2
$\begingroup$

I asked 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.

$\endgroup$
4
  • 1
    $\begingroup$ 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 $z\notin K(A)$. $\endgroup$ May 1, 2011 at 7:43
  • $\begingroup$ 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? $\endgroup$
    – David
    May 1, 2011 at 9:51
  • $\begingroup$ It's the ring of the valuation on $\mathbb{F}_p(t,z^p)$ induced by the $t$-adic valuation on $\mathbb{F}_p((t))$. $\endgroup$ May 1, 2011 at 10:22
  • $\begingroup$ @Laurent: Thanks for your instructive example. I have made a revision (see below). $\endgroup$
    – David
    May 9, 2011 at 6:09

1 Answer 1

2
$\begingroup$

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 won'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 (because $A\to B$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.