Let $F$ be a nonarchimedean local field. Does there exist a semisimple algebraic group over $F$ which is not split over a maximal unramified extension of $F$ ?
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$\begingroup$ Have you looked at the papers of BruhatTits or the 1979 AMS lecture notes by Tits on structure and buildings? To see more recent online resources, check MO questions with the tags localfields and algebraicgroups (which should also be added tags for your question). If I recall correctly, the answer to your question is no; but exact information can be found in the references I've mentioned. $\endgroup$– Jim HumphreysJul 17, 2013 at 17:29

3$\begingroup$ @Jim: You are probably remembering that for a connected semisimple group $G$ of adjoint type over a field $k$ with cd$(k)\le 1$, ${\rm{H}}^1(k,G)=1$ (since all $k$forms of $G$ are quasisplit by Steinberg, so for a Borel $k$subgroup $B$ of $G$ the map ${\rm{H}}^1(k,B)\rightarrow {\rm{H}}^1(k,G)$ is surjective, and ${\rm{H}}^1(k,B)=1$ because in the adjoint case maximal tori of $B$ are "induced", hence cohomologically trivial). Hence, for connected semisimple $G$ over $F$, ${\rm{H}}^1(F^{\rm{un}},G^{\rm{ad}})=1$. But for types A, D, and E$_6$ the Autscheme is disconnected! $\endgroup$– user36938Jul 18, 2013 at 1:20

$\begingroup$ @user36938: Yes, my remark overlooked the nature of the automorphisms. Thanks for the reminders. $\endgroup$– Jim HumphreysJul 18, 2013 at 15:39

$\begingroup$ @user36938, isn't the vanishing of cohomology that you mention true for any connected reductive group (maybe even connected group), by Steinberg? $\endgroup$– LSpiceJul 14, 2019 at 13:12
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The Galois group of $F$ maps onto $S_3$, the symmetric group on three letters. The group $S_3$ is also the group of diagram automorphisms of the Dynkin diagram of $SO(8)$. That means that there is a quasisplit group $G$ defined over $F$, which over an extension $E/F$ with Galois group $S_3$, becomes the split group $SO(8)$ over $E$. The group $G$ cannot split over an unramified extension $L/F$, because its Galois group would be abelian, and cannot cover all the diagram automorphisms.

4$\begingroup$ This example has a serious error, but there is a nice fix. The mistake is that automorphisms of the simply connected central cover might not preserve the kernel of the covering map, and so may not descend; equivalently, a diagram automorphism might not preserve the based root datum. In fact, for $m>1$, ${\rm{SO}}_{2m}$ always has outer automorphism group $\mathbf{Z}/(2)$, even if $m=4$! So use a ramified quadratic Galois $F'/F$ to make a quasisplit outer form $G$ of ${\rm{SO}}_{2m}$ arising from the ${\rm{Gal}}(F'/F)$action through the standard diagram involution for ${\rm{D}}_m$. $\endgroup$ Jul 18, 2013 at 1:05

4$\begingroup$ Alternatively, one can fix the example by working with the spin group instead of the special orthogonal group to exploit triality. But it seems simpler to use quadratic ramified extensions to make examples at the level of SO in every even rank $\ge 4$. $\endgroup$ Jul 18, 2013 at 1:06

2$\begingroup$ I understand the error you pointed out and that one can fix it by replacing $SO(8)$ by $Spin(8)$, but I don't understand why constructing nonsplit groups via totally ramified extensions is of any help. I mean a group arising as a twist by a nontrivial Galois cocycle of a ramified quadratic extension can (and in general will be) split by an unramified extension. $\endgroup$ Jul 18, 2013 at 12:46

4$\begingroup$ @Nicolas: It's not twisted by any old Autvalued 1cocycle: it is twisting by one whose projection into the outer automorphism scheme is a ramified cohomology class, so by canonicity of the outer automorphism scheme we can conclude  the logic is literally identical to what is needed to justify that your Spin(8) construction doesn't split over an unramified extension. That is why the reasoning doesn't apply to inner forms of type A (as we know it cannot); see my comment to Jim Humphreys. $\endgroup$ Jul 18, 2013 at 13:06

2$\begingroup$ It is also true that for any simply connected absolutely almost simple group $G$ defined over $F$, there exists a finite unramified extension over which $G$ is quasisplit. So the obstruction to being split really comes from homomorphisms of $Gal(F)$ into the group of automorphisms of the Dynkin diagram. $\endgroup$ Jul 19, 2013 at 0:04