# Does $0\to I\to\mathrm{Gal}_K\to\mathrm{Gal}_k\to 0$ always split?

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 $$V$$.)

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.
• 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! Mar 11, 2021 at 16:46
• 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-$p$). Mar 11, 2021 at 21:24
• @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