# Algebraic group scheme is a torsor under kernel of a group homomorphism.

In the book by 'Algebraic Groups' by Milne, a right $$G$$- torsor over $$S_0$$ is a scheme $$S$$ faithfully flat over $$S_0$$ together with an action $$S\times_{S_0}G\to S$$ of $$G$$ on $$S$$ such that the map

$$(s,g)\mapsto(s,gs): S\times_{S_0}G\to S\times_{S_0}S$$

is an isomorphism of $$S_0$$-schemes. then There is an exercise that says, let $$G\to Q$$ be a faithfully flat homomorphism of algebraic groups with kernel $$N$$. The action $$G\times_{Q}N\to G$$ of $$N$$ on $$G$$ induces an isomorphism $$G\times_{Q}G \to G\times N$$ and so $$G$$ is a torsor under $$N$$ over $$Q$$.

Why does $$G\times N \simeq G\times_{Q}G$$ say its $$N$$ torsor when it should be $$G\times_{Q}N \simeq G\times_{Q}G$$?

Thank you for the help.

• Stupid question: didn’t the author just forget a subscript $Q$ to the product $G \times N$? Commented Jan 24, 2022 at 22:34
• Thank you. How are the sets isomorphic then? Commented Jan 25, 2022 at 6:53

The homomorphism $$G\rightarrow Q$$ and its kernel $$N$$ are both over a field $$k$$. We have $$G\times_{Q}G\simeq G\times N\simeq G\times_{Q}N_{Q}$$, which says that $$G$$ is a torsor under $$N_{Q}:=N\times Q$$ over $$Q$$ (unadorned products are over Spec$$(k)$$). By an abuse of language, we just say that $$G$$ is a torsor under $$N$$ over $$Q$$.