By Grégory Berhuy

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version 26 may possibly 2010

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

**Example text**

AN INTRODUCTION TO GALOIS COHOMOLOGY 35 1 We may then define a cohomology set Hcont (GΩ , G(Ω)) as in the finite case, but we ask for continuous cocycles, where GΩ is endowed with the Krull topology and G(Ω) is endowed with the discrete topology. The nice thing is that we have 1 (GΩ , G(Ω)) Hcont 1 (GΩ , G(Ω)). Hind Therefore, we have achieved what we wanted, that is finding a way to patch a family of cohomology classes together. Moreover, as long as (1) and (2) are satisfied, all the previous results generalize to arbitrary Galois extensions.

The field Ω = k(i, d) (where i is a square root of −1) contains all the roots of χ and Ω/k is Galois, with Galois group isomorphic to Z/2Z or (Z/2Z)2 (depending on the fact that −d is whether or not a square in k × ). i 0 Moreover, we have QM Q−1 = M0 ∈ M2 (Ω), with Q = , so 0 −i M and M0 are conjugate by an element of SL2 (Ω). However, they are not conjugate by an element of SL2 (k) in general. To see this, let us compute the class in k × /NL/k (L× ) corresponding AN INTRODUCTION TO GALOIS COHOMOLOGY 33 to the conjugacy class of M .

This implies that α = e σ·e−1 for all σ ∈ GΩ , meaning that α is trivial. Let us go back to the general case. If L = L1 × · · · × Lr , then we have an isomorphism of GΩ -modules, (L ⊗k Ω)× (L1 ⊗k Ω)× × · · · × (Lr ⊗k Ω)× . 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).

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

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