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Let $k$ be a field of characteristic zero and let $G$ be an adjoint semi-simple algebraic group over $k$.

On p34 of the paper "Sansuc - Groupe de Brauer et arithmétique des groupes algébriques lineaires sur um corps de nombres", it is claimed that there exists a collection of finite field extensions $k \subset k_i$ such that $$G \cong \prod_i \mathrm{R}_{k_i/k}(G_i) \quad (*)$$ where the $G_i$ are absolutely simple adjoint groups over $k_i$ and $\mathrm{R}_{k_i/k}$ denotes the Weil restriction.

I was quite surprised when I saw this, as it certainly seems to be something special about adjoint groups. However, Sansuc unfortunately gives no explanation nor reference why this holds.

Why does the stated isomorphism (*) exist?

I would be happy with either a proof or a reference.

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    $\begingroup$ This result goes back to Borel and Tits: see 6.21(ii) in their IHES 27 paper (same proof for the adjoint case). Their proof is as in zeno's answer, since they work throughout with groups over fields (i.e., the idea to introduce the finite etale $k$-algebra $k' = \prod k_i$ and the $k'$-group $G' = \coprod G_i$ so as to write the right side as ${\rm{R}}_{k'/k}(G')$ is something that comes very naturally when one is accustomed to considering groups over rings but might look weird otherwise; treating $k'$ as a single package is very handy when contemplating scalar extension on $k$). $\endgroup$
    – nfdc23
    Commented May 17, 2016 at 2:33

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See Proposition 6.4.4 and Remark 6.4.5 in Brian Conrad's article "Reductive groups schemes" in "Autour des schemas en groupes, Vol. I" (alternatively, http://math.stanford.edu/~conrad/papers/luminysga3smf.pdf) for a proof (in a more general setting).

As you mention, this is special for adjoint groups (but would also work for simply connected groups).

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    $\begingroup$ I guess the characteristic 0 hypothesis in the question is irrelevant here? $\endgroup$
    – Jim Humphreys
    Commented May 16, 2016 at 17:07
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    $\begingroup$ @JimHumphreys: Good point. Also, the cited result shows that in positive characteristic the appearing finite extensions $k_i$ may be chosen to be separable. $\endgroup$
    – Kestutis Cesnavicius
    Commented May 16, 2016 at 17:30
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    $\begingroup$ Not only can the $k_i/k$ be chosen to be separable: they must be separable. That is, if $k'/k$ is a finite extension of fields that is not separable and $G'$ is a nontrivial connected reductive $k'$-group then ${\rm{R}}_{k'/k}(G')$ is a smooth connected affine $k$-group that is never reductive (in brief, nonzero nilpotents in $k' \otimes_k \overline{k}$ give rise to a nontrivial geometric unipotent radical, as is seen concretely for $G' = {\rm{SL}}_n$). $\endgroup$
    – nfdc23
    Commented May 17, 2016 at 2:23
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The proof is quite easy. Let $G$ be a simply connected (or adjoint) semisimple group over a field $k$. Over the separable closure $k^{s}$ of $k$, $G$ is a product $G=G_{1}\times\cdots\times G_{n}$ of almost-simple groups $G_{i}$. The Galois group $\Gamma$ of $k^{s}/k$ acts on the set $\{G_{1},\ldots,G_{n}\}$ and the product of the groups in an orbit is stable under $\Gamma$, and hence defined over $k$. In this way, $G$ is a product of quasi-simple groups over $k$. Thus, we may suppose that $G$ itself is quasi-simple. Now $\Gamma$ acts transitively on the set $\{G_{1},\ldots,G_{n}\}$. Let $\Delta$ be the stabilizer of $G_{1}$, and let $K$ be the subfield of $k^{s}$ fixed by $\Delta$. Then $Res_{K/k}(G_{1})$ and $G$ are isomorphic over $k^s$, by an isomorphism invariant under the action of of $\Gamma$, and so they are isomorphic over $k$.

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  • $\begingroup$ Thanks. Though can you please emphasise at which point in the argument you use that $G$ simply connected or adjoint? $\endgroup$
    – Daniel Loughran
    Commented May 16, 2016 at 19:27
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    $\begingroup$ The second sentence of the proof: in general $G$ will only be isogenous to such a product. $\endgroup$
    – zeno
    Commented May 16, 2016 at 20:36
  • $\begingroup$ Curiously, the proof becomes a bit easier if one doesn't pass to the quasi-simple case, as one can formulate the result with a strong uniqueness aspect (pinning down $(k'/k, G')$ uniquely up to unique isomorphism, with $k'$ a nonzero finite etale $k$-algebra and $G'$ a smooth affine $k'$-group whose fibers are connected semisimple, absolutely simple, and simply connected -- or adjoint); that handles the Galois descent by pure thought. See Proposition A.5.14 in the book "Pseudo-reductive groups" for such a proof, the version over fields for the reference given by Cesnavicius over rings. $\endgroup$
    – nfdc23
    Commented May 17, 2016 at 2:28

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