Riccati Differential Equation for Hypergeometric Differential Equation
TAKAHIRONAKAGAWA
ABSTRACT- In this paper, we study the solutions of the Riccati differential equation collesponding to ap-adic differential equation which is solvable on the generic disc. As an application, we consider the Grothendieck conjecture for Riccati differential equations. We see that the Riccati differential equation for some globally nilpotent differential equation with coefficients inQ(t) have, for almost all prime number, a solution inFp(t), butdo nothave any algebraic solutions.
1. Introduction
Letpbe a prime andK a complete non-archimedean valued field with mixed characteristics. In this paper, we consider the solutions of the Ric- cati differential equation for ap-adic differential equationLwhich is sol- vable on the generic discD(t;1 ) (see the Definition 2.5).
We attach the Riccati differential equationC(L)1for each linear ordinary differential operatorLin section 3. LetEbe the completion ofK(t) under the Gauss norm. We see that a solution of the Riccati differential equationC(L)1 forLonEis connected with a bounded solution ofLat the generic point.
LetbddKert(L) be the vector space consisting of the bounded solutions of L. If its dimension is one, then C(L)1 has a unique solution on E. In particular if we can reduce Lmodulop, thenC(L)1 has a solution on the rational function fieldK(t) over the residue field K ofK.
Let L be an n-th order linear ordinary differential operator with coefficients in Q(t). The Grothendieck conjecture (in its simplest form) asserts the following statements are equivalent (see [Put]):
(i) the equation L(y)0 has n linear independent(over Qalg) solutions which are algebraic overQ(t).
(ii) For almostallp, the equationL(y)0 hasnlinear independent solutions over the fieldFp(tp) in the fieldFp(t).
(*) Indirizzo dell'A.: Mathematical Institute CHIBA University, 263-8522 Japan.
E-mail: [email protected]
Van der Put studied the Grothendieck conjecture for non-linear differ- ential equations. He proved that the differential equationy0(1=t2)y1 has no algebraic solutions but there exists a solution inFp(t) for all primesp. He also proved that the conjecture is true for some Risch equations [Put], [Put2].
We will prove the following theorem.
THEOREM 1.1. Let L be a monic differential operator with coeffi- cients in Q(t). We assume that when we regard L as a p-adic differ- ential equation, L is solvable on the generic disc D(t;1) for almost all prime p, and the dimension of the vector space which consists of the bounded solutions of L is one. If L has no algebraic solutions, then C1(L)modulo p has a solution onFp(t)for almost all prime p and no algebraic solutions.
In particular some hypergeometric differential equations satisfy the condition of Theorem 1.1.
The author is grateful to Professor S. Matsuda for his helpful advice and thank referee for his suggestion.
2. Notation
2.1 ±Analytic elements
Let K be a complete non-archimedean valued field with mixed char- acteristics and K the residue field ofK. Let K(t) be the field of rational functions. For a polynomialf P
aitiwith coefficients inK, we set jfj0maxjaij:
(2:1)
The field of rational functionsK(t) has the Gauss norm j j0 defined by jfj0 jgj0=jhj0 if f g=h and g;h2K[t]. Let E be a completion ofK(t) under the Gauss norm. An element ofEis called an analytic element. The derivationd=dtofK(t) is extended toE. We regard (E;d=dt) as a differ- ential field.
Fora2Eandr2R>0, let
D(a;r ) fz2Ej jz aj05rg (2:2)
and
D(a;r) fz2Ej jz aj0rg:
(2:3)
We define the ring of analytic functions on the discD(a;r ) t o be A(a;r) X1
n0
an(x a)n2E[[x a]]an2E;n!1lim janjsn0 for all 0s5r
( )
:
We define the subringE[[x a]]0of the ring of analytic functionsA(a;1) by E[[x a]]0 X1
n0
an(x a)n 2E[[x a]] sup
0n51janj51
( )
:
PROPOSITION2.1 ([Christol, 2.5.1]). We define a ring homomorphism t:E!E[[x t]]0 so that, for f 2E,
t:E!E[[x t]]0 f 7!X1
n0
(d=dt)n(f)
n! (x t)n:
Thentis an isometric isomorphism from E to the subfield of E[[x t]]0 consisting of elements f which satisfy:
t((d=dt)(f))(d=dx)(t(f)):
(2:4)
2.2 ±Differential modules
Let(H;D) be a differential field of characteristic zero and CH the constant field of H. We say differential field (L;D0) is a differential ex- tensionof (H;D) ifLHandD0jHD. AnH-differential module (M;r) is a free H-module M of finite type endowed with a CH-linear homo- morphismr(D) which satisfiesr(D)(am)Da:mar(D)(m) fora2H andm2M.
Let(M;r) be anH-differential module andLa differential extension of H. AnH-linear maps:M!Lis saidhorizontal ifs(rm)D(s(m)) for m2M. We denote the CH-vector space by Homr(M;L):
fs2HomCH(M;K)js is horizontalg.
PROPOSITION 2.2. Let (M;r) be an H-differential module. For any differential extension L=H,
dimCH(Homr(M;L))dimH(M) (2:5)
REMARK2.3. By the existence of Picard-Vessiot field of HCHCalgH , there exists a differential extension L=H such that equality (2.5) holds (see [PS]).
PROPOSITION2.4. Let G be a matrix in Mn(E). There exists a unique matrix U of GLn(E[[x t]])such that
U(t)In; (d=dx)(U)t(G)U:
(2:6)
LetfGigi2Nbe the sequence of matrix in Mn(E)defined by G0In; Gi1(d=dt)(Gi)GiG:
Then we have
UX1
i0
Gi(x t)i i! :
DEFINITION2.5. Let(M;r) be anE-differential module of rankn. Let e1; ;en be a basis of M and G(gij) the matrix such that r(D)eiPn
j0gijej. We say (M;r) is solvableover D(t;r) if the matrix U which is defined by (2.6) is an elementof GLn(A(t;r)).
For a differential field (H;D) which satisfiesK(t)HE, we call an H-differential module (M;r) is solvable overD(t;r) if t heE-differential module (MHE;r) is solvable overD(t;r).
3. Riccati differential equation
Let(H;D) be a differential field of characteristic zero andCHa constant field ofH. LetLDn gnDn 1 g1be a differential operator with coefficients inH. Then (M;r)H[D]=H[D]Lis anH-differential module of rankn. LetGbe the matrix of representation ofr(D) with respect to the basis 1; ;Dn 1of (M;r), i.e,
G
0 1 0 0
0 0 1 0
0... 1
g1 g2 gn
0 BB BB
@
1 CC CC A: (3:1)
We suppose thatLhas a decomposition
LL1R; L1;R2H[D] andR is monic.
We write RDk ckDk 1 c1. Since the canonical morphism MH[D]=H[D]L!H[D]=H[D]R is surjective, the induced morphism (H[D]=H[D]R)_!M_is injective.
Let f1; ;fn (resp. l1; ;lk) be the dual basis of the basis 1;D; ;Dn 1ofM(resp. (H[D]=H[D]R)). If we regard (H[D]=H[D]R)_as a submodule ofM_then eachliis a linear combination offi. In fact, by the definition of dual basis, we have
l1
l2
l...k 0 BB B@
1 CC CAC
f1 f2 f...n 0 BB B@
1 CC CA (3:2)
such thatC fci;jgis theknmatrix defined by
Dm 1ck;mDk 1ck 1;mDk 2 c1;m (modR):
(3:3)
In particularcj;k1cjand ifjkthenci;jdi;j.
Let F=Hbe a differential extension such that dimCH(Homr(M;F)) dimH(M) (see Remark 2.3). Lets1;. . .;snbe a basis of Homr(M;F). By the equality (3.2), we have
s1(l1) s2(l1) sn(l1) s1(l2) s2(l2) sn(l2)
... ...
s1(lk) s2(lk) sn(lk) 0
BB B@
1 CC CAC
s1(f1) s2(f1) sn(f1) s1(f2) s2(f2) sn(f2)
... ...
s1(fn) s2(fn) sn(fn) 0
BB B@
1 CC CA: (3:4)
OperatingDon both sides the equation (3.4), we have 0 c1
1 ...
1 ck 0
BB BB
@
1 CC CC A
s1(l1) s2(l1) sn(l1) s1(l2) s2(l2) sn(l2)
... ...
s1(lk) s2(lk) sn(lk) 0
BB BB
@
1 CC CC A
D(C) C
0 g1
1 g2
... ...
0 1 gn
0 BB BB
@
1 CC CC A 0
BB BB
@
1 CC CC A
s1(f1) s2(f1) sn(f1) s1(f2) s2(f2) sn(f2)
... ...
s1(fn) s2(fn) sn(fn) 0
BB BB
@
1 CC CC A
Since the matrix (si(fj))i;jis invertible, we have 0 c1
1 ...
1 ck 0
BB B@
1 CC
CACD(C) C
0 g1
1 g2
... ...
0 1 gn
0 BB B@
1 CC CA:
Therefore we have the equations
c1ck;nD(c1;n) g1 c1;k1gk1 c1;ngn0;
c1;nc2ck;nD(c2;n) g2 c2;k1gk1 c2;ngn0;
ck 1;nckck;nD(ck;n) gk ...ck;k1gk1 ck;ngn 0:
(3:5)
We define the differential systemC(L)kink-variabley1;. . .;ykby y1yk;nD(y1;n) g1 y1;k1gk1 y1;ngn0;
y1;ny2yk;nD(y2;n) g2 y2;k1gk1 y2;ngn 0;
yk 1;nykyk;nD(yk;n) gk ... yk;k1gk1 yk;ngn0 (3:6)
whereyi;j (1ik; j1) is defined inductively by y1;11;yj;10 (2jk);
y1;m1D(y1;m)yk;m:y1; yj;m1D(yj;m)yk;m:yjyj 1;m (2jk):
By definition, the vector (c1; ;ck) is solutions of differential systemC(L)k. In particular,C(L)1is a differential equation
y1y1;nD(y1;n) g1 y1;2g2 y1;ngn0 (3:7)
wherey1;jis defined inductively by
y1;1 1; y1;m1 D(y1;m)y1;m:y1:
Conversely we assume differential system C(L)k has a solution (c1; ;ck)2Hk. Let F=H be a differential extension which satisfies dimCH(fy2Fj(Dk ckDk 1 c1)y0g)k (see Remark 2.3). If y2Fsatisfies the equality (Dk ckDk 1 c1)y0 then we have
G y Dy
...
Dn 1y 0 BB BB
@
1 CC CC AG:tC
y Dy
...
Dk 1y 0 BB BB
@
1 CC CC A
D(tC) y Dy
...
Dk 1y 0 BB BB
@
1 CC CC AtC:
0 1 0
0 1 0
...
c1 c2 ck
0 BB BB
@
1 CC CC A
y Dy
...
Dk 1y 0 BB BB
@
1 CC CC A
D(tC) y Dy
...
Dk 1y 0 BB BB
@
1 CC CC AtC:
Dy D2y ...
Dky 0 BB BB
@ 1 CC CC A
Dy D2y
...
Dny 0 BB BB
@ 1 CC CC A:
Thereforeysatisfies a solution of differential equation DnygnDn 1ygn 1Dn 2y g1y:
(3:8)
By proposition 2.2, we have a decomposition
Dn gnDn 1 g12H[D](Dk ckDk 1 c1):
(3:9)
As a result, we have the following proposition.
PROPOSITION3.1. Let H be a differential field such that the character- istic of H is zero. Let LDn gnDn 1 . . . g1 be a differential poly- nomial with coefficients in H.
There is a one to one correspondence between a k-th order differential operator R2H[D] which satisfies L2H[D]R and a solution of differ- ential system C(L)k in Hk.
If there exists a differential polynomial RDk ckDk 1 c1
such that L2H[D]R then (c1;c2;. . .;ck)2Hk is a solution of C(L)k. Conversely, if (c1;c2;. . .;ck)2Hk is a solution of C(L)k then the differ- ential polynomial RDk ckDk 1 c1satisfies L2H[D]R.
By induction, we have the following lemma.
LEMMA 3.2. We assume HE. Let LDn gnDn 1 . . . g1 be a differential polynomial with coefficients in E. Ifjgjj01for j1; ;n, then, for any solution y2E of C1(L), we havejyj01.
EXAMPLE3.3. For a differential operatorLD2 g2D g1of rank 2, C(L)1 is a Riccati differential equation c2c0 g1 g2c0. Young con- sidered in this case [Young].
4. Analytic element solution of Riccati differential equation
LetL be a differential operator with coefficients inEandDd=dt. LetC(L)k be the differential system defined byL. We suppose thatLis solvable onD(t;1) (i.e.,E[D]=E[D]Lis solvable onD(t;1)). We denote the vector space which is generated by the solutions of (resp. the bounded solutions) ofLby
Kert(L) fy2 A(t;1)jt(L)y0g;
(resp. Kerbddt(L) fy2E[[x t]]0jt(L)y0g):
In general, sinceLis solvable onD(t;1), we have dimK(Kerbddt(L))1 by [Robba, Theorem 3.5]. If dimK(Kerbddt(L))n 1, then there exists a monic differential operatorR2E[D] such thatL2E[D]Rand Kert(R) Kerbddt(L) by [Robba, Theorem 2.6].
PROPOSITION 4.1. Let L2E[D] be a monic n-th order differential operator which is solvable on D(t;1).
(i) I f kdimK(Kerbddt(L))n 1, then C(L)khas a solution on E.
(ii) If1dimK(Kerbddt(L)), C(L)1has a unique solution on E.
(iii) If there exists fundamental system of solutions y1;. . .ynof L at t such that yiis exactly of log-growth i 1(see the definition of log-growth in [C-T]) then, for each j, C(L)j has a unique solution over E.
PROOF. There exists a monic differential operatorR2E[D] such that L2E[D]Rand Kert(R) fy2E[[x t]]jt Ly0 andy is log-growthig by [Robba, Theorem 2.6]. By Proposition 3.1, there exists a solution of C(L)konEwherekdeg(R). The uniqueness of case (ii),(iii) is confirmed
by [Dwork1, Theorem 1]. p
By [D-R, Theorem 2.4.3 (iii)], we have the following proposition.
PROPOSITION4.2. Let L2E[D]be a monic n-th differential operator which is solvable on D(t;1). Assume kdimK(Kerbddt(L))n 1. Let S be the setfa2K jL has no pole on D(a;1 )g. Then there exists a solution
(y1;. . .;yk)2Ek of C(L)k such that
fa2Sjdim(Kerbddt(L))>dim(Kerbdda(L))g fa2Sjyiis notanalytic on D(a;1 ) for someig
5. Hypergeometric differential equation and Riccati differential equation modp
Letpbe a prime number andDd=dx. Fora;b;c2Q\Zp, we setthe hypergeometric differential operator
L(a;b;c)D2 c (1ab)x
x(1 x) D ab
x(1 x): (5:1)
We assume the following condition:
c a;c b;b;a all lie outsideZ:
(5:2)
c2Q f0; 1; 2;. . .g:
(5:3)
By [Dwork2, p. 11, Remark], we see the Riccati differential equation y2y0c (1ab)x
x(1 x) y ab
x(1 x)0 (5:4)
has no rational solutions.
For a rational number a2Q\Zp, we set ma2[0;p 1] such that ma a (modp). According t o Dwork, we say (a;b;c)2Qissplit(resp.
unsplit) if mc5minfma;mbg or mc >maxfma;mbg (resp. ma5mc5mb or mb5mc5ma). There exists an integer msuch that
F:pm Xp 1
n0
(a;n)(b;n)
(c;n)n! xn2Zp[x] pZp[x];
(5:5)
where (a;n) is defined
(a;n) 1 (n0)
a(a1) (an 1) (n1):
If (a;b;c) is split , t henFis uniquely determined. SinceFis a solution of hypergeometric differential equation
D2c (1ab)x
x(1 x) D ab
x(1 x)
y0 (modp);
(5:6)
by Lemma 5.2,F0=Fis a solution of the Riccati differential equation y2y0c (1ab)x
x(1 x) y ab
x(1 x)0 (modp):
(5:7)
Under the special condition, Dwork proved that if (a;b;c) is split, then L(a;b;c) has a unbounded solution and if (a;b;c) is unsplit, thenL(a;b;c) has two independent bounded solutions (see [Dwork2, Theorem 6.6, 9.4]).
We consider solutions of the Riccati differential equation (5.7).
LEMMA 5.1. We assume that(a;b;c)2(Q\Zp)3 is unsplit. Then the Riccati differential equation(5.7)has infinite solutions.
PROOF. In this case, the hypergeometric differential equation (5.6) has solution
uminfmXa;mbg
n0
(a;n)(b;n)
(c;n)n! xnvpmmaxfmXa;mbg
nmc1
(a;n)(b;n)
(c;n)n! xn (u;v2Fp(xp)) (5:8)
with m2Z which satisfies pmmaxfmPa;mbg
nmc1
(a;n)(b;n)
(c;n)n! xn2Zp[x] pZp[x]. By Lemma 5.2, the Riccati differential equation (5.7) has infinite solutions. p LEMMA 5.2. Let RDn gnDn 1 gn 1Dn 2 . . . g1 be a differ- ential operator with coefficients Fp(x). Then R has a decomposition RQ(D c)where Q2Fp(x)[D]and c2Fp(x)if and only if c2Fp(x) is a solution of the Riccati differential equation C(R)1.
PROOF. Assume that R(Dn 1bn 1Dn 2 b1)(D c) where b1; ;bn 1;c2Fp(x). Comparing the both sides, we have the equalities
gnc bn 1
gn 1 n 1 1
c(1)cbn 1 bn 2
...
gi n 1 n i
c(n i) n 2 n i 1
c(n i 1)bn 1 c0 i
1 bi1cbi bi 1
...
g2 n 1 n 2
c(n 2) n 2 n 3
c(n 3)bn 1 c0 2
1 b3cb2 b1
g1c(n 1)c(n 2)bn 1 c(1)b2cb1:
We setcm;Cmby
cm1c0mccm; c11 CmXm
i1
m 1
i 1
cm i1c(i 1) cm1:
We have the equality
g1 c2g2 cngncn1CnXn 1
i1
Cibi:
To prove thatcis a solution ofC(R)1, we have only to proveCm 0 for allm.
We use induction onm. Form1,C1c1c(0) c20. Therefore the case m1 is true.
We assume that the assertion is true formk 1.
c0k Xk 1
i1
k 2 i 1
ck ic(i 1)
!0
Xk 1
i1
k 2 i 1
(c0k ic(i 1)ck ic(i))
Xk 1
i1
k 2 i 1
((ck i1 cck i)c(i 1)ck ic(i))
Xk 1
i1
k 2 i 1
(ck i1c(i 1)ck ic(i)) cck
Xk 1
i2
k 2 i 1
k 2 i 2
ck i1c(i 1)c(k 1)
Xk 1
i2
k 1 i 1
ck i1c(i 1)c(k 1)Ck cckck1
This completes the induction.
Conversely let c2Fp(x) be a solution of C(R)1. We define
b1;. . .;bn 12Fp(x) as above. Then we have the decomposition
R(Dn 1bn 1Dn 2 b1)(D c). p
LEMMA5.3. We assume that(a;b;c)2 Q\Zp3
is split. The Riccati differential equation(5.7)has a unique solution if and only if
xmc(1 x) 1mamb mc F2
!(p 1)
60 (modp) (5:9)
PROOF. Substituting yzF0=F for y in the Riccati differential equation (5.7), we have
z2z0 2 F0
Fc (ab1)x x(1 x)
z0 (modp):
(5:10)
Substitutingzhxmc(1 x) 1mamb mc=F2forzin the Riccati differential equation (5.7), we have
xmc(1 x) 1mamb mch2
F2h00 (modp):
(5:11)
We supposeh60. By this equality, 1
h
0xmc(1 x) 1mamb mc
F2 (modp):
(5:12)
The differential equation (5.12) has a solution if and only if xmc(1 x) 1mamb mc
F2 2Fp(xp)xFp(xp). . .xp 2Fp(xp)
(5:13) i.e.
xmc(1 x) 1mamb mc F2
!(p 1)
0 (modp):
(5:14)
p PROPOSITION5.4. We assume thatmcp 1and(a;b;c)is split. Then the Riccati differential equation(5.7)has a unique solution F0=F.
PROOF. Let vx be a valuation ofFp(x) with respect to a prime ideal xF[x] of F[x]. Since p 1mc>maxfma;mbg by assumption, vx(F)0.
Therefore we have vx 1 h 0
vx xmc(1 x) 1mamb mc F2
!
p 1:
(5:15)
Since there exists noh2F(x) which satisfies (5.15), we haveh0. p
6. Algebraic solution of Riccati differential equation
Let L be a monic differential operator with coefficients in Q(t) and Dd=dt. For almostall primep, we can reduce L modulo p. If we can reduceLmodulop, then we denoteLmodulopbyLp.
THEOREM 6.1. (i) If L has an algebraic solution, then C(L)1 has an algebraic solution.
(ii) We assume that for almost all prime p, L is solvable on the generic disc D(t;1) as a p-adic differential equation. If C(L)1 has a ra- tional solution then L has an algebraic solution.
(iii) We assume that for almost all prime p, L is solvable on the generic disc D(t;1) and dim(bddKert(L))1 as a p-adic differential equation. Then for almost all prime p, C(L)1has a solution inFp(t). And if C(L)1 has an algebraic solution, then L has an algebraic solution.
PROOF. IfLhas an algebraic solution then there exists an algebraic function h such that L2Q(t)alg[D](D h). Hence the Proposition 3.1 implies (i).
Suppose that C(L)1 has a rational solution y. By Proposition 3.1, we haveL2Q(t)[D](D y). Letpbe a prime such that when we regardLas a p-adic differential equation,Lis solvable on the generic disc D(t;1) and that we can reduceLandD ymodulop. By [DGS, 3. Proposition 5.1],Lp is nilpotent. And by [DGS, 3. Theorem 2.1], (D y)pis also nilpotent and therefore has a solution ofFp(t). Since for all almostprimep, (D y)phas a solution ofFp(t), by Honda's theorem([Put2, Proposition 2.2]),D yhas an algebraic solution. This implies (ii).
We will prove (iii). By Lemma 3.2 and Proposition 4.1(ii), for all almost prime p, C(L)1 has a solution on Fp(t). We assume that C(L)1 has an al- gebraic solution y which is not a rational function. Let G be the Galois group which corresponds to a finite Galois extension ofQ(t) containsy. For alls2G,s(y) is a solution ofC(L)1. Sinceyis nota rational function,C(L)1 has at least two algebraic solutions. By Proposition 3.1, this implies dim(bddKert(L))>1. This is a contradiction. p COROLLARY 6.2. Let L be a monic differential operator with coeffi- cients inQ(t). We assume that for almost all primep, when we regardLas ap-adic differential equation,Lis solvable on the generic discD(t;1)and dim(bddKert(L))1. I f L has no algebraic solutions, then C(L)1 has a solution onFp(t)for almost all primepand no algebraic solutions.
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Manoscritto pervenuto in redazione il 20 maggio 2011.