I am concerned with valuations of $\mathbb Q_p$ distinct from the $p$-adic valuation.

It is of course well-known that the residue field of the $p$-adic valuation is finite with $p$ elements. Also, by Chevalley's Extension Theorem, one can extend all other $q$-adic valuations to $\mathbb Q_p$. As is folklore, none of these extensions will be discrete valuations (in fact $v_p$ is the unique discrete valuation on $\mathbb Q_p$). But what about the residue fields?

Is there a valuation domain of $\mathbb Q_p$ independent from $\mathbb Z_p$ that has a finite residue field?

Of course, we could try to use the fundamental equation linking the degrees of the field extension and the residue field extension for finite subextensions. But that gives us nothing, if the indices of the value groups grow infinitely large.

Thank you for your help!

  • $\begingroup$ Well any such valuation ring will have to contain the integers. Restricting the valuation to the integers, we see that it’s either trivial or $q$-adic for some prime in $\mathbb Z$, no? $\endgroup$
    – Arkady
    Aug 3, 2022 at 10:45
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    $\begingroup$ Yes. I can't see the relation to my question though. Sorry... $\endgroup$
    – Daniel W.
    Aug 3, 2022 at 11:03
  • $\begingroup$ So by Chevalley's extension argument you've made doesn't this show that the only discrete one will come from $q=p$? I guess it's not clear if the original valuation on $\mathbb Q_p$ matches the $p$-adic valuation on it. $\endgroup$
    – Arkady
    Aug 3, 2022 at 14:28
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    $\begingroup$ The only discrete one comes from $p$, as you say. But what about non-discrete valuations? $\endgroup$
    – Daniel W.
    Aug 3, 2022 at 14:32

1 Answer 1


No, if $v$ is a valuation on $\Bbb{Q}_p$ and $O_v/\mathfrak{m}_v$ has finite characteristic $\ell\ne p$ then it contains the algebraic closure of $\Bbb{F}_\ell$.

Proof: take $f\in \Bbb{F}_\ell[x]$, monic irreducible. As $\gcd(p,\ell)=1$ there is $g\in \Bbb{Z}[x]$ monic such that $g\equiv f\bmod \ell, g\equiv x^{\deg(f)}-1\bmod p$.

(if $p$ divides $\deg(f)$ then take instead $g\equiv f\bmod \ell, g\equiv x^{\deg(f)}-x\bmod p$)

By Hensel lemma $g$ has a root in $\Bbb{Q}_p$. This will be a root of $f$ in $O_v/\mathfrak{m}_v$.

Whence $O_v/\mathfrak{m}_v$ contains $\overline{\Bbb{F}}_\ell$.


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