# On function fields all absolute values give discrete valuations

Let $K$ be an algebraic function field in one variable over a field $k$. So, $K$ is a finite extension of $k(t)$, and here I allow $k$ to be any field.

Let $|\cdot|:K\to\mathbb R^+$ be an absolute value such that $|k^\times|=1$. It induces a valuation $v: K\to\mathbb R$ given by

$$v(x)=-\log|x|$$ which is trivial on $k^\times$. Mi question is the following:

Is $v$ always a discrete valuation? This means that $v(K^\times)=t\mathbb Z$ for $t\in\mathbb R_+$

I think that the answer should be yes, because on each subspace of the form $kt^i:=\{at^i\colon a\in k\}$ the valuation take constant value.

• Since your absolute value is non-arquimidian it will be discrete iff the maximal ideal $\mathfrak{m}= \{ x \in K : | x |<1\}$ of $\mathcal{O}= \{ x \in K : |x | \leq 1\}$ is principal see proposition 7.6 here jmilne.org/math/CourseNotes/ANT.pdf . But in this case this is true see for example the theorem of page 5 in here www1.spms.ntu.edu.sg/~weil0005/mas720/lecture2.pdf Commented Apr 22, 2017 at 1:34
• Thank you for the references. Commented Apr 22, 2017 at 17:58

The absolute values on $k(t)$ whose restrictions to $k$ are trivial are described explicitly here (where $k$ is assumed to be finite, but all that is actually needed in the proof is $|k^{\times}| = 1$). In particular, they are all discrete. Now an extension of a discrete value to a finite field extension is discrete again, so you are done.