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Let $k$ be a field of characteristic zero (I'm only interested in number fields), and let $\mathbb{G}_{/k}$ be a linear algebraic group defined over $k$ which is almost $k$-simple (all normal subgroups defined over $k$ are finite). This group need not be absolutely almost simple, and one way this situation can arise is when $\mathbb{G}$ is the Weil restriction of scalars of an almost-simple $K$-group defined over a finite extension $K/k$. Is that the only possibility?

In other words, suppose $\mathbb{G}$ is $k$-almost simple. Is there a finite extension $K$ and an absolutely almost simple group $\mathbb{H}_{/K}$ such that $R^K_k\mathbb{H}$ is $k$-isogenous to $\mathbb{G}$?

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No. Algebraic groups geometrically isomorphic to $\mathbb G_m^n$ are classified by homomorphisms from the Galois group to $\operatorname{Aut}( \mathbb G_m^n) = GL_n(\mathbb Z)$. Almost simple ones correspond to irreducible representations. You are asking whether the irreducible representations are induced representations of trivial representations - in fact induced representations are never irreducible, so the answer is no.

All other cases are fine, though:

Consider the group $G$ over an algebraically closed field. It has no proper canonical normal subgroups - ones definable using just the structure of the group, which would be definable over $k$. Hence it is either reductive or unipotent.

If unipotent, it must be abelian, hence of the form $\mathbb G_a^n$. Then it must be of the form $\mathbb G_a^n$ over $k$ by Hilbert 90, hence $k=1$ and we are done.

If reductive, it must be either a torus or semisimple. If semisimple it is isogenous to a product of copies of simple groups. The semisimple case is fine - there is a transitive Galois action on the set of simple factors, which you can use to write it as a Weil restriction from the field corresponding to the stabilizer of one of the factors. The torus case, as we saw earlier, is problematic.

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  • $\begingroup$ I was thinking about semisimple groups: thanks for confirming my intuition. It didn't occur to me that a torus could be almost simple, but of course it can be. $\endgroup$ Mar 6, 2015 at 2:41
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    $\begingroup$ Not that it matters here, but as with finite groups (where "simple" usually includes "non-abelian") the usual convention is that "almost simple" in the context of linear algebraic groups is meant to include a semisimplicity hypothesis (as Lior intended). A literature reference for that case (with the same idea of proof as above) is 6.21(ii) in the Borel-Tits IHES paper on reductive groups, and this is proved there over any field at all (though Lior only wants char. 0). $\endgroup$
    – user74230
    Mar 6, 2015 at 2:43
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    $\begingroup$ More refined: if $G$ is a connected semisimple $k$-group that is simply connected (or adjoint) with $k$ any field then there is a pair $(k'/k,G')$ consisting of a finite etale $k$-algebra $k'$ and semisimple $k'$-group $G'$ with connected fibers that are absolutely simple and simply connected (or adjoint) so that ${\rm{R}}_{k'/k}(G') \simeq G$, and moreover that this pair is unique up to unique isomorphism (i.e., isom between such Weil restrictions uniquely arises from $k$-algebra isom and group isom over that). By uniqueness and Galois descent we can assume $k=k_s$, which is easy! $\endgroup$
    – user74230
    Mar 6, 2015 at 2:47
  • $\begingroup$ @user74230 Does simple really mean non-abelian in the finite group case? Almost every reference I've seen includes $\mathbb Z/p$ among them. $\endgroup$
    – Will Sawin
    Mar 6, 2015 at 3:08
  • $\begingroup$ Of course the Jordan-Holder theorem would be awkward to state if $\mathbf Z/(p)$ were not among the finite simple groups. I am sure user74230 only had in mind that when you talk about classifying the finite simple groups you may make a convention about them being non-abelian to avoid having to say that a lot. For instance, on en.wikipedia.org/wiki/Feit%E2%80%93Thompson_theorem I find the following quote from Burnside: "The contrast that these results show between groups of odd and even order suggests inevitably that simple groups of odd order do not exist." $\endgroup$
    – KConrad
    Mar 6, 2015 at 3:36

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