HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER
EQUATIONSTHOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

Abstract. The last years have seen a growing interest from mathematiciansin Mahler functions. This class of functions includes the generating series of theautomatic sequences. The present paper is concerned with the following problem, which is omnipresent in combinatorics: a set of Mahler functions u1 , ..., unbeing given, are u1 , ..., un and their successive derivatives algebraically independent? In this paper, we give general criteria ensuring an affirmative answer
to this question. We apply our main results to the generating series attachedto the so-called Baum-Sweet and Rudin-Shapiro automatic sequences. In particular, we show that these series are hyperalgebraically independent i.e. thatthese series and their successive derivatives are algebraically independent. Our
approach relies of the parametrized difference Galois theory (in this context,
the algebro-differential relations between the solutions of a given Mahler equation are reflected by a linear differential algebraic group).

Contents
Introduction
1. Mahler equations and difference Galois theory
1.1. Difference Galois theory
1.2. More specific results about Mahler equations
2. Parametrized difference Galois theory
2.1. Differential algebra
2.2. Difference-differential algebra
2.3. Parametrized difference Galois theory
2.4. Hypertranscendency criteria for equations of order one
2.5. Isomonodromy and projective isomonodromy
3. Hypertranscendence of solutions of Mahler equations
3.1. Homogeneous Mahler equations of order one
3.2. Mahler equations of higher order with large classical difference Galois
group
4. Applications
4.1. User-friendly hypertranscendence criteria
4.2. The Baum-Sweet sequence
4.3. The Rudin-Shapiro sequence
4.4. Direct sum of the Baum-Sweet and of the Rudin Shapiro equations
References

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Date: July 9, 2015.
2010 Mathematics Subject Classification. 39A06,12H10.
Key words and phrases. Mahler functions, Automatic sequences, Difference Galois theory,
Parametrized difference Galois theory.
The first (resp. the second) author would like to thank the ANR-11-LABX-0040-CIMI within
the program ANR-11-IDEX-0002-0 for its total (resp. partial) support. The second authors work
is also supported by ANR Iso-Galois.
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THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

Introduction
This paper grew out of an attempt to understand the algebraic relations between
classical Mahler functions and their successive derivatives. By Mahler function, wemean a function f (z) such that(1)nan (z)f (z p ) + an−1 (z)f (z pn−1

) + · · · + a0 (z)f (z) = 0

for some integers p ≥ 2, n ≥ 1, and some a0 (z), . . . , an (z) ∈ C(z) witha0 (z)an (z) 6= 0.
The study of this class of functions was originally motivated by the work of
Mahler in [Mah29, Mah30a, Mah30b] about the algebraic relations between special
values at algebraic points of Mahler functions. This arithmetic aspect of the theoryof the Mahler functions was developed further by several authors, e.g. Becker,
Kubota, Loxton, van der Poorten, Masser, Nishioka, Töfer. We refer to Nishioka’sbook [Nis96] and Pellarin’s paper [Pel09] for more informations and references. We
shall simply mention that, quite recently, Philippon [Phi15] proved a refinementof Nishioka’s analogue of the Siegel-Shidlovski theorem, in the spirit of Beukers’
refinement of the Siegel-Shidlovski theorem [Beu06]. Roughly speaking, it saysthat the algebraic relations over Q between the above-mentioned special values
come from algebraic relations over Q(z) between the functions themselves. These
functional relations are at the hearth of the present paper.
The attractiveness of the theory of Mahler functions comes from (to alarge extend) its close connection with automata theory: the generating seriesPNf (z) = k≥0 sk z k of any p-automatic sequence (sk )k≥0 ∈ Q (and, actually, of any
p-regular sequence) is a Mahler function; see Mendès France’s [MF80], Adamczewski
and Bell’s [AB13], and Becker’s [Bec94], and the references therein, especially to theworks of Dumas and Randé. The famoust examples are the generating series of theThue-Morse, the paper-folding, the Baum-Sweet and the Rudin-Shapiro sequences(see Allouche and Shallit’s book [AS03]). Note that all of these classical series areassociated to Mahler equations of order n ≤ 2.
The Mahler functions also appear in many other circumstances such as the combinatorics of partitions, the enumeration of words and the analysis of algorithms ofthe type divide and conquer; see for instance [DF96] and the references therein.
It is a classical problem (in combinatorics in particular) to determine whether
or not a given generating series is transcendental or even hypertranscendental overC(z)1.
The hypertranscendence over C(z) of Mahler functions solutions of inhomogeneous Mahler equations of order one can be studied by using the work of Nishioka [Nis96]; see also the work of Nguyen [Ngu11, Ngu12] via difference Galois
theory. This can be applied to the paper-folding generating series for instance.
The present work started with the observation that, besides this case, very few
was known. For instance, the hypertranscendence of the Baum-Sweet or of the
Rudin-Shapiro generating series was not known (the corresponding Mahler equations are of order two). The main objective of the present work is to develop an
approach, as systematic as possible, in order to prove the hypertranscendence ofsuch series.
To give an idea of the content of this paper, we mention the following result(see Theorem 4.2), which is a consequence of one of our main hypertranscendence
criteria. In what follows, we consider the field K = ∪j≥1 C(z 1/j ) endowed with thefield automorphism φ given by φ(f (z)) = f (z p ). We obtain in this way a difference
1We say that a series f (z) ∈ C((z)) is hypertranscendental over C(z) if f (z) and all its derivatives are algebraically independent over C(z).

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

3field with field of constants Kφ = C, and we have at our disposal a difference Galois
theory over K (see Section 1.1).
Theorem. Assume that the difference Galois group over K of the Mahler equation(1) contains SLn (C) and that an (z)/a0 (z) is a monomial. Let f (z) ∈ C((z)) ben−1
a nonzero solution of (1). Then, the series f (z), f (z p ), . . . , f (z p ) and all theirsuccessive derivatives are algebraically independent over C(z). In particular, f (z)is hypertranscendental over C(z).
The hypothesis that an (z)/a0 (z) is a monomial is satisfied in any of the abovementioned cases. Moreover, in the case n = 2, there is an algorithm to determine
whether or not the difference Galois group over K of equation (1) contains SL2 (C);see [Roq15]. It turns out that the difference Galois groups involved in the BaumSweet and in the Rudin-Shapiro cases both contain SL2 (C) (see [Roq15, Section 9]).
Therefore, we have the following consequences of the above theorem (see Theorems
4.3 and 4.4). In what follows, we denote by fBS (z) and fRS (z) the generating seriesof the Baum-Sweet and of the Rudin-Shapiro sequences.
Corollary. The series fBS (z), fBS (z 2 ) and all their successive derivatives are
algebraically independent over C(z). In particular, fBS (z) is hypertranscendental
over C(z).
Corollary. The series fRS (z), fRS (−z) and all their successive derivatives are
algebraically independent over C(z). In particular, fRS (z) is hypertranscendental
over C(z).
Actually, our methods also allow to study the relations between these series. We
prove the following result (see Theorem 4.6).
Corollary. The series fBS (z), fBS (z 2 ), fRS (z), fRS (−z) and all their successive
derivatives are algebraically independent over C(z).
We shall now say a few words about the proofs of these results. Our approach relies on the parametrized difference Galois theory developed by Hardouin and Singer
in [HS08]. Roughly speaking, to the difference equation (1), they attach a linear dife of C – called the parametrized
ferential algebraic group over a differential closure C
difference Galois group – which reflects the algebro-differential relations between
the solutions of the equation. The above theorem is actually a consequence ofthe following purely Galois theoretic statement (see Section 3.2 for more general
results).
Theorem. Assume that the difference Galois group over K of the Mahler equation(1) contains SLn (C) and that an (z)/a0 (z) is a monomial. Then, the parametrized
e and C× SLn (C).
edifference Galois group of equation (1) is caught between SLn (C)Roughly speaking, the fact that the parametrized difference Galois group ofe says that the algebro-differential relations between
equation (1) contains SLn (C)the elements of a basis f1 , . . . , fn of solutions (in a suitable sense) of the equation(1) are generated by the relations satisfied by the determinant of the associatedi−1
Wronskian matrix (fj (z p ))1≤i,j≤n . In particular, there is no nontrivial algebrodifferential relations between the entries of a given column of this matrix, and this isexactly the conclusion of the first theorem stated in this introduction (with f1 = f ).
Note that, in order to use the parametrized difference Galois theory developed
by Hardouin and Singer, one cannot work with the base field K endowed withthe automorphism φ and the usual derivation d/dz because φ and d/dz do notcommute. In order to solve this problem, we follow an idea due to Michael Singer,by introducing the field K(log(z)) and the derivation z log(z)d/dz. The use of

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THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

this derivation, the field K(log(z)), Lemma 2.3 and Proposition 2.6 below appear
in
proof2 due to M. Singer of the fact that the Mahler function
P an unpublishedpn
is hypertranscendental.
n≥0 zThis paper is organized as follows. Section 1 contains reminders and complements on difference Galois theory. Section 2 starts with reminders andcomplements on parametrized difference Galois theory. Then, we state and proveuser-friendly hypertranscendence criteria for general difference equations of orderone. We finish this section with complements on (projective) isomonodromy for
general difference equations from a Galoisian point of view. In Section 3, we firststudy the hypertranscendence of the solutions of Mahler equations of order 1. We
then come to higher order equations and give our main hypertranscendence criteriafor Mahler equations. Section 4 provides user-friendly hypertrancendence criteria
and is mainly devoted to applications of our main results to the generating seriesof classical automatic sequences.
Acknowledgements. We would like to thank Michael Singer for discussions andsupport of this work.
General conventions. All rings are commutative with identity and contain thefield of rational numbers. In particular, all fields are of characteristic zero.
1. Mahler equations and difference Galois theory
1.1. Difference Galois theory. For details on what follows, we refer to [vdPS97,
Chapter 1].
A φ-ring (R, φ) is a ring R together with a ring automorphism φ : R → R. An
ideal of R stabilized by φ is called a φ-ideal of (R, φ). If R is a field, then (R, φ)is called a φ-field. To simplify the notations, we will, most of the time, write R
instead of (R, φ).
The ring of constants of the φ-ring R is defined by
Rφ := {f ∈ R | φ(f ) = f }.
If Rφ is a field, it is called the field of constants.
eee φ)A φ-morphism (resp. φ-isomorphism) from the φ-ring (R, φ) to the φ-ring(R,ee
is a ring morphism (resp. ring isomorphism) ϕ : R → R such that ϕ ◦ φ = φ ◦ ϕ.
ee is a R-φ-algebra if Re φ)e is a ring extension of R
Given a φ-ring (R, φ), a φ-ring(R,
e1 , φe1 )and φe|R = φ; in this case, we will often denote φe by φ. Two R-φ-algebras (Reeeand (R2 , φ2 ) are isomorphic if there exists a φ-isomorphism ϕ from (R1 , φe1 ) toe2 , φe2 ) such that ϕ|R = IdR .
(R
We fix a φ-field K such that k := Kφ is algebraically closed. We consider thefollowing linear difference system(2)

φ(Y ) = AY, with A ∈ GLn (K), n ∈ N∗ .

By [vdPS97, §1.1], there exists a K-φ-algebra R such that
1) there exists U ∈ GLn (R) such that φ(U ) = AU (such a U is called afundamental matrix of solutions of (2));2) R is generated, as a K-algebra, by the entries of U and det(U )−1 ;3) the only φ-ideals of R are {0} and R.
2Letter from Michael Singer to the second author (February 25, 2010).

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

5

Such a R is called a Picard-Vessiot ring, or PV ring for short, for (2) over K.
By [vdPS97, Lemma 1.8], we have Rφ = k. Two PV rings are isomorphic asK-φ-algebras. A PV ring R is not always an integral domain. However, there
exist idempotents elements e1 , . . . , es of R such that R = R1 ⊕ · · · ⊕ Rs where theRi := Rei are integral domains which are transitively permuted by φ. In particular,
R has no nilpotent element and one can consider its total ring of quotients QR , i.e.
the localization of R with respect to the set of its nonzero divisors, which can be
decomposed as the direct sum QR = K1 ⊕ · · · ⊕ Ks of the fields of fractions Ki ofthe Ri . The ring QR has a natural structure of R-φ-algebra and we have QR φ = k.
Moreover, the Ki are transitively permuted by φ. We call the φ-ring QR a total
PV ring for (2) over K.
The following lemma gives a characterization of the PV rings.
Lemma 1.1 ([HS08, Proposition 6.17]). Let S be a K-φ-algebra with no nilpotent
element and let QS be its total ring of quotients. If the following properties hold:(1) there exists V ∈ GLn (S) such that φ(V )V −1 = B ∈ GLn (K) and such that
S is generated, as a K-algebra, by the entries of V and by det(V )−1 ,(2) QS φ = k,then S is a PV ring for the difference system φ(Y ) = BY over K.
As a corollary of the above lemma, we findLemma 1.2. Let R be a PV ring over K and let S be a K-φ-subalgebra of R. If
there exists V ∈ GLn (S) such that φ(V )V −1 = B ∈ GLn (K) and such that S is
generated, as a K-algebra, by the entries of V and by det(V )−1 then S is a PV ringfor φ(Y ) = BY over K.
Proof. Since R has no nilpotent element, S has no nilpotent element. By [HS08,
Corollary 6.9], the total ring of quotients QS of S embed into the total ring of
quotients QR of R. Since QR φ = k, we have QS φ = k. Lemma 1.1 yields thedesired result.

The difference Galois group Gal(QR /K) of R over K is the group of K-φautomorphisms of QR commuting with φ:Gal(QR /K) := {σ ∈ Aut(QR /K) | φ ◦ σ = σ ◦ φ}.
Abusing notations, we shall sometimes denote by Gal(QR /F ) the group{σ ∈ Aut(QR /F ) | φ ◦ σ = σ ◦ φ} for F a K-φ-subalgebra of QR .
An easy computation shows that, for any σ ∈ Gal(QR /K), there exists a uniqueC(σ) ∈ GLn (k) such that σ(U ) = U C(σ). By [vdPS97, Theorem 1.13], the faithfulrepresentation
Gal(QR /K) → GLn (k)σ

7→ C(σ)identifies Gal(QR /K) with a linear algebraic subgroup of GLn (k). If we choose
another fundamental matrix of solutions U , we find a conjugate representation.
A fundamental theorem of difference Galois theory ([vdPS97, Theorem 1.13])
says that R is the coordinate ring of a G-torsor over K. In particular, the dimensionof Gal(QR /K) as a linear algebraic group over k coincides with the transcendence
degree of the Ki over K. Thereby, the difference Galois group controls the algebraic
relations satisfied by the solutions.
The following proposition gives a characterization of the normal algebraic subgroups of Gal(QR /K).

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THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

Proposition 1.3. An algebraic subgroup H of Gal(QR /K) is normal if and only
if the φ-ring QR H := {g ∈ QR | ∀σ ∈ H, σ(g) = g} is stable under the action ofGal(QR /K). In this case, the K-φ-algebra QR H is a total PV ring over K and thefollowing sequence of group morphisms is exact
0

/H

ι

/ Gal(QR /K)/ Gal(QR H /K)

π

/ 0,

where ι is the inclusion of H in Gal(QR /K) and π denotes the restriction of theelements of Gal(QR /K) to QR H .
Proof. Assume that H is normal in Gal(QR /K). For all τ ∈ Gal(QR /K), g ∈ QR H ,and σ ∈ H, we haveσ(τ (g)) = τ ((τ −1 στ )(g)) = τ (g).
This shows that QR H is stable under the action of Gal(QR /K). Conversely, assume
that QR H is stable under the action of Gal(QR /K). Then, we can consider the
restriction morphism
π : Gal(QR /K) →σ

Gal(QR H /K)

7→ σ|QR H .

By Galois correspondence (see [HS08, Theorem 6.20]), we have ker(π) = H and,hence, H is normal in Gal(QR /K). The rest of the proof is [vdPS97, Corollary1.30].

Corollary 1.4. Let f be an invertible element of R such that φ(f ) = af for somea ∈ K. Let Qf ⊂ QR be the total ring of quotients of K[f, f −1 ]; this is a total PV
ring for φ(y) = ay over K. Then, Gal(QR /Qf ) is a solvable algebraic group if andonly if Gal(QR /K) is a solvable algebraic group.
Proof. We have QR Gal(QR /Qf ) = Qf (in virtue of the Galois correspondence[vdPS97, Theorem 1.29.3]) and Qf is stable under the action of Gal(QR /K) (because, for all σ ∈ Gal(QR /K), we have σ(f )f −1 ∈ QR φ = k). By Proposition 1.3,
Gal(QR /Qf ) is normal in Gal(QR /K) and the sequence0/ Gal(QR /Qf )ι/ Gal(QR /K)π/ Gal(Qf /K)/0

is exact. Since Gal(Qf /K) ⊂ GL1 (k) is abelian, the group Gal(QR /K) is solvable
if and only if the same holds for Gal(QR /Qf ).

1.2. More specific results about Mahler equations. Now, we restrict ourselves to the Mahlerian context.
We let p ≥ 2 be an integer.
We consider the field
K := ∪j≥1 C z 1/j .
The field automorphism
φ:K

→ K

f (z) 7→ f (z p )gives a structure of φ-field on K such that Kφ = C.
We also consider the field K0 := K(log(z)). The field automorphism
φ : K0f (z, log(z))→ K0
7→ f (z p , p log(z))

gives a structure of φ-field on K0 such that K0φ = C.

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

7

In the sequel, we shall consider Mahler equations above the φ-field K and also
above its φ-field extension K0 . We shall now study the effect of the base extension
for K to K0 on the difference Galois groups.
We first state and prove a lemma.
Lemma 1.5. Let L be a φ-subfield of K0 that contains K. There exists an integerk ≥ 0 such that L = K(log(z)k ).
Proof. The case L = K is obvious (take k = 0). We shall now assume that L 6= K.
Lemma 1.1 ensures that K0 is a total PV ring over K for the equation φ(y) = py.
The action of Gal(K0 /K) on log(z) allows to see Gal(K0 /K) as an algebraic subgroup of C× . Since log(z) is transcendental over K, we have Gal(K0 /K) = C× .
Since L 6= K, the group Gal(K0 /L) is a proper algebraic subgroup of C× and,hence, is a group of roots of the unity. Let k ≥ 1 be an integer such that
Gal(K0 /L) = µk := {c ∈ C× | ck = 1}. Then, log(z)k is fixed by Gal(K0 /L) and,hence, belongs to L by Galois correspondence. Since Gal(K0 /K(log(z)k )) ⊂ µk , we
get that L = K(log(z)k ).

We consider the difference system(3)

φ(Y ) = AY

with A ∈ GLn (C(z)). Let R0 be a PV ring for (3) over K0 ; then QR0 is a total PV
ring for (3) over K0 . Let U ∈ GLn (R0 ) be a fundamental matrix of solutions of(3). Let R be the K-subalgebra of R0 generated by the entries of U and det(U )−1 .
By [HS08, Corollary 6.9], we have QR ⊂ QR0 . Since QR0 φ = K0φ = C, we have
QR φ = C and Lemma 1.1 allows to conclude that R is a PV ring for (3) over K
and QR is a total PV ring for (3) over K.
The restriction morphismι : Gal(QR0 /K0 ) → Gal(QR /K)
is a closed immersion; we will freely identify Gal(QR0 /K0 ) with the subgroupι(Gal(QR0 /K0 )) of Gal(QR /K).
Proposition 1.6. The difference Galois group Gal(QR0 /K0 ) is a normal subgroupof Gal(QR /K) and the quotient Gal(QR /K)/Gal(QR0 /K0 ) is either trivial or isomorphic to C× .
Proof. We set G0 := ι(Gal(QR0 /K0 )) and G := Gal(QR /K). Let us consider00
F := (QR )G = (QR0 )G ∩ QR = K0 ∩ QR . The Galois correspondence [HS08, Theorem 6.20] ensures that G0 = Gal(QR /F ). Since F/K is a φ-subfield extension of K0 /K, Lemma 1.5 ensures that there exists an integer k ≥ 0 such that
F = K(log(z)k ). Since F φ = C, Lemma 1.1 shows that F is a total PV ring over
K for φ(y) = pk y. Using Proposition 1.3, we see that G0 is a normal subgroup of G
and that G/G0 is isomorphic to the difference Galois group over K of φ(y) = pk y,which is trivial if k = 0 and equal to C× otherwise.
Corollary 1.7. If SLn (C) ⊂ Gal(QR /K) then SLn (C) ⊂ Gal(QR0 /K0 ).

2. Parametrized difference Galois theory
We will use standard notions and notations of difference and differential algebra
which can be found in [Coh65] and [vdPS97].

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THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

2.1. Differential algebra. A δ-ring (R, δ) is a ring R endowed with a derivation δ : R → R (this means that δ is additive and satisfies the Leibniz ruleδ(ab) = δ(a)b + aδ(b), for all a, b ∈ R). If R is a field, then (R, δ) is called a δfield. To simplify the notations, we will, most of the time, write R instead of(R, δ).
We denote by Rδ the ring of δ-constants of the δ-ring R, i.e.
Rδ := {c ∈ R | δ(c) = 0}.
If Rδ is a field, it is called the field of δ-constants.
e
e is a R-δ-algebra if Re δ)e is a ring extension of
Given a δ-ring (R, δ), a δ-ring(R,ee
R and δ|R = δ; in this case, we will often denote δ by δ. Let K be a δ-field. If
L is a K-δ-algebra and a field, we say that L/K is a δ-field extension. Let R bea K-δ-algebra and let a1 , . . . , an ∈ R. We denote by K{a1 , . . . , an } the smallest
K-δ-subalgebra of R containing a1 , . . . , an . Let L/K be a δ-field extension andlet a1 , . . . , an ∈ L. We denote by Kha1 , . . . , an i the smallest K-δ-subfield of L
containing a1 , . . . , an .
The ring of δ-polynomials in the differential indeterminates y1 , . . . , yn and withcoefficients in a differential field (K, δ), denoted by K{y1 , . . . , yn }, is the ring ofpolynomials in the indeterminates {δ j yi | j ∈ N, 1 ≤ i ≤ n} with coefficients in K.
Let R be be a K-δ-algebra and let a1 , . . . , an ∈ R. If there exists a nonzero
δ-polynomial P ∈ K{y1 , . . . , yn } such that P (a1 , . . . , an ) = 0, then we say that
a1 , . . . , an are hyperalgebraically dependent over K. Otherwise, we say that
a1 , . . . , an are hyperalgebraically independent over K.
A δ-field k is called differentially closed if, for every (finite) set of δ-polynomials
F, if the system of differential equations F = 0 has a solution with entries in someδ-field extension L, then it has a solution with entries in k. Note that the fieldof δ-constants kδ of any differentially closed δ-field k is algebraically closed. Anye i.e. a differentially closed δ-field extension,
δ-field k has a differential closure k,δe = k.
and we have k
From now on, we consider a differentially closed δ-field k.
A subset W ⊂ kn is Kolchin-closed (or δ-closed, for short) if there exists
S ⊂ k{y1 , . . . , yn } such that
W = {a ∈ kn | ∀f ∈ S, f (a) = 0} .
The Kochin-closed subsets of kn are the closed sets of a topology on kn , called theKolchin topology. The Kolchin-closure of W ⊂ kn is the closure of W in kn for the
Kolchin topology.
Following Cassidy in [Cas72, Chapter II, Section 1, p. 905], we say that a subgroup G ⊂ GLn (k) ⊂ kn×n is a linear differential algebraic group (LDAG) if G
2is the intersection of a Kolchin-closed subset of kn×n (identified with kn ) withGLn (k).
A δ-closed subgroup, or δ-subgroup for short, of an LDAG is a subgroup that is
Kolchin-closed. The Zariski-closure of a LDAG G ⊂ GLn (k) is denoted by G andis a linear algebraic group.
We will use the following fundamental result.
Proposition 2.1 ([Cas72], Proposition 42). Let k be a differentially closed field.
Let C := kδ . A Zariski-dense δ-closed subgroup of SLn (k) is either conjugate toSLn (C) or equal to SLn (k).
We will also use the following results.
Lemma 2.2. Let k be a differentially closed field. Let C := kδ . The normalizer ofSLn (C) in GLn (k) is k× SLn (C).

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

9Proof. Let M ∈ GLn (k) be in the normalizer of SLn (C). Consider N ∈ SLn (C).
We have M N M −1 ∈ SLn (C). In particular, δ(M N M −1 ) = 0, i.e. δ(M )N M −1 −M N M −1 δ(M )M −1 = 0 so M −1 δ(M ) commutes with N . It follows thatM −1 δ(M ) = cIn for some c ∈ k× . So, the entries of M = (mi,j )1≤i,j≤n are
solutions of δ(y) = cy. Let i0 , j0 be such that mi0 ,j0 6= 0. Then, M = mi0 ,j0 M 0with M 0 = mi1,j M ∈ GLn (kδ ) = GLn (C). Hence, the normalizer of SLn (C) in
0 0
GLn (k) is included in k× SLn (C). The other inclusion is obvious.

2.2. Difference-differential algebra. A (φ, δ)-ring (R, φ, δ) is a ring R endowedwith a ring automorphism φ and a derivation δ : R → R (in other words, (R, φ) isa φ-ring and (R, δ) is a δ-ring) such that φ commutes with δ. If R is a field, then(R, φ, δ) is called a (φ, δ)-field. If there is no possible confusion, we will write R
instead of (R, φ, δ).
We have straightforward notions of (φ, δ)-ideals, (φ, δ)-morphisms, (φ, δ)algebras, etc, similar to the notions recalled in Sections 1 and 2.1. We omit thedetails and refer for instance to [HS08, Section 6.2], and to the references therein,for details.
In order to use the parametrized difference Galois theory developed in [HS08],
we will need to work with a base (φ, δ)-field K such that k := Kφ is differentiallyclosed. Most of the common function fields do not satisfy this condition. The
following result shows that we can embed any (φ, δ)-field with algebraically closed
field of constants into a (φ, δ)-field with differentially closed field of constants.
e be aLemma 2.3. Let F be a (φ, δ)-field with k := F φ algebraically closed. Let kedifferentially closed field containing k. Then, the ring k ⊗k F is an integral domainewhose fraction field K is a (φ, δ)-field extension of F such that Kφ = k.
Proof. The first assertion follows from the fact that, since k is algebraically closed,
e is regular.
the extension k/ke ⊗k F and ke in ke ⊗k F via the maps
In what follows, we see F in k
F
f

→
7→

e ⊗k F
k
1⊗f

and

e
k
a

e ⊗k F→ k7→
a ⊗ 1.

The mapse ⊗k F →φ:k(a, b) 7→e ⊗k Fka ⊗ φ(b)ande ⊗k F → ke ⊗k F
δ:k(a, b) 7→ δ(a) ⊗ b + a ⊗ δ(b)e ⊗k F with a structure of F -(φ, δ)-algebra.
are well-defined and endow k
e ⊗k F is
To prove the second statement, we first show that any φ-ideal of keetrivial.
PnLet (ci )i∈I be a k-basis of k. Let I be a nonzero φ-ideal of k ⊗k F and let
w = i=1 ci ⊗ fi be a nonzero element of I with fi ∈ F and nPminimal. Withoutnloss of generality, we can assume that f1 = 1. Since φ(w)−w = i=2 ci ⊗(φ(fi )−fi )is an element of I with fewer terms than w, it must be equal toP
0. This impliesnthat, for all i ∈ {1, . . . , n}, φ(fi ) = fi , i.e. fi ∈ k. Then, w = ( i=1 ci fi ) ⊗ 1 ise ⊗k F and, hence, I = ke ⊗k F .
invertible in k
φee k F } is a nonzero φ-ideal of k⊗e kF,
Let c ∈ K . Since I := {d ∈ k⊗k F | dc ∈ k⊗ee
we mustP have I = k ⊗k F . In particular, 1 ∈ I and, hence, c ∈ k ⊗k F . Writingc = i∈I ci ⊗ fi , we see that φ(c) = c implies φ(fi ) = fi for all i ∈ {1, . . . , n}.
eTherefore, the fi are in k and, hence, c belongs to k.

2.3. Parametrized difference Galois theory. For details on what follows, werefer to [HS08].

10

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUESLet K be a (φ, δ)-field with k := Kφ differentially closed. We consider thefollowing linear difference systemφ(Y ) = AY

(4)

with A ∈ GLn (K) for some integer n ≥ 1.
By [HS08, § 6.2.1], there exists a K-(φ, δ)-algebra S such that
1) there exists U ∈ GLn (S) such that φ(U ) = AU (such a U is called afundamental matrix of solutions of (4));2) S is generated, as K-δ-algebra, by the entries of U and det(U )−1 ;3) the only (φ, δ)-ideals of S are {0} and S.
Such a S is called a parametrized Picard-Vessiot ring, or PPV ring for short, for(4) over K. It is unique up to isomorphism of K-(φ, δ)-algebras. A PPV ring is notalways an integral domain. However, there exist idempotent elements e1 , . . . , es of
R such that R = R1 ⊕ · · · ⊕ Rs where the Ri := Rei are integral domains stable byδ and transitively permuted by φ. In particular, S has no nilpotent element andone can consider its total ring of quotients QS . It can be decomposed as the direct
sum QS = K1 ⊕ · · · ⊕ Ks of the fields of fractions Ki of the Ri . The ring QS has
a natural structure of R-(φ, δ)-algebra and we have QS φ = k. Moreover, the Ki
are transitively permuted by φ. We call the (φ, δ)-ring QS a total PPV ring for (4)over K.
The parametrized difference Galois group Galδ (QS /K) of S over (K, φ, δ) is thegroup of K-(φ, δ)-automorphisms of QS :Galδ (QS /K) := {σ ∈ Aut(QS /K) | φ ◦ σ = σ ◦ φ and δ ◦ σ = σ ◦ δ}.
Note that, if δ = 0, then we recover the difference Galois groups considered in
Section 1.1.
A straightforward computation shows that, for any σ ∈ Galδ (QS /K), there exists
a unique C(σ) ∈ GLn (k) such that σ(U ) = U C(σ). By [HS08, Proposition 6.18],
the faithful representationGalδ (QS /K) →σ

7→GLn (k)C(σ)δ

identifies Gal (QS /K) with a linear differential algebraic subgroup of GLn (k). If
we choose another fundamental matrix of solutions U , we find a conjugate representation.
The parametrized difference Galois group Galδ (QS /K) of (4) reflects the differential algebraic relations between the solutions of (4). In particular, the δ-dimensionof Galδ (QS /K) coincides with the δ-transcendence degree of the Ki over K (see[HS08, Proposition 6.26] for definitions and details).
A parametrized Galois correspondence holds between the δ-closed subgroups ofGalδ (QS /K) and the K-(φ, δ)-subalgebras F of QS such that every nonzero divisorof F is a unit of F (see for instance [HS08, Theorem 6.20]). Abusing notations,
we still denote by Galδ (QS /F ) the group of F -(φ, δ)-automorphisms of QS . The
following proposition is at the heart of the parametrized Galois correspondence.
Proposition 2.4 ([HS08, Theorem 6.20]). Let S be a PPV ring over K. Let F bea K-(φ, δ)-subalgebra of QS such that every nonzero divisor of F is a unit of F .
Let H be a δ-closed subgroup of Galδ (QS /K). Then, the following hold:Galδ (Q /F )S
• QS:= {f ∈ QS | ∀τ ∈ Galδ (QS /F ), τ (f ) = f } = F ;• Galδ (QS /QHS ) = H.

Let S be a PPV ring over K for (4) and let U ∈ GLn (S) be a fundamental matrix
of solutions. Then, the K-φ-algebra R generated by the entries of U and det(U )−1

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

11

is a PV ring for (4) over K and we have QR ⊂ QS . One can identify Galδ (QS /K)
with a subgroup of Gal(QR /K) by restricting the elements of Galδ (QS /K) to QR .
Proposition 2.5 ([HS08], Proposition 2.8). The group Galδ (QS /K) is a Zariskidense subgroup of Gal(QR /K).
2.4. Hypertranscendency criteria for equations of order one. The hypertranscendence criteria contained in [HS08] are stated for (φ, δ)-fields K such thatthe δ-field k := Kφ is differentially closed. Recently some schematic versions (seefor instance [Wib12] or [DVH12]) of [HS08] have been developed and allow to workover (φ, δ)-fields with algebraically closed field of constants. One could use thisschematic approach to show that the hypertranscendence criteria of [HS08] still
hold over (φ, δ)-fields with algebraically closed field of constants (not necessarily
differentially closed). However, for sake of clarity and simplicity of exposition, we
prefer to show that one can deduce these criteria directly from the ones containedin [HS08], via simple descent arguments.
Proposition 2.6. Let K be a (φ, δ)-field with k := Kφ algebraically closed and let(a, b) ∈ K× × K . Let R be a K-(φ, δ)-algebra and let v ∈ R \ {0}.
• If φ(v) − v = b and v is hyperalgebraic over K, then there exist a nonzero
linear homogeneous δ-polynomial L(y) ∈ k{y} and an element f ∈ K suchthat
L(b) = φ(f ) − f.
• Assume moreover that v is invertible in R. If φ(v) = av and if v is hyperalgebraic over K, then there exist a nonzero linear homogeneous δ-polynomial
L(y) ∈ k{y} and an element f ∈ K such that


δ(a)
L
= φ(f ) − f.
a
The converse of either statement is true if Rφ = k.
e be a δ-closure of k. Lemma 2.3
Proof. Let us prove the first statement. Let keassures that L := Frac(k⊗k K) is a (φ, δ)-field extension of K such that Lφ = k̃. Let
L{y} be the ring of δ-polynomials in one variable over L endowed with the structureof L-(φ, δ)-algebra induced by setting φ(y) := y + b. Without loss of generality, we
can assume that R = K{v}. We identify R with K{y}/I for some (φ, δ)-ideal Iof K{y}. Since v is hyperalgebraic over K, we have I 6= {0}. Moreover, we have
I 6= K{y} because R 6= {0}. We claim that (I) ∩ K{y} = I where (I) denotes
the (φ, δ)-ideal generated by I in L{y}. Indeed, choose a K-basis (ci )i∈I of L with
ci0 = 1 for some i0 ∈ I. Note that (ci )i∈I isPalso a basis of the K{y}-module L{y}.
Then, (I) consists of the sums of the formai ci with ai ∈ I. It follows easily that(I) ∩ K{y} = I, as claimed. In particular, (I) is a proper ideal of L{y} and, hence,is contained in some maximal (φ, δ)-ideal M of L{y}. The ring S := L{y}/M is a
PPV ring over L for φ(y) = y + b. The image u of y in S is hyperalgebraic overL (because M 6= {0}) and is a solution of φ(y) = y + b. By [HS08, Propositione3.1], there exist a nonzero linear homogenous δ-polynomial L0 (y) ∈ k{y}and g ∈ L
such that(5)

L0 (b) = φ(g) − g.

Let (hi )i∈I be a k-basis of K. Without loss of generality, we can assume thatPrnXai ⊗ hi
n+1i
L0 (y) = δ(y) +ci δ (y) and g := Pi=1s
i=1 bi ⊗ hi
i=0

12

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUESe and b1 = 1. It is clear that the equation (5) can be rewritten as
where ai , bi , ci ∈ k
an equation of the formX
Pj ((ai )i∈{1,...,r} , (bi )i∈{2,...,s} , (ci )i∈{1,...,n} ) ⊗ hj = 0
jwhere the Pj are polynomials with coefficients in k. Thus, for all j,Pj ((ai )i∈{1,...,r} , (bi )i∈{2,...,s} , (ci )i∈{1,...,n} ) = 0.
Since k is algebraically closed, there exist αi , βi , γi ∈ k such that, for all j, we have
Pj ((αi )i∈{1,...,r} , (βi )i∈{2,...,s} , (γi )i∈{1,...,n} ) = 0.
Set β1 := 1. Then, we see that
L(y) := δ

n+1

n
X

P
αi ⊗ hi(y) +γi δ (y) and f := Pii βi ⊗ hi
i=0i

satisfy the conclusion of the first part of the proposition.
Conversely, if Rφ = k and if there exist a nonzero linear homogeneous δpolynomial L(y) ∈ k{y} and an element f ∈ K such that L(b) = φ(f ) − f , thenL(v) − f belongs to Rφ = k. Since L(y) is nonzero, v is differentially algebraic overK.
The proof of the second statement is similar. It can also be deduced from theδv
δafirst statement by noticing that, if φ(v) = av then φ( δvv ) = v + a and by using
the fact that v is hyperalgebraic over K if and only if the same holds for δvv .
Remark 2.7. In Proposition 2.6, we require that v is invertible in R. This assumption is automatically satisfied if we assume that R is similar to a total PPV ring.
More precisely, assume that R = ⊕x∈Z/sZ Kx , where the Kx are δ-field extensionsof K, such that φ(Kx ) = Kx+1 . Then, any nonzero solution v ∈ R of φ(y) = ayPfor a ∈ K× is invertible. Indeed, v = x∈Z/sZ vx for some vx ∈ Kx . Since v 6= 0,
there exists x0 ∈ Z/sZ such that vx0 6= 0. From the equation φ(v) = av, we getφ(vx0 −1 ) = avx0 . So, vx0 −1 6= 0. Iterating this argument, we see that, for allx ∈ Z/sZ, vx 6= 0. Hence, v is invertible in R.
2.5. Isomonodromy and projective isomonodromy. Let K be a (φ, δ)-field
e be a δ-closure of k. Let C := keδ be thewith k := Kφ algebraically closed. Let ke Lemma 2.3 ensures that k
e ⊗k K is an(algebraically closed) field of constants of k.
eintegral domain and that L := Frac(k ⊗k K) is a (φ, δ)-field extension of K suche We let QS be the total ring of quotients of a PPV ring S over L ofthat Lφ = k.
the difference systemφ(Y ) = AY
where A ∈ GLn (K).
Proposition 2.8. The following statements are equivalent:(1) the group Galδ (QS /L) is conjugate to a subgroup of GLn (C);e ∈ Ln×n such that(2) there exists B(6)

e = ABAe −1 + δ(A)A−1 ;φ(B)(3) there exists B ∈ Kn×n such that
φ(B) = ABA−1 + δ(A)A−1 .

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

13

Proof. The equivalence between (1) and (2) is [HS08, Proposition 2.9]. In order tocomplete the proof of the proposition, it remains to prove that, if the equation (6)e in Ln×n , then it has a solution in Kn×n . This follows from an
has a solution B
argument similar to the descent argument used in the proof of Proposition 2.6. 
We now consider a “projective isomonodromic” situation, in the spirit of [MS13].
Let U ∈ GLn (S) be a fundamental matrix of solutions of φ(Y ) = AY and let
d := det(U ) ∈ S × .
Proposition 2.9. Assume that the difference Galois group of φ(Y ) = AY overthe φ-field K contains SLn (k) and that the parametrized difference Galois groupof φ(y) = det(A)y over the (φ, δ)-field L is included in C× . Then, we have thefollowing alternative:(1) Galδ (QS /L) is conjugate to a subgroup of GLn (C) that contains SLn (C);e that contains SLn (k).
e(2) Galδ (QS /L) is equal to a subgroup of C× SLn (k)n×nMoreover, the first case holds if and only if there exists B ∈ Ksuch that(7)

φ(B) = ABA−1 + δ(A)A−1 .

Proof. Let R be the L-φ-algebra generated by the entries of U and by det(U )−1 ;this is a PV ring for φ(Y ) = AY over the φ-field L. Using [CHS08, Corollary2.5], we see that the hypothesis that the difference Galois group of φ(Y ) = AYe So,over the φ-field K contains SLn (k) implies that Gal(QR /L) contains SLn (k).
δdereGal(QR /L)= SLn (k). Since Gal (QS /L) is Zariski-dense in Gal(QR /L), we
have that Galδ (QS /L)derδ (this is the Kolchin-closure of the derived subgroup ofe ByGalδ (QS /L); see Section 4.4.1) is Zariski-dense in Gal(QR /L)der = SLn (k).
δderδeProposition 2.1, Gal (QS /L)
is either conjugate to SLn (C) or equal to SLn (k).
δδderδ
Since Gal (QS /L)
is a normal subgroup of Gal (QS /L), Lemma 2.2 ensures thate× SLn (C) containing SLn (C)Galδ (QS /L) is either conjugate to a subgroup of ke containing SLn (k).
e But, the parametrized
or is equal to a subgroup of GLn (k)difference Galois group of φ(y) = det(A)y over L, which can be identified withdet(Galδ (QS /L)), is contained in C× . Therefore, Galδ (QS /L) is either containede Whence the first part of the proposition.
in C× SLn (C) = GLn (C) or in C× SLn (k).
The second part of the proposition follows from Proposition 2.8.
Proposition 2.10. Assume that the difference Galois group of φ(Y ) = AY overthe φ-field K contains SLn (k) and that the parametrized difference Galois group ofe× . We have the following alternative:φ(y) = det(A)y over the (φ, δ)-field L is k
e× SLn (C);(1) Galδ (QS /L) is conjugate to kδe(2) Gal (QS /L) is equal to GLn (k).
Moreover, the first case holds if and only if there exists B ∈ Kn×n such that1
(8)
φ(B) = ABA−1 + δ(A)A−1 − δ(det(A)) det(A)−1 In .
n
Proof. Arguing as for the proof of Proposition 2.9, we see that Galδ (QS /L) is either
e× SLn (C) containing SLn (C) or equal to a subgroupconjugate to a subgroup of ke containing SLn (k).
e Now, the first part of the proposition follows fromof GLn (k)the fact that the parametrized difference Galois group of φ(y) = det(A)y over L,
e× .
which can be identified with det(Galδ (QS /L)), is equal to k
We shall now prove that the first case holds if and only if there exists B ∈ Ln×n
such that1(9)φ(B) = ABA−1 + δ(A)A−1 − δ(det(A)) det(A)−1 In .
n

14

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

e× SLn (C). So, there exists
Let us first assume that Galδ (QS /L) is conjugate to ka fundamental matrix of solutions U ∈ GLn (S) of φ(Y ) = AY such that, for all
e× and Mσ ∈ SLn (C) such that σ(U ) = U ρσ Mσ .
σ ∈ Galδ (QS /L), there exist ρσ ∈ knNote that σ(d) = dρσ . Easy calculations show that the matrix
B := δ(U )U −1 −1
δ(d)d−1 In ∈ S n×nn

is left invariant by the action of Galδ (QS /L), and, hence, belongs to Ln×n in virtueof Proposition 2.4, and that B satisfies equation (9).
Conversely, assume that there exists B ∈ Ln×n satisfying equation (9). Consider
B1 = B +1
δ(d)d−1 In ∈ S n×n .
nNote thatφ(B1 ) = AB1 A−1 + δ(A)A−1 .
Let U ∈ GLn (S) be a fundamental matrix of solutions of φY = AY . We
en×n such thathave φ(δ(U ) − B1 U ) = A(δ(U ) − B1 U ). So, there exists C ∈ ke is differentially closed, we can find D ∈ GLn (k)e suchδ(U ) − B1 U = U C. Since kthat δ(D) + CD = 0. Then, V := U D is a fundamental matrix of solutions ofe
φY = AY such that δ(V ) = B1 V. Consider σ ∈ Galδ (QS /L) and let Mσ ∈ GLn (k)be such that σ(V ) = V Mσ ; note that σ(d) = dρσ where ρσ = det(Mσ ). On
the one hand, we have σ(δ(V )) = σ(B1 V ) = (B1 + n1 δ(ρσ )ρ−1
σ In )V Mσ . On theother hand, we have σ(δ(V )) = δ(σ(V )) = δ(V Mσ ) = B1 V Mσ + V δ(Mσ ). So,1−1n δ(ρσ )ρσ Mσ = δ(Mσ ). So, the entries of Mσ = (mi,j )1≤i,j≤n are solutions of
10
δ(y) = n δ(ρσ )ρ−1
σ y. Let i0 , j0 be such that mi0 ,j0 6= 0. Then, Mσ = mi0 ,j0 M witheδ ) = GLn (C), whence the desired result.
M 0 = mi1,j Mσ ∈ GLn (k0 0
To conclude the proof, we have to show that if (9) has a solution B in Ln×n
then it has a solution in Kn×n . This can be proved by using an argument similar
to the descent argument used in the proof of Proposition 2.6.


3. Hypertranscendence of solutions of Mahler equations
Now, we focus our attention on Mahler equations.

We use the notations of Section 1.2: p ≥ 2 is an integer, K := ∪j≥1 C z 1/j
and K0 := K(log(z)). We endow K with the structure of φ-field given byφ(f (z)) := f (z p ). We endow K0 := K(log(z)) with the structure of φ-field given byφ(f (z, log(z))) := f (z p , p log(z)). We have Kφ = K0φ = C.
The derivationdδ := z log(z)dzgives a structure of (φ, δ)-field over K0 (so, δ commutes with φ, and this is thereason why we work with δ instead of a simplest derivation). We also setϑ := zd.
dz

e a differential closure of (C, δ). We have Ce δ = C. As in Lemma
We denote by Ce ⊗C K0 ) = ∪j≥1 C
e z 1/j (log(z)), which is a (φ, δ)-field
2.3, we consider L = Frac(C0φeextension of K such that L = C.

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

15

3.1. Homogeneous Mahler equations of order one. In this section, we consider the difference equation of order one
φ(y) = ay(10)
×where a ∈ C(z) . We let S be a PPV ring over L for the equation (10).
Since S is an L-(φ, δ)-algebra, it can be seen as a C(z)-ϑ-algebra, (i.e. over the1
differential field (C(z), ϑ)) by letting ϑ acts as log(z)δ.
Proposition 3.1. Let R be a K0 -(φ, δ)-algebra such that Rφ = C. Let u be an invertible element of R such that φ(u) = au. The following statements are equivalent:(1) u is hyperalgebraic over (C(z), ϑ)3;
(2) Galδ (QS /L) is conjugate to a subgroup of C× ;(3) there exists d ∈ C(z) such that ϑ(a) = a(pφ(d) − d);)(4) there exist c ∈ C× , m ∈ Z and f ∈ C(z)× such that a = cz m φ(ff .
Proof. We first prove the implication (3) ⇒ (2). Assume that there existsd ∈ C(z) such that ϑ(a) = a(pφ(d) − d). Then, d1 := d log(z) ∈ K0 satisfies
δ(a) = a(φ(d1 ) − d1 ) and, hence, Galδ (QS /L) is conjugate to a subgroup of C× invirtue of Proposition 2.8.
We now prove (2) ⇒ (3). We assume that Galδ (QS /L) is conjugate to a subgroupof C× . By Proposition 2.8, there exists d1 ∈ K0 such that δ(a) = a(φ(d1 ) − d1 ).
Therefore, we haveϑ(a) = a(pφ(d2 ) − d2 )

(11)

d1
with d2 := log(z)∈ K0 . We shall now prove that there exists d3 ∈ K such that
ϑ(a) = a(pφ(d3 ) − d3 ). Indeed, let u(X, Y ) ∈ C(X, Y ), k ≥ 1 and v ∈ C(X) be suchthat d2 = u(z 1/k , log(z)) and ϑ(a)a = v(z). The equation (11) can be rewritten as
followsv(z) = pu(z p/k , p log(z)) − u(z 1/k , log(z)).
Since z 1/k and log(z) are algebraically independent over C, we getv(X k ) = pu(X p , pY ) − u(X, Y ).
We see u(X,P Y ) has an element of C(X)(Y ) ⊂ C(X)((Y )) as follows:u(X, Y ) = j≥−N uj (X)Y j , for some N ∈ Z. We haveXv(X k ) = pu(X p , pY ) − u(X, Y ) =(pj+1 uj (X p ) − uj (X))Y j .
j≥−NEquating the coefficients of Y

0

in this equality, we obtain

pu0 (X p ) − u0 (X) = v(X k ).
Hence, d3 := u0 (z 1/k ) has the required property.
We claim that d3 belongs to C(z). Indeed, suppose to the contrary thatd3 6∈ C(z).
Let k ≥ 2 be such that d3 ∈ C(z 1/k ). We see d3 in C((z 1/k )):P
d3 = j≥−N d3,j z j/k for some N ∈ Z. Let j0 ∈ Z be such that k 6 |j0 and d3,j0 6= 0,
with |j0 | minimal for this property. Then, the coefficient of z j0 /k in pφ(d3 ) − d3 isnonzero, and this contradicts the fact that pφ(d3 ) − d3 belongs to C(z). This proves(3).
We now prove (3) ⇒ (1). Assume that there exists d ∈ C(z) such that
ϑ(a) = a(pφ(d) − d). Then, d1 := d log(z) ∈ K0 satisfies δ(a) = a(φ(d1 ) − d1 )and, hence, Proposition 2.6 ensures that u is hyperalgebraic over (K0 , δ). Therefore, u is hyperalgebraic over (K0 , ϑ) and the conclusion follows from the fact that
(K0 , ϑ) is hyperalgebraic over (C(z), ϑ).
3Of course, u is hyperalgebraic over (C(z), ϑ) if and only if u is hyperalgebraic over (C(z), d/dz).

16

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

We now prove (1) ⇒ (3). Proposition 2.6, applied to the differencePν equation
φ(y) = ay over the (φ, δ)-field K0 , ensures that there exist L1 := i=1 βi δ i withcoefficients β1 , . . . , βν = 1 in C and g1 ∈ C(z 1/k , log(z)) such that


δ(a)
(12)
L1
= φ(g1 ) − g1 .
a
We shall now prove that there exists g2 ∈ C(z 1/k ) such that


ϑ(a)
ϑν
= pν+1 φ(g2 ) − g2 .
aIndeed,it is easily seen that there exists v(X, Y ) ∈ C(X)[Y ] such that
δ(a)
L1 a
= v(z, log(z)). Using the fact that δ i = log(z)i ϑi + terms of lower degree in log(z), we see that






δ(a)
ϑ(a)
ϑ(a)
L1
= L1 log(z)= ϑν(log(z))ν+1aaa+ terms of lower degree in log(z).
On the other hand, let u(X, Y ) ∈ C(X, Y ) and k ≥ 1 be such that
g1 = u(z 1/k , log(z)). The equation (12) can be rewritten as followsv(z, log(z)) = u(z p/k , p log(z)) − u(z 1/k , log(z)).
Since z 1/k and log(z) are algebraically independent over C, we getv(X k , Y ) = u(X p , pY ) − u(X, Y ).
We see u(X,P Y ) has an element of C(X)(Y ) ⊂ C(X)((Y )) as follows:u(X, Y ) = j≥−N uj (X)Y j for some N ∈ Z. So,
Xv(X k , Y ) = u(X p , pY ) − u(X, Y ) =(pj uj (X p ) − uj (X))Y j .
j≥−Nin this equality, and letting X = z 1/k , we obtain,


ϑ(a)
.
pν+1 uν+1 (z p/k ) − uν+1 (z 1/k ) = ϑνaEquating the coefficients of Y

ν+1

Therefore, g2 = uν+1 (z 1/k ) ∈ C(z 1/k ) has the required property. One can showthat g2 belongs to C(z) by arguing as for the proof of the fact that d3 ∈ C(z) in
the proof of (2) ⇒ (3) above. We now claim that there exists g3 ∈ C(z) such thatϑ(a)= pφ(g3 ) − g3 .
a
If ν =R 0, then g3 := g2 has the expected property. Assume that ν > 0. Let
G2 = gz2 be some primitive of gz2 that we see as a function on some interval (0, ), > 0. We haveϑ(a)
ϑ (pν φ(G2 ) − G2 ) = pν+1 φ(g2 ) − g2 = ϑν,aso there exists C ∈ C such thatpν φ(G2 ) − G2 = ϑν−1



ϑ(a)
a


+ C.

Hence, G3 := G2 − pνC−1 satisfiesνp φ(G3 ) − G3 = ϑ

ν−1



ϑ(a)
a


.

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

17

But G3 = G4 + ` where G4 ∈ C(z) and ` is a C-linear combination of log(z) and offunctions of the form log(1 − zξ)4 with ξ ∈ C× . Using the C-linear independenceof any C-linear combination of log(z) and of functions of the form log(1 − zξ) withξ ∈ C× with any element of C(z), we see that the equality


ϑ(a)
ν
ν
ν
ν−1
p φ(G3 ) − G3 = (p φ(G4 ) − G4 ) + (p φ(`) − `) = ϑaimplies thatϑ(a)pν φ(G4 ) − G4 = ϑν−1.
aIterating this argument, we find g3 ∈ C(z) with the expected property. This proves(3).
We shall now prove (3) ⇒ (4). We assume that there exists d ∈ C(z) such that
ϑ(a) = a(pφ(d) − d). We write a = cz m l with c ∈ C× , m ∈ Z and l ∈ C(z) withoutϑ(c−1 a)pole at 0 and such that l(0) = 1. Since ϑ(a)a = c−1 a , we can assume that c = 1.
A fundamental solution of φ(y) = ay is given by
m Y
m1
f0 = z p−1
φk (l)−1 ∈ z p−1 C[[z]] ⊂ C((z p−1 )).
k≥0−1e − de with de = log(z)d. This is the integrability condition
We have δ(a)a = φ(d)for the system of equations(
φ(y) = aye i.e. ϑ(y) = dy.
δ(y) = dy,e 0 is a solution of φ(y) = ay so
A straightforward calculation shows that δ(f0 ) − df
e 0 , i.e. log(z)ϑ(f0 ) = (q + log(z)d)f0
there exists q ∈ C such that δ(f0 ) = (q + d)f1
p−1))(log(z)) and we have used the fact that
(here, we work in the (φ, δ)-field C((z1the field of φ-constants of C((z p−1 ))(log(z)) is equal to C, so that the solutions of
1
φ(y) = ay in C((z p−1 ))(log(z)) are of the form λf0 for some λ ∈ C). Therefore,ϑ(f0 ) = df0 . So, f0 satisfies a nonzero linear differential equation with coefficients
in K, and also a nonzero linear Mahler equation with coefficients in K. It follows
1mfrom [Béz94, Theorem 1.3] that f0 ∈ C(z p−1 ). Therefore, f0 = z p−1 h for someh ∈ C(z), and, hence, a = φ(f0 )f0−1 = z m φ(h)h−1 .
We shall now prove (4) ⇒ (3). We assume that there exists c ∈ C× , m ∈ Z)
and f ∈ C(z)× such that a = cz m φ(fThen, ϑ(a)/a = pφ(d) − d withf .
)d = m/(p − 1) + ϑ(ff ∈ C(z). Whence the desired result.

Remark 3.2. The technics employed above could also be used in order to recover a
famous result of Nishioka about the hypertranscendence of solutions of inhomogeneous Mahler equations of order one [Nis84]. A Galoisian approach (but withoutparametrized Picard-Vessiot theory) of the work of Nishioka has been proposed by
Nguyen in [Ngu11].
3.2. Mahler equations of higher order with large classical difference Galois group. Consider the difference system(13)

φ(Y ) = AY

with A ∈ GLn (C(z)). We let S be a PPV ring for (13) over L. The aim of thepresent section is to study the parametrized difference Galois group Galδ (QS /L) of(13) over L under the following assumption.
4Here, log(z) is the principal determination of the logarithm, and log(1 − zξ) is such thatlog(1 − 0ξ) = 0

18

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUESAssumption 3.3. In the rest of this section, we assume that the difference Galois
group of (13) over the φ-field K contains SLn (C).
Note the following result.
Lemma 3.4. Assume that the assumption (3.3) holds. Then, the difference Galoisegroup of (13) over the φ-field L contains SLn (C).
Proof. Corollary 1.7 ensures that the difference Galois group of (13) over the φfield K0 contains SLn (C). The fact that the difference Galois group of (13) overe is now a direct consequence of [CHS08, Corollary
the φ-field L contains SLn (C)2.5].
Let U ∈ GLn (S) be a fundamental matrix of solutions of (13) and set
d := det(U ) ∈ S × .
Then, d is a fundamental solution of φ(y) = det(A)y in S. We split our study ofGalδ (QS /L) in two cases, depending on whether d is hyperalgebraic or hypertranscendental over (L, δ). Note that Proposition 3.1 may be used to check whether dis hyperalgebraic or not.
3.2.1. Hyperalgebraic determinant. This section is devoted to the proof of the following result.
Theorem 3.5. Assume that the assumption (3.3) holds and that d is hyperalgebraic over (C(z), ϑ) (or, equivalently, that the parametrized difference Galois groupof φ(y) = det(A)y over L is included in C× ; see Proposition 3.1). Then, thee conparametrized difference Galois group Galδ (QS /L) is a subgroup of C× SLn (C)etaining SLn (C).
Before proceeding with the proof of this theorem, we give some lemmas.
Lemma 3.6. Assume that the assumption (3.3) holds and that d is hyperalgebraic
over (C(z), ϑ) (or, equivalently, that the parametrized difference Galois group of
φ(y) = det(A)y over L is included in C× ; see Proposition 3.1). Then, we have thefollowing alternative:(1) Galδ (QS /L) is conjugate to a subgroup of GLn (C) containing SLn (C);e containing SLn (C).
e(2) Galδ (QS /L) is equal to a subgroup of C× SLn (C)Moreover, the first case holds if and only if there exists B ∈ Kn×n such that(14)pφ(B) = ABA−1 + ϑ(A)A−1 .

Proof. Using Proposition 2.9, we are reduced to prove that the equation(15)

φ(B) = ABA−1 + δ(A)A−1

has a solution B ∈ K0n×n if and only if the equation (14) has a solution B ∈ Kn×n .
Assume that the equation (15) has a solution B ∈ K0n×n .
Letu(X, Y ) ∈ C(X, Y )n×n , k ≥ 1, v(X) ∈ GLn (C(X)) and w(X) ∈ C(X)n×n be suchthat
B = u(z 1/k , log(z)), A = v(z) and δ(A)A−1 = log(z)w(z).
The equation (15) can be rewritten as followsu(z p/k , p log(z)) = v(z)u(z 1/k , log(z))v(z)−1 + log(z)w(z).
Since z 1/k and log(z) are algebraically independent over C, we getu(X p , pY ) = v(X k )u(X, Y )v(X k )−1 + Y w(X k ).

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

19

n×n
We
⊂ C(X)((Y ))n×n : u(X, Y ) =P see u(X, Y )j has an element of C(X)(Y )j≥−N uj (X)Y for some N ∈ Z. We haveXX
uj (X p )pj Y j = v(X k )uj (X)v(X k )−1 Y j  + Y w(X k ).
j≥−N

j≥−N

Equating the terms of degree 1 in Y , we get
pu1 (X p ) = v(X k )u1 (X)v(X k )−1 + w(X k ).
Therefore, B1 := u1 (z 1/k ) ∈ K is a solution of (14).
Conversely, assume that the equation (14) has a solution B ∈ Kn×n . Then
B1 := B log(z) ∈ K0n×n satisfies
φ(B1 ) = AB1 A−1 + δ(A)A−1 .
0Lemma 3.7. Assume that the system φ(Y ) = BY , with B ∈ GLn (K ), has asolution u = (u1 , . . . , un )t with coefficients in C((z 1/k )) for some integer k ≥ 1.
Then, there exists a PPV ring T over L of φ(Y ) = BY that contains the L-δalgebra L{u1 , . . . , un }.
Proof. The result is obvious if u = (0, . . . , 0)t . We shall now assume thatc0 := ∪j≥1 C((z 1/j ))(log(z)). We equip
u 6= (0, . . . , 0)t . We consider the field K
c0 with the structure of (φ, δ)-field given by φ(f (z, log(z))) = f (z p , p log(z)) andKφdc0 = C. One can see K0 as a (φ, δ)-subfield. It is easily seen that K
δ = log(z)z dz
c0 . We let F = K0 hu1 , . . . , un i be the δ-subfield of K
c0 generated over K0 byof Ke ⊗C Fc0 such that F φ = C. By Lemma 2.3, C
u1 , . . . , un ; this is a (φ, δ)-subfield of K
is an integral domain and its field of fractions L1 = Lhu1 , . . . , un i is a (φ, δ)-fielde We consider a PPV ring S1 for φ(Y ) = BY over L1 and we letsuch that Lφ1 = C.
U ∈ GLn (S1 ) be a fundamental matrix of solutions of this difference system. We
can assume that the first column of U is u. Then, the L-(φ, δ)-algebra T generated
by the entries of U and by det(U )−1 contains L{u1 , . . . , un } and is a PPV ring forφ(Y ) = BY over L. Whence the result.
Lemma 3.8. Let us consider a vector u = (u1 , . . . , un )t with coefficients in
b := ∪j≥1 C((z 1/j )) such that φ(u) = Bu for some B ∈ GLn (K). Assume moreK
over that each ui satisfies some nonzero linear differential equation with coefficientse 1/j ), with respect to the derivation ϑ. Then, the ui actually belong toin ∪j≥1 C(zK.
Proof. According to the cyclic vector lemma, there exists P ∈ GLn (K) suchb which is a solution of athat P u = (f, φ(f ), . . . , φn−1 (f ))t for some f ∈ K
nonzero linear Mahler equation (i.e. a φ-difference equation) of order n with coefficients in K. Moreover, f satisfies a nonzero linear differential equation with
e 1/j ), with respect to the derivation ϑ, because it is a Kcoefficients in ∪j≥1 C(zlinear combination of the ui and the ui themselves satisfy such equations. It
follows from [Béz94, Theorem 1.3] that f belongs to K. Hence, the entries ofu = P −1 (P u) = P −1 (f, φ(f ), . . . , φn−1 (f ))t actually belong to K, as expected. 
Lemma 3.9. There exists c ∈ C× such that the difference system φ(Y ) = c−1 AY
b := ∪j≥1 C((z 1/j )).
has a nonzero solution u = (u1 , . . . , un )t with coefficients in K
Proof. According to [Roq15, Section 4], the system φ(Y ) = AY is triangularizableb i.e. there exists Pb ∈ GLn (K)b such that φ(Pb)−1 APb =: (vi,j )1≤i,j≤n isover K,×
upper-triangular. Let c ∈ C , m ∈ Z and l ∈ 1 + z 1/k C[[z 1/k ]] be such that

20

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUES

v1,1 = cz m l. We consider A1 = c−1 A ∈ GLn (C(z)). Then, the system φ(Y ) = A1 Y
b namely u = (u1 , . . . , un )t := Pb(f, 0, . . . , 0)t
has a nonzero solution with entries in K,
m Q
j
−1
p−1
with f := z.
j≥0 φ (l)Proof of Theorem 3.5. We let c ∈ C× and u = (u1 , . . . , un )t be as in Lemma 3.9,
and we set A1 := c−1 A ∈ GLn (C(z)). Thanks to Lemma 3.7, we can consider a PPV
ring S1 for φ(Y ) = A1 Y over L that contains L{u1 , . . . , un }. We let U1 ∈ GLn (S1 )be a fundamental matrix of solutions of φ(Y ) = A1 Y whose first column is u.
We denote by G the difference Galois group of φ(Y ) = AY over the φ-field K,
and by Gδ its parametrized difference Galois group over the (φ, δ)-field L. Similarly,
we denote by G1 the difference Galois group of φ(Y ) = A1 Y over the φ-ring K,
and by Gδ1 its parametrized difference Galois group over the (φ, δ)-field L.
We have Gder
= Gder = SLn (C), so G1 contains SLn (C). Moreover, the1parametrized difference Galois group of φ(y) = det(A1 )y = c−n det(A)y over
L is a subgroup of C× (because the parametrized difference Galois group of
φ(y) = det(A)y over L is a subgroup of C× by hypothesis, and the parametrized
difference Galois group of φ(y) = c−n y over L satisfies the same property).
e that contains SLn (C).
e Indeed,
We claim that Gδ1 is a subgroup of C× SLn (C)according to Lemma 3.6, it is sufficient to prove that there is no B ∈ Kn×n
such that ϑ(A1 ) = pφ(B)A1 − A1 B. Assume at the contrary that such a B
exists. The equation ϑ(A1 ) = pφ(B)A1 − A1 B, which can be rewritten as
δ(A1 ) = φ(log(z)B)A1 − A1 (log(z)B), ensures the integrability of the system of
equations(φ(Y ) = A1 Y
δ(Y ) = (log(z)B)Y.
e such that V := U1 D ∈ GLn (S1 ) satisfies
So, there exists D ∈ GLn (C)(φ(V ) = A1 V
δ(V ) = (log(z)B)V, i.e. ϑ(V ) = BV.
Hence, we have the equalities ϑ(U1 )D + U1 ϑ(D) = ϑ(U1 D) = ϑ(V ) = BV = BU1 D
so ϑ(U1 ) = BU1 − U1 ϑ(D)D−1 . This formula implies that the (finite dimensional)
e 1/j )-vector space generated by the entries of U1 is stable by ϑ. In particular,∪j≥1 C(zany ui (recall that the ui are the entries of the first column of U ) satisfies a nonzero
e 1/j ), with respect to thelinear differential equation with coefficients in ∪j≥1 C(zderivation ϑ. It follows from Lemma 3.8 that the ui belong to K. Hence, the firstcolumn of U1 is fixed by the Galois group G1 and this contradicts the fact that G1
contains SLn (C).
e Now, the theorem follows from
Therefore, (Gδ )der = (Gδ1 )der contains SLn (C).
Lemma 3.6.

3.2.2. Hypertranscendental determinant. In the case of an hypertranscendental determinant, we can reduce the computation of the parametrized difference Galois
group to a question concerning the existence of a rational solution of a given Mahler
equation as follows.
Lemma 3.10. Assume that the assumption (3.3) holds and that d is hypertranscendental over (C(z), θ) (or, equivalently, that the parametrized difference Galoise × ). We have the following alternative:
group of φ(y) = det(A)y over L is equal to C
e × SLn (C);(1) Galδ (QS /L) is conjugate to Cδe(2) Gal (QS /L) is equal to a GLn (C).

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

21

Moreover, the first case holds if and only if there exists B ∈ Kn×n such that(16)

pφ(B) = ABA−1 + ϑ(A)A−1 −1ϑ(det(A)) det(A)−1 In .
n

Proof. Note that d is hypertranscendental over (L, δ). Using Proposition 2.10, it
remains to prove that the equation(17)

φ(B) = ABA−1 + δ(A)A−1 −1δ(det(A)) det(A)−1 Innhas a solution B ∈ K0n×n if and only if the equation (16) has a solution B ∈ Kn×n .
The proof of this fact is similar to the proof of Lemma 3.6.

Unlike to the situation of Section 3.2.1, it is not completely obvious that we canbypass the search of rational solutions of (16) to decide which of the two optionsof Lemma 3.10 is satisfied. However, we can still get directly some informations on
the hypertranscendence of solutions in ∪j≥1 C(z 1/j ) as follows.
Theorem 3.11. Assume that the assumption (3.3) holds and that d is hypertranscendental over (C(z), ϑ). Assume that the difference system φ(Y ) = AY admits
a nonzero solution u = (u1 , . . . , un )t with coefficients in C((z 1/k )) for some integerk ≥ 1. Then, at least one of the ui is hypertranscendental over (C(z), ϑ).
Note the following immediate corollary, which is particularly interesting when
one works with difference equations rather than with difference systems.
Corollary 3.12. Assume that the assumption (3.3) holds and that d is hypertranscendental over (C(z), ϑ). Assume that the difference system φ(Y ) = AY admits anonzero solution u = (f, φ(f ), . . . , φn−1 (f ))t for some f ∈ C((z 1/k )) and some integer k ≥ 1. Then, f (and, hence, any φi (f )) is hypertranscendental over (C(z), ϑ).
The arguments employed in the proof Theorem 3.11 given below are very similar
to the ones used in the hyperalgebraic case. But, we need a new descent argument,that is contained in the following lemma.
Lemma 3.13. Let L be a δ-field and let Lhai and Lhb1 , . . . , bn i be two δ-field
extensions of L, both contained in a same δ-field extension of L. Assume that ais hypertranscendental over L and that any bi is hyperalgebraic over L. Then, thefield extensions Lhai and Lhb1 , . . . , bn i are linearly disjoint over L.
Proof. If Lhai and Lhb1 , . . . , bn i are not linearly disjoint over L then a is hyperalgebraic over Lhb1 , . . . , bn i. This implies that the differential transcendence degreeof the field Lha, b1 , . . . , bn i over Lhb1 , . . . , bn i is zero. Since the differential transcendence degree of Lhb1 , . . . , bn i over L is also zero, by hypothesis, we find thatthe differential transcendence degree of Lha, b1 , . . . , bn i over L is zero by classicalproperties of the transcendence degree. This implies that a is hyperalgebraic overL.

Proof of Theorem 3.11. Thanks to Lemma 3.7, we can assume that the PPV ring
S for φ(Y ) = AY over L contains L{u1 , . . . , un }. We can assume that the firstcolumn of the fundamental matrix of solutions U ∈ GLn (S) of φ(Y ) = AY is u.
We denote by G the difference Galois group of φ(Y ) = AY over the φ-field K,
and by Gδ its parametrized difference Galois group over the (φ, δ)-ring L. Since d
is hypertranscendental over L, the parametrized difference Galois group of φ(y) =e ×.
det(A)y over L is C
Note that Lemma 3.10 implies that Gδ is Kolchin-connected. So, S is an integral
domain.

22

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUESWe claim that at least one of the ui is hypertranscendental over L. Supposeat the contrary that all of them are hyperalgebraic. In particular, Gδ is a stricte Lemma 3.10 ensures that there exists B ∈ Kn×n such that
subgroup of GLn (C).
(18)

pφ(B) = ABA−1 + ϑ(A)A−1 −1ϑ(det(A)) det(A)−1 In .
n

This equation can be rewritten as
φ(B0 ) = AB0 A−1 + δ(A)A−1 −1δ(det(A)) det(A)−1 In ,nwhere B0 = log(z)B. Set B1 := B0 + δ(d)nd . Note thatφ(B1 ) = AB1 A−1 + δ(A)A−1 .
This equation ensures the integrability of the system of equations(φ(Y ) = AY
δ(Y ) = B1 Y.
e such that V := U D ∈ GLn (S) satisfies
So, there exists D ∈ GLn (C)(φ(V ) = AV
δ(V ) = B1 V i.e. ϑ(V ) = (B + ϑ(d)nd )V.
In particular, we have ϑ(U )D + U ϑ(D) = ϑ(U1 D) = ϑ(V ) = (B + ϑ(d)nd )U D soϑ(U ) = (B +ϑ(d))U − U ϑ(D)D−1 .
nde 1/j ), the previous formula implies that the F hdi5-vector subIf we set F = ∪j≥1 C(zspace of QS generated by the entries of U and all their successive ϑ-derivatives is offinite dimension. In particular, any ui satisfies a nonzero linear differential equation
Li (y) = 0 with coefficients in F hdi, with respect to the derivation ϑ.PWe can assume that the coefficients of Li (y) belong to F {d}. We write Li (y) = α Li,α (y)dαwhere Li,α (y) is a linear differential operator with coefficients in F , with respect tothe derivation ϑ, and dα is a monomial in the ϑi (d)’s. By Lemma 3.13, the ϑ-fields
F hdi and F hu1 , ..., un i are linearly disjoint over F . It follows easily that there exists
some nonzero Li,α (y) such that Li,α (ui ) = 0. Therefore, any ui satisfies a nonzero
linear differential equation with coefficients in F , with respect to the derivation ϑ.
It follows from Lemma 3.8 that the ui belong to K. Hence, the first column of U is
fixed by the difference Galois group G and this contradicts the fact that G contains
SLn (C).

4. Applications
In this section, we will use the notations introduced at the beginning of Section 3.
4.1. User-friendly hypertranscendence criteria. Consider the Mahler system(19)

φ(Y ) = AY, with A ∈ GLn (C(z)).

Theorem 4.1. Assume that the difference Galois group of the Mahler system (19)over the φ-field K contains SLn (C) and that det A(z) is a monomial. Then, thefollowing properties hold:(1) The parametrized difference Galois group of the Mahler system (19) over L
e containing SLn (C).
e
is a subgroup of C× SLn (C)5Here, F hdi denotes the ϑ-field extension generated by d over F .

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

23(2) Let u = (u1 , ..., un )t be a nonzero solution of (19) with entries in C((z)).
Then, the series u1 , u2, . . . , un and all their successive derivatives are algebraically independent over C(z). In particular, any ui is hypertranscendental over C(z).
Proof. The fact that det A(z) is a monomial ensures, in virtue of Proposition 3.1,
that the parametrized difference Galois group of φ(y) = det(A)y is included in C× .
Theorem 3.5 yields the first assertion of the theorem.
We claim that u1 , u2, . . . , un are hyperalgebraically independent over C(z). Assume at the contrary that they are hyperalgebraically dependent over C(z). Thanks
to Lemma 3.7, there exists a PPV ring S for the system (19) over L containing
K0 {u1 , u2 , . . . , un }. Let U ∈ GLn (S) be a fundamental matrix of solutions of thesystem (19) whose first column is u. Then, det(U ) is hyperalgebraic over L andthe elements of the first column of U are hyperalgebraically dependent over L. It
follows easily that the δ-transcendence degree of S over L is lower than or equal ton2 −2. This contradicts the fact that the δ-dimension of the parametrized difference
Galois group of equation (19) over L, namely n2 −1, is equal to the δ-transcendence
degree of S over L (see [HS08, Proposition 6.26]).
We shall now state a variant of the last theorem for Mahler equations. Consider
the following Mahler equation(20)nan (z)y(z p ) + an−1 (z)y(z pn−1) + · · · + a0 (z)y(z) = 0

for some integers p ≥ 2, n ≥ 1, and some a0 (z), . . . , an (z) ∈ C(z) with a0 (z)an (z) 6=
0. In what follows, by “difference Galois group of equation (20)”, we mean the
difference Galois group of the associated system


0
1
0 ···
0
.. 

..
 0
.
0
1
. 


 ∈ GLn (C(z)).

..
..
..
(21) φ(Y ) = AY, with A =  ..
.
.
0 
.

 .
 0
0
···
0
1 
− aan0 − aan1 · · · · · · − an−1
anTheorem 4.2. Assume that the difference Galois group over the φ-field K of theMahler equation (20) contains SLn (C) and that an (z)/a0 (z) is a monomial. Then,the following properties hold:(1) The parametrized difference Galois group of equation (20) over L is a sube containing SLn (C).
egroup of C× SLn (C)(2) Let f (z) ∈ C((z)) be a nonzero solution of (20). Then, the sen−1
ries f (z), f (z p ), . . . , f (z p ) and all their successive derivatives are algebraically independent over C(z). In particular, f (z) is hypertranscendental
over C(z).
Proof. Using the fact that the determinant of the matrix A given by formula (21)is equal to a0 /an and the fact that, if f (z) ∈ C((z)) is a nonzero solution of (20),n−1then (f (z), f (z p ), . . . , f (z p ))t is a nonzero solution of (21) with entries in C((z)),
we see that this theorem is a consequence of Theorem 4.1.

4.2. The Baum-Sweet sequence. The Baum-Sweet sequence (an )n≥0 is the automatic sequence defined by an = 1 if the binary representation of n contains noblock of consecutive 0 of odd length, and an = 0 otherwise. It is characterized by
the following recursive equations:a0 = 1, a2n+1 = an , a4n = an , a4n+2 = 0.

24

THOMAS DREYFUS, CHARLOTTE HARDOUIN, AND JULIEN ROQUESP
Let fBS (z) = n≥0 an z n be the corresponding generating series. The above recursive equations show thatfBS (z)
Y (z) =fBS (z 2 )satisfies the following Mahler system:


0 1
(22)
φ(Y ) = AY where A =∈ GL2 (K).
1 −z
We have used the following notations: p = 2, K = ∪j≥1 C(z 1/j ) and φ is the field
automorphism of K such that φ(z) = z 2 .
Theorem 4.3. The parametrized difference Galois group of (22) over L is equal toe where µ4 ⊂ C× is the group of 4th roots of the unity. The series fBS (z),µ4 SL2 (C),2fBS (z ) and all their successive derivatives are algebraically independent over C(z).
Proof. According to [Roq15, Theorem 50], the difference Galois group of (22) overthe φ-field K is equal to µ4 SL2 (C). Now, the result is a direct consequence ofTheorem 4.2.

4.3. The Rudin-Shapiro sequence. The Rudin-Shapiro sequence (an )n≥0 is theautomatic sequence defined by an = (−1)bn where bn is the number of pairs ofconsecutive 1 in the binary representation of n. It is the characterized by thefollowing recurrence relations:a0 = 1, a2n = an , a2n+1 = (−1)n an .
Let fRS (z) = n≥0 an z n be the corresponding generating series. The above recursive equations show thatfRS (z)
Y (z) =fRS (−z)satisfies the following Mahler system:


1 1
1
∈ GL2 (K).
(23)
φ(Y ) = AY where A =2 z1 − z1P

We have used the following notations: p = 2, K = ∪j≥1 C(z 1/j ) and φ is the field
automorphism of K such that φ(z) = z 2 .
Theorem 4.4. The parametrized difference Galois group of (23) over L is equaleto GL2 (C).
The series fRS (z), fRS (−z) and all their successive derivatives are
algebraically independent over C(z).
Proof. According to [Roq15, Theorem 54], the difference Galois group of (23) overthe φ-field K is equal to GL2 (C). Now, the result is a direct consequence of Theorem 4.1.

4.4. Direct sum of the Baum-Sweet and of the Rudin Shapiro equations.
The aim of this section is to illustrate how one can use the results of this paper
in order to prove the hyperalgebraic independence of Mahler functions solutions ofdistinct equations.
4.4.1. A differential group theoretic preliminary result. In what follows, we denoteby G◦ the neutral component of the linear algebraic group G, and by Gder its
derived subgroup. We recall that G◦ and Gder are Zariski-closed in G.
We denote by G◦δ the neutral component of the differential algebraic group
G (so, here, we consider Kolchin’s topology), and Gder its derived subgroup. In
general, Gder is not Kolchin-closed. We denote by Gderδ its Kolchin-closure in G.

HYPERTRANSCENDENCE OF SOLUTIONS OF MAHLER EQUATIONS

25

Theorem 4.5. Let k be a differentially closed δ-field. Let r ≥ 2 be an integer and,for any i ∈ {1, . . . , r}, let GQ
i be an algebraic subgroup of GLni (k). We consider the
linear algebraic group G = i∈{1,...,r} Gi . We assume that, for any i ∈ {1, . . . , r},Q
G◦,deris quasi-simple and that G◦,der = i∈{1,...,r} G◦,der. Let H be a Zariski-denseiidifferential algebraic subgroup of G. Let Hi be the projection of H in Gi ⊂ G. Then,(1) for all i ∈ {1, . . . , r}, Hi◦δ ,derδ is Zariski-dense in G◦,der;i(2) we have:YYH ◦δ ,derδ =Hi◦δ ,derδ ⊂G◦,der.
ii∈{1,...,r}i∈{1,...,r}

Proof. By hypothesis, H is Zariski-dense in G and, hence, Hi is Zariski-dense inGi (because the projection pi : G → Gi is continuous for the Zariski topology
and, hence, Gi = pi (G) = pi (H) ⊂ pi (H)). Therefore, H ◦δ ,derδ is Zariski-dense inQG◦,der = i∈{1,...,r} G◦,derand Hi◦δ ,derδ is Zariski-dense in G◦,der. Recall that theiiG◦,derare quasi-simple by hypothesis. It follows from [Cas89, Theorem 15] thatiY
H ◦δ ,derδ =Ki
i∈{1,...,r}for some δ-closed subgroups Ki of G◦,der. (With the terminologies of [Cas89, Thei
orem 15], the simple components Ai of G◦,der are the {1}i−1 × G◦,der× {1}r−i−1 ).
i◦δ ,derδWe necessarily have Ki = Hi.

4.4.2. Baum-Sweet and Rudin-Shapiro.
Theorem 4.6. The parametrized difference Galois group of the direct sum of thesystems (22) and (23) is equal to the direct product of the parametrized differencee × GL2 (C).
e
Galois groups of the systems (22) and (23), namely µ4 SL2 (C)The
22series fBS (z), fBS (z ), fRS (z), fRS (z ) and all their successive derivatives are algebraically independent over C(z).
Proof. We denote by MBS and MRS the φ-modules associated to the systems (22)and (23).
It is proved in [Roq15, Section 9.3] that the difference Galois group over K of thedirect sum MBS ⊕ MRS is the direct product of the difference Galois groups, i.e.
µ4 SL(C)×GL2 (C). If follows from Theorems 4.5, 4.3, and 4.4, that the parametrizede × SL2 (C).
e The fact thatdifference Galois group of MBS ⊕ MRS contains SL2 (C)e × GL2 (C)e is
the parametrized difference Galois group of MBS ⊕ MRS is µ4 SL2 (C)now clear.
The proof of the last assertion is similar to the proof of the last statement ofTheorem 4.2.

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Université Paul Sabatier - Institut de Mathématiques de Toulouse, 118 route deNarbonne, 31062 Toulouse.
E-mail address: tdreyfus@math.univ-toulouse.fr
Université Paul Sabatier - Institut de Mathématiques de Toulouse, 118 route deNarbonne, 31062 Toulouse.
E-mail address: hardouin@math.univ-toulouse.fr
Université Grenoble Alpes, Institut Fourier, CNRS UMR 5582, 100 rue desMaths, BP 74, 38402 St Martin d’Hères
E-mail address: Julien.Roques@ujf-grenoble.fr