By Grégory Berhuy

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1) Via the isomorphism H 1 (GΩ , Gm,L (Ω)) H 1 (GΩ , ZSLn (M0 )(Ω)) induced by f∗ , the cohomology class [α(Q) ] correspond to the cohomology class of the cocycle (1) β (Q) : GΩ → Gm,L (Ω), σ → (1 ⊗ i)−1 σ·1 ⊗ i. Now NL⊗Ω/Ω (1 ⊗ i) = (1 ⊗ i)2 = −1, and thus the conjugacy class of M corresponds to the class of −1 in k × /NL/k (L× ). In particular, M and M0 are conjugate over k if and only if −1 ∈ NL/k (L× ). Therefore, to produce counterexamples, one may take for k any subfield of R and d < 0, as we did in the introduction.

These two conditions say that an element g ∈ G(Ω) ‘comes from’ an element of G(L) for some finite Galois subextension L/k of Ω/k, and that the action of GΩ is in fact the same as the action of GL on g when viewed as an element of G(L). One can show that any functor G defined by a finite set of polynomial equations with coefficients in k satisfy these assumptions. For example, this is exactly what happens in the case of matrices. 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.

I will be extremely vague here, since it can become very quickly quite technical. 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).

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