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Is the Lie algebra of the solvable radical of an algebraic group G always the solvable radical of Lie(G)?
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Assume everything is defined oder the field k. If $char(k)=p>0$, then there is a counter-example:
It is well known that $SL_p(k)$ is semisimple, i.e. the radical is trivial. So the Lie algebra of its radical will also be trivial, but the radical of $sl(2,p)$ is not trivial as the scalar matrices $k^times I_p$ are then contained in $sl(2,p)$ and they form an abelian (hence solvable) ideal.
Does the result work on characteristic $0$ or is there also a counterexample?
algebraic-geometry
asked Jul 25 at 14:05
Yann
11
add a comment |Â
up vote
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down vote
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Assume everything is defined oder the field k. If $char(k)=p>0$, then there is a counter-example:
It is well known that $SL_p(k)$ is semisimple, i.e. the radical is trivial. So the Lie algebra of its radical will also be trivial, but the radical of $sl(2,p)$ is not trivial as the scalar matrices $k^times I_p$ are then contained in $sl(2,p)$ and they form an abelian (hence solvable) ideal.
Does the result work on characteristic $0$ or is there also a counterexample?
algebraic-geometry
asked Jul 25 at 14:05
Yann
11
I meant $sl(p,k)$ instead of $sl(2,p)$
– Yann
Jul 25 at 14:10
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume everything is defined oder the field k. If $char(k)=p>0$, then there is a counter-example:
It is well known that $SL_p(k)$ is semisimple, i.e. the radical is trivial. So the Lie algebra of its radical will also be trivial, but the radical of $sl(2,p)$ is not trivial as the scalar matrices $k^times I_p$ are then contained in $sl(2,p)$ and they form an abelian (hence solvable) ideal.
Does the result work on characteristic $0$ or is there also a counterexample?
algebraic-geometry
asked Jul 25 at 14:05
Yann
11
Assume everything is defined oder the field k. If $char(k)=p>0$, then there is a counter-example:
It is well known that $SL_p(k)$ is semisimple, i.e. the radical is trivial. So the Lie algebra of its radical will also be trivial, but the radical of $sl(2,p)$ is not trivial as the scalar matrices $k^times I_p$ are then contained in $sl(2,p)$ and they form an abelian (hence solvable) ideal.
Does the result work on characteristic $0$ or is there also a counterexample?
algebraic-geometry
asked Jul 25 at 14:05
Yann
11
asked Jul 25 at 14:05
Yann
11
asked Jul 25 at 14:05
Yann
11
asked Jul 25 at 14:05
Yann
11
11
I meant $sl(p,k)$ instead of $sl(2,p)$
– Yann
Jul 25 at 14:10
add a comment |Â
I meant $sl(p,k)$ instead of $sl(2,p)$
– Yann
Jul 25 at 14:10
I meant $sl(p,k)$ instead of $sl(2,p)$
– Yann
Jul 25 at 14:10
I meant $sl(p,k)$ instead of $sl(2,p)$
– Yann
Jul 25 at 14:10
add a comment |Â
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I meant $sl(p,k)$ instead of $sl(2,p)$
– Yann
Jul 25 at 14:10