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By Grégory Berhuy

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

Example text

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 , . . , 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.

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.

But the fact that M is conjugate or not to M0 by an element of SLn (k) is an intrisic property of M and of the field k, and should certainly not depend on the chosen Galois extension L/k. Therefore, we need to find a way to patch these local obstructions together. There are two ways to proceed. First of all, notice that if L1 /k and L2 /k are two finite Galois extensions such that L1 ⊂ L2 , then the maps GL2 → GL1 , σ2 → (σ2 )|L1 and ZSLn (M0 )(L1 ) → ZSLn (M0 )(L2 ) are compatible, so we have a well-defined map inf L1 ,L2 : H 1 (GL1 , ZSLn (M0 )(L1 )) → H 1 (GL2 , ZSLn (M0 )(L2 )), 34 ´ GREGORY BERHUY which sends [α] to the class of the cocycle GL2 → ZSLn (M0 )(L2 ), σ2 → α(σ2 )|L .

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