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By Grégory Berhuy

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We then have H 1 (GΩ , L ⊗k Ω) H 1 (GΩ , (L1 ⊗k Ω)× ) × · · · × H 1 (GΩ , (Lr ⊗k Ω)× ), and we use the previous case. If E is a finite dimensional algebra over a field F , we denote by NE/F (x) the determinant of left multiplication by x (considered as an endomorphism of the F -vector space E). 2. Let E = k n , n ≥ 1. If x = (x1 , . . , xn ), then we have NE/k (x) = x1 · · · xn , since the representative matrix of x in the canonical basis of E is simply the diagonal matrix whose diagonal entries are x1 , .

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 .

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 . Now it is obvious from the properties of the determinant that we have NL⊗k Ω/Ω (x ⊗ 1) = NL/k (x) for all x ∈ L.

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