because $\mathbb{C}[x]$ is an integral domain. The only elements of $\mathbb{C}(x)$ not in $\mathbb{C}[x]_{(x-a)}$ are those with a factor of $(x-a)$ in the denominator. Thus $\mathbb{C}[x]_{(x-a)}$ is a maximal proper subring of $\mathbb{C}(x)$. Since $A$ was assumed to be proper, $\overline{j}$ is surjective, hence an isomorphism.

Derive what the valuation must be for this description of $A$.

The ring $\mathbb{C}[x]_{(x-a)}$, for any $a\in\mathbb{C}$, is easily seen to be a valuation ring of $\mathbb{C}(x)$. The valuation is $v:\mathbb{C}(x)^{\times}\rightarrow\mathbb{C}(x)^{\times}/\mathbb{C}[x]_{(x-a)}^{\times}=G$. Note that the elements of $G$ are uniquely identified by coset representatives of the form $(x-a)^n$, where $n\in\mathbb{Z}$. So $G$ is isomorphic to $\mathbb{Z}$. Choose the identification of $(x-a)\in G$ with $1\in \mathbb{Z}$ (there are two possible isomorphisms, this picks one). The ordering on $G$ is just the ordering of the integers (powers of $(x-a)$). Identifying $G$ and $\mathbb{Z}$, we get $$v((x-a)^mp(x)+(x-a)^nq(x))=v((x-a)^k[(x-a)^{m-k}p(x)+(x-a)^{n-k}q(x)])\geq v((x-a)^k)=k,$$ where $p(x),q(x)$ are assumed to have no factors of $(x-a)$ in numerator or denominator, and $k=\min(m,n)$, $m,n\in\mathbb{Z}$. Similarly, $$v((x-a)^mp(x)\cdot(x-a)^nq(x))=v((x-a)^{m+n}p(x)q(x))=m+n.$$ So $v$ is in fact a discrete valuation of $\mathbb{C}(x)$.

Now assume that $x^{-1}\in A$ and $x\notin A$. Find a valuation ring and its valuation that were not found above.

We also have that the ring $\mathbb{C}[x^{-1}]_{(x^{-1})}$ is a valuation ring of $\mathbb{C}(x)$. One way to see this is that $x\mapsto x^{-1}$ is an automorphism of $\mathbb{C}(x)$, and $\mathbb{C}[x]_{(x)}$ is one of the valuation rings from the previous paragraph. Repeating the same construction as before yields a discrete valuation.

Finally, show that there are no other discrete valuation rings of $\mathbb{C}(x)$.

Our characterization in the case that $x\in A$ was forced. So there cannot be any others in that case. Suppose that $x\notin A$. If we run through the characterization of $A$ analogously to the case $x\in A$ we end up with $A=\mathbb{C}[x^{-1}]_{(x^{-1}-a)}$ for some $a\in\mathbb{C}$. If $a\neq0$, then $x^{-1}$ is a unit of $A$ because it is not in the maximal ideal of $A$. This is a contradiction, so we are forced to have $a=0$ and we end up with the valuation ring from before. Therefore, the only wiggle room on valuations and valuation rings of $\mathbb{C}(x)$ comes from dropping the assumption that $A\cap\mathbb{C}$ is a closed subset of $\mathbb{C}$. This will lead to considerations of possibly non-measurable subsets of $\mathbb{C}$ and other set theoretic difficulties that I cannot at present address. Furthermore, it should not yield any more discrete valuations, but I cannot prove that right now either.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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