Let $K$ be a henselian valuation field with residue field $k$, then the decomposition group surjects onto Galois group of the residue field, with kernel the inertia subgroup, namely we have short exact sequence:$$0\to I\to\mathrm{Gal}_K\to\mathrm{Gal}_k\to 0$$

When $K$ is a local field, we can split the sequence by lifting the Frobenius; when $K=k((t))$, we can split the sequence by lifting the Galois action (with trivial action on $t$). But in general, do we know if the sequence always split?

(The splitting of the sequence is mentioned as a well-known fact in Proposition A5 in "Exposant et indice d'algèbres simples centrales non ramifiées", but I couldn't find a reference.. Splitting of the sequence would imply that the restriction morphism $H^i_{et}(\mathrm{Spec}(V),M)\to H^i_{et}(\mathrm{Spec}(K),M)$ is injective for any locally constant sheave $M$ on the valuation ring $V$.)


1 Answer 1


Good question! Let me try to guess what Gabber had in mind there. (Note that he only says "known" (to him), not "well-known"...)

The claim is that the extension splits. Note that to prove this, we are free to replace $K$ by any (algebraic) extension $K'$ whose residue field $k'$ is purely inseparable over $k$. By Zorn's lemma, we can choose $K$ so that it admits no further such extensions. In particular, $K$ is perfect and the value group is divisible. Let $p$ be the characteristic of $k$. Then it follows that $I$ must be pro-$p$, as the maximal unramified extension $K^{\mathrm{ur}}$ of $K$ will still have divisible value group, and so have no nontrivial tame extensions.

Now, if $I$ was nonzero, there is a map $I\to \overline{I}$ where $\overline{I}$ is a nonzero $\mathbb F_p$-vector space (the quotient by the Frattini subgroup). By maximality of $K$, the induced sequence $$0\to \overline{I}\to \overline{G}\to \mathrm{Gal}_k\to 0$$ is nonsplit, so gives a nonzero class in $H^2(\mathrm{Gal}_k,\overline{I})$. But Galois groups in characteristic $p$ have $p$-cohomological dimension $\leq 1$ by Artin-Schreier theory.

  • $\begingroup$ Is the only reason to enlarge K to ensure that I is pro-p? That's a really interesting technique! $\endgroup$
    – Asvin
    Mar 11, 2021 at 12:30
  • $\begingroup$ One thing I am a bit confused about: given that $K^{ur}$ has a divisible value group, how do we know that its tame extensions are trivial? Would you explain a bit more? Thanks! $\endgroup$
    – Qixiao
    Mar 11, 2021 at 16:46
  • $\begingroup$ The book of Gabber-Ramero on Almost ring theory has a nice Section 6.2 on ramification theory. See in particular 6.2.17 (and 6.2.12 for the assertion that the non-tame part is pro-$p$). $\endgroup$ Mar 11, 2021 at 21:24
  • 1
    $\begingroup$ @Asvin I think enlarging $K$ is also essential for the last part of the argument. If $K$ is not maximal but only have divisible value group then the last exact sequence could spilt $\endgroup$
    – ali
    Mar 14, 2021 at 20:19

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