Let $k$ be a field. By a simple algebraic group over $k$ I mean an affine group scheme $G$ of finite type over $k$ such that $G$ is connected, non-commutative and every normal closed subgroup of $G$ is trivial. I would like to know an example of a simple algebraic group such that the base extension $G_{\overline{k}}$ of $G$ to the algebraic closure $\overline{k}$ of $k$ is not simple anymore.

If $G$ is connected and non-commutative then also $G_{\overline{k}}$ is connected and non-commutative. So the problem is really about normal subgroups of $G_{\overline{k}}$ not being defined over $k$.

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    $\begingroup$ Your definition of simple does not quite agree with the usual one (which also has the subgroups be connected). Is that on purpose? $\endgroup$ – Tobias Kildetoft Aug 9 '13 at 9:46
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    $\begingroup$ Yes this is on purpose. The ones you are referring to I would call "almost simple". $\endgroup$ – Peter Aug 9 '13 at 9:49
  • $\begingroup$ Btw, are you identifying the group with its points over the field, or are you considering it as a scheme? Because in the latter case, I can't think of any examples of such a group (at least not in positive characteristic). (By such a group, I mean one that is simple). $\endgroup$ – Tobias Kildetoft Aug 9 '13 at 11:07
  • $\begingroup$ I am considering it as a scheme. I am mainly interested in characteristic zero. You can get a simple group by taking an almost simple group and then taking the quotient modulo the center. $\endgroup$ – Peter Aug 9 '13 at 11:19
  • $\begingroup$ I guess it is just not obvious to me why that results in a connected group, why the center is maximal among normal closed subgroups, or why this all works in characteristic $0$, when it fails in positive characteristic. $\endgroup$ – Tobias Kildetoft Aug 9 '13 at 11:22

Consider $SL(d,\mathbb{C})$ as a real algebraic group (ie, replace each matrix entry with a $2 \times 2$ matrix representing a complex number with real entries). Then the complexification has the structure of $SL(d,\mathbb{C}) \times SL(d,\mathbb{C})$ and is hence $SL(d,\mathbb{C})$, considered as a group over $\mathbb{R}$, is not absolutely simple.


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