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I remember reading (quite a while ago, and I can't remember where!) that linear algebraic groups of multiplicative type over a field of characteristic zero are closed under extensions. This is certainly true if we replace "groups of multiplicative type" by "tori" (and the proof is not hard, but it uses connectedness), but I can't prove it in the case of groups of multiplicative type. As I'm starting doubting my memory, does anyone know if

1) the result is true, and 2) if true, a reference for it?

Quick proofs instead of a reference are also welcomed.

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    $\begingroup$ I don't think this is true. If $\sigma$ is the nontrivial automorphism $x \mapsto x^{-1}$ of $\mathbb{G}_m$, then the semidirect product $\mathbb{G}_m \rtimes \langle \sigma \rangle$ seems to be a counterexample. (Groups of multiplicative type are abelian.) $\endgroup$
    – Dave Witte Morris
    Commented Oct 11, 2015 at 17:06
  • $\begingroup$ @DaveWitteMorris That works indeed, thanks. $\endgroup$
    – user81354
    Commented Oct 11, 2015 at 17:50
  • $\begingroup$ OK, I'll make my comment into an answer, then. $\endgroup$
    – Dave Witte Morris
    Commented Oct 11, 2015 at 18:41

2 Answers 2

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This is not true. If $\sigma$ is the nontrivial automorphism $x \mapsto x^{-1}$ of $\mathbb{G}_m$, then the semidirect product $\mathbb{G}_m \rtimes \langle \sigma \rangle$ is a counterexample. (Groups of multiplicative type are abelian.)

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    $\begingroup$ If I understand the question correctly, both $\langle\sigma\rangle$ and the semidirect product must be realized as linear algebraic groups, the former - of multiplicative type, and the maps in the extension as morphisms of algebraic groups, no? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Oct 11, 2015 at 19:52
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    $\begingroup$ That is correct. If we let $T$ be the group of diagonal matrices in $\mathrm{SL}_2(F)$, then the semidirect product is the subgroup $T \cdot \left\langle \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\right\rangle$ of $\mathrm{GL}_2(F)$. Everything is algebraic. $\endgroup$
    – Dave Witte Morris
    Commented Oct 11, 2015 at 23:57
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The answer above is correct, but apparently the non-abelianness of the extension is the only obstruction. In particular, if you assume that the middle group in the extension is abelian then it will be of multiplicative type as soon as the other two are, see chapter IV, §1, Proposition 4.5 in the book:

M. Demazure, P. Gabriel, Groupes algébriques, Tome I : Géométrie algébrique, généralités, groupes commutatifs.

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