# Basic question about affine group schemes

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.

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## 2 Answers

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.

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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.

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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