# An introduction to Galois cohomology and its applications by Grégory Berhuy

By Grégory Berhuy

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There are two ways to proceed. First of all, notice that if L1 /k and L2 /k are two finite Galois extensions such that L1 ⊂ L2 , then the maps GL2 → GL1 , σ2 → (σ2 )|L1 and ZSLn (M0 )(L1 ) → ZSLn (M0 )(L2 ) are compatible, so we have a well-defined map inf L1 ,L2 : H 1 (GL1 , ZSLn (M0 )(L1 )) → H 1 (GL2 , ZSLn (M0 )(L2 )), 34 ´ GREGORY BERHUY which sends [α] to the class of the cocycle GL2 → ZSLn (M0 )(L2 ), σ2 → α(σ2 )|L . 1 (L) (L ) If we examine the classes [α ] and [α ] above, one can check that they are both equal to [α(LL ) ] when we ”push them in LL ”.

The desired equality then follows immediately. Now for λ ∈ Ω× , set xλ = ϕ−1 ((λ, 1, . . , 1)). 2 then yield NL⊗k Ω/Ω (xλ ) = NΩn /Ω ((λ, 1, . . , 1)) = λ. Therefore NL⊗k Ω/Ω is surjective and we have an exact sequence of GΩ modules 1 / / (1) Gm,L (Ω) / (L ⊗k Ω)× Ω× / 1, where the last map is given by the norm NL⊗Ω/Ω . It is known that the condition on L implies in particuliar that L is the direct product of finitely many finite field extensions of k. 1 yield the exact sequence (1) (L ⊗k 1)× → k × → H 1 (GΩ , Gm,L (Ω)) → 1, the first map being NL⊗k Ω/Ω .

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 ), .