By Grégory Berhuy

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version 26 could 2010

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**Extra info for An introduction to Galois cohomology and its applications [Lecture notes]**

**Sample text**

13. The case of infinite Galois extensions In this ultimate paragraph, we would like to indicate quickly how to generalize all this machinery to arbitrary Galois extensions, even infinite ones. I will be extremely vague here, since it can become very quickly quite technical. Let us come back to the conjugacy problem of matrices one last time, but assuming that Ω/k is completely arbitrary, possibly of infinite degree. The main idea is that the problem locally boils down to the previous case. Let us fix M0 ∈ Mn (k) and let us consider a specific matrix M ∈ Mn (k) such that QM Q−1 = M0 for some Q ∈ SLn (Ω).

Therefore, for this particular matrix M , the descent problem may be solved by examining the corresponding element [α(L) ] ∈ H 1 (GL , ZSLn (M0 )(L)). Now if we take another finite Galois subextension L /k such that M ∈ Mn (L ) and Q ∈ SL(L ), we obtain an obstruction [α(L ) ] ∈ H 1 (GL , ZSLn (M0 )(L )). But the fact that M is conjugate or not to M0 by an element of SLn (k) is an intrisic property of M and of the field k, and should certainly not depend on the chosen Galois extension L/k. Therefore, we need to find a way to patch these local obstructions together.

Assume that L ⊗k Ω Ωn for some n ≥ 1. Then we have (1) H 1 (GΩ , Gm,L (Ω)) k × /NL/k (L× ). (1) Proof. The idea of course is to fit Gm,L (Ω) into an exact sequence of GΩ -modules. We first prove that the norm map NL⊗k Ω/Ω : (L ⊗k Ω)× → Ω× ∼ is surjective. For, let ϕ : L ⊗k Ω → Ωn be an isomorphism of Ωalgebras. We claim that we have NL⊗k Ω (x) = NΩn /k (ϕ(x)) for all x ∈ AN INTRODUCTION TO GALOIS COHOMOLOGY 31 L ⊗k Ω. Indeed, if e = (e1 , . . , en ) is a Ω-basis of L ⊗k Ω, then ϕ(e) = (ϕ(e1 ), .

### An introduction to Galois cohomology and its applications [Lecture notes] by Grégory Berhuy

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