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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?

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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. – B R Oct 27 '11 at 3:16
Another key word is Amitsur's (co)homology. – Mikhail Bondarko Oct 27 '11 at 17:30

1 Answer 1

I cannot answer your question, but point to the right algebraic framework in my opinion:

There is a well worked out classical (but somewhat underestimated) theory of inseparable Galois extensions. It is Jacobson 1964 I think and is also in Jacobsons book. One just has to get used to the fact that the Galois group is replaced by a (p-restricted) Lie algebra. In my opinion it gets much more conceptual nowadays by looking at Hopf Galois theory, which contains both, but this goes to far for a first answer.

Example: Let your $k=\bar{\mathbb{F}}_2[x]$ and $K=\bar{\mathbb{F}}_2[\sqrt{x}]$ be a quadratic extension. It is purely inseperable. But there is a "Galois derivation" $\partial:K\to K$ with the properties $$\partial\sqrt{x}=1$$ $$\partial(ab)=a\cdot \partial(b)+\partial(a)\cdot b$$ $$char(k)=2\Rightarrow\;\partial x= \sqrt{x}\cdot 1+1\cdot\sqrt{x}=0$$ So it acts trivial on $k$ (meaning for Lie algebras that it act as $0$ ) and is therefor $k$-linear map $\partial(ka)=k\partial(a)$. There is also a normal basis theorem, fairily trivial in this quadratic case.

Hence the "Galois Lie algebra" is the very simple abelian Lie algebra $\partial k$, p-restricted by $\partial^2=0$; respectively by the Hopf algebra $k[\partial]/(\partial^2), \Delta(\partial)=1\otimes \partial+\partial\otimes 1$ which exists as it is only in characteristic 2.

For your question, I suggest at look at cohomology of this Lie algebra with coefficients in a module. For the beginning, one should be able to describe twisted objects over $K$ as you explained.

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Since Jacobson's useful older work has dropped off the radar screen for most people (including myself), it would help to give more precise references. – Jim Humphreys Feb 4 at 14:09
P.S. Jacobson's volume III (1964) in his old algebra series is one reference, but his interest in inseparability goes back to much earlier papers such as and (the first freely available online, the second via JSTOR). – Jim Humphreys Feb 4 at 14:37

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