Sign up ×

I've been reading Waterhouse's book "Introduction to affine group schemes", in part to help prepare myself for an (oral) advanced topic exam in algebraic geometry. There is one exercise in chapter 1 that has been giving me trouble. Let $G$ be an affine group scheme with associated Hopf algebra $A$. The exercise says that I should prove the following Hopf-algebraic fact about $A$ by translating it to a basic fact about the group theory of $G$ : "The map $A \otimes A \rightarrow A \otimes A$ sending $a \otimes b$ to $(a \otimes 1)(\Delta(b))$ is an algebra isomorphism".

The other parts of the exercise give Hopf-algebraic facts corresponding to really basic group theory facts, like $(x^{-1})^{-1} = x$ and $(xy)^{-1} = y^{-1} x^{-1}$ and $1^{-1} = 1$. However, I can't figure out which group-theoretic fact the above corresponds to. It almost seems like it is saying that there is some automorphism of the group corresponding to the above Hopf-algebra isomorphism; however, the only group automorphisms I know that exist in general are the inner ones, and those don't seem to do the job.

share|cite|improve this question

2 Answers 2

up vote 9 down vote accepted

You are confusing algebra isomorphisms with Hopf algebra isomorphisms. The map $A\otimes A \to A\otimes A$ given by $a\otimes b\mapsto \left(a\otimes 1\right)\left(\Delta\left(b\right)\right)$ is an algebra isomorphism but not a Hopf algebra isomorphism in general. So it corresponds not to a group automorphism of $G\times G$, but to an automorphism of the affine scheme $G\times G$. This automorphism is the one that sends $\left(x,y\right)$ to $\left(x,xy\right)$ in terms of points.

share|cite|improve this answer

Consider the map of groups $G\times G \to G\times G$, $(g_1,g_2) \mapsto (g_1,g_1\cdot g_2)$. Clearly, it is an isomorphism, hence induces an isomorphism of the corresponding Hopf algebras, which is given by precisely the map you are asking about.

share|cite|improve this answer
Thank you very much! – Catherine Dec 11 '12 at 5:45
It's not a map of groups, though, unless you assume $G$ to be abelian. – darij grinberg Dec 11 '12 at 5:46
@darij grinberg : Yes, I figured that out an instant ago, so I switched the answer to yours. – Catherine Dec 11 '12 at 5:47
Of course it is not an isomorphism of groups, but still it gives an isomorphism of algebras. – Sasha Dec 11 '12 at 5:56

Your Answer


