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Then e = (1 ⊗ 1, α ⊗ 1, . . , αn−1 ⊗ 1) is a n−1 n−1 i Ω-basis of L ⊗k Ω. Let x = λi X i . α ⊗ λi ∈ L ⊗k Ω, and let P = i=0 i=0 Clearly, we have x = P ( α⊗1 ). Now the matrix of α⊗1 in the basis e is easily seen to be Cχ = M0 , and so the matrix of x in the basis e is P (M0 ) = f (x). Therefore det( x ) = det(f (x)), and we are done. We then get ZSLn (M0 )(Ω) (1) Gm,L (Ω) as a Galois module. e. χ has only simple roots in an algebraic closure of k) and that Ω/k is a Galois extension containing all the roots of χ.

1, we then get H 1 (GΩ , ZGLn (M0 )(Ω)) = 1, as expected. Now let us identify ZSLn (M0 )(Ω). Claim: We have det(f (x)) = NL⊗k Ω/Ω (x) for all x ∈ L ⊗k Ω. To see this, set α = X ∈ L. Then e = (1 ⊗ 1, α ⊗ 1, . . , αn−1 ⊗ 1) is a n−1 n−1 i Ω-basis of L ⊗k Ω. Let x = λi X i . α ⊗ λi ∈ L ⊗k Ω, and let P = i=0 i=0 Clearly, we have x = P ( α⊗1 ). Now the matrix of α⊗1 in the basis e is easily seen to be Cχ = M0 , and so the matrix of x in the basis e is P (M0 ) = f (x). Therefore det( x ) = det(f (x)), and we are done.

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. 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).

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