Automorphism of tensor product of Hopf algebras

Question

Let $A$ be a commutative Hopf algebra over the commutative ring $R$, that is we have comultiplication $\Delta:A\to A\otimes_{R}A$, counit $\epsilon: A\to R$ and coinverse (or antipode) $S: A\to A$. I want to prove that the map $a\otimes b\mapsto (a\otimes 1)\cdot \Delta(b)$ is an automorphism of $R$ algebras of $A\otimes_{R}A$. I found a similar exercise on Milne's notes on affine group schemes, though my context is just that of commutative algebra. My first idea was to prove directly injectivity and surjectivity, but I'm completely stuck. Can anyone give me some hints on which is the correct way to follow?

Answer 1

You want to write down an inverse explicitly using the other operations.

Since $A$ is commutative $G = \text{Spec } A$ is an affine group scheme so we can think about this map in terms of the dual map $G \times G \to G \times G$ of affine group schemes, which is easier to think about. Its action on a pair of $S$-points $g, h : A \to S$ of $G(S)$, for any $R$-algebra $S$, sends $(g, h) \in G(S) \times G(S)$ to some other pair of points in $G(S) \times G(S)$, which we can calculate by restricting the given map to the first and then the second copy of $A$. Restricting to the first copy gives

$$a \mapsto (a \otimes 1) \Delta(1) = a \otimes 1$$

which reveals that the first component of our map is $(g, h) \mapsto g$. Restricting to the second copy gives

$$b \mapsto (1 \otimes 1) \Delta(b) = \Delta(b)$$

which reveals that the second component of our map is $(g, h) \mapsto gh$. Altogether, our map is

$$G(S) \times G(S) \ni (g, h) \mapsto (g, gh) \in G(S) \times G(S).$$

This is much easier to think about: its inverse is just $(g, h) \mapsto (g, g^{-1} h)$.