Timeline for Field of definition of a normal subgroup of Reductive group
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2017 at 14:07 | comment | added | nfdc23 | A big class of examples is the unipotent radical $H$ when $G$ is non-reductive but pseudo-reductive. For example, if $k'/k$ is a nontrivial non-separable finite extension of fields (take $k=k_s$ if you wish; not necessary) and $G'$ is a connected reductive $k'$-group $\ne 1$ then the Weil restriction $G={\rm{R}}_{k'/k}(G')$ is smooth connected affine and not reductive (nonzero nilpotents in $k'\otimes_k \overline{k}$ force $\mathscr{R}_u(G_{\overline{k}})\ne 1$) but the so-called $k$-unipotent radical of $G$ is trivial. See Prop. 1.1.10 and Example 1.6.1 in the book Pseudo-reductive Groups. | |
| Sep 19, 2017 at 13:38 | comment | added | random123 | I mean example of a non reductive group with a normal subgroup not defined over a finite separable extension. | |
| Sep 19, 2017 at 13:13 | comment | added | nfdc23 | What do you mean by asking for a counterexample? Both answers are explaining that your question admits no counterexample. | |
| Sep 19, 2017 at 6:32 | comment | added | random123 | Thanks. What would be a possible counter-example to the statement? | |
| Sep 19, 2017 at 6:25 | vote | accept | random123 | ||
| S Sep 19, 2017 at 5:34 | history | answered | nfdc23 | CC BY-SA 3.0 | |
| S Sep 19, 2017 at 5:34 | history | made wiki | Post Made Community Wiki by nfdc23 |