By V. Bargmann
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Yn ) is invertible. Furthermore, the matrix A = W W −1 is left invariant by G/H and so has entries in k. 22 implies that LH is a Picard-Vessiot ﬁeld. 41 Let G be a connected solvable linear algebraic group. In this exercise the fact that any G-torsor over k is trivial, will be used. 51. 1. Picard-Vessiot extensions with Galois group (Ga )r . Suppose that K is a Picard-Vessiot extension of k with Galois group (Ga )r . Show that there exist t1 , . . , tr ∈ K with ti ∈ k such that K = k(t1 , . .
Let Z ⊂ Hk ⊂ Let q ⊃ (I) be a maximal diﬀerential ideal of k[Xi,j , det GLn,k be the reduced, irreducible subspace deﬁned by q. For any M in the diﬀerential Galois group and any B ∈ Z(k) one has BM ∈ Z(k) and thus M ∈ H(k). Further H(k) ∩ GLn (C) = H(C). (2) If A˜ ∈ g(k), then the proof of part (1) yields that Gk is its torsor and BGk is the torsor of y = Ay. If Z is a trivial torsor, then Z = BGk for some B ∈ Z(k). The equivalent ˜ obtained by the substitution y = Bv, has the diﬀerential equation v = Av, 1 ] of Gk , where (Xi,j ) = B(Zi,j ), is a property that the ideal q˜ ⊂ k[Zi,j , det maximal diﬀerential ideal.
S y ∈ V are linearly independent over C. The vector space W ⊂ V spanned by σ1 y, . . , σs y is clearly invariant under the action of Gal(K/k). Let f (s) +as−1 f (s−1) +· · ·+a0 f by the unique diﬀerential equation M over K with M (σi y) = 0 for i = 1, . . , s. For any σ ∈ Gal(K/k), the transformed equation σM has the same space W as solution space. Hence σM = M and we conclude that M has coeﬃcients in k. We replace now L by M . Consider the liouvillian ﬁeld extension k(t, u1 , . . , us , σ1 y, .