I have question which arises from the answer of my former thread: https://mathoverflow.net/questions/324887/why-is-for-a-group-scheme-of-finite-type-smooth-resp-irreducible-equivale
Let $G/K$ be a group scheme of finite type. Fix the algebraic closure $\bar{K}$ of $K$ and consider the resulting Galois group ${\rm Gal}(\bar K/K)$.
As @Piotr Achinger stated there exist an "canonical" action on $\bar G = G \otimes \bar{K}$ by ${\rm Gal}(\bar K/K)$.
My question is what kind of action is here meant?
My two possible guesses/ candidates:
- The action on "points" $\bar{G}(\bar{K})= Hom(Spec(\bar{K}), \bar{G})$ via composing $\phi \mapsto \phi \circ g$ for a $g \in {\rm Gal}(\bar K/K)$
or
- the action via "base change": namely that $g$ induces automorphism on $\bar{G}$ via $id⊗g$?
Which of these two is here considered?