TITLE: Are simple algebraic groups absolutely simple?
QUESTION [6 upvotes]: 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 $k$.

REPLY [3 votes]: 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.