By Dung Trang Le

Blending basic effects and complicated equipment, Algebraic method of Differential Equations goals to accustom differential equation experts to algebraic equipment during this niche. It offers fabric from a faculty geared up through The Abdus Salam foreign Centre for Theoretical Physics (ICTP), the Bibliotheca Alexandrina, and the overseas Centre for natural and utilized arithmetic (CIMPA).

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2) If M = D/DP , to describe R HomD (M, N ) we take the free resolution of M −1 ·P 0 0 → D −→ D → M = D/DP → 0, −1 ·P 0 M • = · · · → 0 → D −→ D → 0 → · · · , and R HomD (M, N ) = HomD (M • , N ) = 0 HomD (·P,N ) 1 · · · → 0 → HomD (D, N ) −−−−−−−−→ HomD (D, N ) → 0 → · · · = 0 P 1 ··· → 0 → N − → N → 0 → ··· . In particular, HomD (M, N ) = Ext0D (M, N ) = h0 R HomD (M, N ) = ker(P : N → N ), Ext1D (M, N ) = h1 R HomD (M, N ) = coker(P : N → N ) and ExtiD (M, N ) = 0 for all i = 0, 1. e. ExtiD (M, N ) = 0 for all i = 0.

2) For any injection ι : O → U, the left D-module FO is generated by ι. (3) Let ι : O → U be an injection and (E, F, U, V ) = im ι ⊂ U. Let us define I = {P ∈ D | P f = 0, ∀f ∈ E, P g = 0, ∀g ∈ F }. Prove that I is a left maximal ideal of D and E(I) = E, F (I) = F . (4) FO is a simple regular holonomic left D-module. 2. With the above notations, FO is a regular holonomic left D-module for any object O of C. Moreover, lg FO ≤ lg O. Proof. The proof goes easily by induction on the length of the object O.

A . +        Fq−2   ∂Fq−2  0 z αq−1 −αq Fq Fq−1 ∂Fq−1 with A a matrix with entries in O. By applying ψ we find       ∂gp gp 0 ∂gp+1  gp+1    0        ..   ..    .  .  = A . +        ∂gq−2  gq−2    0 ∂gq−1 gq−1 z αq−1 −αq gq or         fp fp 0 hp fp+1  fp+1    hp+1  0        d   ..   ..     ..  .  .  = A . +  +  . ,        dz  fq−2  fq−2    hq−2  0 fq−1 z αq−1 −αq fq fq−1 hq−1 where hd ∈ O for d = p, .

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