Let $A$ be a characteristic $p>0$ commutative ring. Let $B$ be a finitely presented etale $A$ algebra i.e. $$ B=A[x_1,\ldots,x_n]/(f_1,\ldots,f_n), $$ with $det(\frac{\partial f_i}{\partial x_j})\in B^{\times}$. Consider the multiplication map \begin{align*} m:A\otimes_A B\rightarrow B\\\ a\otimes b\mapsto ab^p \end{align*} Note here that for $a\in A$ we have the tensor relation $1\otimes ab=a^p\otimes b$.
Q: How does one prove directly that $m$ is an isomorphism of rings?
Ok, I think one can quickly reduce to the case where $A$ is local and also that $B$ is $A$-free of finite rank (I need to think about this latter assumption a little, but probably by completion we can do this).
But granting these assumptions, we can proceed as follows (I think):
Set $A \to A'$ to be the absolute Frobenius on $A$ and likewise $B \to B'$ the Frobenius on $B$. The primes will just help me distinguish source and target.
Then we want to show that the obvious map: $$\Psi : A' \otimes_A B \to B'$$ is an isomorphism.
Now, $B'$ is a free $A'$ module, likewise so is $A' \otimes_A B$. So we are trying to establish that a certain map of free $A'$-modules is an isomorphism. First we show surjectivity, so we mod out by the maximal ideal $m' \subseteq A'$. Then we obtain: $$\Phi : A'/m' \otimes_{A/m} B/mB \cong A'/m' \otimes_A B \to B'/m'B'.$$ Here the isomorphism comes from the fact that $Fr^{-1}(m') = m$.
Now, $A/m$ is a field, and $B/mB$ is a separable extension (by the etale hypothesis). $A'/m'$ is a purely inseparable extension and so we easily see (by using that the extensions are linearly disjoint) that $B'/m'B'$ is identified naturally with $A'/m' \otimes_{A/m} B/mB$.
It follows that $\Phi$ is an isomorphism. Thus $A' \otimes_A B \to B$ is surjective by Nakayama's Lemma. But $A' \otimes_A B$ and $B'$ have the same rank as $A'$-modules by the above argument. It follows that then $\Psi$ is a surjection between finite free modules of the same rank. Thus $\Psi$ is an isomorphism.
