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For an element $g$ of a connected reductive group $G$ over a field $F$,

$g$ is called $regular$ if the dimension of the centralizer of $g$ is equal to the rank of the algebraic group $G$,

$g$ is called $elliptic$ if it is semisimple and the maximal split subtorus of the center of the centralizer of $g$ is equal to the maximal split subtorus of the center of $G$.

My question : how can one prove the following statements?

Let $D$ be a central division algebra over $F$ of dimension $n^{2}$, and set $G=D^{\ast}$ (multiplicative group).

Statements:

(1) any element of $G$ is elliptic

(2) an element $g$ of $G$ is regular if and only if $F[g] \subset D$ is a finite field extension of degree $n$.

Please give me any advice.

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    $\begingroup$ What do you mean by "This is a related question as before"? Related to what? $\endgroup$
    – Tobias Kildetoft
    Commented Jan 29, 2014 at 9:10
  • $\begingroup$ Sorry, I asked a similar question in the case G is GL_n before. But this question is independent of it. $\endgroup$
    – Hiro
    Commented Jan 29, 2014 at 10:32

1 Answer 1

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Here you have to know some basic facts about central division algebras.

(1): every element $g$ of $D$ is contained in a maximal subfield $L$; then $L^*$ is contained in the center of the centralizer of $g$, hence is equal to it (because $L$ is maximal). The maximal split subtorus of $L^*$ is $F^*$.

(2): Let $F[g]'$ be the commuting algebra of $F[g]$ in $D$; the centralizer $Z(g)$ is $(F[g]')^*$. It is a fact that $[F[g]:F].[F[g]':F]=n^2$, thus $[F[g]:F]=n$ is equivalent to $[F[g]':F]=n$, hence to $\dim Z(g)=n= \mathrm{rk}(D^*)$.

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  • $\begingroup$ Thank you for providing a proof again! Where can one find the proof of the fact you used? $\endgroup$
    – Hiro
    Commented Jan 29, 2014 at 11:40
  • $\begingroup$ Any book on division algebra... The book by Rowen, Ring Theory -- student edition has a chapter on central simple algebras which should be accessible. Or Bourbaki, Algebra 8. $\endgroup$
    – abx
    Commented Jan 29, 2014 at 12:02
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    $\begingroup$ But statement (1) in the question posed is generally false when the field $F$ is not perfect (e.g., counterexamples arise over every global or local function field over a finite field) because $L$ can fail to be separable over $F$: in such cases "$L^{\ast}$ as an $F$-group", which is to say ${\rm{R}}_{L/F}({\rm{GL}}_1)$, is not an $F$-torus. This is related to the fact that Jordan decomposition of an $F$-point need not be $F$-rational when $F$ is not perfect. $\endgroup$
    – user76758
    Commented Jan 29, 2014 at 14:01
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    $\begingroup$ I think the OP was assuming characteristic 0 -- my answer certainly does. $\endgroup$
    – abx
    Commented Jan 29, 2014 at 14:13

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