Algebraic Aspects of Linear Differential and Difference by Hendriks P.A.

By Hendriks P.A.

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In this case there are exactly two solutions u; u~ 2 k((t)) of the Riccati equation. The rst solution u satis es v(au) = v(u(u)) < v(b) and v(u) = v(a) and the other solution u~ satis es v(au~) = v(b) < v(~u(~u)) and v(~u) = v(b) v(a). We can compute succesively the coecients of u and u~ by simply solving linear equations with coecients in k. Hence no eld extension of k is needed. (b) v(b)  2v(a). If v(b) is odd then there is clearly no solution of the Riccati equation. If v(b) is even and u 2 k((t)) satis es the Riccati equation then we have v(b) = v(u(u))  v(au) and v(u) = 21 v(b).

2 The main results in [BBH88] are the next two theorems. 7 Let M be a D-module. Suppose dimK M = n  2. Further let G = DGal(M ). Then the following statements are equivalent: 1. M is simple and homogeneous algebraic Siegel normal. 2. 8 Let M be a D-module and let G = DGal(M ). Then the following statements are equivalent: 36 1. M is homogeneous algebraic Siegel normal. 2. M = ( L Mi ) L( L Nj ), where the Mi are are non-cogredient and noni=1 j =1 contragredient simple linear Siegel normal D-modules with dimK Mi  2 and the Nj are one dimensional D-modules satisfying the following condition: r s S k N1    S ks Ns ' Ntriv Ps implies either k ; : : : ; k = 0 or k 6= 0.

If Mi = Mk + Ml with 0 < k; l < i, then de ne Ei := Ei 1. Obviously, Ei satis es the conditions i) and ii) if Ei 1 satis es these conditions. If Mi 6= Ml +Mk for all k; l with 0 < k; l < i, then there exists a unique h 2 f0; : : : ; ig such that Mh  Mi and Mi=Mh is a simple D-module. Suppose that EMi = EMh [ E~i (disjunct union) is a K -base of Mi , where EMh  Ei 1 is a K -base of Mh. De ne Ei := Ei 1 [ E~i . Of course Ei satis es condition ii). Now 40 we assume that Ei doesn't satisfy condition i) and derive a contradiction.

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