Criterion for the integrality of the Taylor coefficients of mirror maps in several variables
aa r X i v : . [ m a t h . N T ] A ug CRITERION FOR THE INTEGRALITY OF THE TAYLORCOEFFICIENTS OF MIRROR MAPS IN SEVERAL VARIABLES
E. DELAYGUE
Abstract.
We give a necessary and sufficient condition for the integrality of the Taylorcoefficients at the origin of formal power series q i ( z ) = z i exp( G i ( z ) /F ( z )) , with z =( z , . . . , z d ) and where F ( z ) and G i ( z ) + log( z i ) F ( z ) , i = 1 , . . . , d are particular solutionsof certain A -systems of differential equations. This criterion is based on the analyticalproperties of Landau’s function (which is classically associated with the sequences offactorial ratios) and it generalizes the criterion in the case of one variable presented in“Critère pour l’intégralité des coefficients de Taylor des applications miroir” [J. ReineAngew. Math.]. One of the techniques used to prove this criterion is a generalization ofa version of a theorem of Dwork on the formal congruences between formal series, provedby Krattenthaler and Rivoal in “Multivariate p -adic formal congruences and integrality ofTaylor coefficients of mirror maps” [arXiv:0804.3049v3, math.NT]. This criterion involvesthe integrality of the Taylor coefficients of new univariate mirror maps listed in “Tables ofCalabi–Yau equations” [arXiv:math/0507430v2, math.AG] by Almkvist, van Enckevort,van Straten and Zudilin. Introduction
The mirror maps considered in this article are formal series of d variables z i ( x , . . . , x d ) , i = 1 , . . . , d , such that the map ( x , . . . , x d ) ( z ( x , . . . , x d ) , . . . , z d ( x , . . . , x d )) is the compositional inverse of the map ( y , . . . , y d ) ( q ( y , . . . , y d ) , . . . , q d ( y , . . . , y d )) , with, writing y = ( y , . . . , y d ) , q i ( y ) = y i exp( G i ( y ) /F ( y )) for i = 1 , . . . , d and where F ( y ) and G i ( y ) + log( y i ) F ( y ) are particular solutions of a certain A -system of lineardifferential equations. These objects are geometric in nature because the series F ( y ) are A -hypergeometric functions ( ) which can be viewed as the period of certain multi-parameterfamilies of algebraic varieties in a product of weighted projective spaces (see [5] for details).A classic example of multivariate mirror maps, studied in [2], [13] and [8] is related tothe series F ( z , z ) = X m,n ≥ (3 m + 3 n )! m ! n ! z m z n (1.1) The A -hypergeometric series are also called GKZ hypergeometric series. See [13] for an introductionto these series, which generalize the classic hypergeometric series in the multivariate case. hich is solution of the system of differential equations (cid:26) D y − z (3 D + 3 D + 1) (3 D + 3 D + 2) (3 D + 3 D + 3) y = 0 ,D y − z (3 D + 3 D + 1) (3 D + 3 D + 2) (3 D + 3 D + 3) y = 0 , where D = z ddz and D = z ddz . We find two other solutions of this system G ( z , z ) +log( z ) F ( z , z ) and G ( z , z ) + log( z ) F ( z , z ) where G ( z , z ) = X m,n ≥ (3 m + 3 n )! m ! n ! (3 H m +3 n − H m ) z m z n and G ( z , z ) = X m,n ≥ (3 m + 3 n )! m ! n ! (3 H m +3 n − H n ) z m z n . This set of solutions enables us to define two canonical coordinates q ( z , z ) = z exp( G ( z , z ) /F ( z , z )) and q ( z , z ) = z exp( G ( z , z ) /F ( z , z )) . The associated mirror maps are defined by the formal series z ( q , q ) and z ( q , q ) suchthat the map ( q , q ) ( z ( q , q ) , z ( q , q )) is the compositional inverse of the map ( z , z ) ( q ( z , z ) , q ( z , z )) .According to the Corollary from [8], the series q ( z , z ) , q ( z , z ) , z ( q , q ) and z ( q , q ) have integral Taylor coefficients.Mirror maps are of interest in Mathematical Physics and Algebraic Geometry. Particu-larly, within Mirror Symmetry Theory, it has been observed that the Taylor coefficients ofmirror maps are integers. This surprising observation has led to the study of these objectswithin Number Theory, which has led to its proof in many cases (see further down in theintroduction). The aim of this article is to establish a necessary and sufficient condition forthe integrality of all the Taylor coefficients of mirror maps defined by ratios of factorialsof linear forms.1.1. Definition of mirror maps.
In order to define the mirror maps involved in thisarticle, we introduce some standard multi-index notation, which we use throughout thearticle. Namely, given a positive integer d , k ∈ { , . . . , d } and vectors m := ( m , . . . , m d ) and n := ( n , . . . , n d ) in R d , we write m · n for the scalar product m n + · · · + m d n d and m ( k ) for m k . We write m ≥ n if and only if m i ≥ n i for all i ∈ { , . . . , d } . In addition, if z := ( z , . . . , z d ) is a vector of variables and if n := ( n , . . . , n d ) ∈ Z d , then we write z n forthe product z n · · · z n d d . Finally, we write for the vector (0 , . . . , ∈ Z d .Given two sequences of vectors in N d e := ( e , . . . , e q ) and f := ( f , . . . , f q ) , wewrite | e | := P q i =1 e i and | f | := P q i =1 f i ∈ N d so that, for all k ∈ { , . . . , d } , we have | e | ( k ) = P q i =1 e ( k ) i and | f | ( k ) = P q i =1 f ( k ) i . For all n ∈ N d , we write Q e,f ( n ) := ( e · n )! · · · ( e q · n )!( f · n )! · · · ( f q · n )! . e define the formal series F e,f ( z ) := X n ≥ ( e · n )! · · · ( e q · n )!( f · n )! · · · ( f q · n )! z n and G e,f,k ( z ) := X n ≥ ( e · n )! · · · ( e q · n )!( f · n )! · · · ( f q · n )! q X i =1 e ( k ) i H e i · n − q X j =1 f ( k ) j H f j · n ! z n , (1.2)where k ∈ { , . . . , d } and, for all m ∈ N , H m := P mi =1 1 i is the m -th harmonic number.The series F e,f ( z ) is a A -hypergeometric series and is therefore a solution of a A -system oflinear differential equations. In some cases, we find d additional solutions of this systemtogether with at most logarithmic singularities at the origin, the G e,f,k ( z ) + log( z k ) F ( z ) for k ∈ { , . . . , d } .In the context of mirror symmetry, when | e | = | f | , the d functions q e,f,k ( z ) := z k exp( G e,f,k ( z ) /F e,f ( z )) , k ∈ { , . . . , d } , are canonical coordinates . The compositional inverse of the map z ( q e,f, ( z ) , . . . , q e,f,d ( z )) defines the vector ( z e,f, ( q ) , . . . , z e,f,d ( q )) of mirror maps .The aim of this article is to establish a necessary and sufficient condition for the in-tegrality of the coefficients of the d mirror maps z e,f,k ( q ) , that is, to determine underwhich conditions, for all k ∈ { , . . . , d } , we have z e,f,k ( q ) ∈ Z [[ q ]] . In the context ofNumber Theory of this article, the mirror map z e,f,k ( q ) and the corresponding canonicalcoordinate q e,f,k ( z ) play strictly the same role because, for all k ∈ { , . . . , d } , we have q e,f,k ( z ) ∈ z k Z [[ z ]] if and only if, for all k ∈ { , . . . , d } , we have z e,f,k ( q ) ∈ q k Z [[ q ]] (see [8,Partie 1.2]). Therefore, we shall formulate the criterion exclusively for canonical coordinatebut it also holds for the corresponding mirror maps.1.2. Statement of the criterion.
Before stating the criterion for the integrality of theTaylor coefficients of q e,f,k ( z ) , we recall the definition of Landau’s function associated witha ratio of factorials of linear forms. Given two sequences of vectors in N d e := ( e , . . . , e q ) and f := ( f , . . . , f q ) , we write ∆ e,f the Landau’s function associated with Q e,f , which isdefined, for all x ∈ R d , by ∆ e,f ( x ) := q X i =1 ⌊ e i · x ⌋ − q X j =1 ⌊ f j · x ⌋ , where ⌊·⌋ denotes the floor function. We also write {·} for the fractional part function. Westill write ⌊·⌋ , respectively {·} , for the function defined, for all x = ( x , · · · , x d ) ∈ R d , by ⌊ x ⌋ := ( ⌊ x ⌋ , · · · , ⌊ x d ⌋ ) , respectively by { x } := ( { x } , · · · , { x d } ) . For all c ∈ N d , we have ⌊ c · x ⌋ = ⌊ c · { x }⌋ + c · ⌊ x ⌋ and therefore ∆ e,f ( x ) = ∆ e,f ( { x } ) + ( | e | − | f | ) · ⌊ x ⌋ . So, we have | e | = | f | if and only if ∆ e,f is -periodic in each of its variables. We write D e,f for the semi-algebraic set of all x ∈ [0 , d such that there exists d ∈ { e , · · · , e q , f , · · · , f q } verifying d · x ≥ . The set [0 , d \D e,f is nonempty and the function ∆ e,f vanishes on [0 , d \D e,f . he following proposition shows that the Landau’s function provides a characterization ofthe sequences e and f such that, for all n ∈ N d , Q e,f ( n ) is an integer. Landau’s criterion.
Let e and f be two sequences of vectors in N d . We have the followingdichotomy. ( i ) If, for all x ∈ [0 , d , we have ∆ e,f ( x ) ≥ , then, for all n ∈ N d , we have Q e,f ( n ) ∈ N . ( ii ) If there exists x ∈ [0 , d such that ∆ e,f ( x ) ≤ − , then there are only finitely manyprime numbers p such that all terms of the family Q e,f are in Z p .Remark. Assertion ( i ) is a result of Landau from [10]: he has proved that it is in fact anecessary and sufficient condition. We prove Landau’s criterion assertion ( ii ) in Section 2.In literature, one can distinguish several results proving the integrality of the Taylorcoefficients of univariate mirror maps ( i.e. d = 1 ) when | e | = | f | . One can find them, in anincreasing order of generality, in [11], [15], [7] and [3]. Refer to the introduction from [3]for a detailed statement of all these results. In the univariate case, the most general resultbuilds up a criterion for the integrality of the Taylor coefficients of mirror maps definedby sequences of ratios of factorials. According to the notations of this article, it reads asfollows: Criterion for univariate mirror maps (Theorem 1 from [3]) . Let e and f be two disjointsequences of positive integers such that Q e,f is a sequence of integers (which is equivalentto ∆ e,f ≥ on [0 , ) and which satisfy | e | = | f | . Then, we have the following dichotomy. ( i ) If, for all x ∈ D e,f , we have ∆ e,f ( x ) ≥ , then q e,f, ( z ) ∈ z Z [[ z ]] . ( ii ) If there exists x ∈ D e,f such that ∆ e,f ( x ) = 0 , then there are only finitely manyprime numbers p such that q e,f, ( z ) ∈ z Z p [[ z ]] . In the multivariate case, Krattenthaler and Rivoal proved in [8] the integrality of theTaylor coefficients of mirror maps belonging to large infinite families. In order to statethis result, for all k ∈ { , . . . , d } , we write k for the vector in N d , all coordinates of whichequal to zero except the k -th which is equal to . Theorem (Corollary 1 from [8]) . Let e and f be two sequences of vectors in N d verifying | e | = | f | and such that f is only composed of vectors of the form k with k ∈ { , . . . , d } .Then, for all k ∈ { , . . . , d } , we have q e,f,k ( z ) ∈ z k Z [[ z ]] . The purpose of this article is to prove the following theorems, which provide a characteri-zation of the multivariate mirror maps, associated with integral ratios of factorials of linearforms and all the Taylor coefficients of which are integers. We prove in Section 1.3 thatthey contain the results of other authors who worked on this subject previously. First, weconsider the case | e | = | f | and then we state the results when there exists k ∈ { , . . . , d } such that | e | ( k ) > | f | ( k ) . When there exists k ∈ { , . . . , d } such that | e | ( k ) < | f | ( k ) , thefamily Q e,f has a term that is not an integer and the question of the integrality of theTaylor coefficients of q e,f,k ( z ) is still open. heorem 1. Let e and f be two disjoint sequences of nonzero vectors in N d such that Q e,f is a family of integers (equivalent to ∆ e,f ≥ on [0 , d ) and which satisfy | e | = | f | . Thenwe have the following dichotomy. ( i ) If, for all x ∈ D e,f , we have ∆ e,f ( x ) ≥ , then, for all k ∈ { , . . . , d } , we have q e,f,k ( z ) ∈ z k Z [[ z ]] . ( ii ) If there exists x ∈ D e,f such that ∆ e,f ( x ) = 0 , then there exists k ∈ { , . . . , d } suchthat there are only finitely many prime numbers p such that q e,f,k ( z ) ∈ z k Z p [[ z ]] .Remarks. • Note the similarity between Landau’s criterion and Theorem 1. • We assume that the terms of the sequences e and f are nonzero and that thesesequences are disjoint in order to rule out the possibility that ∆ e,f vanish identically,which corresponds to the formal series F e,f ( z ) = (1 − z ) − · · · (1 − z d ) − , G e,f,k ( z ) =0 and q e,f,k ( z ) = z k . • Assertion ( ii ) of Theorem 1 is optimal as, if ∆ e,f vanishes on D e,f and if d ≥ ,then there may exist k ∈ { , . . . , d } such that q e,f,k ( z ) ∈ z k Z [[ z ]] . Indeed, if onechooses d = 2 , e = ((3 , and f = ((2 , , (1 , . Then we have D e,f = { ( x , x ) ∈ [0 , : x ≥ / } , ∆ e,f ((1 / , and q e,f, ( z ) = z . • Theorem 1 generalizes the criterion for univariate mirror maps and Corollary 1 from[8] (see Section 1.3).We will now state a criterion for the integrality of the Taylor coefficients of mirror-type maps q L ,e,f defined, for all L ∈ N d , by q L ,e,f ( z ) := exp( G L ,e,f ( z ) /F e,f ( z )) , where G L ,e,f isthe formal power series G L ,e,f ( z ) := X n ≥ ( e · n )! · · · ( e q · n )!( f · n )! · · · ( f q · n )! H L · n z n . (1.3)We write E e,f for the set of all L ∈ N d \{ } such that there is a d ∈ { e , . . . , e q , f , . . . , f q } satisfying L ≤ d . We have q L ,e,f ( z ) ∈ P dj =1 z j Q [[ z ]] and z − k q e,f,k ( z ) = q Y i =1 (cid:0) q e i ,e,f ( z ) (cid:1) e ( k ) i ! / q Y j =1 (cid:0) q f j ,e,f ( z ) (cid:1) f ( k ) j ! , (1.4)so that if, for all L ∈ E e,f , we have q L ,e,f ( z ) ∈ Z [[ z ]] , then, for all k ∈ { , . . . , d } , we have q e,f,k ( z ) ∈ z k Z [[ z ]] . Thus, assertion ( i ) of Theorem 2 implies assertion ( i ) of Theorem 1.Assertion ( ii ) of Theorem 2 adds details to assertion ( ii ) of Theorem 1. To be moreprecise, it proves that there exists k ∈ { , . . . , d } such that q e,f,k ( z ) / ∈ z k Z [[ z ]] and that allthe mirror-type maps indeed involved in (1.4) have at least one Taylor coefficient which isnot an integer. Thus Theorem 1 can be seen as a corollary of Theorem 2. Theorem 2.
Let e and f be two disjoint sequences of nonzero vectors in N d such that Q e,f is a family of integers (which is equivalent to ∆ e,f ≥ on [0 , d ) and which satisfy | e | = | f | . Then we have the following dichotomy. ( i ) If, for all x ∈ D e,f , we have ∆ e,f ( x ) ≥ , then, for all L ∈ E e,f , we have q L ,e,f ( z ) ∈ Z [[ z ]] . ii ) If there exists x ∈ D e,f such that ∆ e,f ( x ) = 0 , then there exists k ∈ { , . . . , d } suchthat, if L ∈ E e,f verifies L ( k ) ≥ , then there are only finitely many prime numbers p such that q L ,e,f ( z ) ∈ Z p [[ z ]] . Furthermore, there are only finitely many primenumbers p such that q e,f,k ( z ) ∈ z k Z p [[ z ]] . Theorem 2 generalizes Theorem 2 from [3] and Theorem 2 from [8] (see Section 1.3). Ifthere exists k ∈ { , . . . , d } such that | e | ( k ) > | f | ( k ) , we have the following theorem whichgeneralizes Theorem 3 from [3]. Theorem 3.
Let e and f be two disjoint sequences of nonzero vectors in N d such that Q e,f is a family of integers (which is equivalent to ∆ e,f ≥ on [0 , d ) and such that there exists k ∈ { , . . . , d } verifying | e | ( k ) > | f | ( k ) . Then, ( a ) there are only finitely many prime numbers p such that q e,f,k ( z ) ∈ z k Z p [[ z ]] ; ( b ) for all L ∈ E e,f verifying L ( k ) ≥ , there are only finitely many prime numbers p such that q L ,e,f ( z ) ∈ Z p [[ z ]] . Comparison of Theorems 1, 2 and 3 with previous results.
First, we provethat Theorems 1 and 2 generalize Corollary 1 and Theorem 2 from [8]. We only have toprove that, if e and f are two disjoint sequences of nonzero vectors in N d , verifying | e | = | f | and such that f is only constituted by vectors k with k ∈ { , . . . , d } , then, for all x ∈ D e,f ,we have ∆ e,f ( x ) ≥ . Indeed, if x ∈ D e,f , then x ∈ [0 , d and, for all k ∈ { , . . . , d } , wehave k · x = 0 . Thus, there exists an element d in e such that d · x ≥ and we have ∆ e,f ( x ) = q X i =1 ⌊ e i · x ⌋ − q X j =1 ⌊ f j · x ⌋ = q X i =1 ⌊ e i · x ⌋ ≥ . Let us now prove that Theorems 2 and 3 generalize Theorems 2 and 3 from [3]. Itis sufficient to note that if d = 1 , then e and f are two sequences of positive integersand, writing M e,f for the greatest element in the sequences e and f , we obtain E e,f = { , . . . , M e,f } and D e,f = [1 /M e,f , .1.4. Structure of proofs.
First, we prove assertion ( ii ) of Landau’s criterion in Section 2.Section 3 is dedicated to the statement and the proof of Theorem 4, which generalizescriteria of formal congruences proved by Dwork and by Krattenthaler and Rivoal. Thesecriteria were crucial for the previous results about the integrality of the Taylor coefficientsof mirror maps. Theorem 4 is central to the proofs of Theorems 1 and 2.In Section 4, we reduce the proofs of Theorems 1, 2 and 3 to the proofs of p -adic relations.Section 5 is dedicated to the statement and the proof of a technical lemma which we willuse to prove both assertions of Theorems 1 and 2.We prove assertions ( i ) of Theorems 1 and 2 in Section 6, this is by far the longest andthe most technical part of this article. Particularly, we have to prove certain number ofdelicate p -adic estimations in order to be able to apply Theorem 4.In Sections 7 and 8, we prove assertions ( ii ) of Theorems 1 and 2 and the Theorem 3,which ensue rather fast from reformulations of these theorems established in Section 4. inally in Section 9, we prove that Theorems 1 and 2 enable us to obtain the integralityof the Taylor coefficients of new univariate mirror maps listed in [1] by Almkvist, vanEnckevort, van Straten and Zudilin.2. Proof of assertion ( ii ) of Landau’s criterion First, let us introduce some additional notations which we will use throughout thisarticle. Given d ∈ N , d ≥ , λ ∈ R , k ∈ { , . . . , d } and vectors m := ( m , . . . , m d ) and n := ( n , . . . , n d ) in R d , we write m + n for ( m + n , . . . , m d + n d ) , λ m or m λ for ( λm , . . . , λm d ) , and m /λ for ( m /λ, . . . , m d /λ ) when λ is nonzero.To prove assertion ( ii ) of Landau’s criterion, we will use the fact that, for all prime p andall n ∈ N d , we have v p ( Q e,f ( n )) = P ∞ ℓ =1 ∆( n /p ℓ ) . Indeed, we recall that, for all m ∈ N , wehave the formula v p ( m !) = P ∞ ℓ =1 ⌊ m/p ℓ ⌋ . Thereby, we get v p ( Q e,f ( n )) = v p (cid:18) ( e · n )! · · · ( e q · n )!( f · n )! · · · ( f q · n )! (cid:19) = ∞ X ℓ =1 q X i =1 ⌊ e i · n /p ℓ ⌋ − q X j =1 ⌊ f j · n /p ℓ ⌋ ! = ∞ X ℓ =1 ∆ (cid:18) n p ℓ (cid:19) . We will need the following lemma, which we will also use for the proofs of assertions ( ii ) of Theorems 1 and 2. In the rest of the article, we write for the vector (1 , . . . , ∈ N d . Lemma 1.
Let u := ( u , . . . , u n ) be a sequence of vectors in N d and x ∈ R d . Then, thereexists µ > such that, for all x ∈ R d satisfying ≤ x ≤ µ and all i ∈ { , . . . , n } , wehave ⌊ u i · ( x + x ) ⌋ = ⌊ u i · x ⌋ .Proof. For all y > , there exists ν y > such that ⌊ y + ν y ⌋ = ⌊ y ⌋ . Thus, writing ν :=min { ν u i · x : 1 ≤ i ≤ n } > , we obtain that, for all i ∈ { , . . . , n } , we have ⌊ u i . x + ν ⌋ = ⌊ u i . x ⌋ . Therefore, writing µ := min { ν/ | u i | : 1 ≤ i ≤ n, u i = } > , we get that, for all ≤ x ≤ µ and all i ∈ { , . . . , n } , we have u i · x ≤ µ | u i | ≤ ν so ⌊ u i · ( x + x ) ⌋ = ⌊ u i · x ⌋ .This completes the proof of the lemma. (cid:3) Proof of assertion ( ii ) of Landau’s criterion. Given x ∈ [0 , d satisfying ∆ e,f ( x ) ≤ − and applying Lemma 1 with, instead of u , the sequence constituted by the elements of e and f , we obtain that there exists µ > such that, for all x ∈ R d verifying ≤ x ≤ µ ,we have ∆ e,f ( x + x ) = ∆ e,f ( x ) ≤ − . We write U := { x + x : ≤ x ≤ µ } during theproof.There exists a constant N such that, for all prime p ≥ N , there is n p ∈ N d suchthat n p /p ∈ U . There exists a constant N such that, for all prime p ≥ N and all d ∈ { e , . . . , e q , f , . . . , f q } , we have | d | ( µ + 1) /p < .Thus, for all prime number p ≥ N := max( N , N ) and all integer ℓ ≥ , we have ∆ e,f ( n p /p ) ≤ − and, as n p /p ∈ U , we have n p /p ≤ (1 + µ ) and n p /p ℓ ≤ n p /p ≤ ( µ +1) /p . As a result, for all d ∈ { e , . . . , e q , f , . . . , f q } , we obtain d · n p /p ℓ ≤ | d | ( µ + 1) /p < , which leads to n p /p ℓ ∈ [0 , d \D e,f and so ∆ e,f ( n p /p ℓ ) = 0 . hus, for all prime p ≥ N , we have v p ( Q e,f ( n p )) = P ∞ ℓ =1 ∆ e,f ( n p /p ℓ ) ≤ − , whichfinishes the proof of Landau’s criterion. (cid:3) Formal congruences
The proof of assertion ( i ) of Theorem 2 is essentially based on the generalization (The-orem 4 below) of a theorem of Krattenthaler and Rivoal [8, Theorem 1, p. 3] which is amultivariate adaptation of a Dwork’s theorem [4, Theorem 1, p. 296].Before stating the Theorem 4, we introduce some notations. Let p be a prime numberand d ∈ N , d ≥ . We write Ω for the completion of the algebraic closure of Q p and O forthe ring of integers of Ω .If N is a subset of S t ≥ (cid:0) { , . . . , p t − } d × { t } (cid:1) , then, for all s ∈ N , we write Ψ s ( N ) for the set of all u ∈ { , . . . , p s − } d such that, for all ( n , t ) ∈ N , with t ≤ s , and all j ∈ { , . . . , p s − t − } d , we have u = j + p s − t n .Given u ∈ { , . . . , p s − } d , u := P s − k =0 u k p k with u k ∈ { , . . . , p − } d , we write M s ( u ) for the word u · · · u s − of length s on the alphabet { , . . . , p − } d . According to thisdefinition, we have u ∈ Ψ s ( N ) if and only if none of the words M t ( n ) , ( n , t ) ∈ N , is asuffix of M s ( u ) .For example, let us take N := { ( , t ) : t ≥ } . In this case, Ψ s ( N ) is the set of all u = P s − k =0 u k p k such that u s − = . We observe that Ψ s ( N ) = Ψ s ( N ′ ) with N ′ = { ( , } . Theorem 4.
Let us fix a prime number p . Let ( A r ) r ≥ be a sequence of maps from N d to Ω \ { } and ( g r ) r ≥ be a sequence of maps from N d to O \ { } . We assume that there exists N ⊂ S t ≥ (cid:0) { , . . . , p t − } d × { t } (cid:1) such that, for all r ≥ , we have ( i ) | A r ( ) | p = 1 ; ( ii ) for all m ∈ N d , we have A r ( m ) ∈ g r ( m ) O ; ( iii ) for all s ∈ N and m ∈ N d , we have: ( a ) for all u ∈ Ψ s ( N ) and v ∈ { , . . . , p − } d , we have A r ( v + u p + m p s +1 ) A r ( v + u p ) − A r +1 ( u + m p s ) A r +1 ( u ) ∈ p s +1 g r + s +1 ( m ) A r ( v + u p ) O ;( a ) furthermore, if v + p u ∈ Ψ s +1 ( N ) , then we have A r ( v + u p + m p s +1 ) A r ( v + u p ) − A r +1 ( u + m p s ) A r +1 ( u ) ∈ p s +1 g r + s +1 ( m ) g r ( v + u p ) O ;( a ) on the other hand, if v + p u / ∈ Ψ s +1 ( N ) , then we have A r +1 ( u + p s m ) A r +1 ( u ) ∈ p s +1 g s + r +1 ( m ) g r ( v + p u ) O ;( b ) For all ( n , t ) ∈ N , we have g r ( n + p t m ) ∈ p t g r + t ( m ) O ; hen, for all a ∈ { , . . . , p − } d , m ∈ N d , s, r ∈ N and K ∈ Z d , we have S r ( a , K , s, p, m ) := X m p s ≤ j ≤ ( m + ) p s − (cid:0) A r ( a + p ( K − j )) A r +1 ( j ) − A r +1 ( K − j ) A r ( a + j p ) (cid:1) ∈ p s +1 g s + r +1 ( m ) O , (3.1) where we extend A r to Z d by A r ( n ) = 0 if there is an i ∈ { , . . . , d } such that n i < . This theorem generalizes Theorem 1 from [8]. Indeed, let A : N d Z p \ { } and g : N d Z p \ { } be two maps verifying conditions ( i ) , ( ii ) and ( iii ) of Theorem 1 from [8].Let ( A r ) r ≥ be the constant sequence of value A and ( g r ) r ≥ be the constant sequence ofvalue g . These two sequences verify conditions ( i ) and ( ii ) of Theorem 4. Let us choose N := ∅ so that, for all s ∈ N , we have Ψ s ( N ) = { , . . . , p s − } d . In particular, conditions ( a ) and ( b ) of Theorem 4 are empty. Thus we only have to prove that ( A r ) r ≥ and ( g r ) r ≥ verify assertions ( a ) and ( a ) of Theorem 4. The equality Ψ s +1 ( N ) = { , . . . , p s +1 − } d ,associated with assertion ( ii ) , proves that condition ( a ) implies assertion ( a ) . But assertion ( a ) corresponds to no other assertion than ( iii ) of Theorem 1 from [8]. Thus the conditionsof Theorem 4 are valid and we have the conclusion of Theorem 1 from [8].The aim of the end of this section is to prove Theorem 4.3.1. Proof of Theorem 4.
The structure of the proof is based on those of the theoremsof Dwork and Krattenthaler and Rivoal, but it rather appreciably differs in details.For all s ∈ N , s ≥ , we write α s for the following assertion: for all a ∈ { , . . . , p − } d , u ∈ { , . . . , s − } , m ∈ N d , r ≥ and K ∈ Z d , we have the congruence S r ( a , K , u, p, m ) ∈ p u +1 g u + r +1 ( m ) O . For all s ∈ N , s ≥ and t ∈ { , . . . , s } , we write β t,s for the following assertion: for all a ∈ { , . . . , p − } d , m ∈ N d , r ≥ and K ∈ Z d , we have the congruence S r ( a , K + m p s , s, p, m ) ≡ X j ∈ Ψ s − t ( N ) A t + r +1 ( j + m p s − t ) A t + r +1 ( j ) S r ( a , K , t, p, j ) mod p s +1 g s + r +1 ( m ) O . For all a ∈ { , . . . , p − } d , K ∈ Z d , r ∈ N and j ∈ N d , we set U r ( a , K , p, j ) := A r ( a + p ( K − j )) A r +1 ( j ) − A r +1 ( K − j ) A r ( a + j p ) . Then we have S r ( a , K , s, p, m ) = X ≤ j ≤ ( p s − U r ( a , K , p, j + m p s ) . We state now four lemmas enabling us to prove (3.1).
Lemma 2.
Assertion α is true. emma 3. For all s, r ∈ N , m ∈ N d , a ∈ { , . . . , p − } d , j ∈ Ψ s ( N ) and K ∈ Z d , we have U r ( a , K + m p s , p, j + m p s ) ≡ A r +1 ( j + m p s ) A r +1 ( j ) U r ( a , K , p, j ) mod p s +1 g s + r +1 ( m ) O . Lemma 4.
For all s ∈ N , s ≥ , if α s is true, then, for all a ∈ { , . . . , p − } d , K ∈ Z d , r ≥ and m ∈ N d , we have S r ( a , K , s, p, m ) ≡ X j ∈ Ψ s ( N ) U r ( a , K , p, j + m p s ) mod p s +1 g s + r +1 ( m ) O . Lemma 5.
For all s ∈ N , s ≥ , and all t ∈ { , . . . , s − } , assertions α s and β t,s implyassertion β t +1 ,s . Before proving these lemmas, we check that their validity implies (3.1). We prove byinduction on s that α s is true for all s ≥ , which leads to the conclusion of Theorem 4.According to Lemma 2, α is true. Let us assume that α s is true for a fixed s ≥ . Wenote that β ,s is the assertion β ,s : S r ( a , K + m p s , s, p, m ) ≡ X j ∈ Ψ s ( N ) A r +1 ( j + m p s ) A r +1 ( j ) S r ( a , K , , p, j ) mod p s +1 g s + r +1 ( m ) O . As S r ( a , K , , p, j ) = U r ( a , K , p, j ) , we have X j ∈ Ψ s ( N ) A r +1 ( j + m p s ) A r +1 ( j ) S r ( a , K , , p, j ) = X j ∈ Ψ s ( N ) A r +1 ( j + m p s ) A r +1 ( j ) U r ( a , K , p, j ) and, according to Lemma 3, we get X j ∈ Ψ s ( N ) A r +1 ( j + m p s ) A r +1 ( j ) U r ( a , K , p, j ) ≡ X j ∈ Ψ s ( N ) U r ( a , K + m p s , p, j + m p s ) mod p s +1 g s + r +1 ( m ) O≡ S r ( a , K + m p s , s, p, m ) mod p s +1 g s + r +1 ( m ) O , (3.2)where (3.2) is obtained via Lemma 4.Hence, assertion β ,s is true. Then we get, according to Lemma 5, the validity of β ,s .By iteration of Lemma 5, we finally obtain β s,s which is S r ( a , K + m p s , s, p, m ) ≡ X j ∈ Ψ ( N ) A s + r +1 ( j + m ) A s + r +1 ( j ) S r ( a , K , s, p, j ) mod p s +1 g s + r +1 ( m ) O≡ A s + r +1 ( m ) A s + r +1 ( ) S r ( a , K , s, p, ) mod p s +1 g s + r +1 ( m ) O , (3.3)where we used the fact that Ψ ( N ) = { } for (3.3). e will now prove that, for all a ∈ { , . . . , p − } d , r ∈ N and K ∈ Z d , we have S r ( a , K , s, p, ) ∈ p s +1 O . For all N ∈ Z d , we write P N for the assertion: “for all a ∈{ , . . . , p − } d and r ∈ N , we have S r ( a , N , s, p, ) ∈ p s +1 O ”. If there exists i ∈ { , . . . , d } such that N i < , then, for all j ∈ { , . . . , p s − } d , we have A r ( a + p ( N − j )) = 0 and A r +1 ( N − j ) = 0 so that S r ( a , N , s, p, ) = 0 ∈ p s +1 O . First, we prove by contradictionthat, for all N ∈ Z d , P N is true. Let us assume that there is a minimal element N ∈ N d such that P N is false. Given m ∈ N d \ { } and N ′ := N − m p s and applying (3.3) with N ′ instead of K , we get S r ( a , N , s, p, m ) ≡ A s + r +1 ( m ) A s + r +1 ( ) S r ( a , N ′ , s, p, ) mod p s +1 g s + r +1 ( m ) O . As m ∈ N d \ { } , we have N ′ < N , which, according to the definition of N , leads to S r ( a , N ′ , s, p, ) ∈ p s +1 O . According to conditions ( i ) and ( ii ) , we have | A s + r +1 ( ) | p = 1 and A s + r +1 ( m ) ∈ g s + r +1 ( m ) O , so we get S r ( a , N , s, p, m ) ∈ p s +1 g s + r +1 ( m ) O ⊂ p s +1 O .Thereby, for all m ∈ N d \ { } , we have S r ( a , N , s, p, m ) ∈ p s +1 O . Given T ∈ N d such that,for all i ∈ { , . . . , d } we have ( T i + 1) p s > N i , we get X ≤ m ≤ T S r ( a , N , s, p, m )= X ≤ m ≤ T X m p s ≤ j ≤ ( m + ) p s − ( A r ( a + p ( N − j )) A r +1 ( j ) − A r +1 ( N − j ) A r ( a + j p ))= X ≤ j ≤ N ( A r ( a + p ( N − j )) A r +1 ( j ) − A r +1 ( N − j ) A r ( a + j p )) (3.4) = 0 , (3.5)where we used the fact that A r ( n ) = 0 when there is an i ∈ { , . . . , d } such that n i < for (3.4), and (3.5) occurs because the term of sum (3.4) is changed into its opposite whenchanging the index j in N − j . So we obtain S r ( a , N , s, p, ) = − P < m ≤ T S r ( a , N , s, p, m ) ∈ p s +1 O , which is contradictory to the status of N . Thus, for all N ∈ Z d , P N is true.Furthermore, conditions ( i ) and ( ii ) respectively lead to | A s + r +1 ( ) | p = 1 and A s + r +1 ( m ) ∈ g s + r +1 ( m ) O . Then we obtain, according to (3.3), that S r ( a , K + m p s , s, p, m ) ∈ p s +1 g s + r +1 ( m ) O . Thislatest congruence is valid for all a ∈ { , . . . , p − } d , K ∈ Z d , m ∈ N d and r ≥ , whichproves that the assertion α s +1 is true and completes the induction on s . We now have toprove Lemmas 2, 3, 4 and 5.3.1.1. Proof of Lemma 2.
Given a ∈ { , . . . , p − } d , K ∈ Z d , m ∈ N d and r ≥ , we have S r ( a , K , , p, m ) = A r ( a + p ( K − m )) A r +1 ( m ) − A r +1 ( K − m ) A r ( a + p m ) . (3.6)If K − m / ∈ N d , then we have A r ( a + p ( K − m )) = 0 and A r +1 ( K − m ) = 0 so that S r ( a , K , , p, m ) = 0 ∈ p g r +1 ( m ) O , as expected. Thus we can assume that K − m ∈ N d . e rewrite (3.6) as follows. S r ( a , K , , p, m ) = A r ( a ) A r +1 ( m ) (cid:18) A r ( a + p ( K − m )) A r ( a ) − A r +1 ( K − m ) A r +1 ( ) (cid:19) − A r +1 ( K − m ) (cid:18) A r ( a + m p ) A r ( a ) − A r +1 ( m ) A r +1 ( ) (cid:19) ! . (3.7)As Ψ ( N ) = { } , we can use ( a ) , with instead of u and a instead of v , to obtain A r ( a + p ( K − m )) A r ( a ) − A r +1 ( K − m ) A r +1 ( ) ∈ p g r +1 ( K − m ) A r ( a ) O and A r ( a + m p ) A r ( a ) − A r +1 ( m ) A r +1 ( ) ∈ p g r +1 ( m ) A r ( a ) O . This leads to A r ( a ) A r +1 ( m ) (cid:18) A r ( a + p ( K − m )) A r ( a ) − A r +1 ( K − m ) A r +1 ( ) (cid:19) ∈ p g r +1 ( K − m ) A r +1 ( m ) O⊂ p g r +1 ( m ) O (3.8)and A r ( a ) A r +1 ( K − m ) (cid:18) A r ( a + m p ) A r ( a ) − A r +1 ( m ) A r +1 ( ) (cid:19) ∈ p g r +1 ( m ) A r +1 ( K − m ) O⊂ p g r +1 ( m ) O , (3.9)where we used condition ( ii ) for (3.8) and (3.9), which leads to A r +1 ( m ) ∈ g r +1 ( m ) O and A r +1 ( K − m ) ∈ g r +1 ( K − m ) O ⊂ O . Applying (3.8) and (3.9) to (3.7), we obtain S r ( a , K , , p, m ) ∈ p g r +1 ( m ) , which finishes the proof of the lemma.3.1.2. Proof of Lemma 3.
We have U r ( a , K + m p s , p, j + m p s ) − A r +1 ( j + m p s ) A r +1 ( j ) U r ( a , K , p, j )= − A r +1 ( K − j ) A r ( a + j p ) (cid:18) A r ( a + j p + m p s +1 ) A r ( a + j p ) − A r +1 ( j + m p s ) A r +1 ( j ) (cid:19) . (3.10)As j ∈ Ψ s ( N ) , hypothesis ( a ) implies that the right-hand side of equality (3.10) lies in A r +1 ( K − j ) A r ( a + j p ) p s +1 g s + r +1 ( m ) A r ( a + j p ) O . Furthermore, according to condition ( ii ) , we have A r +1 ( K − j ) ∈ g r +1 ( K − j ) O ⊂ O . Theseestimates prove that the left-hand side of (3.10) lies in p s +1 g s + r +1 ( m ) O , which completesthe proof of the lemma. .1.3. Proof of Lemma 4.
Let us fix r, s ∈ N , s ≥ , such that α s is true.For all u ∈ { , . . . , s } , we write A u for the assertion: for all n ∈ { , . . . , p s − u − } d , wehave X ≤ j ≤ ( p u − U r ( a , K , p, j + n p u + m p s ) = S r ( a , K , u, p, n + m p s − u ) . We will prove by induction on u that, for all u ∈ { , . . . , s } , the assertion A u is true.If u = 0 , then there is nothing to prove so A is true. Let u ∈ { , . . . , s − } such that A u is true. Let us prove that A u +1 is true. For all n ∈ { , . . . , p s − u − − } d , we have S r ( a , K , u + 1 ,p, n + m p s − u − ) = X ≤ v ≤ ( p − S r ( a , K , u, p, v + n p + m p s − u )= X ≤ v ≤ ( p − X ≤ j ≤ ( p u − U r ( a , K , p, j + v p u + n p u +1 + m p s ) (3.11) = X ≤ j ≤ ( p u +1 − U r ( a , K , p, j + n p u +1 + m p s ) , (3.12)where we used assertion A u for (3.11). Equality (3.12) proves that A u +1 is true, whichfinishes the induction on u .If Ψ s ( N ) = { , . . . , p s − } d , then Lemma 4 is trivial. In the sequel of this proof, weassume that Ψ s ( N ) = { , . . . , p s − } d . We have u ∈ { , . . . , p s − } d \ Ψ s ( N ) if and onlyif there exists ( n , t ) ∈ N , t ≤ s , and j ∈ { , . . . , p s − t − } d such that u = j + p s − t n . Wewrite N s the set of all ( n , t ) ∈ N with t ≤ s . So we have { , . . . , p s − } d \ Ψ s ( N ) = [ ( n ,t ) ∈N s { j + p s − t n : j ∈ { , . . . , p s − t − } d } . In particular, the set N s is nonempty.We will prove that there exists k ∈ N , k ≥ , and ( n , t ) , . . . , ( n k , t k ) ∈ N s suchthat the sets J ( n i , t i ) := { j + p s − t i n i : j ∈ { , . . . , p s − t i − } d } induce a partition of { , . . . , p s − } d \ Ψ s ( N ) . We observe that N s ⊂ S st =1 ( { , . . . , p t − } × { t } ) and thus N s is finite. Therefore, we only have to prove that if ( n , t ) , ( n ′ , t ′ ) ∈ N s , j ∈ { , . . . , p s − t − } d and j ′ ∈ { , . . . , p s − t ′ − } d verify j + p s − t n = j ′ + p s − t ′ n ′ , then we have J ( n , t ) ⊂ J ( n ′ , t ′ ) or J ( n ′ , t ′ ) ⊂ J ( n , t ) . Let us assume, for example, that t ≤ t ′ . Then there exists j ∈{ , . . . , p t ′ − t − } d such that j = j ′ + p s − t ′ j , so that p s − t ′ n ′ = p s − t n + p s − t ′ j and thus J ( n ′ , t ′ ) ⊂ J ( n , t ) . Also, if t ≥ t ′ , then we have J ( n , t ) ⊂ J ( n ′ , t ′ ) . Thus, we get S r ( a , K , s, p, m ) = X j ∈ Ψ s ( N ) U r ( a , K , p, j + m p s ) + X j ∈{ ,...,p s − } d \ Ψ s ( N ) U r ( a , K , p, j + m p s ) , (3.13)with X j ∈{ ,...,p s − } d \ Ψ s ( N ) U r ( a , K , p, j + m p s ) = k X i =1 X j ∈{ ,...,p s − ti − } d U r ( a , K , p, j + p s − t i n i + m p s ) . (3.14) e now prove that, for all i ∈ { , . . . , k } , we have X j ∈{ ,...,p s − ti − } d U r ( a , K , p, j + p s − t i n i + m p s ) ∈ p s +1 g s + r +1 ( m ) O . (3.15)Given i ∈ { , . . . , k } , assertion A s − t i leads to X ≤ j ≤ ( p s − ti − U r ( a , K , p, j + p s − t i n i + m p s ) = S r ( a , K , s − t i , p, n i + m p t i ) . As t i ≥ , we get, via α s , that S r ( a , K , s − t i , p, n i + m p t i ) ∈ p s − t i +1 g s − t i + r +1 ( n i + m p t i ) O . Applying assertion ( b ) with t i instead of t and r + s − t i + 1 instead of r , we obtain p s − t i +1 g s − t i + r +1 ( n i + m p t i ) ∈ p s − t i +1 p t i g s + r +1 ( m ) O = p s +1 g s + r +1 ( m ) O . Thus, for all i ∈ { , . . . , k } , we have (3.15).Congruence (3.15), associated with (3.14) and (3.13), proves that S r ( a , K , s, p, m ) ≡ X j ∈ Ψ s ( N ) U r ( a , K , p, j + m p s ) mod p s +1 g s + r +1 ( m ) O , which completes the proof of Lemma 4.3.1.4. Proof of Lemma 5.
During this proof, i indicates an element of { , . . . , p − } d and u indicates an element of { , . . . , p s − t − − } d . For t < s , we write β t,s as follows S r ( a , K + m p s , s, p, m ) ≡ X i + u p ∈ Ψ s − t ( N ) A t + r +1 ( i + u p + m p s − t ) A t + r +1 ( i + u p ) S r ( a , K , t, p, i + u p ) mod p s +1 g s + r +1 ( m ) O . (3.16)We want to prove β t +1 ,s , which is S r ( a , K + m p s , s, p, m ) ≡ X u ∈ Ψ s − t − ( N ) A t + r +2 ( u + m p s − t − ) A t + r +2 ( u ) S r ( a , K , t + 1 , p, u ) mod p s +1 g s + r +1 ( m ) O . We note that S r ( a , K , t + 1 , p, u ) = P ≤ i ≤ ( p − S r ( a , K , t, p, i + u p ) . Thus, writing X := S r ( a , K + m p s , s, p, m ) − X ≤ i ≤ ( p − X u ∈ Ψ s − t − ( N ) A t + r +2 ( u + m p s − t − ) A t + r +2 ( u ) S r ( a , K , t, p, i + u p ) , we only have to prove that X ∈ p s +1 g s + r +1 ( m ) O . We have i + u p ∈ Ψ s − t ( N ) ⇒ u ∈ Ψ s − t − ( N ) . (3.17) ndeed, if u / ∈ Ψ s − t − ( N ) , then there exists ( n , k ) ∈ N , k ≤ s − t − , and j ∈{ , . . . , p s − t − − k − } d such that u = j + p s − t − − k n . Thus we have i + u p = i + j p + p s − t − k n ,which leads to i + u p / ∈ Ψ s − t ( N ) . Hence, according to β t,s written as (3.16), we obtain X ≡ X i + u p ∈ Ψ s − t ( N ) S r ( a , K , t, p, i + u p ) (cid:18) A t + r +1 ( i + u p + m p s − t ) A t + r +1 ( i + u p ) − A t + r +2 ( u + m p s − t − ) A t + r +2 ( u ) (cid:19) + X u ∈ Ψ s − t − ( N ) i + u p / ∈ Ψ s − t ( N ) A t + r +2 ( u + m p s − t − ) A t + r +2 ( u ) S r ( a , K , t, p, i + u p ) mod p s +1 g s + r +1 ( m ) O . Furthermore, applying ( a ) with s − t − instead of s and t + r + 1 instead of r , we get A t + r +1 ( i + u p + m p s − t ) A t + r +1 ( i + u p ) − A t + r +2 ( u + m p s − t − ) A t + r +2 ( u ) ∈ p s − t g s + r +1 ( m ) g t + r +1 ( i + u p ) O . In addition, as t < s and since α s is true, we have S r ( a , K , t, p, i + u p ) ∈ p t +1 g t + r +1 ( i + u p ) O (3.18)and we obtain X ≡ X u ∈ Ψ s − t − ( N ) i + u p / ∈ Ψ s − t ( N ) A t + r +2 ( u + m p s − t − ) A t + r +2 ( u ) S r ( a , K , t, p, i + u p ) mod p s +1 g s + r +1 ( m ) O . (3.19)Finally, when i + u p / ∈ Ψ s − t ( N ) , we can apply condition ( a ) with s − t − instead of s , i instead of v and r + t + 1 instead of r , which leads to A t + r +2 ( u + m p s − t − ) A t + r +2 ( u ) ∈ p s − t g s + r +1 ( m ) g t + r +1 ( i + u p ) O . (3.20)Applying (3.18) and (3.20) to (3.19), we obtain X ∈ p s +1 g s + r +1 ( m ) O . This finishes theproof of Lemma 5 and thus the one of Theorem 4.4. A p -adic reformulation of Theorems 1, 2 and 3 Let e and f be two disjoint sequences of nonzero vectors in N d such that Q e,f is a familyof integers. We fix L ∈ E e,f throughout this section. We recall that, for all k ∈ { , . . . , d } ,we have q e,f,k ( z ) ∈ z k Z [[ z ]] , respectively q L ,e,f ( z ) ∈ Z [[ z ]] , if and only if, for all primenumber p , we have q e,f,k ( z ) ∈ z k Z p [[ z ]] , respectively q L ,e,f ( z ) ∈ Z p [[ z ]] .We will define, for all prime number p , two elements Φ p,k ( a + p K ) and Φ L ,p ( a + p K ) of Q p , where a ∈ { , . . . , p − } d and K ∈ N d , and we will prove that q e,f,k ( z ) ∈ z k Z p [[ z ]] ,respectively q L ,e,f ( z ) ∈ Z p [[ z ]] , if and only if, for all a ∈ { , . . . , p − } d and all K ∈ N d , wehave Φ p,k ( a + p K ) ∈ p Z p , respectively Φ L ,p ( a + p K ) ∈ p Z p .To simplify notations, we will write E := E e,f , D := D e,f , ∆ := ∆ e,f , Q := Q e,f , F := F e,f , G k := G e,f,k , G L := G L ,e,f , q k := q e,f,k and q L := q L ,e,f , as throughout the rest of thearticle. We fix a prime number p in this section. efore proving Theorems 1, 2 and 3, we will reformulate them. The following result isdue to Krattenthaler and Rivoal’s Lemma [8, Lemma 2, p. 7]; it is the analogue in severalvariables of a lemma of Dieudonné and Dwork [6, Chap. IV, Sec. 2, Lemma 3]; [9, Chap.14, Sec. 2]. Lemma 6.
Given two formal power series F ( z ) ∈ P di =1 z i Z [[ z ]] and G ( z ) ∈ P di =1 z i Q [[ z ]] ,we define q ( z ) := exp( G ( z ) /F ( z )) . Then we have q ( z ) ∈ P di =1 z i Z p [[ z ]] if and only if F ( z ) G ( z p ) − pF ( z p ) G ( z ) ∈ p P di =1 z i Z p [[ z ]] . Lemma 6 will enable us to “eliminate” the exponential in the formulas q k ( z ) = z k exp( G k ( z ) /F ( z )) and q L ( z ) = exp( G L ( z ) /F ( z )) . Since ∆ ≥ on [0 , d , we obtain, according to Landau’s criterion, Q as a family ofintegers and thus F ( z ) ∈ P di =1 z i Z [[ z ]] . Furthermore, according to identities (1.2)and (1.3) defining the power series G k and G L , we have G k ( ) = G L ( ) = 0 and so G k ( z ) and G L ( z ) lie in P di =1 z i Q [[ z ]] . Thereby, following Lemma 6, we have q k ( z ) ∈ z k Z p [[ z ]] , respectively q L ( z ) ∈ Z p [[ z ]] , if and only if we have F ( z ) G k ( z p ) − pF ( z p ) G k ( z ) ∈ p P di =1 z i Z p [[ z ]] , respectively F ( z ) G L ( z p ) − pF ( z p ) G L ( z ) ∈ p P di =1 z i Z p [[ z ]] .According to identity (1.2) which defines G k , the coefficient of z a + p K in F ( z ) G k ( z p ) − pF ( z p ) G k ( z ) is Φ p,k ( a + p K ) := X ≤ j ≤ K Q ( K − j ) Q ( a + p j ) q X i =1 e ( k ) i ( H ( K − j ) · e i − pH ( a + p j ) · e i ) − q X i =1 f ( k ) i ( H ( K − j ) · f i − pH ( a + p j ) · f i ) ! and, according to identity (1.3) defining G L , the coefficient of z a + p K in F ( z ) G L ( z p ) − pF ( z p ) G L ( z ) is Φ L ,p ( a + K p ) := X ≤ j ≤ K Q ( K − j ) Q ( a + j p )( H L · ( K − j ) − pH L · ( a + j p ) ) . Thus we have q k ( z ) ∈ z k Z p [[ z ]] , respectively q L ( z ) ∈ Z p [[ z ]] , if and only if, for all a ∈{ , . . . , p − } d and K ∈ N d , we have Φ p,k ( a + p K ) ∈ p Z p , respectively Φ L ,p ( a + p K ) ∈ p Z p .5. A technical lemma
The aim of this section is to prove the following lemma which we will use for the proofsof assertions ( i ) and ( ii ) of Theorem 2. Lemma 7.
Let e and f be two sequences of vectors in N d such that | e | = | f | . Then, forall s ∈ N , c ∈ { , . . . , p s − } d and m ∈ N d , we have Q e,f ( c ) Q e,f ( c p ) Q e,f ( c p + m p s +1 ) Q e,f ( c + m p s ) ∈ p s +1 Z p . o prove Lemma 7, we will use certain properties of the p -adic gamma function definedas follows, Γ p ( n ) := ( − n γ p ( n ) , where γ p ( n ) := Q n − k =1( k,p )=1 k . The function Γ p can be extend tothe whole set Z p but we shall not need it here. Lemma 8. ( i ) For all n ∈ N , we have the formula ( np )! n ! = p n γ p (1 + np ) . ( ii ) For all k, n, s ∈ N , we have Γ p ( k + np s ) ≡ Γ p ( k ) mod p s . Assertion ( i ) of Lemma 8 is obtained by observing that γ p (1 + np ) = ( np )! n ! p n . Assertion ( ii ) of Lemma 8 is Lemma 1.1 from [9]. We are now able to prove Lemma 7. Proof of Lemma 7.
We have Q e,f ( c p + m p s +1 ) Q e,f ( c + m p s ) = q Y i =1 ( e i · ( c p + m p s +1 ))!( e i · ( c + m p s ))! q Y i =1 ( f i · ( c + m p s ))!( f i · ( c p + m p s +1 ))!= Q q i =1 p e i · ( c + m p s ) γ p (1 + p e i · ( c + m p s )) Q q i =1 p f i · ( c + m p s ) γ p (1 + p f i · ( c + m p s ))= p ( | e |−| f | ) · m p s Q q i =1 ( p e i · c ( − p e i · ( c + m p s ) Γ p (1 + p e i · ( c + m p s ))) Q q i =1 ( p f i · c ( − p f i · ( c + m p s ) Γ p (1 + p f i · ( c + m p s )))= ( − ( | e |−| f | ) · m p s +1 Q q i =1 ( p e i · c ( − e i · c p Γ p (1 + p e i · ( c + m p s ))) Q q i =1 ( p f i · c ( − f i · c p Γ p (1 + p f i · ( c + m p s ))) (5.1) = Q q i =1 p e i · c ( − e i · c p Q q i =1 p f i · c ( − f i · c p · Q q i =1 Γ p (1 + p e i · ( c + m p s )) Q q i =1 Γ p (1 + p f i · ( c + m p s )) , (5.2)where we used the identity | e | − | f | = for (5.1) and (5.2). According to assertion ( ii ) ofLemma 8, for all n ∈ N d , we have Γ p (1 + n · c p + n · m p s +1 ) ≡ Γ p (1 + n · c p ) mod p s +1 .So we get Q q i =1 Γ p (1 + e i · c p + e i · m p s +1 ) Q q i =1 Γ p (1 + f i · c p + f i · m p s +1 ) = Q q i =1 (Γ p (1 + e i · c p ) + O ( p s +1 )) Q q i =1 (Γ p (1 + f i · c p ) + O ( p s +1 )) , where we write x = O ( p k ) when x ∈ p k Z p . Furthermore, according to the definition of Γ p ,for all n ∈ N d , we have Γ p (1 + n · c p ) ∈ Z × p . Then we obtain Q q i =1 (Γ p (1 + e i · c p ) + O ( p s +1 )) Q q i =1 (Γ p (1 + f i · c p ) + O ( p s +1 )) = Q q i =1 Γ p (1 + e i · c p ) Q q i =1 Γ p (1 + f i · c p ) (1 + O ( p s +1 )) nd thus, Q e,f ( c p + m p s +1 ) Q e,f ( c + m p s ) = Q q i =1 p e i · c ( − e i · c p Q q i =1 p f i · c ( − f i · c p · Q q i =1 Γ p (1 + e i · c p ) Q q i =1 Γ p (1 + f i · c p ) (1 + O ( p s +1 ))= Q q i =1 p e i · c Q q i =1 p f i · c · Q q i =1 γ p (1 + e i · c p ) Q q i =1 γ p (1 + f i · c p ) (1 + O ( p s +1 ))= Q e,f ( c p ) Q e,f ( c ) (1 + O ( p s +1 )) . This completes the proof of the lemma. (cid:3) Proof of assertions ( i ) of Theorems 1 and 2 We assume the hypothesis of Theorems 1 and 2. Furthermore, we assume that, for all x ∈ D , we have ∆( x ) ≥ . As we said in Section 1.2, assertion ( i ) of Theorem 2 impliesassertion ( i ) of Theorem 1. So the aim of this section is to prove that, for all L ∈ E , wehave q L ( z ) ∈ Z [[ z ]] . Following Section 4, we only have to prove that, for all L ∈ E , allprime number p , all a ∈ { , . . . , p − } d and K ∈ N d , we have Φ L ,p ( a + p K ) ∈ p Z p . We fixa L ∈ E in this section.6.1. New reformulation of the problem.
For all prime number p , all s ∈ N , a ∈{ , . . . , p − } d and K , m ∈ N d , we define S ( a , K , s, p, m ) := X m p s ≤ j ≤ ( m + ) p s − ( Q ( a + j p ) Q ( K − j ) − Q ( j ) Q ( a + ( K − j ) p )) , where we extend Q to Z d by Q ( n ) = 0 if there is an i ∈ { , . . . , d } such that n i < .The aim of this section is to produce, for all prime number p , a function g p from N d to Z p such that: if, for all prime p , all s ∈ N , a ∈ { , . . . , p − } d and K , m ∈ N d , wehave S ( a , K , s, p, m ) ∈ p s +1 g p ( m ) Z p , then we have Φ L ,p ( a + K p ) ∈ p Z p . Thus the proofof assertion ( i ) of Theorem 2 will amount to finding a suitable lower bound of the p -adicvaluation of S ( a , K , s, p, m ) for all prime p . This reduction method is an adaptation of theapproach to the problem made by Dwork in [4].6.1.1. A reformulation of Φ L ,p ( a + K p ) modulo p Z p . This step is the analogue of a refor-mulation made by Krattenthaler and Rivoal in Section from [7]. We fix a prime number p . We will prove that Φ L ,p ( a + K p ) ≡ − X ≤ j ≤ K H L · j ( Q ( a + j p ) Q ( K − j ) − Q ( j ) Q ( a + ( K − j ) p )) mod p Z p . (6.1) or all a ∈ { , . . . , p − } d and j ∈ N d , we have pH L · ( a + j p ) = p L · j p X i =1 i + L · a X i =1 L · j p + i ! ≡ p L · j X i =1 ip + ⌊ L · a /p ⌋ X i =1 L · j p + ip mod p Z p ≡ H L · j + ⌊ L · a /p ⌋ X i =1 L · j + i mod p Z p . (6.2)We need a result that we shall prove further by means of Lemma 10 stated in Section6.1.2:For all L ∈ E , a ∈ { , . . . , p − } d and j ∈ N d , we have Q ( a + j p ) ⌊ L · a /p ⌋ X i =1 L · j + i ∈ p Z p . (6.3)Applying (6.3) to (6.2) and with the fact that Q ( a + j p ) ∈ Z p and Q ( K − j ) ∈ Z p , weobtain Q ( K − j ) Q ( a + j p ) pH L · ( a + j p ) ≡ Q ( K − j ) Q ( a + j p ) H L · j mod p Z p . This leads to Φ L ,p ( a + K p ) = X ≤ j ≤ K Q ( K − j ) Q ( a + j p )( H L · ( K − j ) − pH L · ( a + j p ) ) ≡ X ≤ j ≤ K Q ( K − j ) Q ( a + j p )( H L · ( K − j ) − H L · j ) mod p Z p ≡ − X ≤ j ≤ K H L · j ( Q ( a + j p ) Q ( K − j ) − Q ( j ) Q ( a + ( K − j ) p )) mod p Z p , which is the expected equation (6.1).We now use a Krattenthaler and Rivoal’s combinatorial lemma (see [8, Lemma 5, p. 14])which enables us to write X ≤ j ≤ K H L · j ( Q ( a + j p ) Q ( K − j ) − Q ( j ) Q ( a + ( K − j ) p ))= r − X s =0 X ≤ m ≤ ( p r − s − W L ( a , K , s, p, m ) , where r is such that p r − > max( K , . . . , K d ) and W L ( a , K , s, p, m ) := S ( a , K , s, p, m )( H L · m p s − H L ·⌊ m /p ⌋ p s +1 ) . If we prove that, for all s ∈ N and m ∈ N d , we have W L ( a , K , s, p, m ) ∈ p Z p , then we willhave Φ L ,p ( a + K p ) ∈ p Z p , as expected. or all m ∈ N d , we set µ p ( m ) := P ∞ ℓ =1 D ( { m /p ℓ } ) and g p ( m ) := p µ p ( m ) , where D isthe characteristic function of D . We now use the following lemma which we will prove inSection 6.1.2. Lemma 9.
For all prime number p , all L ∈ E , m ∈ N d and s ∈ N , we have p s +1 g p ( m ) (cid:0) H L · m p s − H L ·⌊ m /p ⌋ p s +1 (cid:1) ∈ p Z p . According to Lemma 9, if we prove that, for all a ∈ { , . . . , p − } d , K , m ∈ N d and s ∈ N , we have S ( a , K , s, p, m ) ∈ p s +1 g p ( m ) Z p , then we will have q L ( z ) ∈ Z p [[ z ]] , which isthe announced reformulation.6.1.2. Proofs of (6.3) and Lemma 9.
We state a result which enables us to prove (6.3) andLemma 9.
Lemma 10.
Given s ∈ N , s ≥ , a ∈ { , . . . , p s − } d , m ∈ N d and L ∈ E . If we have ⌊ L · a /p s ⌋ ≥ , then, for all u ∈ { , . . . , ⌊ L · a /p s ⌋} and ℓ ∈ { s, . . . , s + v p ( L · m + u ) } , wehave (cid:26) a + m p s p ℓ (cid:27) ∈ D . Proof.
We recall that D is the set of all x ∈ [0 , d such that there exists an element d of e or f satisfying d · x ≥ . We have n a + m p s p ℓ o ∈ [0 , d , so we only have to prove that L · n a + m p s p ℓ o ≥ . Indeed, as L ∈ E , there exists d ∈ { e , . . . , e q , f , . . . , f q } such that d ≥ L , which leads to L · (cid:26) a + p s m p ℓ (cid:27) ≥ ⇒ d · (cid:26) a + p s m p ℓ (cid:27) ≥ ⇒ (cid:26) a + p s m p ℓ (cid:27) ∈ D . We write m = P ∞ j =0 m j p j with m j ∈ { , . . . , p − } d . We have (cid:26) a + m p s p ℓ (cid:27) = a + p s P ℓ − s − j =0 m j p j p ℓ . We have p ℓ − s divide ( u + L · m ) and so p ℓ − s divides u + L · m − L · ∞ X j = ℓ − s m j p j ! = u + L · ℓ − s − X j =0 m j p j ! . Thus, we obtain p ℓ − s ≤ u + L · (cid:16)P ℓ − s − j =0 m j p j (cid:17) ≤ p s L · a + L · (cid:16)P ℓ − s − j =0 m j p j (cid:17) and we have ≤ L · a + p s L · (cid:16)P ℓ − s − j =0 m j p j (cid:17) p ℓ = L · (cid:26) a + m p s p ℓ (cid:27) . (cid:3) We will now apply Lemma 10 to prove (6.3). roof of (6.3) . Given L ∈ E , a ∈ { , . . . , p − } d and j ∈ N d , we have to prove that Q ( a + j p ) P ⌊ L · a /p ⌋ i =1 1 L · j + i ∈ p Z p . If ⌊ L · a /p ⌋ = 0 , it is evident. Thus let us assume that ⌊ L · a /p ⌋ ≥ . Applying Lemma 10 with s = 1 and m = j , we obtain that, for all i ∈ { , . . . , ⌊ L · a /p ⌋} and ℓ ∈ { , . . . , v p ( i + L · j ) } , we have { ( a + j p ) /p ℓ } ∈ D and so ∆(( a + j p ) /p ℓ ) ≥ . Since ∆ ≥ on R d , we get v p ( Q ( a + j p )) = ∞ X ℓ =1 ∆ (cid:18)(cid:26) a + j pp ℓ (cid:27)(cid:19) ≥ v p ( L · j + i ) X ℓ =1 ∆ (cid:18)(cid:26) a + j pp ℓ (cid:27)(cid:19) ≥ v p ( L · j + i ) , which finishes the proof of (6.3). (cid:3) Proof of Lemma 9.
Given L ∈ E , m ∈ N d and s ∈ N , we have to prove that p s +1 g p ( m )( H L · m p s − H L ·⌊ m /p ⌋ p s +1 ) ∈ p Z p . We write m = b + q p where b ∈ { , . . . , p − } d and q ∈ N d . Then we have L · m p s = L · b p s + L · q p s +1 and L · ⌊ m /p ⌋ p s +1 = L · q p s +1 . Therefore, we get H L · m p s − H L ·⌊ m /p ⌋ p s +1 = L · b p s X j =1 L · q p s +1 + j ≡ ⌊ L · b /p ⌋ X i =1 L · q p s +1 + ip s +1 mod 1 p s Z p and so p s +1 g p ( m )( H L · m p s − H L ·⌊ m /p ⌋ p s +1 ) ≡ g p ( b + q p ) P ⌊ L · b /p ⌋ i =1 1 L · q + i mod p Z p . We nowhave to prove that g p ( b + q p ) P ⌊ L · b /p ⌋ i =1 1 L · q + i ∈ p Z p . If ⌊ L · b /p ⌋ = 0 , it is evident. Letus assume that ⌊ L · b /p ⌋ ≥ . Applying Lemma 10 with s = 1 and q instead of m , weobtain that, for all i ∈ { , . . . , ⌊ L · b /p ⌋} and all ℓ ∈ { , . . . , v p ( i + L · q ) } , we have { ( b + q p ) /p ℓ } ∈ D and thus v p ( g p ( b + q p )) = µ p ( b + q p ) = ∞ X ℓ =1 D (cid:18)(cid:26) b + q pp ℓ (cid:27)(cid:19) ≥ v p ( L · q + i ) X ℓ =1 D (cid:18)(cid:26) b + q pp ℓ (cid:27)(cid:19) ≥ v p ( L · q + i ) , which completes the proof of Lemma 9. (cid:3) Application of Theorem 4.
We will use Theorem 4 to finish the proof of assertions ( i ) of Theorems 1 and 2. In the following sections, we will prove that, setting A r = Q and g r = g p for all r ≥ , then there exists N ⊂ S t ≥ (cid:0) { , . . . , p t − } d × { t } (cid:1) such that thesequences ( A r ) r ≥ and ( g r ) r ≥ satisfy assertions ( i ) , ( ii ) and ( iii ) of Theorem 4. Thus, wewill obtain S ( a , K , s, p, m ) ∈ p s +1 g p ( m ) Z p , as expected.In the following sections, we check the assumptions for the application of Theorem 4. .3. Verification of assertions ( i ) and ( ii ) of Theorem 4. We fix a prime number p and we write g := g p and µ := µ p . For all r ≥ , we set A r = Q and g r = g . In thissection, we will prove that the sequences ( A r ) r ≥ and ( g r ) r ≥ verify assertions ( i ) and ( ii ) of Theorem 4.For all r ≥ , we have | A r ( ) | p = |Q ( ) | p = 1 . Furthermore, for all m ∈ N d , wehave v p ( g ( m )) = µ ( m ) ≥ , so we get g ( m ) ∈ Z p \ { } . We now have to prove that A ( m ) ∈ g ( m ) Z p , which amounts to proving that µ p ( m ) ≤ v p ( Q ( m )) . This is true because,for all ℓ ∈ N , ℓ ≥ , we have ∆( m /p ℓ ) = ∆( { m /p ℓ } ) ≥ D ( { m /p ℓ } ) , because ∆( x ) ≥ for x ∈ D .6.4. Verification of assertion ( iii ) of Theorem 4. We fix a prime number p and weset N := [ t ≥ (cid:18)(cid:26) n ∈ { , . . . , p t − } d : ∀ ℓ ∈ { , . . . , t } , (cid:26) n p ℓ (cid:27) ∈ D (cid:27) × { t } (cid:19) . Verification of assertion ( b ) . Let ( n , t ) ∈ N and m ∈ N d . We have to prove that g ( n + p t m ) ∈ p t g ( m ) Z p . We have v p ( g ( n + p t m )) = ∞ X ℓ =1 D (cid:18)(cid:26) n + p t m p ℓ (cid:27)(cid:19) = t X ℓ =1 D (cid:18)(cid:26) n p ℓ (cid:27)(cid:19) + ∞ X ℓ = t +1 D (cid:18)(cid:26) n + p t m p ℓ (cid:27)(cid:19) = t + ∞ X ℓ = t +1 D (cid:18)(cid:26) n + p t m p ℓ (cid:27)(cid:19) . (6.4)Let us write m = P ∞ k =0 m k p k , where the m k ∈ { , . . . , p − } d are zero except for a finitenumber of k . For all ℓ ≥ t + 1 , we have (cid:26) n + p t m p ℓ (cid:27) = n + p t (cid:16)P ℓ − t − k =0 m k p k (cid:17) p ℓ ≥ p t (cid:16)P ℓ − t − k =0 m k p k (cid:17) p ℓ = (cid:26) m p ℓ − t (cid:27) . Thus, for all ℓ ≥ t + 1 , if n m p ℓ − t o ∈ D , then there exists L ∈ E such that ≤ L · (cid:26) m p ℓ − t (cid:27) ≤ L · (cid:26) n + p t m p ℓ (cid:27) , which gives us n n + p t m p ℓ o ∈ D . We get ∞ X ℓ = t +1 D (cid:18)(cid:26) n + p t m p ℓ (cid:27)(cid:19) ≥ ∞ X ℓ = t +1 D (cid:18)(cid:26) m p ℓ − t (cid:27)(cid:19) = ∞ X ℓ =1 D (cid:18)(cid:26) m p ℓ (cid:27)(cid:19) = v p ( g ( m )) , which, associated with (6.4), leads to v p ( g ( n + p t m )) ≥ t + v p ( g ( m )) , i.e. g ( n + p t m ) ∈ p t g ( m ) Z p , as expected. .4.2. Verification of assertion ( a ) . Given s ∈ N , u ∈ Ψ s ( N ) and v ∈ { , . . . , p − } d suchthat v + p u / ∈ Ψ s +1 ( N ) , we have to prove that Q ( u + p s m ) Q ( u ) ∈ p s +1 g ( m ) g ( v + p u ) Z p . (6.5)First, we give another expression for Ψ s ( N ) = { u ∈ { , . . . , p s − } d : ∀ ( n , t ) ∈ N , t ≤ s, ∀ j ∈ { , . . . , p s − t − } d , u = j + p s − t n } . For that purpose, we need the following lemma.
Lemma 11.
Given s ∈ N , s ≥ , and u ∈ { , . . . , p s − } d , we write u = P s − k =0 u k p k , with u k ∈ { , . . . , p − } d . Then, the following assertions are equivalent. (1) We have { u /p s } ∈ D . (2) There exists ( n , t ) ∈ N , t ≤ s and j ∈ { , . . . , p s − t − } d such that u = j + p s − t n .Proof of Lemma 11. (1) ⇒ (2) : For all s ≥ , u ∈ { , . . . , p s − } d such that { u /p s } ∈ D and all i ∈ { , . . . , s − } , we write A s,i ( u ) for the assertion: for all ℓ ∈ { , . . . , s − i } , wehave (cid:8)(cid:0)P s − k = i u k p k − i (cid:1) /p ℓ (cid:9) ∈ D .For all s ≥ , we write B s for the assertion: for all u ∈ { , . . . , p s − } d such that { u /p s } ∈ D , there exists i ∈ { , . . . , s − } , such that A s,i ( u ) is true.First, we will prove by induction on s that, for all s ≥ , B s is true.If s = 1 , then, for all u ∈ { , . . . , p − } d such that { u /p } ∈ D , assertion A , ( u ) corresponds to no other assertion than { u /p } ∈ D and thus is true. Hence, B is true.Given s ≥ such that B , . . . , B s − are true, and u ∈ { , . . . , p s − } d verifying { u /p s } ∈D such that A s, ( u ) , . . . , A s,s − ( u ) are false, we will prove that assertion A s, ( u ) is true.This will imply the validity of B s and will finish the induction on s .Let us give a proof by contradiction, assuming that there exists ℓ ∈ { , . . . , s } such that a ℓ := P ℓ − k =0 u k p k p ℓ = ( P s − k =0 u k p k p ℓ ) / ∈ D . We actually have ℓ ∈ { , . . . , s − } because { u /p s } ∈ D . For all L ∈ { e , . . . , e q , f , . . . , f q } ,we have L · a ℓ < . We write (cid:26) u p s (cid:27) = u p s = p ℓ a ℓ + p ℓ P s − k = ℓ u k p k − ℓ p s = a ℓ p s − ℓ + P s − k = ℓ u k p k − ℓ p s − ℓ . Since { u /p s } ∈ D , there exists L ∈ { e , . . . , e q , f , . . . , f q } such that ≤ L · (cid:26) u p s (cid:27) = L · a ℓ p s − ℓ + L · P s − k = ℓ u k p k − ℓ p s − ℓ < p s − ℓ + L · P s − k = ℓ u k p k − ℓ p s − ℓ , which leads to L · (cid:0)P s − k = ℓ u k p k − ℓ (cid:1) > p s − ℓ − . Since L · (cid:0)P s − k = ℓ u k p k − ℓ (cid:1) is an integer, we get L · (cid:0)P s − k = ℓ u k p k − ℓ (cid:1) ≥ p s − ℓ , i.e. (cid:8)(cid:0)P s − k = ℓ u k p k − ℓ (cid:1) /p s − ℓ (cid:9) ∈ D . We write v := P s − k = ℓ u k p k − ℓ ∈{ , . . . , p s − ℓ − } d . Thus we have { v /p s − ℓ } ∈ D and, applying B s − ℓ , we obtain that thereexists i ∈ { , . . . , s − ℓ − } such that A s − ℓ,i ( v ) is true, i.e. , for all r ∈ { , . . . , s − ℓ − i } , e have n(cid:16)P s − ℓ − k = i v k p k − i (cid:17) /p r o ∈ D . Furthermore, for all k , we have v k = u ℓ + k andtherefore P s − ℓ − k = i v k p k − i = P s − k = i + ℓ u k p k − i − ℓ . Thereby, assertion A s − ℓ,i ( v ) becomes: for all r ∈ { , . . . , s − ℓ − i } , we have (cid:8)(cid:0)P s − k = i + ℓ u k p k − i − ℓ (cid:1) /p r (cid:9) ∈ D ; which corresponds to noother assertion than A s,i + ℓ ( u ) . Since we assumed that A s, ( u ) , . . . , A s,s − ( u ) are false, weget a contradiction. Hence A s, ( u ) is true and B s is also true, which finishes the inductionon s .As { u /p s } ∈ D , assertion B s tells us that an i ∈ { , . . . , s − } exists such that A s,i ( u ) is true, i.e. for all ℓ ∈ { , . . . , s − i } , we have (cid:8)(cid:0)P s − k = i p k − i u k (cid:1) /p ℓ (cid:9) ∈ D . Thus we have (cid:0)P s − k = i p k − i u k , s − i (cid:1) ∈ N and u = P i − k =0 p k u k + p i P s − k = i p k − i u k . Therefore, the assertion (2) is valid with s − i instead of t , P s − k = i p k − i u k instead of n and P i − k =0 p k u k instead of j . (2) ⇒ (1) : We have (cid:26) u p s (cid:27) = u p s = j + p s − t n p s ≥ n p t = (cid:26) n p t (cid:27) ∈ D and so { u /p s } ∈ D , as expected. (cid:3) According to Lemma 11, we obtain Ψ s ( N ) = { u ∈ { , . . . , p s − } d : { u /p s } / ∈ D} . (6.6)Thus, for all u ∈ Ψ s ( N ) and ℓ ≥ s , we have { u /p ℓ } = u /p ℓ ≤ u /p s = { u /p s } , which givesus that, for all L ∈ { e , . . . , e q , f , . . . , f q } and ℓ ≥ s , we have L · { u /p ℓ } ≤ L · { u /p s } < and so { u /p ℓ } / ∈ D . As a result, for all ℓ ≥ s , we have ∆( { u /p ℓ } ) = 0 and thus v p ( Q ( u )) = ∞ X ℓ =1 ∆ (cid:18)(cid:26) u p ℓ (cid:27)(cid:19) = s X ℓ =1 ∆ (cid:18)(cid:26) u p ℓ (cid:27)(cid:19) . Furthermore, we have v p ( Q ( u + p s m )) = ∞ X ℓ =1 ∆ (cid:18)(cid:26) u + p s m p ℓ (cid:27)(cid:19) = s X ℓ =1 ∆ (cid:18)(cid:26) u p ℓ (cid:27)(cid:19) + ∞ X ℓ = s +1 ∆ (cid:18)(cid:26) u + p s m p ℓ (cid:27)(cid:19) , which leads to v p (cid:18) Q ( u + p s m ) Q ( u ) (cid:19) = ∞ X ℓ = s +1 ∆ (cid:18)(cid:26) u + p s m p ℓ (cid:27)(cid:19) . (6.7)We write m = P ∞ k =0 p k m k , with m k ∈ { , . . . , p − } d . For all ℓ ≥ s + 1 , we have (cid:26) u + p s m p ℓ (cid:27) = u + p s P ℓ − − sk =0 p k m k p ℓ ≥ P ℓ − − sk =0 p k m k p ℓ − s = (cid:26) m p ℓ − s (cid:27) nd thus ∞ X ℓ = s +1 ∆ (cid:18)(cid:26) u + p s m p ℓ (cid:27)(cid:19) ≥ ∞ X ℓ = s +1 D (cid:18)(cid:26) u + p s m p ℓ (cid:27)(cid:19) (6.8) ≥ ∞ X ℓ = s +1 D (cid:18)(cid:26) m p ℓ − s (cid:27)(cid:19) = ∞ X ℓ =1 D (cid:18)(cid:26) m p ℓ (cid:27)(cid:19) = v p ( g ( m )) , (6.9)where inequality (6.8) is true because, for all x ∈ D , we have ∆( x ) ≥ . Applying (6.9) to(6.7), we get v p (cid:18) Q ( u + p s m ) Q ( u ) (cid:19) ≥ v p ( g ( m )) . Thus, to verify assertion ( a ) , we only have to prove that, for all u ∈ Ψ s ( N ) and v ∈ { , . . . , p − } d such that v + p u / ∈ Ψ s +1 ( N ) , we have g ( v + p u ) ∈ p s +1 Z p .We write u = P s − k =0 p k u k , with u k ∈ { , . . . , p − } d . We have { ( v + p u ) /p } = v /p and,for all ℓ ≥ , we have { ( v + p u ) /p ℓ } = (cid:16) v + p P ℓ − k =0 p k u k (cid:17) /p ℓ . We get v p ( g ( v + p u )) = ∞ X ℓ =1 D (cid:18)(cid:26) v + p u p ℓ (cid:27)(cid:19) ≥ D (cid:18) v p (cid:19) + s +1 X ℓ =2 D v + p P ℓ − k =0 p k u k p ℓ ! . Thus, if we prove that v /p ∈ D and that (cid:16) v + p P ℓ − k =0 p k u k (cid:17) /p ℓ ∈ D for all ℓ ∈{ , . . . , s + 1 } , then we would have v p ( g ( v + p u )) ≥ s + 1 . • Let us prove that v /p ∈ D .As v + p u / ∈ Ψ s +1 ( N ) , we obtain, according to (6.6), that { ( v + p u ) /p s +1 } ∈ D . Thusthere exists L ∈ { e , . . . , e q , f , . . . , f q } such that L · { ( v + p u ) /p s +1 } ≥ . We get ≤ L · v + p P s − k =0 p k u k p s +1 = L · v p s +1 + L · P s − k =0 p k u k p s = L · v p s +1 + L · (cid:26) u p s (cid:27) . (6.10)As u ∈ Ψ s ( N ) , we have { u /p s } / ∈ D and so L · { u /p s } < . We have L · { u /p s } ∈ p s N thus L · { u /p s } ≤ ( p s − /p s and we get, via inequality (6.10), that L · v /p s +1 ≥ /p s , i.e. L · v /p ≥ . Thereby, we have v /p ∈ D . • Let us prove that, for all ℓ ∈ { , . . . , s + 1 } , we have (cid:16) v + p P ℓ − k =0 p k u k (cid:17) /p ℓ ∈ D .We assume that s ≥ . Given ℓ ∈ { , . . . , s + 1 } , we have ≤ L · v + p P s − k =0 p k u k p s +1 = L · v + p P ℓ − k =0 p k u k p s +1 + L · p P s − k = ℓ − p k u k p s +1 . (6.11) e have u ∈ Ψ s ( N ) and u = u + p P s − k =1 p k − u k . Thus, applying (3.17) with t = 0 , weobtain P s − k =1 p k − u k ∈ Ψ s − ( N ) . Iterating (3.17), we finally get that P s − k = ℓ − p k − ℓ +1 u k ∈ Ψ s − ℓ +1 ( N ) . Following Lemma 11, we get p P s − k = ℓ − p k u k p s +1 = P s − k = ℓ − p k − ℓ +1 u k p s − ℓ +1 = ( P s − k = ℓ − p k − ℓ +1 u k p s − ℓ +1 ) / ∈ D . In particular, we obtain > L · (cid:0)P s − k = ℓ − p k − ℓ +1 u k (cid:1) /p s − ℓ +1 ∈ p s − ℓ +1 N . Thus we have L · (cid:0)P s − k = ℓ − p k − ℓ +1 u k (cid:1) /p s − ℓ +1 ≤ ( p s − ℓ +1 − /p s − ℓ +1 . Using this latest inequality in (6.11),we get L · v + p P ℓ − k =0 p k u k p s +1 ≥ p s − ℓ +1 . Therefore, for all ℓ ∈ { , . . . , s + 1 } , we have L · (cid:26) v + p u p ℓ (cid:27) = L · v + p P ℓ − k =0 p k u k p ℓ ≥ (6.12)and, for all ℓ ∈ { , . . . , s + 1 } , we obtain (cid:8) ( v + p u ) /p ℓ (cid:9) ∈ D . This completes the verifica-tion of assertion ( a ) .6.4.3. Verification of assertions ( a ) and ( a ) . For all s ∈ N , v ∈ { , . . . , p − } d and u ∈ Ψ s ( N ) , we set θ s ( v + u p ) := Q ( v + u p ) if v + u p / ∈ Ψ s +1 ( N ) , and θ s ( v + u p ) := g ( v + u p ) if v + u p ∈ Ψ s +1 ( N ) .The aim of this section is to prove the following assertion: for all s ∈ N , v ∈ { , . . . , p − } d , u ∈ Ψ s ( N ) and m ∈ N d , we have Q ( v + u p + m p s +1 ) Q ( v + u p ) − Q ( u + m p s ) Q ( u ) ∈ p s +1 g ( m ) θ s ( v + u p ) Z p , (6.13)which will prove assertions ( a ) and ( a ) of Theorem 4. Indeed, for all v ∈ { , . . . , p − } d and u ∈ Ψ s ( N ) , we have Q ( v + u p ) ∈ g ( v + u p ) Z p so that p s +1 g ( m ) θ s ( v + u p ) ∈ p s +1 g ( m ) Q ( v + u p ) Z p and (6.13) implies ( a ) . Furthermore, according to the definition of θ s , when v + u p ∈ Ψ s +1 ( N ) , congruence (6.13) implies ( a ) .Congruence (6.13) is valid if and only if, for all v ∈ { , . . . , p − } d , u ∈ Ψ s ( N ) and m ∈ N d , we have (cid:18) − Q ( v + u p ) Q ( u ) Q ( u + m p s ) Q ( v + u p + m p s +1 ) (cid:19) Q ( v + u p + m p s +1 ) Q ( v + u p ) ∈ p s +1 g ( m ) θ s ( v + u p ) Z p . In the sequel of the proof, we set X s ( v , u , m ) := Q ( v + u p ) Q ( u ) Q ( u + m p s ) Q ( v + u p + m p s +1 ) . hus, to prove (6.13), we only have to prove that ( X s ( v , u , m ) − Q ( v + u p + m p s +1 ) g ( m ) ∈ p s +1 Q ( v + u p ) θ s ( v + u p ) Z p . (6.14)In order to estimate the valuation of X s ( v , u , m ) − , let us set, for all v ∈ { , . . . , p − } d , u ∈ { , . . . , p s − } d , s ∈ N and m ∈ N d , Y s ( v , u , m ) := Q q i =1 Q ⌊ f i · v /p ⌋ j =1 (cid:16) f i · m p s f i · u + j (cid:17)Q q i =1 Q ⌊ e i · v /p ⌋ j =1 (cid:16) e i · m p s e i · u + j (cid:17) . Given s ∈ N , m ∈ N d and a ∈ { , . . . , p s − } d , we write η s ( a , m ) := P ∞ ℓ = s +1 ∆ (cid:16)n a + m p s p ℓ o(cid:17) .We state four lemmas, which we prove in Section 6.4.4. Lemma 12.
For all s ∈ N , v ∈ { , . . . , p − } d , u ∈ Ψ s ( N ) and m ∈ N d , we have X s ( v , u , m ) ∈ Y s ( v , u , m ) (1 + p s +1 Z p ) and v p ( Y s ( v , u , m )) ≥ η s ( u , m ) − η s +1 ( v + u p, m ) . Lemma 13.
Given s ∈ N , v ∈ { , . . . , p − } d and u ∈ { , . . . , p s − } d , if there exists j ∈ { , . . . , s + 1 } such that { ( v + u p ) /p j } / ∈ D , then we have Y s ( v , u , m ) ∈ p s − j +2 Z p . Lemma 14.
For all s ∈ N , a ∈ { , . . . , p s +1 − } d and m ∈ N d , we have η s +1 ( a , m ) ≥ µ ( m ) (6.15) and v p (cid:18) Q ( a + m p s +1 ) g ( m ) (cid:19) ≥ s +1 X ℓ =1 ∆ (cid:18)(cid:26) a p ℓ (cid:27)(cid:19) . (6.16) Lemma 15.
Given s ∈ N and a ∈ Ψ s ( N ) , we have v p ( Q ( a )) = P sℓ =1 ∆ (cid:16)n a p ℓ o(cid:17) . In order to prove (6.14), we will now distinguish two cases. • Case
1: Let us assume that there exists j ∈ { , . . . , s + 1 } such that (cid:26) v + u pp j (cid:27) / ∈ D . (6.17)Let j be the smaller j ∈ { , . . . , s + 1 } verifying (6.17). According to Lemma 13 appliedwith j , we get Y s ( v , u , m ) ∈ p s − j +2 Z p and thus, following Lemma 12, v p ( X s ( v , u , m ) − ≥ s − j + 2 . According to (6.16), we get v p (cid:18) ( X s ( v , u , m ) − Q ( v + u p + m p s +1 ) g ( m ) (cid:19) ≥ v p ( X s ( v , u , m ) −
1) + s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) ≥ s − j + 2 + s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) . (6.18) or all ℓ ∈ { , . . . , j − } , we have { ( v + u p ) /p ℓ } ∈ D and so ∆( { ( v + u p ) /p ℓ } ) ≥ .We get P s +1 ℓ =1 ∆( { ( v + u p ) /p ℓ } ) ≥ j − which, associated with (6.18), leads to v p (cid:18) ( X s ( v , u , m ) − Q ( v + u p + m p s +1 ) g ( m ) (cid:19) ≥ s + 1 . (6.19)If v + u p / ∈ Ψ s +1 ( N ) , then we have θ s ( v + u p ) = Q ( v + u p ) and p s +1 Q ( v + u p ) θ s ( v + u p ) = p s +1 .Hence, when v + u p / ∈ Ψ s +1 ( N ) , inequality (6.19) implies (6.14).We assume, throughout the end of the proof of Case , that v + u p ∈ Ψ s +1 ( N ) , thus θ s ( v + u p ) = g ( v + u p ) . Let us prove that we have v p ( g ( v + u p )) ≥ j − . Indeed, for all ℓ ∈ { , . . . , j − } , we have { ( v + u p ) /p ℓ } ∈ D and therefore v p ( g ( v + u p )) = ∞ X ℓ =1 D (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) ≥ j − . Following (6.18), we get v p ( X s ( v , u , m ) − Q ( v + u p + m p s +1 ) g ( m ) ! ≥ s − j + 2 + v p ( g ( v + u p )) + s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) − v p ( g ( v + u p )) ! ≥ ( s − j + 2) + j − v p (cid:18) Q ( v + u p ) g ( v + u p ) (cid:19) (6.20) ≥ s + 1 + v p (cid:18) Q ( v + u p ) g ( v + u p ) (cid:19) , where (6.20) is valid because, applying Lemma 15 with s + 1 instead of s and v + u p insteadof a , we get v p ( Q ( v + u p )) = P s +1 ℓ =1 ∆( { v + u pp ℓ } ) . Thus we have (6.14) in this case. • Case
2: Let us assume that, for all j ∈ { , . . . , s + 1 } , we have { ( v + u p ) /p j } ∈ D .In particular, we have v + u p / ∈ Ψ s +1 ( N ) and thus θ s ( v + u p ) = Q ( v + u p ) . Furthermore,we obtain P s +1 ℓ =1 ∆( { ( v + u p ) /p ℓ } ) ≥ s + 1 .If v p ( Y s ( v , u , m )) ≥ , then, following Lemma 12, v p ( X s ( v , u , m ) − ≥ and, accordingto (6.16), we have v p (cid:18) Q ( v + u p + m p s +1 ) g ( m ) (cid:19) ≥ s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) ≥ s + 1 . thus we have (6.14).Let us now assume that v p ( Y s ( v , u , m )) < . In this case, according to Lemma 12, wehave v p ( X s ( v , u , m ) −
1) = v p ( Y s ( v , u , m )) ≥ η s ( u , m ) − η s +1 ( v + u p, m ) . urthermore, v p ( Q ( v + u p + m p s +1 )) = ∞ X ℓ =1 ∆ (cid:18)(cid:26) v + u p + m p s +1 p ℓ (cid:27)(cid:19) = s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) + ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) v + u p + m p s +1 p ℓ (cid:27)(cid:19) = s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) + η s +1 ( v + u p, m ) . Thereby, we get v p ( X s ( v , u , m ) − Q ( v + u p + m p s +1 ) g ( m ) ! ≥ η s ( u , m ) − η s +1 ( v + u p, m ) + s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) + η s +1 ( v + u p, m ) − µ ( m ) ≥ s + 1 + η s ( u , m ) − µ ( m ) . If s = 0 , then we have u = and η ( , m ) = P ∞ ℓ =1 ∆( { m p ℓ } ) ≥ P ∞ ℓ =1 D ( { m p ℓ } ) = µ ( m ) and we have (6.14). On the other hand, if s ≥ then, applying Lemma 14 with s − instead of s and a = u , we get η s ( u , m ) ≥ µ ( m ) , which implies (6.14). This finishes theproof of equation (6.13) modulo those of the various lemmas.6.4.4. Proof of Lemmas 12, 13, 14 and 15.Proof of Lemma 12.
We have to prove that X s ( v , u , m ) ∈ Y s ( v , u , m )(1 + p s +1 Z p ) .We have X s ( v , u , m ) = Q ( v + u p ) Q ( u p ) Q ( u p + m p s +1 ) Q ( v + u p + m p s +1 ) · Q ( u p ) Q ( u ) Q ( u + m p s ) Q ( u p + m p s +1 ) . Applying Lemma 7 with c = u we obtain Q ( u p ) Q ( u ) Q ( u + m p s ) Q ( u p + m p s +1 ) ∈ p s +1 Z p , so that X s ( v , u , m ) ∈ Q ( v + u p ) Q ( u p ) Q ( u p + m p s +1 ) Q ( v + u p + m p s +1 ) (1 + p s +1 Z p ) . (6.21) urthermore, we have Q ( v + u p ) Q ( u p ) · Q ( u p + m p s +1 ) Q ( v + u p + m p s +1 )= (cid:16) Q q i =1 Q e i · v k =1 ( e i · u p + k ) (cid:17) (cid:16)Q q i =1 Q f i · v k =1 ( f i · ( u p + m p s +1 ) + k ) (cid:17)(cid:16)Q q i =1 Q f i · v k =1 ( f i · u p + k ) (cid:17) (cid:16) Q q i =1 Q e i · v k =1 ( e i · ( u p + m p s +1 ) + k ) (cid:17) = Q q i =1 Q f i · v k =1 (cid:16) f i · m p s +1 f i · u p + k (cid:17)Q q i =1 Q e i · v k =1 (cid:16) e i · m p s +1 e i · u p + k (cid:17) . If d ∈ { e , . . . , e q , f , . . . , f q } and k ∈ { , . . . , d · v } , then p divides d · u p + k if and onlyif there exists j ∈ { , . . . , ⌊ d · v /p ⌋} such that k = jp . Thus we have d · v Y k =1 (cid:18) d · m p s +1 d · u p + k (cid:19) = ⌊ d · v /p ⌋ Y j =1 (cid:18) d · m p s d · u + j (cid:19) (1 + O ( p s +1 )) . Therefore Q ( v + u p ) Q ( u p ) · Q ( u p + m p s +1 ) Q ( v + u p + m p s +1 ) = Q q i =1 Q ⌊ f i · v /p ⌋ j =1 (cid:16) f i · m p s f i · u + j (cid:17)Q q i =1 Q ⌊ e i · v /p ⌋ j =1 (cid:16) e i · m p s e i · u + j (cid:17) (1 + O ( p s +1 ))= Y s ( v , u , m )(1 + O ( p s +1 )) and so X s ( v , u , m ) ∈ Y s ( v , u , m )(1 + p s +1 Z p ) , as expected.We will now prove that we also have v p ( Y s ( v , u , m )) ≥ η s ( u , m ) − η s +1 ( v + u p, m ) . We have seen above that v p ( Y s ( v , u , m )) = v p ( X s ( v , u , m )) . Furthermore, according to(6.21), we also have v p ( X s ( v , u , m )) = v p (cid:18) Q ( v + u p ) Q ( u p ) · Q ( u p + m p s +1 ) Q ( v + u p + m p s +1 ) (cid:19) = v p ( Q ( v + u p )) − v p ( Q ( u p )) + v p ( Q ( u p + m p s +1 )) − v p ( Q ( v + u p + m p s +1 ))= ∞ X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) − ∞ X ℓ =1 ∆ (cid:18)(cid:26) u pp ℓ (cid:27)(cid:19) + ∞ X ℓ =1 ∆ (cid:18)(cid:26) u p + m p s +1 ) p ℓ (cid:27)(cid:19) − ∞ X ℓ =1 ∆ (cid:18)(cid:26) v + u p + m p s +1 p ℓ (cid:27)(cid:19) . e have ∞ X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) − ∞ X ℓ =1 ∆ (cid:18)(cid:26) v + u p + m p s +1 p ℓ (cid:27)(cid:19) = ∞ X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) − s +1 X ℓ =1 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) − ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) v + u p + m p s +1 p ℓ (cid:27)(cid:19) = ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) v + u pp ℓ (cid:27)(cid:19) − ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) v + u p + m p s +1 p ℓ (cid:27)(cid:19) = η s +1 ( v + u p, ) − η s +1 ( v + u p, m ) , and ∞ X ℓ =1 ∆ (cid:18)(cid:26) u pp ℓ (cid:27)(cid:19) − ∞ X ℓ =1 ∆ (cid:18)(cid:26) u p + m p s +1 p ℓ (cid:27)(cid:19) = ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) u pp ℓ (cid:27)(cid:19) − ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) u p + m p s +1 p ℓ (cid:27)(cid:19) = ∞ X ℓ = s +1 ∆ (cid:18)(cid:26) u p ℓ (cid:27)(cid:19) − ∞ X ℓ = s +1 ∆ (cid:18)(cid:26) u + m p s p ℓ (cid:27)(cid:19) = η s ( u , ) − η s ( u , m ) . Thus, v p ( Y s ( v , u , m )) = η s +1 ( v + u p, ) − η s ( u , )+ η s ( u , m ) − η s +1 ( v + u p, m ) . We now haveto prove that if u ∈ Ψ s ( N ) , then we have η s +1 ( v + u p, ) − η s ( u , ) ≥ . As u ∈ Ψ s ( N ) , wehave { u /p s } / ∈ D . Hence, for all ℓ ≥ s + 1 and all L ∈ { e , . . . , e q , f , . . . , f q } , we obtain L · (cid:26) u p ℓ (cid:27) = L · u p ℓ ≤ L · u p s = L · (cid:26) u p s (cid:27) < , i.e. , for all ℓ ≥ s + 1 , { u /p ℓ } / ∈ D . Then we have η s ( u , ) = P ∞ ℓ = s +1 ∆ (cid:16)n u p ℓ o(cid:17) = 0 and η s +1 ( v + u p, ) − η s ( u , ) = η s +1 ( v + u p, ) ≥ , which completes the proof of Lemma 12. (cid:3) Proof of Lemma 13.
Given s ∈ N , v ∈ { , . . . , p − } d and u ∈ { , . . . , p s − } d , we write u = P ∞ k =0 u k p k , where u k ∈ { , . . . , p − } d . Given L ∈ { e , . . . , e q , f , . . . , f q } , we define s + 1 non-negative integers by the formulas b L , := ⌊ L · v /p ⌋ and b L ,k +1 := ⌊ ( L · u k + b L ,k ) /p ⌋ for k ∈ { , . . . , s − } . For all x ∈ R , we write ⌈ x ⌉ the smallest integer greater than x and wedefine s +1 non-negative integers by the formulas a L , := 1 and a L ,k +1 := ⌈ ( L · u k + a L ,k ) /p ⌉ .First, we will prove by induction on r that assertion A r : ⌊ L · v /p ⌋ Y n =1 (cid:18) L · m p s L · u + n (cid:19) = b L ,r Y n = a L ,r (cid:18) L · m p s − r L · ( P ∞ k = r u k p k − r ) + n (cid:19) (cid:0) O ( p s − r +1 ) (cid:1) is true for all r ∈ { , . . . , s } . e have b L , = ⌊ L · v /p ⌋ and a L , = 1 , thus A is true.Given r ≥ , let us assume that A r is true and prove A r +1 . If a L ,r > b L ,r then a L ,r +1 >b L ,r +1 and A r implies A r +1 . Thus we can assume that a L ,r ≤ b L ,r . If n ∈ { a L ,r , . . . , b L ,r } ,then p divides L · (cid:0)P ∞ k = r u k p k − r (cid:1) + n if and only if p divides L · u r + n , i.e. if and only if an i ∈ {⌈ ( L · u r + a L ,r ) /p ⌉ , . . . , ⌊ ( L · u r + b L ,r ) /p ⌋} exists such that L · u r + n = ip . So we get b L ,r Y n = a L ,r (cid:18) L · m p s − r L · ( P ∞ k = r u k p k − r ) + n (cid:19) = b L ,r +1 Y i = a L ,r +1 L · m p s − r L · (cid:0)P ∞ k = r +1 u k p k − r (cid:1) + ip ! (1 + O ( p s − r ))= b L ,r +1 Y i = a L ,r +1 L · m p s − r − L · (cid:0)P ∞ k = r +1 u k p k − r − (cid:1) + i ! (1 + O ( p s − r )) . (6.22)According to A r and (6.22), we have A r +1 , which finishes the induction on r .Given L ∈ { e , . . . , e q , f , . . . , f q } , we will prove by induction on k that assertion B k : a L ,k ≥ and b L ,k ≤ ⌊ L · { ( v + u p ) /p k +1 }⌋ is true for all k ∈ { , . . . , s } .We have a L , = 1 and b L , = ⌊ L · v /p ⌋ = ⌊ L · { ( v + u p ) /p }⌋ , so B is true.Given k ≥ , let us assume that B k is true and let us prove B k +1 . We have a L ,k +1 = ⌈ ( L · u k + a L ,k ) /p ⌉ and b L ,k +1 = ⌊ ( L · u k + b L ,k ) /p ⌋ , thus a L ,k +1 ≥ ⌈ ( L · u k + 1) /p ⌉ ≥ and b L ,k +1 ≤ (cid:22) L · u k p + L p · (cid:26) v + u pp k +1 (cid:27)(cid:23) = $ L · u k p k +1 p k +2 + v + p P k − i =0 u i p i p k +2 !% = (cid:22) L · (cid:26) v + u pp k +2 (cid:27)(cid:23) , which completes the induction on k .Given j ∈ { , . . . , s +1 } such that { ( v + u p ) /p j } / ∈ D , for all L ∈ { e , . . . , e q , f , . . . , f q } ,we obtain, via B j − , that a L ,j − ≥ and b L ,j − ≤ ⌊ L · { ( v + u p ) /p j }⌋ = 0 . Hence, following A j − , we get ⌊ L · v /p ⌋ Y n =1 (cid:18) L · m p s L · u + n (cid:19) = 1 + O ( p s − j +2 ) and thus Y s ( v , u , m ) = Q q i =1 Q ⌊ f i · v /p ⌋ n =1 (cid:16) f i · m p s f i · u + n (cid:17)Q q i =1 Q ⌊ e i · v /p ⌋ n =1 (cid:16) e i · m p s e i · u + n (cid:17) = 1 + O ( p s − j +2 )1 + O ( p s − j +2 ) = 1 + O ( p s − j +2 ) , which finishes the proof of Lemma 13. (cid:3) roof of Lemma 14. First, we will prove that we have (6.15). Let us write m = P qk =0 m k p k ,where m k ∈ { , . . . , p − } d . We have η s +1 ( a , m ) − µ ( m ) = ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) a + m p s +1 p ℓ (cid:27)(cid:19) − ∞ X ℓ =1 D (cid:18)(cid:26) m p ℓ (cid:27)(cid:19) = ∞ X ℓ = s +2 (cid:18) ∆ (cid:18)(cid:26) a + m p s +1 p ℓ (cid:27)(cid:19) − D (cid:18)(cid:26) m p s +1 p ℓ (cid:27)(cid:19)(cid:19) = ∞ X ℓ = s +2 ∆ a + P ℓ − s − k =0 m k p k + s +1 p ℓ ! − D P ℓ − s − k =0 m k p k + s +1 p ℓ !! . Furthermore, for all ℓ ≥ s + 2 , we have ≤ P ℓ − s − k =0 m k p k + s +1 p ℓ ≤ a + P ℓ − s − k =0 m k p k + s +1 p ℓ ≤ ( p ℓ − p ℓ ∈ [0 , d . Thus D P ℓ − s − k =0 m k p k + s +1 p ℓ ! = 1 = ⇒ P ℓ − s − k =0 m k p k + s +1 p ℓ ∈ D = ⇒ a + P ℓ − s − k =0 m k p k + s +1 p ℓ ∈ D = ⇒ ∆ a + P ℓ − s − k =0 m k p k + s +1 p ℓ ! ≥ and so η s +1 ( a , m ) − µ ( m ) ≥ . This completes the proof of (6.15).Let us now prove (6.16). We have v p (cid:18) Q ( a + m p s +1 ) g p ( m ) (cid:19) = ∞ X ℓ =1 ∆ (cid:18)(cid:26) a + m p s +1 p ℓ (cid:27)(cid:19) − µ ( m )= s +1 X ℓ =1 ∆ (cid:18)(cid:26) a p ℓ (cid:27)(cid:19) + ∞ X ℓ = s +2 ∆ (cid:18)(cid:26) a + m p s +1 p ℓ (cid:27)(cid:19) − µ ( m )= s +1 X ℓ =1 ∆ (cid:18)(cid:26) a p ℓ (cid:27)(cid:19) + η s +1 ( a , m ) − µ ( m ) , ≥ s +1 X ℓ =1 ∆ (cid:18)(cid:26) a p ℓ (cid:27)(cid:19) . (6.23)where we used inequality (6.15) for (6.23). (cid:3) roof of Lemma 15. We have v p ( Q ( a )) = P ∞ ℓ =1 ∆ (cid:16)n a p ℓ o(cid:17) . As a ∈ Ψ s ( N ) , we have { a /p s } / ∈ D and, for all ℓ ≥ s + 1 and all L ∈ { e , . . . , e q , f , . . . , f q } , we get L · (cid:26) a p ℓ (cid:27) = L · a p ℓ ≤ L · a p s = L · (cid:26) a p s (cid:27) < , i.e. { a /p ℓ } / ∈ D . Thus, for all ℓ ≥ s + 1 , we have ∆ (cid:0)(cid:8) a /p ℓ (cid:9)(cid:1) = 0 . This gives us theexpected result. (cid:3) Proof of assertions ( ii ) of Theorems 1 and 2 We assume the hypothesis of Theorems 1 and 2. Furthermore, we assume that x ∈ D e,f is a zero of ∆ e,f . In Section 7.1, we prove an elementary result of analysis which we will usefor the proofs of assertions ( ii ) of Theorems 1 and 2. We prove assertion ( ii ) of Theorem1 in Section 7.2. We will use certain results from Section 7.2 for the proof of assertion ( ii ) of Theorem 2 which we present in Section 7.3.7.1. Preliminary.
The aim of this section is to prove that there exists a nonempty opensubset U of D e,f such that, for all x ∈ U , i ∈ { , . . . , q } and j ∈ { , . . . , q } , we have ⌊ e i · x ⌋ = ⌊ e i · x ⌋ , e i · x = 0 , ⌊ f j · x ⌋ = ⌊ f j · x ⌋ , f j · x = 0 and e i · x = f j · x .Particularly, for all x ∈ U , we would have ∆ e,f ( x ) = ∆ e,f ( x ) = 0 . We will use this openset U throughout the rest of the proof.Applying Lemma 1 with, instead of u , the sequence constituted by the elements of e and f , we obtain that there exists µ > such that, for all x ∈ [0 , µ ] d and all L ∈{ e , . . . , e q , f , . . . f q } , we have ⌊ L · ( x + x ) ⌋ = ⌊ L · x ⌋ . As x ∈ [0 , d , there exists µ > , µ ≤ µ , such that, for all x ∈ [0 , µ ] d , we have x + x ∈ [0 , d . Since x ∈ D e,f , a L ∈ { e , . . . , e q , f , . . . , f q } exists such that L · x ≥ , which gives us the result that, forall x ∈ [0 , µ ] d , we have L · ( x + x ) ≥ L · x ≥ and thus, as x + x ∈ [0 , d , we get that x + x ∈ D e,f . Thereby, there exists a nonempty open subset U of D e,f such that, for all x ∈ U and L ∈ { e , . . . , e q , f , . . . , f q } , we have ⌊ L · x ⌋ = ⌊ L · x ⌋ .For all i ∈ { , . . . , q } and j ∈ { , . . . , q } , we define the sets H e i := { x ∈ R d : e i · x = 0 } , H f j := { x ∈ R d : f j · x = 0 } and H e i , f j := { x ∈ R d : e i · x = f j · x } . Since e and f aretwo disjoint sequences constituted by nonzero vectors, we obtain that the H e i , H f j and H e i , f j are hyperplanes in R d and are therefore closed subsets of R d with empty interiors.Therefore, their complements are dense open subsets of R d and the complement U of theunion of H e i , H f j and H e i , f j is a dense open subset of R d . As a result, U := U ∩ U is anonempty open subset of D e,f and, for all x ∈ U , i ∈ { , . . . , q } and j ∈ { , . . . , q } , wehave e i · x = 0 , f j · x = 0 , e i · x = f j · x , ⌊ e i · x ⌋ = ⌊ e i · x ⌋ and ⌊ f j · x ⌋ = ⌊ f j · x ⌋ .7.2. Proof of assertion ( ii ) of Theorem 1. The aim of this section is to prove thatthere exists k ∈ { , . . . , d } such that there are only finitely many prime numbers p suchthat q e,f,k ( z ) ∈ z k Z p [[ z ]] . Following Section 4, we only have to prove that there exists k ∈{ , . . . , d } such that, for all large enough prime number p , there exists a ∈ { , . . . , p − } d and K ∈ N d such that Φ p,k ( a + p K ) / ∈ p Z p . We will actually prove that there exists ∈ { , . . . , d } such that, for all large enough prime number p , there is an a ∈ { , . . . , p − } d such that Φ p,k ( a ) / ∈ p Z p . In this case, we have Φ p,k ( a ) = − p Q ( a ) q X i =1 e ( k ) i H a · e i − q X i =1 f ( k ) i H a · f i ! . (7.1)For all d ∈ N d , we have pH d · a = p d · a X i =1 i ≡ p ⌊ d · a /p ⌋ X j =1 jp mod p Z p ≡ ⌊ d · a /p ⌋ X j =1 j mod p Z p . For all k ∈ { , . . . , d } and x ∈ [0 , d , we set Ψ k ( x ) := q X i =1 ⌊ e i · x ⌋ X j =1 e ( k ) i j − q X i =1 ⌊ f i · x ⌋ X j =1 f ( k ) i j . Thus, for all k ∈ { , . . . , d } and a ∈ { , . . . , p − } d , we have Φ p,k ( a ) ≡ −Q ( a )Ψ k ( a /p )mod p Z p . Therefore we now have to prove that there exists k ∈ { , . . . , d } such that, forall large enough prime number p , there exists a ∈ { , . . . , p − } d such that v p ( Q ( a )) = v p (Ψ k ( a /p )) = 0 . We set M := max {| d | : d ∈ { e , . . . , e q , f , . . . , f q }} .A constant P ≥ M exists such that, for all prime number p ≥ P , there exists a p ∈{ , . . . , p − } d such that a p /p ∈ U . For all ℓ ≥ , we have a p /p ℓ ≤ a p /p < /p and thus,for all L ∈ { e , . . . , e q , f , . . . , f q } , we have L · a p /p ℓ < L · /p ≤ M /p ≤ . Hence, for allprime number p ≥ P and all ℓ ≥ , we have a p /p ℓ / ∈ D e,f , which implies that v p ( Q ( a p )) = ∞ X ℓ =1 ∆ e,f (cid:18) a p p ℓ (cid:19) = ∆ e,f (cid:18) a p p (cid:19) = 0 , because ∆ e,f vanishes on U and on [0 , d \D e,f .So we now have to prove that there exists k ∈ { , . . . , d } and a constant P ≥ P suchthat, for all prime number p ≥ P , we have v p (Ψ k ( a p /p )) = 0 .For all prime number p ≥ P , all i ∈ { , . . . , q } and j ∈ { , . . . , q } , we write α i := ⌊ e i · a p /p ⌋ and β j := ⌊ f j · a p /p ⌋ . According to the construction of U and since a p /p ∈ U ,we have ⌊ e i · a p /p ⌋ = ⌊ e i · x ⌋ and ⌊ f j · a p /p ⌋ = ⌊ f j · x ⌋ . Therefore, the α i and β j do notdepend on p . Thus there exists a constant P ≥ P such that, for all prime number p ≥ P and all k ∈ { , . . . , d } , we have Ψ k ( a p /p ) = q X i =1 α i X j =1 e ( k ) i j − q X i =1 β j X j =1 f ( k ) i j ∈ Z × p ∪ { } . Therefore we only have to prove that there exists k ∈ { , . . . , d } such that Ψ k ( a p /p ) = 0 .For this purpose, we will use Lemma from [3] which reads as follows. emma 16. Let E := ( E , . . . , E q ) and F := ( F , . . . , F q ) be two disjoint sequences ofpositive integers. We write A := { E , . . . , E q , F , . . . , F q } and γ < · · · < γ t = 1 forthe rational numbers which satisfy { γ , . . . , γ t } = S a ∈A { a , a , . . . , a − a , } and m i ∈ Z theamplitude of the jump of ∆ E , F in γ i . If there exists i ∈ { , . . . , t } such that ∆ E , F ≥ on [ γ , γ i ] , then we have i X k =1 m k γ k > and i Y k =1 (cid:18) γ k (cid:19) m k > . We will use Lemma 16 with E p := ( e · a p , . . . , e q · a p ) instead of E and F p := ( f · a p , . . . , f q · a p ) instead of F .First, we have to prove that E p and F p are two disjoint sequences of positive integers.Indeed, according to the construction of U , for all i ∈ { , . . . , q } and all j ∈ { , . . . , q } ,we have e i · a p /p = 0 , f j · a p /p = 0 and e i · a p /p = f j · a p /p , thus e i · a p = 0 , f j · a p = 0 and e i · a p = f j · a p , which gives us that E p and F p are two disjoint sequences of positiveintegers.We write A := { e · a p , . . . , e q · a p , f · a p , . . . , f q · a p } and γ < · · · < γ t = 1 for the rationalnumbers which satisfy { γ , . . . , γ t } = S a ∈A { a , a , . . . , a − a , } and m i ∈ Z the amplitude ofthe jump of ∆ E p , F p in γ i . As a p /p ∈ D e , f , there exists a ∈ A such that a ≥ p and so max( A ) ≥ p . Hence, we have γ = 1 / max( A ) ≤ /p . Thus there exists i ∈ { , . . . , t − } such that γ i ≤ /p < γ i +1 . Furthermore, for all x ∈ [0 , , we have ∆ E p , F p ( x ) = q X i =1 ⌊ ( e i · a p ) x ⌋ − q X j =1 ⌊ ( f j · a p ) x ⌋ = ∆ e,f ( x a p ) ≥ , because ∆ e , f ≥ on [0 , d . In particular, ∆ E p , F p ≥ on [ γ , γ i ] .We can therefore apply Lemma 16 which results in < i X i =1 m i γ i = X c ∈ E p ⌊ c/p ⌋ X j =1 cj − X d ∈ F p ⌊ d/p ⌋ X j =1 dj (7.2) = q X i =1 ⌊ a p · e i /p ⌋ X j =1 a p · e i j − q X i =1 ⌊ a p · f i /p ⌋ X j =1 a p · f i j = d X k =1 a ( k ) p q X i =1 α i X j =1 e ( k ) i j − q X i =1 β i X j =1 f ( k ) i j ! = d X k =1 a ( k ) p Ψ k ( a p /p ) , where (7.2) is valid because the abscissas of the jumps of ∆ E p , F p on [0 , /p ] are exactly therational numbers j/a with a ∈ A and j ≤ ⌊ a/p ⌋ , and an abscissa j/a corresponds to ajump with positive amplitude when a ∈ E p and to a jump with negative amplitude when a ∈ F p .Thus there exists k ∈ { , . . . , d } such that Ψ k ( a p /p ) = 0 , which finishes the proof ofassertion ( ii ) of Theorem 1. .3. Proof of assertion ( ii ) of Theorem 2. According to Section 7.2, there exists k ∈{ , . . . , d } such that there are only finitely many prime numbers p such that q e,f,k ( z ) ∈ Z p [[ z ]] . In order to finish the proof of assertion ( ii ) of Theorem 2, we only have to provethat, for all L ∈ E satisfying L ( k ) ≥ , there are only finitely many prime numbers p suchthat q L ,e,f ( z ) ∈ Z p [[ z ]] . During the proof, we fix L ∈ E e,f satisfying L ( k ) ≥ ( ). We willseparate the proof into two cases depending on whether ⌊ L · x ⌋ = 0 or ⌊ L · x ⌋ 6 = 0 .According to Section 7.2, we know that there exists a constant P such that, for all primenumber p ≥ P , there exists a p ∈ { , . . . , p − } d such that a p /p ∈ U and v p ( Q ( a p )) = 0 .7.3.1. When ⌊ L · x ⌋ 6 = 0 . The aim of this section is to prove that there exists a constant
P ≥ P such that, for all prime number p ≥ P , we have Φ L ,p ( a p ) / ∈ p Z p , which, accordingto Section 4, will prove that there are only finitely many prime numbers p such that q L ,e,f ( z ) ∈ Z p [[ z ]] .We recall that, for all a ∈ { , . . . , p − } d , we have Φ L ,p ( a ) = − p Q ( a ) H L · a ≡ −Q ( a ) H ⌊ L · a /p ⌋ mod p Z p . (7.3)For all prime number p ≥ P , we have ⌊ L · a p /p ⌋ = ⌊ L · x ⌋ 6 = 0 therefore H ⌊ L · a p /p ⌋ ∈{ H , . . . , H | L | } . A constant P ≥ P exists such that, for all prime number p ≥ P , we have { H , . . . , H | L | } ⊂ Z × p . Thus, for all prime number p ≥ P , we have Q ( a p ) H ⌊ L · a p /p ⌋ ∈ Z × p and, following (7.3), we obtain Φ L ,p ( a p ) / ∈ p Z p .We observe that in this case, we did not use the hypothesis L ( k ) ≥ .7.3.2. When ⌊ L · x ⌋ = 0 . The aim of this section is to prove that there exists r ∈ N , r ≥ and a constant P ′ ≥ P such that, for all prime number p ≥ P ′ , we have Φ L ,p ( a p + pr k ) / ∈ p Z p . According to Section 4, this will prove that there are only finitely many prime numbers p such that q L ,e,f ( z ) ∈ Z p [[ z ]] .In the sequel, for all k ∈ { , . . . , d } , we write R k for the rational function defined by R k ( X ) := Q q i =1 Q α i j =1 (cid:18) e ( k ) i j X (cid:19)Q q i =1 Q β i j =1 (cid:18) f ( k ) i j X (cid:19) . (7.4)We will use the following lemma, which we will prove at the end of this section. Lemma 17.
For all r ∈ N , r ≥ , there exists a constant P r ≥ P such that, for all primenumber p ≥ P r and all k ∈ { , . . . , d } , we have Φ L ,p ( a p + pr k ) ≡ − r X j =1 H j L ( k ) Q ( a p ) Q ( j k ) Q (( r − j ) k ) (cid:0) R k ( j ) − R k ( r − j ) (cid:1) mod p Z p . According to the end of Section 7.2, we know that q X i =1 α i X j =1 e ( k ) i j − q X i =1 β i X j =1 f ( k ) i j = 0 . (7.5) Such a L exists because q e,f,k ( z ) / ∈ z k Z [[ z ]] . nequality (7.5) proves that R k ( X ) is not a constant equal to . Thus there exists r ∈ N such that R k ( r ) = 1 . Let r be the smallest positive integer satisfying R k ( r ) = 1 .Applying Lemma 17 with k instead of k and r instead of r , we obtain that a constant P r ≥ P exists such that, for all prime number p ≥ P r , we have Φ L ,p ( a p + pr k ) ≡ − r X j =1 H j L ( k Q ( a p ) Q ( j k ) Q (( r − j ) k ) (cid:0) R k ( j ) − R k ( r − j ) (cid:1) mod p Z p ≡ − H r L ( k Q ( a p ) Q ( r k ) (cid:0) R k ( r ) − (cid:1) mod p Z p , (7.6)where (7.6) is valid because, for all j ∈ { , . . . , r − } , we have R k ( j ) = R k ( r − j ) = 1 .Since R k ( r ) = 1 , we obtain that if L ( k ) ≥ , then there exists a constant P ≥ P r suchthat, for all prime number p ≥ P , we have H r L ( k Q ( a p ) Q ( r k ) ( R k ( r ) − ∈ Z × p and therefore Φ L ,p ( a p + pr k ) / ∈ p Z p , which completes the proof of assertion ( ii ) ofTheorem 2 modulo the proof of Lemma 17. Proof of Lemma 17.
According to Section 4, for all prime number p ≥ P and all K ∈ N d ,we have Φ L ,p ( a p + p K ) = X ≤ j ≤ K Q ( K − j ) Q ( a p + p j ) (cid:0) H L · ( K − j ) − pH L · ( a p + p j ) (cid:1) . (7.7)Furthermore, we have pH L · ( a p + p j ) ≡ H j L · a p + p L · j p k mod p Z p with j L · a p + p L · j p k = ⌊ L · a p /p ⌋ + L · j = L · j because ⌊ L · a p /p ⌋ = ⌊ L · x ⌋ = 0 . Thereby, for all K , j ∈ N d , j ≤ K , we obtain Q ( K − j ) Q ( a p + p j ) pH L · ( a p + p j ) ≡ Q ( K − j ) Q ( a p + p j ) H L · j mod p Z p . (7.8)Applying (7.8) to (7.7), we obtain that, for all K ∈ N d , we have Φ L ,p ( a p + p K ) ≡ X ≤ j ≤ K Q ( K − j ) Q ( a p + p j ) (cid:0) H L · ( K − j ) − H L · j (cid:1) mod p Z p ≡ − X ≤ j ≤ K H L · j (cid:0) Q ( a p + p j ) Q ( K − j ) − Q ( j ) Q ( a p + p ( K − j )) (cid:1) mod p Z p ≡ − X ≤ j ≤ K H L · j Q ( a p ) Q ( j ) Q ( K − j ) (cid:18) Q ( a p + p j ) Q ( a p ) Q ( j ) − Q ( a p + p ( K − j )) Q ( a p ) Q ( K − j ) (cid:19) mod p Z p . (7.9) pplying (7.9) with r k instead of K , we finally obtain Φ L ,p ( a p + pr k ) ≡− r X j =0 H j L ( k ) Q ( a p ) Q ( j k ) Q (( r − j ) k ) (cid:18) Q ( a p + pj k ) Q ( a p ) Q ( j k ) − Q ( a p + p ( r − j ) k ) Q ( a p ) Q (( r − j ) k ) (cid:19) mod p Z p . (7.10)We will now prove that, for all n ∈ N and k ∈ { , . . . , d } , we have Q ( a p + pn k ) Q ( a p ) Q ( n k ) = R k ( n )(1 + O ( p )) , (7.11)which will enable us to conclude. We have Q ( a p + pn k ) Q ( a p ) Q ( n k ) = Q ( a p + pn k ) Q ( a p ) Q ( pn k ) Q ( pn k ) Q ( n k )= 1 Q ( a p ) Q q i =1 Q e i · a p j =1 ( pn e ( k ) i + j ) Q q i =1 Q f i · a p j =1 ( pn f ( k ) i + j ) (cid:0) O ( p ) (cid:1) (7.12) = Q q i =1 Q e i · a p j =1 (cid:18) pn e ( k ) i j (cid:19)Q q i =1 Q f i · a p j =1 (cid:18) pn f ( k ) i j (cid:19) (cid:0) O ( p ) (cid:1) = Q q i =1 Q ⌊ e i · a p /p ⌋ j =1 (cid:18) e ( k ) i j n (cid:19)Q q i =1 Q ⌊ f i · a p /p ⌋ j =1 (cid:18) f ( k ) i j n (cid:19) (cid:0) O ( p ) (cid:1) (7.13) = R k ( n ) (cid:0) O ( p ) (cid:1) , where we obtain (7.12) by applying Lemma 7 with s = 0 , c = and n k instead of m ,which leads to Q ( pn k ) / Q ( n k ) = 1 + O ( p ) . Equation (7.13) is valid because, for all d ∈ { e , . . . , e q , f , . . . , f q } and j ∈ { , . . . , d · a p } , if j is not divisible by p then we have pn d ( k ) j = 1 + O ( p ) .There exists a constant P r ≥ P such that, for all prime number p ≥ P r and all n ∈{ , . . . , r } , we have R k ( n ) ∈ Z × p and H n L ( k ) ∈ Z p . Therefore, applying (7.11) to (7.10), weobtain that, for all prime number p ≥ P r , we have Φ L ,p ( a p + pr k ) ≡ − r X j =1 H j L ( k ) Q ( a p ) Q ( j k ) Q (( r − j ) k ) ( R k ( j ) − R k ( r − j )) mod p Z p , which finishes the proof of Lemma 17. (cid:3) . Proof of Theorem 3
We assume the hypothesis of Theorem 3. The aim of this section is to prove that thereare only finitely many prime numbers p such that q e,f,k ( z ) ∈ z k Z p [[ z ]] and that, for all L ∈ E e,f satisfying L ( k ) ≥ , there are only finitely many prime numbers p such that q L ,e,f ( z ) ∈ Z p [[ z ]] . We fix a L ∈ E e,f satisfying L ( k ) ≥ throughout this section.According to Section 4, we only have to prove that, for all large enough prime number p , there exists a ∈ { , . . . , p − } d and K ∈ N d such that Φ p,k ( a + K p ) / ∈ p Z p and Φ L ,p ( a + K p ) / ∈ p Z p . In fact, we will prove that, for all large enough prime number p , we have Φ p,k ( p k ) / ∈ p Z p and Φ L ,p ( p k ) / ∈ p Z p . We have Φ p,k ( p k )= X j =0 Q ((1 − j ) k ) Q ( jp k ) q X i =1 e ( k ) i ( H e ( k ) i (1 − j ) − pH e ( k ) i jp ) − q X i =1 f ( k ) i ( H f ( k ) i (1 − j ) − pH f ( k ) i jp ) ! = Q ( k ) q X i =1 e ( k ) i H e ( k ) i − q X i =1 f ( k ) i H f ( k ) i ! − p Q ( p k ) q X i =1 e ( k ) i H e ( k ) i p − q X i =1 f ( k ) i H f ( k ) i p ! (8.1)and Φ L ,p ( p k ) = X j =0 Q ((1 − j ) k ) Q ( jp k )( H L ( k ) (1 − j ) − pH L ( k ) jp )= Q ( k ) H L ( k ) − p Q ( p k ) H L ( k ) p . (8.2)There exists a constant P such that, for all prime number p ≥ P , we have q X i =1 e ( k ) i H e ( k ) i − q X i =1 f ( k ) i H f ( k ) i ∈ Z × p ∪ { } and H L ( k ) ∈ Z × p because L ( k ) ≥ . In the sequel, we write ∆ k for the Landau’s functionassociated with sequences e ( k ) := ( e ( k )1 , . . . , e ( k ) q ) and f ( k ) := ( f ( k )1 , . . . , f ( k ) q ) . We also write M for the largest element of sequences e ( k ) and f ( k ) . We note that M is nonzero because | e | ( k ) > | f | ( k ) , and that ∆ k vanishes on [0 , /M [ . If p > M , then, for all ℓ ≥ , we have /p ℓ < /M and thus v p ( Q ( k )) = P ∞ ℓ =1 ∆ e,f ( k /p ℓ ) = P ∞ ℓ =1 ∆ k (1 /p ℓ ) = 0 . Hence, for allprime number p > max( P , M ) =: P , we have Q ( k ) q X i =1 e ( k ) i H e ( k ) i − q X i =1 f ( k ) i H f ( k ) i ! ∈ Z × p ∪ { } and Q ( k ) H L ( k ) ∈ Z × p . (8.3)Furthermore, we have pH L ( k ) p = p L ( k ) X i =1 ip + L ( k ) p X j =1 p ∤ j j ≡ H L ( k ) mod p Z p , hich gives us that, for all prime number p > P , we have pH L ( k ) p ∈ Z × p . Similarly, we get p q X i =1 e ( k ) i H e ( k ) i p − q X i =1 f ( k ) i H f ( k ) i p ! ∈ Z p . Finally, for all prime number p > P , we have v p ( Q ( p k )) = ∞ X ℓ =1 ∆ e,f (cid:18) p k p ℓ (cid:19) = ∞ X ℓ =1 ∆ k (cid:18) pp ℓ (cid:19) = ∆ k (1) + ∞ X ℓ =1 ∆ k (cid:18) p ℓ (cid:19) = | e | ( k ) − | f | ( k ) ≥ , from which we obtain that, for all prime number p > P , we have p Q ( p k ) q X i =1 e ( k ) i H e ( k ) i p − q X i =1 f ( k ) i H f ( k ) i p ! ∈ p Z p and p Q ( p k ) H L ( k ) p ∈ p Z p . (8.4)Applying (8.3) and (8.4) to (8.2), we obtain that, for all prime number p > P , we have Φ L ,p ( p k ) / ∈ p Z p .Congruences (8.3) and (8.4) associated with (8.1) prove that it suffices to prove that P q i =1 e ( k ) i H e ( k ) i − P q i =1 f ( k ) i H f ( k ) i = 0 to conclude that, for all prime number p > P , we have Φ p,k ( p k ) / ∈ p Z p .For this purpose, we write E and F the respective subsequences of e ( k ) and f ( k ) obtainedas follows. We remove the zero elements of e ( k ) and f ( k ) and, if e ( k ) and f ( k ) have anelement in common, then we remove it from e ( k ) and f ( k ) once only. This latest step isrepeated until the obtained sequences are disjoint. The sequence F can be empty but thehypothesis | e | ( k ) > | f | ( k ) ensures that the sequence E is nonempty. Thus we have q X i =1 e ( k ) i H e ( k ) i − q X i =1 f ( k ) i H f ( k ) i = X c ∈ E cH c − X d ∈ F dH d and ∆ k = ∆ E , F . (8.5)Particularly, if F is empty then we have q X i =1 e ( k ) i H e ( k ) i − q X i =1 f ( k ) i H f ( k ) i = X c ∈ E cH c > . In the sequel of the proof, we assume that F is nonempty.Since E and F are two disjoint sequences of positive integers, we can apply Lemma 16to the sequences E and F . Using the notations of Lemma 16, we obtain X c ∈ E cH c − X d ∈ F dH d = X c ∈ E c X j =1 cj − X d ∈ F d X j =1 dj = t X i =1 m i γ i . (8.6)Furthermore, for all x ∈ R , we have ∆ E , F ( x ) = ∆ k ( x ) = ∆ e,f ( x k ) ≥ so ∆ E , F ≥ on [ γ , γ t ] and Lemma 16 leads to P ti =1 m i γ i > . This inequality associated with (8.5) and (8.6)proves that P q i =1 e ( k ) i H e ( k ) i − P q i =1 f ( k ) i H f ( k ) i = 0 and completes the proof of Theorem 3. . A consequence of Theorems 1 and 2
Almkvist, van Enckevort, van Straten and Zudilin present in [1] a list of more than fourth order differential equations that they call of Calabi–Yau type. In most ofthe considered equations, they give an explicit formula for the analytic solution F ( z ) normalized by the condition F (0) = 1 . One of the required conditions so that an equationto be of Calabi–Yau type is that its indicial equation at z = 0 should have as its onlysolution (see [1]). In particular, according to Section . of [2], there is a unique powerseries without constant term G ( z ) ∈ C [[ z ]] such that G ( z ) + log( z ) F ( z ) is a solution of thedifferential equation linearly independent of F ( z ) . We can then define the q -parameter forthe equation (following [2]) q ( z ) := z exp( G ( z ) /F ( z )) .In [8], Krattenthaler and Rivoal observe that equations from the list [1] have forsolution F a specialization of a series F e,f ( z ) , where the sequences e and f verify theconditions of Theorem from [8]. We understand by specialization of F e,f ( z ) any seriesobtained by replacing each z i , ≤ i ≤ d , by z i = M i z N i , where M i ∈ Z \ { } and N i ∈ N , N i ≥ . According to Theorem from [8], we see that the Taylor coefficients ofcanonical coordinates and mirror-type maps associated with e and f are all integers. It isthe same for their specializations, which ensures the integrality of the Taylor coefficientsof numerous new univariate mirror-type maps. In particular, we can obtain the integralityof the q -parameter for the differential equation.We found additional equations from the list [1] which have as solution F ( z ) a spe-cialization of a series F e,f ( z ) . Among these new cases, correspond to sequences e and f such that ∆ e,f ≥ on D e,f and thus, according to Theorems 1 and 2, such that the special-izations of canonical coordinates and mirror-type maps lie in z Z [[ z ]] . On the other hand,Cases , and correspond to sequences e and f such that there exists x ∈ D e,f such that ∆ e,f ( x ) = 0 . Thus we know that at least one of the canonical coordinates doesnot lie in z Z [[ z ]] .All in all, we obtain equations which are the cases: – , , ∗ , ∗∗ , ∗∗ , ∗∗ , ˆ1 – ˆ14 , , – , – , , , – , – , – , – , , , – , , , , , , – , , , , , , , – , , , , , , , , , and .Let us, for example, give the details for Case . The differential operator is L := θ − z (4 θ + 1)(4 θ + 3)(8 θ + 8 θ + 3) + 2 z (4 θ + 1)(4 θ + 3)(4 θ + 5)(4 θ + 7) , (9.1)where θ = z ddz . The function F canceled by this operator is F ( z ) = ∞ X n =0 z n (4 n )!( n !) (2 n )! n X k =0 k (cid:18) n − k ) n − k (cid:19) (cid:18) kk (cid:19) . Given e = ((4 , , (2 , , (2 , , (0 , and f = ((2 , , (1 , , (1 , , (1 , , (1 , , (1 , , (1 , , (0 , , (0 , , e obtain | e | = | f | and F e,f ( z, z ) = X k,m ≥ (4 k + 4 m )!(2 k + 2 m )!(( k + m )!) · ((2 k )!) ( k !) · ((2 m )!)( m !) m z k + m = ∞ X n =0 z n X k + m = n (4 n )!(2 n )!( n !) m (cid:18) kk (cid:19) (cid:18) mm (cid:19) = F ( z ) . The solution F ( z ) is therefore a specialization of F e,f ( z, w ) . We will now prove that ∆ e,f ≥ on D e,f .For all ( x, y ) ∈ D e,f , we have ∆ e,f ( x, y ) = ⌊ x + 4 y ⌋ + 2 ⌊ x ⌋ + ⌊ y ⌋ − ⌊ x + 2 y ⌋ − ⌊ x + y ⌋ , (9.2)because x and y lie in [0 , . According to the definition of D e,f , at least one of the floorfunctions in (9.2) must be greater than or equal to . If x + 2 y < , then we have ∆ e,f ( x, y ) ≥ . Let us assume that x + 2 y ≥ . Thus we have ⌊ x + 4 y ⌋ ≥ ⌊ x + 2 y ⌋ ≥ ⌊ x + 2 y ⌋ , so that if x + y < , then ∆ e,f ( x, y ) ≥ . On the other hand, if x + y ≥ , then ⌊ x ⌋ ≥ or ⌊ y ⌋ ≥ , and since ⌊ x + 4 y ⌋ ≥ ⌊ x + 2 y ⌋ + 2 ⌊ x + y ⌋ , we obtain ∆ e,f ( x, y ) ≥ . Thus, according to Theorem 1, we have q e,f, ( z, z ) ∈ z Z [[ z ]] and q e,f, ( z, z ) ∈ z Z [[ z ]] . We will now prove that the q -parameter associated with operator(9.1) is equal to q e,f, ( z, z ) .Let us write G ( z ) as the power series without constant term such that G ( z ) + log( z ) F ( z ) is canceled by operator (9.1). In order to determine the power series G ( z ) we use theFrobenius method presented in [14].For all r ∈ C , | r | ≤ / , and all n, k ∈ N , we set h n,k ( r ) := 2 k (cid:18) Γ(1 + 2( n + r − k ))Γ(1 + n + r − k ) (cid:19) (cid:18) kk (cid:19) if r = 0 ,h n,k (0) := 2 k (cid:18) Γ(1 + 2( n − k ))Γ(1 + n − k ) (cid:19) (cid:18) kk (cid:19) if k ≤ n and h n,k (0) = 0 if k ≥ n + 1 .The function Γ is meromorphic on C \ Z ≤ and has a simple pole in all nonpositive integer.Hence, the functions h n,k are analytic on { r ∈ C : | r | ≤ / } and, when k ≥ n + 1 , thefunctions h n,k have a zero of order in r = 0 .For all n ∈ N and r ∈ C , | r | ≤ / , we set c n ( r ) := Γ(1 + 4 n + 4 r )Γ(1 + n + r ) Γ(1 + 2 n + 2 r ) ∞ X k =0 h n,k ( r ) . et us prove that the series c n ( r ) is well defined. We recall Euler’s reflection formula:for all z ∈ C \ Z , we have Γ(1 − z )Γ( z ) = π sin( πz ) . (9.3)We will also use the property that, for all α ∈ ]0 , π [ , when | arg( z ) | < π − α and z → ∞ ,we have Γ( z ) ∼ e − z z z − / √ π . Particularly, there exists a constant K > such that, if ℜ ( z ) > and | z | > K , then (cid:12)(cid:12)(cid:12) e − z z z − / √ π (cid:12)(cid:12)(cid:12) ≤ | Γ( z ) | ≤ (cid:12)(cid:12)(cid:12) e − z z z − / √ π (cid:12)(cid:12)(cid:12) . (9.4)Let us fix n ∈ N . There exists a constant K ′ > n + 1 such that, for all k ≥ K ′ and r ∈ C , | r | ≤ / , we have | k − n − r | ≥ K . Thereby, for all k ∈ N , k ≥ K ′ and all r ∈ C \ { } , r ≤ / , following (9.3), we get Γ(1 + 2( n + r − k )) = π sin(2 π ( k − n − r ))Γ(2( k − n − r )) (9.5)and Γ(1 + n + r − k ) = π sin( π ( k − n − r ))Γ(( k − n − r )) . (9.6)Applying (9.4) to (9.5) and (9.6), we respectively obtain | Γ(1 + 2( n + r − k )) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) √ πe k − n − r ) sin(2 πr )(2( k − n − r )) k − n − r ) − / (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and | Γ(1 + n + r − k ) | ≥ (cid:12)(cid:12)(cid:12)(cid:12) √ πe k − n − r √ πr )( k − n − r ) k − n − r − / (cid:12)(cid:12)(cid:12)(cid:12) . Thus we have (cid:12)(cid:12)(cid:12)(cid:12)
Γ(1 + 2( n + r − k ))Γ(1 + n + r − k ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) sin ( πr ) π cos ( πr )2 k − n − r ) ( k − n − r ) (cid:12)(cid:12)(cid:12)(cid:12) . (9.7)Furthermore, according to (9.4), for all k ∈ N , k ≥ K ′ , we have (cid:18) kk (cid:19) = 2Γ(2 k ) k Γ( k ) ≤ k √ πk . (9.8)Therefore, following (9.7) and (9.8), for all r ∈ C , | r | ≤ / , and all k ∈ N , k ≥ K ′ , wehave | h n,k ( r ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n + r sin ( πr )cos ( πr ) π ( k − n − r ) √ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . (9.9)There exists a constant C > such that, for all r ∈ C , | r | ≤ / , we have | sin ( πr ) / cos ( πr ) | ≤ C, (cid:12)(cid:12) n + r (cid:12)(cid:12) ≤ · n and | k − n − r | ≥ k − n − . hus, for all r ∈ C , | r | ≤ / , and all k ∈ N , k ≥ K ′ , we have | h n,k ( r ) | ≤ C n π ( k − n − √ k . Thus the series c n ( r ) is uniformly convergent on { r ∈ C : | r | ≤ / } and defines ananalytic function.Let us write P ( X ) := X , P ( X ) := − (4 X + 1)(4 X + 3)(8 X + 8 X + 3) and P ( X ) := 2 (4 X + 1)(4 X + 3)(4 X + 5)(4 X + 7) , so that operator (9.1) is written like L = P ( θ ) + zP ( θ ) + z P ( θ ) .First, we prove that, for all r ∈ C , | r | ≤ / , and all n ∈ N , we have P ( n + r + 2) c n +2 ( r ) + P ( n + r + 1) c n +1 ( r ) + P ( n + r ) c n ( r ) = 0 . (9.10)If r = 0 , then, according to [1], F ( z ) = P ∞ n =0 c n (0) z n is canceled by L thus we have(9.10). Let us assume that r = 0 . In order to prove (9.10), we apply the Zeilberger procedure from Maple to the sequence ( b k ( n, r )) k ≥ := Γ(1 + 4 n + 4 r )Γ(1 + n + r ) Γ(1 + 2 n + 2 r ) 2 k (cid:18) Γ(1 + 2( n + r − k ))Γ(1 + n + r − k ) (cid:19) (cid:18) kk (cid:19)! k ≥ . For a hypergeometric term T ( n, k ) this procedure constructs a sequence ( d k ) k ≥ and anoperator R = P v ( n ) δ v + · · · + P ( n ) δ + P ( n ) such that R T ( n, k ) = d k +1 − d k , where P i ( X ) ∈ C [[ X ]] and δ is the shift operator δT ( n, k ) = T ( n + 1 , k ) . In our case, we obtainan explicit sequence ( d k ) k ≥ such that, for all n, k ∈ N , we have P ( n + r +2) b k ( n +2 , r )+ P ( n + r +1) b k ( n +1 , r )+ P ( n + r ) b k ( n, r ) = d ( k +1) − d ( k ) , (9.11)with d (0) = 0 and (quick calculation) d ( k ) = O (cid:18) k k Γ(2( n + r − k ) + 1) Γ( n + r − k + 1) (cid:18) kk (cid:19)(cid:19) = O (cid:18) √ k (cid:19) , when k → + ∞ . Summing identity (9.11) for k from to + ∞ and using that d (0) = 0 and d ( k ) → k → + ∞ , we get (9.10).Therefore, writing ˜ F ( z, r ) := P ∞ n =0 c n ( r ) z n + r , we have L ˜ F ( z, r ) = P ( r ) c ( r ) z r + (cid:0) P ( r + 1) c ( r ) + P ( r ) c ( r ) (cid:1) z r . (9.12)Let us prove that ∂∂r L ˜ F ( z, r ) (cid:12)(cid:12)(cid:12) r =0 = 0 . The series c n ( r ) is analytic on { r ∈ C : | r | ≤ / } and its derivative is obtained by differentiating term by term. When k ≥ n +1 , the functions h n,k ( r ) are analytic in a neighborhood of and have a zero of order in . For all k ≥ n +1 ,we obtain ∂∂r Γ(1 + 4 n + 4 r )Γ(1 + n + r ) Γ(1 + 2 n + 2 r ) (cid:18) Γ(1 + 2( n + r − k ))Γ(1 + n + r − k ) (cid:19) !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r =0 = 0 . n the other hand, if m ∈ N , m ≥ , then we have Γ ′ ( m ) = Γ( m )( H m − − γ ) , where γ is Euler’s constant. Hence, for all k ∈ { , . . . , n } , we obtain ∂∂r Γ(1 + 4 n + 4 r )Γ(1 + n + r ) Γ(1 + 2 n + 2 r ) (cid:18) Γ(1 + 2( n + r − k ))Γ(1 + n + r − k ) (cid:19) !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r =0 = Γ(1 + 4 n )Γ(1 + n ) Γ(1 + 2 n ) (cid:18) Γ(1 + 2( n − k ))Γ(1 + n − k ) (cid:19) × (cid:0) H n − H n − H n + 4 H n − k ) − H n − k (cid:1) . Thus, for all n ∈ N , we have c ′ n (0) = Γ(1 + 4 n )Γ(1 + n ) Γ(1 + 2 n ) n X k =0 (cid:18) Γ(1 + 2( n − k ))Γ(1 + n − k ) (cid:19) × (cid:0) H n − H n − H n + 4 H n − k ) − H n − k (cid:1) (9.13)Particularly, we obtain ddr ( P ( r ) c ( r ) z r ) (cid:12)(cid:12)(cid:12)(cid:12) r =0 = ddr (cid:0) r c ( r ) z r (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) r =0 = 0 and, a simple calculation via Maple leads to ddr (cid:0) ( P ( r + 1) c ( r ) + P ( r ) c ( r )) z r (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) r =0 = 0 . Therefore, we have ∂∂r (cid:16) L ˜ F ( z, r ) (cid:17)(cid:12)(cid:12)(cid:12) r =0 = 0 .Since the sequence ( c n ( r )) n ≥ satisfies recurrence relation (9.10), we can follow Section . from [14] and we obtain that there exists R > such that, for all r ∈ C , | r | ≤ / ,the power series ˜ F ( z, r ) in z has a radius of convergence at least equal to R . Furthermore,if | z | < R , then F ( z, r ) is derivable with respect to r and ∂∂r ˜ F ( z, r ) (cid:12)(cid:12)(cid:12)(cid:12) r =0 = log( z ) F ( z )+ ∞ X n =0 z n (4 n )!( n !) (2 n )! n X k =0 k (cid:18) n − k ) n − k (cid:19) (cid:18) kk (cid:19) (4 H n − H n − H n + 4 H n − k ) − H n − k ) . As the operator ∂∂r commutes with L , we obtain L ∂∂r ˜ F ( z, r ) (cid:12)(cid:12)(cid:12)(cid:12) r =0 = ∂∂r ( L ˜ F ( z, r )) (cid:12)(cid:12)(cid:12)(cid:12) r =0 = 0 . Thereby, ∂∂r ˜ F ( z, r ) (cid:12)(cid:12)(cid:12) r =0 = G e,f, ( z, z ) + log( z ) F ( z ) is canceled by L and, according tothe uniqueness of G ( z ) , we have G ( z ) = G e,f, ( z, z ) . The q -parameter associated with perator (9.1) is thus q ( z ) = z exp (cid:0) G e,f, ( z, z ) /F e,f ( z, z ) (cid:1) = q e,f, ( z, z ) ∈ z Z [[ z ]] . So in Case , the q -parameter is a specialization of a canonical coordinate. We did notverify in detail the cases cited above but it seems that the presented method forCase can prove in many cases that the q -parameter associated with the operator is aspecialization of a canonical coordinate and thus, when ∆ e,f ≥ on D e,f , all its Taylorcoefficients are integers. It would be interesting to have a more general method to provethis. References [1] G. Almkvist, C. van Enckevort, D. van Straten, W. Zudilin,
Tables of Calabi–Yau equations , preprint(2010), arXiv:math/0507430v2 [math.AG].[2] V. V. Batyrev, D. van Straten,
Generalized Hypergeometric Functions and Rational Curves on Calabi-Yau Complete Intersections in Toric Varieties , Commun Math. Phys. (1995), 493–533.[3] E. Delaygue,
Critère pour l’intégralité des coefficients de Taylor des applications miroir , J. Reine Angew.Math. (to appear), published online at http://dx.doi.org/10.1515/CRELLE.2011.094[4] B. Dwork, On p -adic differential equations IV : generalized hypergeometric functions as p -adic functionsin one variable , Annales scientifiques de l’E. N. S. e série, tome , numéro 3 (1973), p. 295-316.[5] S. Hosono, A. Klemm, S. Theisen and S.-T. Yau, Mirror symmetry, Mirror map and applications tocomplete intersection Calabi–Yau spaces , Nuclear Phys. B , no. 3 (1995), 501–552.[6] N. Kobliz, p -Adic Numbers, p -Adic Analysis, and Zeta-functions , Springer-Verlag, Heidelberg, 1977.[7] C. Krattenthaler, T. Rivoal, On the integrality of the Taylor coefficients of mirror maps , Duke Math.J., .2 (2010), 175–218.[8] C. Krattenthaler, T. Rivoal,
Multivariate p -adic formal congruences and integrality of Taylor coeffi-cients of mirror maps , preprint (2008), arXiv:0804.3049v3 [math.NT], to appear in series Séminaire etCongrès of the SMF. Proceedings of the conference
Théorie galoisiennes et arithmétiques des équationsdifférentielles (CIRM, september 2009).[9] S. Lang,
Cyclotomic Fields, I, II , Combined 2nd edition, vol. , Graduate Texts in Math., Springer-Verlag, New York, 1990.[10] E. Landau,
Sur les conditions de divisibilité d’un produit de factorielles par un autre , collected works, I , page 116. Thales-Verlag (1985).[11] B. H. Lian, S. T. Yau, Mirror Maps, Modular Relations and Hypergeometric Series I .arXiv:hep-th/9507151v1. Paru sous le titre :
Integrality of certain exponential series. Algebra and ge-ometry (Taipei, 1995), 215–227, Lect. Algebra Geom., , Int. Press, Cambridge, MA (1998). (Reviewer: Nobuo Tsuzuki).[12] B. H. Lian, S. T. Yau, Arithmetic properties of mirror map and quantum coupling , Comm. Math.Phys. 176, (1996), 163–191.[13] J. Stienstra, GKZ Hypergeometric Structures , Arithmetic and geometry around hypergeometric func-tions, R.-P. Holzapfel, A. Muhammed Uludaćg and M. Yoshida (eds.), Progr. Math., , Birkhĺauser,Basel (2007), 313–371.[14] M. Yoshida,