ON THE ALGEBRAIC DEPENDENCE OF E-FUNCTIONS
T. RIVOAL AND J. ROQUES
Abstract. Siegel introduced and studied the class of E-functions in 1929. They are
power series, solutions of some linear differential equations, whose Taylor coefficients satisfy certain arithmetic and growth conditions. After the work of Siegel and Shidlovskii,
and its refinement by Beukers, on the algebraic relations between values of E-functions
over Q, it is important to know when the E-functions solutions of a given differential
system of order 1 are algebraically dependent or not over Q(z). In this paper, we give thecomplete classification of the vector solutions of two dimensional differential systems oforder 1 whose components are algebraically dependent E-functions over Q(z).

1. Introduction
An E-function is a power seriesf (z) =

∞
X
an

n!
n=0

z n ∈ Q[[z]]with coefficients in the field of algebraic numbers Q such that(1) f (z) satisfies a nonzero linear differential equation with coefficients in Q(z);(2) there exists C > 0 such that(a) the maximum of the moduli of the galoisian conjuguates of an is bounded by
C n+1 ;(b) there exists a sequence of positive integers dn such that dn ≤ C n+1 and dn am
is an algebraic integers for all m ≤ n.
The prototypical example is the exponential function. The E-functions were first introduced by Siegel (1) to generalize the diophantine properties of ez , in particular the
Lindemann-Weierstrass Theorem. The work of Siegel [13] and Shidlovskii [14] culminatedwith the following theorem, which can be seen as vast generalization of the LindemannWeierstrass Theorem.
Theorem 1 (Siegel-Shidlovskii). Let f1 (z), . . . , fn (z) be E-functions such that(f10 (z), . . . , fn0 (z))t = A(z)(f1 (z), . . . , fn (z))t
Date: December 28, 2015.
2010 Mathematics Subject Classification. 11J91, 34M35, 30D15.
Key words and phrases. E-functions, E-operators, E-values.
1His definition was slightly less restrictive, but it is now believed that both definitions define the sameclass of functions.
1

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for some A(z) ∈ Mn (Q(z)). Denote the common denominator of the entries of A(z) by
T (z). Then, for any ξ ∈ Q such that ξT (ξ) 6= 0, we have
deg trQ Q f1 (ξ), . . . , fn (ξ) = deg trQ(z) Q(z) f1 (z), . . . , fn (z) .
An alternative proof of the Siegel-Shidlovskii Theorem was given by Bertrand in [3] using
Laurent’s determinants.
In the seminal paper [1], André elucidated the structure of “E-operators” by means of
their relations with G-operators. Any E-function is in the kernel of an E-operator. Using
these results, André obtained in [2, Théorème 2.3.1] a completely new proof of the SiegelShidlovskii Theorem. Beukers [6] was even able to deduce from the work of André thefollowing important refinement of a theorem of Nesterenko and Shidlovskii [9], which isitself a refinement of the above-mentioned Siegel-Shidlovskii Theorem.
Theorem 2 (Beukers). With the notations and hypotheses of Theorem 1, let us considerξ ∈ Q such that ξT (ξ) 6= 0. For any polynomial relation P f1 (ξ), . . . , fn (ξ) = 0 withP ∈ Q[X1 , . . . , Xn ], there exists Q ∈ Q[z][X1 , . . . , Xn ] such that Q f1 (z), . . . , fn (z) = 0
and P (X1 , . . . , Xn ) = Q(X1 , . . . , Xn )|z=ξ .
In order to apply the above transcendence results, the first naive question is: when are
f1 (z), . . . , fn (z) algebraically dependent over Q(z)? The main result of this paper gives acomplete answer to this question when n = 2. In what follows, for any γ ∈ Q \ Z≤0 , wedenote by 1 F1 (1; γ; z) the hypergeometric function (which is an E-function) defined by:
1 F1 (1; γ; z) =

∞
X

zk
.
γ(γ + 1) · · · (γ + k − 1)
k=0Note that 1 F1 (1; 1; z) = ez .
Theorem 3. Let f (z), g(z) ∈ Q[[z]] be E-functions such that(f 0 (z), g 0 (z))t = E(z)(f (z), g(z))tfor some E(z) ∈ M2 (Q(z)). If f (z) and g(z) are algebraically dependent over Q(z), thenone of the following cases occurs:(i) There exist a(z), b(z), c(z), d(z) ∈ Q[z, z −1 ] and α, β ∈ Q such thatf (z) = a(z)eαz + b(z)eβz and g(z) = c(z)eαz + d(z)eβz .
(ii) There exist a(z), b(z), c(z), d(z) ∈ Q[z, z −1 ], γ ∈ Q \ Z and α ∈ Q such thatf (z) = a(z)1 F1 (1; γ; αz) + b(z) and g(z) = c(z)1 F1 (1; γ; αz) + d(z).
Remark. Let f (z) be an E-function such that f 0 (z) = u(z)f (z) + v(z) for some u(z) ∈Q(z)× and v(z) ∈ Q(z). In particular, f (z) and f 0 (z) are algebraically dependent overQ(z). Using Theorem 3, it is easily seen that f (z) = a(z)1 F1 (1; γ; αz) + b(z) for somea(z), b(z) ∈ Q[z, z −1 ], γ ∈ {1} ∪ Q \ Z and α ∈ Q. This provides a complete proof of a
result suggested by André in [1, Section 4.5], which answers a question asked by Shidlovskii.

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Actually, we will first prove, in section 3, the following result.
Theorem 4. Let f (z), g(z) ∈ Q[[z]] be E-functions such that(f 0 (z), g 0 (z))t = E(z)(f (z), g(z))tfor some E(z) ∈ M2 (Q(z)). Assume that the differential system Y 0 (z) = E(z)Y (z) is
reducible over Q(z). Then, one of the following cases occurs:(i) There exist a(z), b(z), c(z), d(z) ∈ Q[z, z −1 ] and α, β ∈ Q such thatf (z) = a(z)eαz + b(z)eβz and g(z) = c(z)eαz + d(z)eβz .
(ii) There exist a(z), b(z), c(z), d(z) ∈ Q[z, z −1 ], γ ∈ Q \ Z, α ∈ Q and δ ∈ Q such thatf (z) = (a(z)1 F1 (1; γ; αz) + b(z))eδz and g(z) = (c(z)1 F1 (1; γ; αz) + d(z))eδz .
The proof of Theorem 3, given in section 4, relies on Theorem 4. With the notations andhypotheses of Theorem 3, we actually prove that the differential system Y 0 (z) = E(z)Y (z)is automatically reducible over Q(z) and that the algebraic number δ given by Theorem 4
can be choosen equal to 0.
The present work was mainly motivated by the following question: is the group of units
×of the ring E of values of E-functions at algebraic points equal to Q exp(Q)? We referto [10] for an application of Theorem 3 to this problem. The ring E was introduced in thepaper [7], to which we refer the interested reader.
It would be very interesting to extend Theorem 3 to higher order differential systems.
Note that, in [11, 12], Salikhov studied hypergeometric E-functions solutions of a linear
differential equation of order n ≥ 1 with coefficients in Q(z) and algebraically dependentof their first n − 1 derivatives. See [5] for further results in this direction.
2. Preliminary step for the proofs of Theorem 3 and Theorem 4
Let f (z), g(z) ∈ Q[[z]] be E-functions such that(f 0 (z), g 0 (z))t = E(z)(f (z), g(z))tfor some E(z) ∈ M2 (Q(z)). According to the cyclic vector lemma, there exist a linear
differential operator L of order 2 with coefficients in Q(z), a series h(z) ∈ Q[[z]] such that
L h(z) = 0, and a matrix
p1 (z) p2 (z)∈ GL2 (Q(z))
p3 (z) p4 (z)such thatf (z) = p1 (z)h(z) + p2 (z)h0 (z) and g(z) = p3 (z)h(z) + p4 (z)h0 (z).
Hence,
h(z) = q1 (z)f (z) + q2 (z)g(z) and h0 (z) = q3 (z)f (z) + q4 (z)g(z)
where


 
−1
q1 (z) q2 (z)p1 (z) p2 (z)=∈ GL2 (Q(z)).
q3 (z) q4 (z)p3 (z) p4 (z)4Let ∆(z) ∈ Q[z] be a common denominator of the qi (z). Then,k(z) := ∆(z)h(z) = ∆(z)q1 (z)f (z) + ∆(z)q2 (z)g(z)is an E-function (we recall that the set of E-functions is a sub-Q[z]-algebra of Q[[z]]), andis a solution of a linear differential operator with coefficients in Q(z) of order 2, namely
L ∆−1 . By André’s [1, Theorem 4.3], k(z) is solution of a monic linear differential operator
M with coefficients in Q(z) of order ν = 1 or 2 which is a right factor of an E-operator.
Let us first assume that ν = 1. Then, it is well-known that k(z) = q(z)eαz for someq(z) ∈ Q[z] and α ∈ Q. Therefore, there exist a(z), c(z) ∈ Q(z) such that f (z) = a(z)eαz ,and g(z) = c(z)eαz . Since f (z) and g(z) are entire functions, we have a(z), c(z) ∈ Q[z],whence the desired result.
We shall now assume that ν = 2. By André’s [1, Theorem 4.3], the differential operator
M has the following properties, that will be freely used in the rest of this paper:
Proposition 1. We have:(1) M has only apparent singularities on C× ;(2) M is regular singular at 0, and its exponents at 0 are rational;(3) M admits a basis of formal solutions at ∞ of the formba1 (z)eα1 z , ba2 (z)eα2 z = f1 (z), f2 (z) z Γ∞ e∆z
where the fi (z) ∈ Q[[ z1 ]] are Gevrey-1 series of arithmetic type, Γ∞ ∈ M2 (Q) is
upper-triangular and ∆ = diag(α1 , α2 ) ∈ M2 (Q).
3. Proof of Theorem 4
So, we assume that the differential system Y 0 (z) = E(z)Y (z) is reducible over Q(z).
This implies that the differential operator M is reducible over Q(z), i.e., that there exist
η(z), ω(z) ∈ Q(z) such that



d
d
− η(z)− ω(z) .
M =dzdz
Since M (k(z)) = 0, the function u(z) := k 0 (z) − ω(z)k(z) satisfies u0 (z) = η(z)u(z).
We claim that u(z) = r(z)eδz for some r(z) ∈ Q(z)× and δ ∈ Q. Indeed, let $(z) ∈ Q[z]be a denominator of ω(z). Then, $(z)u(z) ∈ Q[[z]] satisfies an homogeneous differential
equation of order 1 with coefficients in Q(z) and is a E-function (because the set of Efunctions is a sub-Q[z]-differential algebra of Q[[z]]). So, $(z)u(z) = r1 (z)eδz for somer1 (z) ∈ Q[z] and δ ∈ Q. So u(z) = r(z)eδz with r(z) = r1 (z)/$(z). Note that r(z) 6= 0
because k(z) is not a solution of a linear differential equation of order 1 with coefficientsin Q(z) by hypothesis.
Then, it is easily seen that l(z) := e−δz k(z) is an E-function solution of an inhomogeneous
linear differential equation of order 1 with coefficients in Q(z), say y 0 + a1 (z)y = v(z), andthat e−δz ζ(z) is a solution of the corresponding homogeneous equation y 0 + a1 (z)y = 0.
Note that e−δz ζ(z) = z γ p(z)eαz for some γ ∈ Q, p(z) ∈ C[z] and α ∈ {α1 −δ, α2 −δ} ⊂ Q.
Indeed, this follows from the fact that ζ 0 (z) = ω(z)ζ(z) together with Proposition 1.

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Using the variation of constants method, we get that there exists C ∈ C× such that
Z z

−γ
−1
−αx
γ
αz
x p(x) v(x)e dx + C ,(3.1)
l(z) = z p(z)e
z0
×

−1

where z0 ∈ C is not a pole of p(z) v(z). We shall now express (3.1) by means ofhypergeometric series. For this, we will use the following lemmas.
Lemma 1. For all γ 6∈ Z, for all Q(z) ∈ C(z), for all z0 ∈ C× which is not a pole of Q(z),there exists R(z) ∈ C(z) with at most simple poles on C× and whose set of poles in C× is
included in the set of poles in C× of Q(z), and there exist λ(z), µ(z) ∈ C(z) and ν ∈ Csuch that
Z z
Z z
−γ
−αx
−γ −αzx−γ R(x)e−αx dx + ν.
x Q(x)e dx = λ(z)z e+ µ(z)z0

z0

Proof. Using the decomposition in partial fractions of Q(z), we see that it is sufficient to1
×prove the lemma for Q(z) = Qn (z) := (z−ξ)
and n ∈ N∗ . We proceed byn with ξ ∈ C
induction on n. The result is obvious for n = 1. Assume that the result is true for somen ∈ N∗ . An integration by parts shows that
Z z
−1
−1
Qn (z) − z0−γ e−αz0
Qn (z0 )x−γ Qn+1 (x)e−αx dx = z −γ e−αz
n
n
z0
Z
Z
α z
γ z
−γ −αx−Qn (x)x e dx −Qn (x)x−γ−1 e−αx dx.
n z0
n z0
The induction hypothesis leads to the desired result.

Lemma 2. For all Q(z) ∈ C(z), for all z0 ∈ C× which is not a pole of Q(z), there exists
R(z) ∈ C(z) with at most simple poles on C and whose set of poles in C× is included in
the set of poles in C× of Q(z), and there exist λ(z), µ(z) ∈ C(z) and ν ∈ C such that
Z z
Z z
−αx
−αz
Q(x)e dx = λ(z)e+ µ(z)
R(x)e−αx dx + ν.
z0z0

Proof. Similar to the proof of Lemma 1.

In what follows, for any γ ∈ C \ Z and α ∈ C, we set:
Z z∞X(−α)n z n+1γ−γ −αxEγ,α (z) = z
x e dx =(n − γ + 1)n!
0
n=0and, for γ ∈ Z, we set Eγ,α (z) = e−αz . If γ ∈ Q and α ∈ Q, then Eγ,α (z) is an E-function.
Lemma 3. Considerγ ∈ C, R(z) ∈ C(z), z0 ∈ C× which is not a pole of R(z), C ∈ C,Rz
and ϕ(z) := z γ ( z0 x−γ R(x)e−αx dx + C). Assume that ϕ(z) is meromorphic over C. Then,
there exist λ(z), µ(z) ∈ C(z) such thatϕ(z) = λ(z)e−αz + µ(z)Eγ,α (z) if γ 6∈ Z

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andϕ(z) = λ(z)e−αz + µ(z) if γ ∈ Z.
Proof. Let us first assume that γ 6∈ Z. By Lemma 1, there exists R(z) ∈ C(z) with at most
simple poles on C× and whose set of poles in C× is included in the set of poles in C× of
Q(z), and there exist λ(z), µ(z) ∈ C(z) and C 0 ∈ C such that
Z z
−αz
γ
x−γ R(x)e−αx dx + C 0 z γ .
ϕ(z) = λ(z)e+ µ(z)zz00

If µ(z) = 0, then we must have C = 0 because ϕ(z) − λ(z)e−αz is meromorphic over C,
and, hence, the result is proved. We now assume Rthat µ(z) 6= 0. If ξ ∈ C× is a (simple)zpole of R(z), then ξ is a logarithmic singularity of z0 x−γ R(x)e−αx dx and this contradicts
the fact that ϕ(z) is meromorphic over C. Therefore, R(z) = z −n S(z) for some integern ≥ 0 and some S(z) ∈ C[z]. Hence,
Z z
−αz
γ
ϕ(z) = λ(z)e+ µ(z)zx−γ−n S(x)e−αx dx + C 0 z γ .
z0RzBut, z0 x−γ−n S(x)e−αx dx is a linear combination with coefficients in C of functions ofRz
the form z0 x−γ−n+k e−αx dx for k ∈ N. Using integrations by parts, we conclude thatR z −γ−nxS(x)e−αx dx is, up to an additive constant in C, a linear combination with coefz0Rzeficients in C(z) of z −γ e−αz and z0 x−γ e−αx dx. Hence, there exist λ(z),µe(z), νe(z) ∈ C(z)such that−αzeϕ(z) = λ(z)e+µ
e(z)Eγ,α (z) + νe(z)z γ .
−αze
We must have νe(z) = 0 because ϕ(z) − λ(z)e−µ
e(z)Eγ,α (z) is meromorphic over C. This
yields the desired result.
Let us assume that γ ∈ Z. By Lemma 2, there exists R(z) ∈ C(z) with at most simple
poles on C and whose set of poles in C× is included in the set of poles in C× of Q(z), andthere exist λ(z), µ(z) ∈ C(z) and C 0 ∈ C such thatZ z−αz
R(x)e−αx dx + C 0 z γ .
ϕ(z) = λ(z)e+ µ(z)z0

If µ(z) = 0, the result is proved. We now assumeR that µ(z) 6= 0. If ξ ∈ C is a (simple)zpole of R(z), then ξ is a logarithmic singularity of z0 R(x)e−αx dx and this contradicts thefact thatRϕ(z) is meromorphic over C. Hence, R(z) ∈ C[z]. Using integration by parts, wezsee that z0 R(x)e−αx dx is, up to an additive constant in C, of the form η(z)e−αz for someη(z) ∈ C[z], and this gives the desired result.

We are now able to express (3.1) in terms of hypergeometric functions.
Let us first assume that γ 6∈ Z. Using (3.1) and Lemma 3, we see that there exist
λ(z), µ(z) ∈ C(z) such thatk(z) = l(z)eδz = λ(z)eδz + µ(z)Eγ,α (z)e(α+δ)z .

(3.2)

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A simple linear algebra argument shows that we can choose λ(z) and µ(z) in Q(z). Furthermore, we have the following relation


αz
γα2
αz
;
Eγ,α (z)e =1 F1 (1; γ; −αz) − 1 +zγthis is an easy consequence of the fact that both members of this equality satisfy the same
nonhomogeneous differential equation of order one, namely
zy 0 (z) − (γ + αz)y(z) = z.

(3.3)

Therefore, there exist a(z), b(z), c(z), d(z) ∈ Q(z) such thatf (z) = a(z)1 F1 (1; γ; −αz)eδz + b(z)eδz and g(z) = c(z)1 F1 (1; γ; −αz)eδz + d(z)eδz . (3.4)
It remains to prove that a(z), b(z), c(z), d(z) belong to Q[z, z −1 ]. Assume that ξ(∈ Q)
is a non-zero pole of a(z) or b(z). Let us denote by n the order of ξ as a pole of a(z).
Let us denote by m the order of ξ as a pole of b(z). Let us first assume that m > n.
mThen, multiplying (the first) equation (3.4) by (z − ξ) and letting z = ξ, we get 0 =mδz(z − ξ) b(z)e |z=ξ and this is a contradiction. So, we have n ≤ m. Then, multiplying(the first) equation (3.4) by (z − ξ)n and letting z = ξ, we obtain that 1 F1 (1; γ; −αξ)
belongs to Q, and this is a contradiction by [14, p. 192, Theorem 3]. Hence, a(z) and b(z)do not have poles on C× and, hence, belong to Q[z, z −1 ]. A similar argument shows thatc(z) and d(z) belong to Q[z, z −1 ].
We shall now assume that γ ∈ Z. Using (3.1) and Lemma 3, we see that there exist
λ(z), µ(z) ∈ C(z) such thatk(z) = λ(z)eδz + µ(z)e(α+δ)z .

(3.5)

It is easily seen that one can choose λ(z), µ(z) ∈ Q(z).
a(z), b(z), c(z), d(z) ∈ Q(z) such that

It follows that there exist

f (z) = a(z)e(α+δ)z + b(z)eδz and g(z) = c(z)e(α+δ)z + d(z)eδz .
The proof of the fact that one can choose a(z), b(z), c(z) and d(z) in Q[z, z −1 ] is analogous
to the proof of the similar result in the case γ 6∈ Z, using Lindemann’s Theorem.
4. Proof of Theorem 3
So, we assume that f (z) and g(z) are algebraically dependent over Q(z). It follows
that the functions k(z) and k 0 (z) are algebraically dependent over Q(z). Therefore, thedifferential Galois group of M over Q(z) does not contain SL2 (Q) (for an introduction todifferential Galois theory, we refer to [4, 15] and the references therein). It follows from
Kovacic’s [8, §1.2, Theorem] that one of the following cases hold:
Case 1: M has a nonzero solution ζ(z) such that ζ 0 (z) = ω(z)ζ(z) for some ω(z) ∈ Q(z).
0Case 2: M has a basis of solutions (ζ(z), ζ(z)) such that ζ 0 (z) = ω(z)ζ(z) and ζ (z) =ω(z)ζ(z) where ω(z) is an algebraic function of degree 2 on Q(z), and ω(z) is its
Galois conjugate.

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Case 3: M has a basis of solutions of the form (ζ(z)ζ1 (z), ζ(z)ζ2 (z)) where ζ(z) satisfies
ζ 0 (z) = ω(z)ζ(z) for some ω(z) ∈ Q(z) and where ζ1 (z) and ζ2 (z) are algebraic
functions over Q(z).
We shall now consider each of these cases (we will prove that the last two cases are impossible: the first case is automatically satisfied and, hence, M is automatically reducible
over Q(z)).
4.1. Case 1. In this case, the operator M is reducible over Q(z) (indeed, by euclideandddivision, there exists η(z) ∈ Q(z) such that M = dz − η(z) dz − ω(z) ). We have seen
in section 3 that one of the following properties holds true:(i) There exist a(z), b(z), c(z), d(z) ∈ Q[z, z −1 ] and α, β ∈ Q such thatf (z) = a(z)eαz + b(z)eβz and g(z) = c(z)eαz + d(z)eβz .
(ii) There exist a(z), b(z), c(z), d(z) ∈ Q[z, z −1 ], γ ∈ Q \ Z, α ∈ Q and δ ∈ Q such thatf (z) = (a(z)1 F1 (1; γ; αz) + b(z))eδz and g(z) = (c(z)1 F1 (1; γ; αz) + d(z))eδz .
If (i) is satisfied, then Theorem 3 is proved. We shall now assume that (ii) is satisfied. The
difference between (ii) and the desired conclusion is the multiplicative factor eδz . Actually,
if α = 0 or µ(z) = 0, then we can easily conclude: in this case, we can take a(z) = c(z) = 0
and, hence, f (z) and g(z) have the expected form. We now assume that α 6= 0 andµ(z) 6= 0. In order to conclude the proof, it is sufficient to show that δ = 0. We will provethat this is indeed the case with the help of the following lemma.
Lemma 4. For all γ ∈ C \ Z, α ∈ C× and δ ∈ C× , the functions Eγ,α (z)e(α+δ)z and eδz
are algebraically independent over C(z).
Proof. We consider the differential system associated with the differential equation (3.3):
γ/z + α 10
Y = A(z)Y with A(z) =.
00
A fundamental matrix of solutions of this system is given by γ αzz eEγ,α (z)eαz.
Y=
01
We let G be the differential Galois group over C(z) of the above differential system, which
is seen as an algebraic subgroup of GL2 (C) via Y. It is easily seen that



a b
×
G=
| a ∈ C ,b ∈ C .
0 1
We now consider the differential Galois group H over C(z) of the differential system
Z 0 = (A(z) + δI2 )Z,
which is seen as an algebraic subgroup of GL2 (C) via the fundamental matrix of solutions
 γ (α+δ)zz eEγ,α (z)e(α+δ)zδz
Z=e Y=.
0eδz

9Since the derived subgroup (G, G) of G is equal to the group of unipotent upper triangular matrices U2 ⊂ GL2 (C), we get that (H, H) = U2 and, hence, U2 ⊂ H. Therefore, the (differential) field extension C(z, z γ e(α+δ)z , Eγ,α (z)e(α+δ)z , eδz )/C(z, z γ e(α+δ)z , eδz )is transcendental. Whence the desired result.

Recall that we want to prove that δ = 0 under the assumptions that α 6= 0 and µ(z) 6= 0.
We know that k(z) and k 0 (z) = u(z) + ω(z)k(z) are algebraically dependent over Q(z).
Therefore, k(z) and u(z) are algebraically dependent over Q(z). Using the equality k(z) =λ(z)eδz + µ(z)Eγ,α (z)e(α+δ)z and the facts that u(z) = r(z)eδz with r(z) 6= 0 and thatµ(z) 6= 0, we deduce that Eγ,α (z)e(α+δ)z and eδz are algebraically dependent over Q(z).
Lemma 4 implies that δ = 0, as expected.
4.2. Case 2. We are going to prove that this case does not occur.
Sincepω(z) has pdegree 2 over Q(z), there exist r(z),ps(z) ∈ Q(z) such that ω(z) =r(z) + s(z) and s(z) 6∈ Q(z). Hence, ω(z) = r(z)− s(z).
RRps(z) some primitive integrals of r(z)In what follows, we will denote by r(z) andRR√pand s(z) respectively
such that ζ(z) = e r(z)+ s(z) . We can and will assume thatRR√ζ(z) = e r(z)− s(z) .
We claim that s(z) = z n q(z)2 for some n ∈ Z \ 2Z and q(z) ∈ Q(z)× . Indeed, writeQ×ni
with c ∈ Q , s1 , . . . , sm√s(z) = c m∈ Q pairwise distinct and n1 , . . . , nm ∈ Z.
i=1 (z − si )Rs(z)If si 6= 0, then ni ∈ 2Z because, otherwise, e2= ζ(z)/ζ(z) would have a non trivial
monodromy around si , and this would be contradiction since both ζ(z) and ζ(z) havetrivial monodromy around si in virtue of Proposition 1. Therefore,s(z) = z n q(z)2 forpsome n ∈ Z and q(z)p∈ Q(z)× . We have n ∈ Z \ 2Z because s(z) 6∈ Q(z).
Rs(z) = z 1/2 h∞ (z) (up to an additive constant, that we take equal
If follows thattoR 0) for some germ of meromorphic function h∞ (z) at ∞. Moreover, near ∞, we havee ω = z γ t∞ (z)ep(z) for some γ ∈ Q, some germ of analytic function t∞ (z) at ∞ and some
1/2p(z) ∈ Q[z]. So, ζ(z) = z γ t∞ (z)ep(z)+z h∞ (z) .
Let λ, µ ∈ C be such that k(z) = λζ(z) + µζ(z). We distinguish two cases:• If h∞ (z) is analytic and vanishes at ∞, then e−p(z) k(z) has moderate growth at∞. Since e−p(z) k(z) is an entire function, it is a polynomial. So, k(z) satisfies afirst order linear differential equation with coefficients in Q(z), which is excluded
by assumption.
1/2• Otherwise, we see that ζ(z) = z γ g∞ (z 1/2 )eq(z ) where g∞ is a germ of meromorphic
function at ∞ and q(z 1/2 ) ∈ Q[z 1/2 ]\Q[z]. But, Proposition 1 implies that q(z 1/2 ) =αi z + β for some i ∈ {1, 2} and β ∈ C, whence a contradiction.

4.3. Case 3. We are going to prove that this case does not occur.
Since k(z) is a C-linear combination of ζ(z)ζ1 (2) and ζ(z)ζ2 (z), we see that k(z) =e where ζ(z)e is an algebraic function over C(z). Since ζ 0 (z) = ζ(z)ω(z) with ω(z) ∈ζ(z)ζ(z)Q(z), we get that ζ(z) = m(z)er(z) for some m(z) algebraic over C(z) and some r(z) ∈ Q(z).

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r(z)
So, k(z) = m(z)eewhere m(z)eis algebraic over C(z). Since k(z) is an entire function, wemust have r(z) ∈ Q[z]. Therefore, the function e−r(z) k(z) is entire and algebraic over C(z)and, hence, e−r(z) k(z) ∈ Q[z]. If follows that k(z) satisfies a linear differential equation oforder 1 with coefficients in Q(z), and this is excluded by hypothesis.

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T. Rivoal, Institut Fourier, Université Grenoble 1, CNRS UMR 5582, 100 rue des Maths,
BP 74, 38402 Saint-Martin d’Hères cedex, France.
Tanguy.Rivoal@ujf-grenoble.fr
J. Roques, Institut Fourier, Université Grenoble 1, CNRS UMR 5582, 100 rue des Maths,
BP 74, 38402 Saint-Martin d’Hères cedex, France.
Julien.Roques@ujf-grenoble.fr