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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]
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. As a final remark, we should point out that we could define the pointed set H 1 (GΩ , G(Ω)) dropping the continuity condition, but the equality above does not hold anymore, and therefore one cannot always do patching, so this definition is not really suitable.
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]. Now let us go back to the conjugacy problem of matrices. Assume that M0 = Cχ ∈ Mn (k) is a companion matrix of some monic polynomial χ ∈ k[X] of degree n ≥ 1. In this case, it is known that every matrix commuting with M0 is a polynomial in M0 , so ZG (M0 )(Ω) = Ω[M0 ] ∩ G(Ω). Moreover, the minimal polynomial and the characteristic polynomial are both equal to χ.
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]. Now let us go back to the conjugacy problem of matrices.