Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

First let me give a general form of my question, and then I'll give some motivation and a more specific version of it. Let $K/k$ be a Galois extension of fields with Galois group $G$, and let $X$ be some object defined over $k$. One can endow the group of automorphisms $\mathrm{Aut}(X)$ with the structure of a $G$-module and then use the Galois cohomology set $H^1(G, \mathrm{Aut}(X) )$ to parameterize the $K/k$-twists of $X$; i.e., the $k$-objects that become isomorphic to $X$ over the field $K$. But this theory seems to be useless if we instead take $K / k$ to be a purely inseparable extension. Does there exist a cohomology theory that allows one to describe twists for inseparable extensions?

Here is the motivation for my question. Arnaud Beauville used Galois cohomology to classify up to $k$-conjugacy all finite subgroups $G$ of $\mathrm{PGL}_2(k)$ such that $\mathrm{char}(k) \nmid |G|$. See here for the arXiv version of his paper, and see here for another thread related to this question.) I am trying to extend his description to the case $\mathrm{char}(k) \mid |G|$. One can work exclusively with separable extensions if the characteristic is different from 2, and so Galois cohomology appears to give a perfectly satisfactory answer. But something odd happens in characteristic 2.

Rather than describe the general picture, let me just give an example. Let $k$ be a separably closed field of characteristic 2. One can parameterize the $k$-conjugacy classes of cyclic groups of order~2 in $\mathrm{PGL}_2(k)$ as follows. Any such group $G$ fixes a unique point of $\mathbb{P}^1(k^{(2)})$, where $k^{(2)}$ denotes the compositum of all (inseparable) quadratic extensions of $k$ (inside some fixed algebraic closure $k^a$). Write $k(G)/k$ for the extension given by adjoining the coordinates of this fixed point. The field $k(G)$ depends only on the $k$-conjugacy class of $G$. Let $\mathbb{E}_2(k)$ denote the set of extensions of $k$ of degree at most 2 (again, inside $k^a$). Then the association $G \mapsto k(G)$ gives a bijection $$ \{k\text{-conjugacy classes of cyclic groups of order 2} \} \rightarrow \mathbb{E}_2(k). $$

We can identify $k^\times / (k^\times)^2$ with the set $\mathbb{E}_2(k)$ via the association $\tau \mapsto k(\sqrt{\tau})$. Under this identification, the inverse to the above bijection $$ k^\times / (k^\times)^2 \rightarrow \{k\text{-conjugacy classes of cyclic groups of order 2} \} $$ can be given by $\tau \mapsto \left\{ \left(\begin{smallmatrix} 1 \\ & 1 \end{smallmatrix} \right), \left(\begin{smallmatrix} & \tau^2 \\ 1 & \end{smallmatrix}\right) \right\}$. (A similar strategy shows that $k$-conjugacy classes of dihedral subgroups of fixed order $2n$ with $n$ odd are in bijection with $k^\times / (k^\times)^2$.)

The above parameterization of $k$-conjugacy classes looks remarkably like the kind of answer one gets from Galois cohomology. So is there a cohomological explanation?

share|improve this question
Yes, it's called flat cohomology. –  anon Oct 26 '11 at 23:49
@anon - Can you point me to a reference? –  Xander Faber Oct 27 '11 at 2:27
I'm only passingly familiar with the book, but Milne's Arithmetic Duality Theorems has a chapter on flat cohomology. jmilne.org/math/Books/index.html –  B R Oct 27 '11 at 3:16
Another key word is Amitsur's (co)homology. –  Mikhail Bondarko Oct 27 '11 at 17:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.