By Hendriks P.A.

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**Extra info for Algebraic Aspects of Linear Differential and Difference Equations**

**Example text**

Then the numbers 1) ;:::;p; ;:::;q ( ); : : : ; (p ;:::; p ; ;:::;q ( ) 1 1 1 1 are algebraic independent with an eective measure of algebraic independence. 44 Proof. Consider the homogeneous linear dierential equation Yq ( + r(i 1)) i=1 Yp r z ( + ri) y i=1 =0 The q q system corresponding to this dierential equation is simple and homogeneous algebraic Siegel normal. 5 in [BBH88]. Hence the (q + 1) (q + 1) system of dierential equations corresponding to (y) is homogeneous algebraic Shidlovskii irreducible.

C CA = BB( ... CA T ; (U ) = B ( . (un1) (unn) un1 unn where T 2 Gl(n; Q ). So the elements of the dierential Galois group act as Q -linear maps on the space of solutions V = fc1 u1 + + c1 un j c1 ; : : : ; cn 2 Q g. (ui = (u1i; : : : ; uni)t 2 Rn). But even a stronger statement holds. 5 G = DGal(R=K^ ) is a linear algebraic group over the eld of . constants Q If G is an algebraic subgroup of Gl(n; Q ), then we denote by G(K^ ) the sub- group of Gl(n; K^ ) which is dened by the same equations.

It seems possible to make a proof of this without the explicit knowledge of P1 . 4, one nds a form (M ) of L over k and a quadratic extension k0 k such that the Galois group of k0=k permutes the corresponding polynomials fP1; P2 g of M . Let k0 = k() with 2 2 k and 62 k. Then one can take for the automorphism given by (x) = 1+22x and suitable (@x). A calculation shows that M has the form 2 2 @x2 + 9(1 3242x2 )2 4(1 342 x2) Order three with a 1-reducible Galois group Let T 3 + pT + q 2 k[T ] denote an irreducible polynomial with Galois group S3.

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

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