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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.

Dubious
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    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 http://jmilne.org/math/CourseNotes/ANT.pdf . But in this case this is true see for example the theorem of page 5 in here http://www1.spms.ntu.edu.sg/~weil0005/mas720/lecture2.pdf – CRKarl Apr 22 '17 at 01:34
  • Thank you for the references. – Dubious Apr 22 '17 at 17:58

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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.

Torsten Schoeneberg
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