By Grégory Berhuy

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version 26 might 2010

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

**Example text**

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 ), . . , ϕ(en )) is a Ω-basis of Ωn , and we have easily Mat( ϕ(x) , ϕ(e)) = Mat( x , e). The desired equality then follows immediately.

The exactness of the sequence above then gives the desired result. 5. The isomorphim above works as follows: If a ∈ k × /NL/k (L× ), pick z ∈ L ⊗k Ω such that a = NL⊗k Ω/Ω (z) (this is possible since NL⊗k Ω/Ω is surjective). Then the corresponding cohomology class is represented by the cocycle (1) α : GΩ → Gm,L (Ω), σ → z −1 σ·z. (1) Conversely, if [α] ∈ H 1 (GΩ , Gm,L (Ω)), pick z ∈ (L ⊗k Ω)× such that ασ = z −1 σ·z for all σ ∈ GΩ . Then a = NL⊗k Ω/ (z) lies in fact in k × , and a ∈ k × /NL/k (L× ) is the class corresponding to [α].

Then e = (1 ⊗ 1, α ⊗ 1, . . , αn−1 ⊗ 1) is a n−1 n−1 i Ω-basis of L ⊗k Ω. Let x = λi X i . α ⊗ λi ∈ L ⊗k Ω, and let P = i=0 i=0 Clearly, we have x = P ( α⊗1 ). Now the matrix of α⊗1 in the basis e is easily seen to be Cχ = M0 , and so the matrix of x in the basis e is P (M0 ) = f (x). Therefore det( x ) = det(f (x)), and we are done. We then get ZSLn (M0 )(Ω) (1) Gm,L (Ω) as a Galois module. e. χ has only simple roots in an algebraic closure of k) and that Ω/k is a Galois extension containing all the roots of χ.