By Grégory Berhuy

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The case of infinite Galois extensions In this ultimate paragraph, we would like to indicate quickly how to generalize all this machinery to arbitrary Galois extensions, even infinite ones. 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).

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. 13. The case of infinite Galois extensions In this ultimate paragraph, we would like to indicate quickly how to generalize all this machinery to arbitrary Galois extensions, even infinite ones.

En )) is a Ω-basis of Ωn , and we have easily Mat( ϕ(x) , ϕ(e)) = Mat( x , e). The desired equality then follows immediately. Now for λ ∈ Ω× , set xλ = ϕ−1 ((λ, 1, . . , 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.

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