I'm currently learning algebraic groups with the help of the book of Milne. He makes the following claim :
"every action (by group automorphisms) of a connected algebraic group on a multiplicative group is trivial"
and the proof essentially boils down to the fact that there can be no non-trivial morphism $G\to \mathrm{Aut}(\mu_n)\subseteq \mathbf{Z}/n\mathbf{Z}$ if $G$ is connected.
But I don't understand what allows us to assume that the above morphism is continuous for the discrete topology on the right.
In particular, in characteristic $p$ $\mu_p$ is both connected and multiplicative, and it acts non-trivially on itself [EDIT : not by group automorphisms, so it is a bad example].
So, is an hypothesis missing (like $G$ being smooth) ?
Here are the definitions used :
- an algebraic group is a group scheme of finite type over a field
- a multiplicative group is a group which becomes diagonalizable (ie, a subgroup of some $\mathbb{G}_m^k$) over a field extension
- $\mu_n = \mathrm{spec}\ k[T,T^{-1}]/(T^n-1)$
Thank you for reading me!