2
$\begingroup$

Let $G$ be an algebraic group over an algebraically closed field. Furthermore, let $G$ be semi-simple, i.e. its radical (viz. its maximal closed, connected, solvable normal subgroup) is trivial. One step in a proof I'm trying to understand seems to use the following fact:

If $G$ is connected and non-trivial, then there is a non trivial maximal torus in $G$.

Why is that true? Thank you!

$\endgroup$

1 Answer 1

3
$\begingroup$

It's a theorem of Grothendieck, even true over any base field. See SGA 3, tome II, Exp. XIV, Theorem 1.1 (it is also available in Linear algebraic groups by Borel, Theorem 18.2).

$\endgroup$
1
  • $\begingroup$ In my copy of Borels book, Theorem 18.2 only says that a connected group contains a maximal torus, but not that it is non-trivial if $G$ is non-trivial. Or am I understanding it wrong? $\endgroup$
    – Sh4pe
    Jan 3, 2013 at 9:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .