The title is meant to be punchy, but also a tongue-in-cheek acknowledgement of the prevalence of ‘reduce’-derived words in this area. (Unfortunately, I overlooked the fact that the question in the title is the opposite of the one in the body. Fortunately, [@LaurentMoret-Bailly's answer](https://mathoverflow.net/a/396667) is quite clear about this: ‘yes’ to the question in the title, ‘no’ to the question below.)


Let $k$ be a separably closed field, let $\overline k/k$ be an algebraic closure, and suppose that $G$ is a connected $k$-group scheme of finite type.


I know that $G\_\text{red}$ need not be geometrically reduced, but the standard examples of this have $G\_\overline k$ unipotent. Is that for convenience, or is it necessary?


To put it more precisely, suppose that $(G\_\overline k)\_\text{red}$ is a *reductive* $\overline k$-group scheme. Does it follow that $G\_\text{red}$ is geometrically reduced?



Can non-geometrically reduced reduced subschemes happen for reductive groups?
No, it does not follow that $G\_\mathrm{red}$ is geometrically reduced (thus the answer to the title question is yes).


Let $p=\mathrm{char}(k)>0$ and let $H=\mathbb{G}\_a \rtimes \mathbb{G}\_m$ be the group of affine transformations of $\mathbb{A}^1\_k$.


Let $(x,y)\in H$ act on $\mathbb{A}^1\_k$ by the "twisted" action $z\mapsto x^p+y^p z$. If $z\in k$ is not a $p$-th power, its stabilizer $G$ is easily seen to be reduced, but $G\_{\overline{k}}$ is isomorphic to the stabilizer of the origin which is $\alpha\_p \rtimes \mathbb{G}\_m$.


In particular, $(G\_{\overline{k}})\_\mathrm{red}$ is isomorphic to $\mathbb{G}\_m$.


