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Hi,

I made this question a couple of weeks ago.

The question arose while looking for a criterion of separability for extensions of fraction fields $K(A)\to K(B)$ induced by a faithfully flat morphism $A\to B$ between algebras over a field of positive characteristic which are domains with $A$ Noetherian, and such that the induced morphism $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$.

G.Leuschke gave me this reference, whose results (concretely, the Theorem 1.8) allows one to conclude that the separability is guaranteed when $B$ is also Noetherian by the argument that follows:

  1. The next theorem is an exercise from Bourbaki's Algebra II:

    Theorem 1 (Bourbaki Alg II, V.15. Ex.11): Let $K$ be a field of characteristic $p>0$ and let $C$ be a $K$-algebra. Then $C$ is separable if and only if for every family of elements $\{k_{i} \}\subset K$ linearly free over $K^{p}$ and every family $\{c_{i}\}\subset C$ (with $c_{i}=0$ except for a finite number of subindices) the equality $$\sum_{i} k_{i}c_{i}^{p}=0$$ implies that $c_{i}=0$ (for every $i$).

  2. Now, it is not hard to see that this exercise gives the following:

    Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $A^{1/p}$ (resp, by $B^{1/p}$) the algebra $A$ (resp. $B$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in A$ of free elements in $A^{1/p}$ is free in $B^{1/p}$. This happens if and only if the canonical map
    $$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ is injective.

  3. Here comes Frankild et al's paper: when $A, B$ are Noetherian, $A\to B$ is faithfully flat and the induced map $A/m\to B/mB$ is bijective for every maximal ideal $m$ of $A$, an easy application of the local-global principle together with Theorem 1.8 in the paper guarantees the separability of $K(A)\to K(B)$.

I thought the criterion was extendible to the case in which $B$ is not Noetherian, but there were mistakes in my argument. Does anybody have an idea on how to prove (or refute) the corresponding affirmation in such a case? (Frankild et al's paper can give you some hints, but I don't write them down in order to avoid bias).

P.S: The argument can be extended easily, I think, to the case in which the localization of $B$ at every maximal ideal $\eta$ is $\eta$-adically separated.

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Something's wrong. There are discrete valuation rings $A$ of char. $p>0$ such that $K(\widehat{A})$ is inseparable over $K(A)$. For instance, choose some $z\in\mathbb{F}_p[[t]]$ which is transcendental over $\mathbb{F}_p(t)$. Then take $A=\mathbb{F}_p[[t]]\cap\mathbb{F}_p(t,z^p)$. Then $A$ is a DVR containing $\mathbb{F}_p[t]$, with completion $\mathbb{F}_p[[t]]$, but we have $z\in\widehat{A}$, $z^p\in A$ and $z\notin K(A)$. –  Laurent Moret-Bailly May 1 '11 at 7:43
    
Thanks. I'll think about it. Right now I have only a question: (I hope this don't be too obvious, but) why is A a DVR? Why is it actually Noetherian? –  David May 1 '11 at 9:51
    
It's the ring of the valuation on $\mathbb{F}_p(t,z^p)$ induced by the $t$-adic valuation on $\mathbb{F}_p((t))$. –  Laurent Moret-Bailly May 1 '11 at 10:22
    
@Laurent: Thanks for your instructive example. I have made a revision (see below). –  David May 9 '11 at 6:09

1 Answer 1

Thanks again to prof. Moret-Bailly for taking me out of this delusion.

I will explain where the mistakes were, for there could be someone curious about it (I hope there don't be new mistakes after my revision):

  1. All the mistakes come from the ``translation'' of Bourbaki's exercise to the case of fraction fields as in the hypotheses of the theorem. The (right) theorem is:

    Theorem 2: Let $A\hookrightarrow B$ be a flat extension of algebras over a field of characteristic $p>0$ which are domains, and denote by $K(A)^{1/p}$ (resp, by $K(B)^{1/p}$) the algebra $K(A)$ (resp. $K(B)$) seen as an $A$ (resp $B$) Module via the Frobenius map. Then the field extension $K(A)\to K(B)$ is separable if and only every finite set $a_{1},\dots, a_{n}\in K(A)$ of free elements in $K(A)^{1/p}$ is free in $K(B)^{1/p}$. This happens if and only if the composite map
    $$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right \rangle_{K(A)^{1/p}}\to B \otimes_{A}\left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}\to \frac{B \otimes_{A}\left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}}{ker(\varphi)}$$

    is injective, where $$ \varphi: B \otimes_{A}\left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}$$

    is the canonical surjection (given by $b\otimes a\mapsto ba$), and the first map is induced by the inclusion.

  2. The first mistake (to take elements in $A$ instead of $K(A)$) is independent from what follows in the argument. The fatal error, concerning to the injectivity stuff, came from confusing the modules $\left\langle a_{1},\dots,a_{n}\right\rangle_{K(A)^{1/p}}$ and $\left\langle a_{1},\dots,a_{n}\right\rangle_{K(B)^{1/p}}$ (this is a dumb mistake. I fell in it, perhaps, by thinking all the time in terms of generators).

  3. Now: the injectivity of the map $\varphi$ in the theorem obviously gives you a sufficience criterion to that of the composite map ('cause $A\to B$ is flat) but this condition is, as you see, only suficcient, and even worst:

  4. If, after these corrections you keep the argument (with the hope it gives still a helpful criterion of separability), Frankild et al's theorem 1.9 implies that the "corrected'' case will hold if and only if $A\to B$ is an isomorphism. This vanishes the hope of rescuing a decent thing from all of this.

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