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By Grégory Berhuy

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

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The exactness of the sequence above then gives the desired result. 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].

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 , . . , xn . 3. If L/k is a finite dimensional commutative k-algebra, (1) we denote by Gm,L the functor defined by (1) Gm,L (K) = {x ∈ (L ⊗k K)× | NL⊗k K/K (x) = 1}, for every field extension K/k. (1) We now compute H 1 (GΩ , Gm,L (Ω)) in a special case. 4. Let L be a finite dimensional commutative k-algebra, and let Ω/k be a finite Galois extension.

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. Set L = k[X]/(χ), so that we have an isomorphism of k-algebras L → k[M0 ], P → P (M0 ), 32 ´ GREGORY BERHUY which induces in turn a Galois equivariant isomorphism of Ω-algebras ∼ f : L ⊗k Ω → Ω[M0 ], X ⊗ λ → λM0 .

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


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