A note on p-curvatures
Julien Roques

Abstract. In this note, we give an arithmetic criterion for the Lie-irreducibility of linear
differential equations based on p-curvatures.

Contents
1 Introduction - Statement of the main result

12 Differential equations, differential systems and p-curvatures2

3 Proof of Theorem 1

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Introduction - Statement of the main result

A. Grothendieck conjectured that a linear differential equation with coefficients in Q(x) has
a full set of algebraic solutions if and only if its p-curvatures are zero for almost all prime p.
This conjecture was reformulated by N. Katz in [1] (Grothendieck-Katz’ conjecture).
In contrast with the original Grothendieck’s conjecture, this note is concerned with the nonvanishing of the p-curvatures. Our main result is :Theorem 1. Let L be a linear differential operator with coefficients in Q(x), irreducible overQ(x). Assume that the order n of L is a prime and that, for infinitely many prime p, thereduction of L mod. p is nilpotent and has non zero p-curvature. Then L is Lie-irreducible.
We recall that an operator L as above is Lie-irreducible if the neutral component of itsdifferential Galois group over Q(x) acts irreducibly (on the solutions).
This note is inspired by J. F. Voloch’s paper [5] which is concerned with second order operators. Indeed, it is easily seen that, in the case that n = 2, Theorem 1 can be paraphrased asfollows : Let L be a linear differential operator of order 2 with coefficients in Q(x), irreducible
over Q(x). Assume that, for infinitely many prime p, the reduction of L mod. p is nilpotent
and has non zero p-curvature. Then the Galois group of L over Q(x) contains SL2 (Q). This
result is proved in [5]. The starting point of this work was a question raised by N. Katz in
the introduction of [2]. It is interesting to note the similarity of Theorem 1 with N. Katz’
Proposition 2.7.2 in [3].
Acknowledgments. Je remercie L. Di Vizio et D. Bertrand.

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Differential equations, differential systems and p-curvatures

We will denote by Q(x)h∂i be the usual non commutative algebra of differential operatorswith coefficients in Q(x) (i.e. the non commutative algebra of non commutative polynomials
with coefficients in Q(x) satisfying to the relation ∂x = x∂ + 1).
Let us consider L = ∂ n + an−1 ∂ n−1 + · · · + a0 ∈ Q(x)h∂i. The corresponding linear homogeneous differential equation is∂ n y + an−1 ∂ n−1 y + · · · + a0 y = 0.

(1)

Setting Y = (y, ∂y, ..., ∂ n−1 y)t , this differential equation is equivalent to the differential system

0
 ..
In−1∂Y = AY ; A =  .
(2) ∈ Mn (Q(x)).
 0−a0 −a1 · · · −an−1We define a sequence (Ak )k∈N∗ of elements of Mn (Q(x)) by A1 = A and, for all k ∈ N∗ ,Ak+1 = ∂Ak + Ak A (in other terms, for all k ∈ N∗ , ∂ k Y = Ak Y ). For all k ∈ N, we will set :Ak = (ak;i,j )1≤i,j≤n .
For almost all prime p, we define the p-curvature of L as Ap mod. p. Note if the first line of thep-curvature is zero then the p-curvature itself is zero (immediate from the formula ∂ p Y = Ap Y ).

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Proof of Theorem 1
We start with some preliminary results.

Lemma 1. Let L be an irreducible element of Q(x)h∂i of order m ∈ N. Let y be a non zero element of some differential field extension of (Q(x), ∂) such that Ly = 0. Then y, ∂y, ∂ 2 y, ..., ∂ m−1 y
are linearly independent over Q(x).
Proof. This lemma means that if L0 ∈ Q(x)h∂i is a differential operator of order < m suchthat L0 y = 0 then L0 = 0. This is a direct consequence of the fact that the left ideal {L0 ∈Q(x)h∂i | L0 y = 0} of Q(x)h∂i is generated by L (by the usual Euclidean division argument).
Lemma 2. Let n be a prime. Let G ⊂ GLn (Q) be a linear algebraic group which acts irreducibly
nnn
on Q . Then either G0 acts irreducibly on Q or there exists a line in Q invariant by theaction of G0 .
nn

Proof. Assume that G0 acts reducibly on Q . Let V 6= {0}, Q be a non trivial subspace
nof Q invariant for the action of G0 and minimal for this property. For all g ∈ G, gV is00invariant undersincePthe action of G because G is normalized by G. So,P G acts irreduciblynnnon Q , Q = g∈G gV . Let E be a finite subset of G such that Q = g∈E gV and minimal
Pfor this property. For any h ∈ E, ( g∈E\{h} gV ) ∩ hV is an invariant subspace for G0 so,2Pby
P the minimality property of the (dimensionPof) V , either ( g∈E\{h} gV ) ∩ hV = {0} or( g∈E\{h} gV ) ∩ hV = hV . The case that ( g∈E\{h} gV ) ∩ hV = hV is excluded by theLnminimality property of E. Therefore Q = g∈E gV . In particular, since n is prime, we getdim V = 1.
Notations. In what follows, we will use the Picard-Vessiot approach for differential Galois
theory ([4]). For any L ∈ Q(x)h∂i, we will denote by KL some Picard-Vessiot extension for
L over Q(x) and by SL = {y ∈ KL | Ly = 0} the corresponding Q-vector space of solutions(whose dimension is the order of L).
Proposition 1. Let n be a prime. Let L be an element of Q(x)h∂i of order n, irreducible overkQ(x). Then either L is Lie-irreducible or there exists y 6= 0 in SL such that, for all k ∈ N, ∂y yis algebraic over Q(x).
Proof. Let G ⊂ GL(SL ) be the differential Galois group of L over Q(x). Since L is irreducible
over Q(x), G acts irreducibly on SL . Assume that L is Lie-reducible. Lemma 2 ensures that
there exists y ∈ SL which spans a Q-line invariant by the action of G0 ; in particular, for anyk (gy)kk ∈ N, g ∂y y = ∂ gydoes not depend on g ∈ G0 . So, we can set (without ambiguity), for any
kkg ∈ G/G0 , g ∂y y = g ∂y y . It is clear that any symmetric polynomial with coefficients in Q(x)kkin (g ∂y y | g ∈ G/G0 ) is fixed by the action of G and hence belongs to Q(x). Therefore, ∂y y isalgebraic over Q(x) of degree at most [G : G0 ].
We now prove our main result.
Proof of Theorem 1. Assume that L is Lie-reducible. Proposition 1 ensures that there exists
ky 6= 0 in SL such that, for all k ∈ N, ∂y y is algebraic over Q(x). For the sake of conciseness, wekset, for all k ∈ N, uk = ∂y y and K = Q(x)(u1 , ..., un−1 ) ⊂ KL . Then K is a finite differential
extension of Q(x).
PLet T be an indeterminate over Q(x) and let F (T ) = nk=0 fk T k be a unitary irreducible element of Q(x)[T ] such that K can be identified with Q(x)[T ]/(F (T ));P
we denote by t the class ofdT in Q(x)[T ]/(F (T )). With this identification ∂ is given by ∂t = −( nk=0 ∂(fk )tk )( dTF (t))−1 .
Let r ∈ Z[x] be some multiple of denominators of the coefficients of F (T ) and of L such
dthat the image R of Z[x][r−1 ][T ] in K = Q(x)[T ]/(F (T )) contains ( dTF (t))−1 and u1 , ..., un−1 .
It is clear that R is a subring of K stable by ∂. Moreover, for all k ∈ N, uk ∈ R as it is easily
seen from the relation Ly = 0.
In what follows by “mod. p” we will mean “in R/pR”.
p kp+kLet us consider k ∈ J0, n − 1K. Using Leibniz formula, we get up+k = ∂ y y = ∂ ∂y y =P∂ p (uk y)= pj=0 pj ∂ j (uk )up−j = uk up + ∂ p uk mod. p, for almost all prime p. But, since uk
yis algebraic over Q(x), ∂ p uk = 0 mod. p, for almost all prime p. Thus, we are lead to thep+kp
fact that, for almost all prime p, up+k = uk up mod. p i.e. ∂ y y = uk ∂y y mod. p and henceap;k+1,1 + ap;k+1,2 u1 + · · · + ap;k+1,n un−1 = uk (ap;1,1 + ap;1,2 u1 + · · · + ap;1,n un−1 ) mod. p (weuse the notations of section 2 for the p-curvatures).
Therefore, for almost all prime p, the vector (1, u1 , ..., un−1 )t mod. p is an eigenvector
with coefficients in R/pR for the p-curvature Ap mod. p associated to the eigenvalue ap;1,1 +3

ap;1,2 u1 + · · · + ap;1,n un−1 mod. p. Since Ap mod. p is nilpotent and non zero for infinitely
many prime p and since the first coordinate of the above eigenvector is equal to 1, we getthat ap;1,1 + ap;1,2 u1 + · · · + ap;1,n un−1 mod. p is a nilpotent element of R/pR for infinitely
many prime p (the fact that the first coordinate is equal to 1 is used because R/pR need notbe entire). Since R/pR is a reduced ring for almost all prime p, we get a non trivial linear
relation ap;1,1 + ap;1,2 u1 + · · · + ap;1,n un−1 = 0 mod. p for infinitely many prime p in theZ[x][r−1 ]/pZ[x][r−1 ]-module R/pR..
So 1 mod. p, u1 mod. p,..., un−1 mod. p are linearly dependent in the Z[x][r−1 ]/pZ[x][r−1 ]module R/pR, for infinitely many prime p. Using the fact that R is a free Z[x][r−1 ]-module offinite type, we get that 1, u1 , ..., un−1 are linearly dependent over Q(x) : this is a contradiction
in virtue of Lemma 1.

References[1] N. M. Katz. A conjecture in the arithmetic theory of differential equations. Bull. Soc.
Math. France, 110(2):203–239, 1982.
[2] N. M. Katz. On the calculation of some differential Galois groups. Invent. Math., 87(1):13–61, 1987.
[3] N. M. Katz. Exponential sums and differential equations, volume 124 of Annals of Mathematics Studies. Princeton University Press, Princeton, NJ, 1990.
[4] M. Van der Put and M. F. Singer. Galois Theory of Linear Differential Equations, volume328 of Grundlehren der mathematischen Wissenschaften. Springer-Verlag, Berlin, 2003.
[5] J. P. Voloch. A note on the arithmetic of differential equations. Indag. Math., 11(4):617–621,2000.
Julien RoquesUniversité Grenoble 1 - CNRS UMR 5582
Institut Fourier
100 rue des Maths
BP 74
38402 St Martin d’Hères cedex (France)
E-mail : Julien.Roques@ujf-grenoble.fr

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