On the Separability of Certain Extensions of Fields. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:17:43Z http://mathoverflow.net/feeds/question/63559 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63559/on-the-separability-of-certain-extensions-of-fields On the Separability of Certain Extensions of Fields. David 2011-04-30T20:01:58Z 2011-12-03T18:25:56Z <p>Hi,</p> <p>I made <a href="http://mathoverflow.net/questions/62206/criteria-for-preservation-of-a-module-structure-under-extension-of-scalars" rel="nofollow">this question</a> a couple of weeks ago. </p> <p>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 $A$.</p> <p>G.Leuschke gave me <a href="http://www.math.unl.edu/~rwiegand1/ExtVanish/paper.pdf" rel="nofollow">this reference</a>, 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:</p> <ol> <li><p>The next theorem is an exercise from Bourbaki's <em>Algebra II</em>: </p> <p><strong>Theorem 1 (Bourbaki Alg II, V.15. Ex.11):</strong> <em>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</em> ${k_{i} }\subset K$ <em>linearly free over</em> $K^{p}$ <em>and every family</em> ${c_{i}}\subset C$ (<em>with</em> $c_{i}=0$ <em>except for a finite number of subindices) the equality</em> $$\sum_{i} k_{i}c_{i}^{p}=0$$ <em>implies that</em> $c_{i}=0$ <em>(for every $i$).</em></p></li> <li><p>Now, it is not hard to see that this exercise gives the following: </p> <p><strong>Theorem 2:</strong> <em>Let</em> $A\hookrightarrow B$ <em>be a flat extension of algebras over a field of characteristic</em> $p>0$ <em>which are domains, and denote by</em> $A^{1/p}$ <em>(resp, by</em> $B^{1/p}$) <em>the algebra</em> $A$ <em>(resp.</em> $B$) <em>seen as an</em> $A$ <em>(resp $B$) Module via the Frobenius map. Then the field extension</em> $K(A)\to K(B)$ <em>is separable if and only every finite set</em> $a_{1},\dots, a_{n}\in A$ <em>of free elements in</em> $A^{1/p}$ <em>is free in</em> $B^{1/p}$. <em>This happens if and only if the canonical map</em><br> $$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{B^{1/p}}$$ <em>is injective.</em></p></li> <li><p>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 $K(A)\to K(B)$.</p></li> </ol> <p>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> <p>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.</p> http://mathoverflow.net/questions/63559/on-the-separability-of-certain-extensions-of-fields/64355#64355 Answer by David for On the Separability of Certain Extensions of Fields. David 2011-05-09T06:08:26Z 2011-12-03T18:25:56Z <p>Thanks again to prof. Moret-Bailly for taking me out of this delusion.</p> <p>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):</p> <ol> <li><p>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:</p> <p><strong>Theorem 2:</strong> <em>Let</em> $A\hookrightarrow B$ <em>be a flat extension of algebras over a field of characteristic</em> $p>0$ <em>which are domains, and denote by</em> $K(A)^{1/p}$ <em>(resp, by</em> $K(B)^{1/p}$) <em>the algebra</em> $K(A)$ <em>(resp.</em> $K(B)$) <em>seen as an</em> $A$ <em>(resp $B$) Module via the Frobenius map. Then the field extension</em> $K(A)\to K(B)$ <em>is separable if and only every finite set</em> $a_{1},\dots, a_{n}\in K(A)$ <em>of free elements in</em> $K(A)^{1/p}$ <em>is free in</em> $K(B)^{1/p}$. <em>This happens if and only if the composite map</em><br> $$B \otimes_{A} \left\langle a_{1},\dots ,a_{n} \right \rangle_{K(A)^{1/p}}\to B \otimes_{A}\left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}\to \frac{B \otimes_{A}\left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}}{ker(\varphi)}$$</p> <p><em>is injective, where</em> $$\varphi: B \otimes_{A}\left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}\to \left\langle a_{1},\dots ,a_{n} \right\rangle_{K(B)^{1/p}}$$</p> <p><em>is the canonical surjection</em> (given by $b\otimes a\mapsto ba$), <em>and the first map is induced by the inclusion.</em></p></li> <li><p>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).</p></li> <li><p>Now: the injectivity of the map $\varphi$ in the theorem obviously gives you a <em>sufficience</em> 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:</p></li> <li><p>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.</p></li> </ol>