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Let $G$ be a smooth affine algebraic group over a field $k$. If $G_{\bar{k}}$ is commutative over $\bar{k}$, is it necessary that $G$ is a commutative algebraic group over $k$?

I think the answer is yes. Since the formation of centralizer commutes with extension of the base field, we have $Z(G)_{\bar{k}}=Z(G)_{\bar{k}}=G_{\bar{k}}$ so $Z(G)=G$. The "smoothness" seems to be unnecessary here so I don't know if my attempt is wrong or not.

vutuanhien
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A straightforward way to see this, is by noting that for two $k$-schemes $X,Y$ with morphisms $$ f:X\to Y \ , \ g:X\to Y$$ $f=g$ iff $f_{\bar{k}}=g_{\bar{k}}$, see for example this question.

As commutativity (by definition) means $m=\sigma\circ m$, where $$\sigma:G\times G\to G$$ is the permutation $\sigma(g,h)=(h,g)$, the result follows.

Notone
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  • Thanks a lot! However I'm still a little bit of confused. Consider the field $k=\mathbb{F}{p}(T)$ and $G=\alpha{p}\rtimes \mu_{p}$, then $G_{\bar{k}}$ is trivial but the group $G$ is not commutative. – vutuanhien Mar 15 '20 at 13:29
  • What do you mean by "it is trivial" and why do you think so? – Notone Mar 15 '20 at 18:19
  • I mean $G_{\bar{k}}=e$. I know that the group of rational points might be trivial but the group still isn't, but here the ground field is algebraically closed so there is an equivalence categories between the set of closed point and the affine scheme (of finite type). Can you help me to point out where I am wrong? Thanks a lot for your help. – vutuanhien Mar 15 '20 at 19:20
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    I see. The problem is that this equivalence of categories is more restrictive. It is not enough to take an affine scheme of finite type, it needs to be also reduced (+maybe more conditions?). Clearly some of the schemes involved here aren"t reduced, so the equivalence doesn't hold – Notone Mar 15 '20 at 20:22