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By Grégory Berhuy

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Example text

Let L be a finite dimensional commutative k-algebra, and let Ω/k be a finite Galois extension. 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.

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). If L/k is any finite Galois subextension of Ω/k with Galois group GL containing all the entries of Q, then Q ∈ SLn (L) and the equality above may be read in Mn (L).

Of course, there is no particular reason to limit ourselves to ZG (M0 ). If G is a group-valued functor, we may consider the disjoint union H 1 (GL , G(L)), L where L/k runs over all finite Galois subextensions of Ω. We now put the following equivalence relation on this set: we say that [α] ∈ H 1 (GL , G(L)) and [α ] ∈ H 1 (GL , G(L )) are equivalent if there exists a finite Galois extension L , L ⊃ L, L ⊃ L such that inf L,L ([α]) = inf L ,L ([α ]). We denote the quotient set by 1 Hind (GΩ , G(Ω)).

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