By V. F. Demianov, Aleksandr Moiseevich Rubinov

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Tn+1 ), y = 0 is a solution of L. The induction hypothesis implies that the algebraic closure k(t), with t = t1 , contains solutions of the Riccati equation of L. It t is algebraic over k, then we are done. 43, the Picard-Vessiot ﬁeld of L over k(t) which is denoted by Kk(t) or K(t). Further Kk(t) denotes the Picard-Vessiot ﬁeld of L over k(t). Let V ⊂ K denote the solution space of L (in K and also in Kk(t)). Let a y ∈ V, y = 0 be given such that yy is algebraic over k(t). For any σ ∈ Gal(K/k) the element σ(y) has the same property.

2) The Lie algebra of G coincides with the Lie algebra of the derivations of L/k that commute with the derivation on L. (3) The ﬁeld LG of G-invariant elements of L is equal to k. Proof. An intuitive proof of (1) and (2). 1 ]/q, where q is a maximal difL is the ﬁeld of fractions of R := k[Xi,j , det ferential ideal. 26 one can identify G with the group of matrices 1 ], given M ∈ GLn (C) such that the automorphism σM of R0 := k[Xi,j , det by (σXi,j ) = (Xi,j )M , has the property σM (q) ⊂ q. One has to verify that the property σM (q) ⊂ q deﬁnes a Zariski closed subset of GLn (C).

Let (Xi,j ) denote an n × n-matrix of indeterminates and let “det” 1 ] denote the determinant of (Xi,j ). t. the ele1 ] with the derivation, ment “det”. Consider the diﬀerential ring R0 = k[Xi,j , det extending the one of k, given by (Xi,j ) = A(Xi,j ). 1 shows the existence and unicity of such a derivation. Let I ⊂ R0 be a maximal diﬀerential ideal. Then R = R0 /I is easily seen to be a Picard-Vessiot ring for the equation. 2. Let R1 , R2 denote two Picard-Vessiot rings for the equation. Let B1 , B2 denote the two fundamental matrices.