# Frobenius base change of etale maps

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?

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Just to clarify, when you write $A \tensor_A B$, you mean $A$ is acting on itself by Frobenius. In other words, if $F : A \to A'$ is the Frobenius on $A$, you are really forming $A' \otimes_A B$, right? Otherwise, if $A$ is acting on itself by the identity, you have $a \otimes b = 1 \otimes(ab)$ which is sent to both $ab^p$ and $a^p b^p$. –  Karl Schwede Jun 14 '12 at 20:25
Yes of course, @Karl –  Hugo Chapdelaine Jun 14 '12 at 20:44

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.

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Dear @Karl, thanks a lot for your proof, it seems to work. Note though that in your proof you are using another definition of etale namely that the map is flat an unramified. You use the flatness (at least) in order to reduce to the case where $B$ is a free $A$ module (a flat finitely presented modules over a local ring is free) and you use the fact that it is unramified to guarantee that $B/mB$ is a separable field extension of $A/mA$. But in my set up, my definition of etale was different since I used the invertibility of the jacobian. –  Hugo Chapdelaine Jun 15 '12 at 18:41
I guess that the equivalence of these 2 definitions is probably not too difficult to prove, what do you think? –  Hugo Chapdelaine Jun 15 '12 at 18:41
@Karl, I'm confused about one point, if $A=F_p[x]/x^p$ and $m=(\bar{x})$ then $Fr_p(m)=m'=0$ and therefore is not a maximal ideal... –  Hugo Chapdelaine Jun 15 '12 at 18:50
Hugo, you are right, but I don't think I'm doing this, or at least I don't intend to (can you point out to me where?). However, $Fr^{−1}_p(m′)=m$ for obvious reasons (this is the only thing I'm intending to use). In terms your first comment, I've seen Mel Hochster give some variant of the above proof, so that's the one I know. I don't see how to get at it directly from the Jacobian criterion. –  Karl Schwede Jun 15 '12 at 21:29
Karl, so I've tried to show that the Jacobian's criterion implies flatness and no ramification but it does not seem as easy as what I first thought... –  Hugo Chapdelaine Jun 16 '12 at 13:44