On the set of zero coefficients of a function satisfying a linear differential equation
aa r X i v : . [ m a t h . N T ] M a y ON THE SET OF ZERO COEFFICIENTS OF A FUNCTION SATISFYING ALINEAR DIFFERENTIAL EQUATION
JASON P. BELL, STANLEY N. BURRIS, AND KAREN YEATS
Abstract.
Let K be a field of characteristic zero and suppose that f : N → K satisfies arecurrence of the form f ( n ) = d X i =1 P i ( n ) f ( n − i ) , for n sufficiently large, where P ( z ) , . . . , P d ( z ) are polynomials in K [ z ]. Given that P d ( z ) isa nonzero constant polynomial, we show that the set of n ∈ N for which f ( n ) = 0 is a unionof finitely many arithmetic progressions and a finite set. This generalizes the Skolem-Mahler-Lech theorem, which assumes that f ( n ) satisfies a linear recurrence. We discuss examples andconnections to the set of zero coefficients of a power series satisfying a homogeneous lineardifferential equation with rational function coefficients. Introduction
The Skolem-Mahler-Lech theorem is a well-known result that describes the set of solutions toan equation f ( n ) = 0, where f ( n ) is a sequence given by a linear recurrence. Throughout thispaper, we take N := { , , , . . . } . Theorem 1.1. (Skolem-Mahler-Lech) Let K be a field of characteristic zero and let f : N → K be a sequence satisfying a linear recurrence over K ; that is, a recurrence of the form f ( n ) = P dj =1 a j f ( n − j ) , for n ≥ d , with a , . . . , a d ∈ K . Then the set (cid:8) n ∈ N : f ( n ) = 0 (cid:9) is a union offinitely many arithmetic progressions and a finite set. This theorem was first proved for linear recurrences over the rational numbers by Skolem [32],and it was then proved for linear recurrences over the algebraic numbers by Mahler [22]. Theversion above was proved first by Lech [20] and later by Mahler [23, 24]. Further backgroundabout linear recurrences and the Skolem-Mahler-Lech theorem can be found in the book byEverest, van der Poorten, Shparlinski, and Ward [13]. There are many different proofs andextensions of the Skolem-Mahler-Lech theorem in the literature [3, 4, 6, 7, 17, 28, 29]—all knownproofs of the Skolem-Mahler-Lech theorem use p -adic methods in some way, albeit sometimes ina disguised manner.Recall that if K is a field, then f : N → K satisfies a linear recurrence over K if and only if ∞ X n =0 f ( n ) z n ∈ K [[ z ]]is the power series expansion of a rational function in K ( z ). Stanley [33, Theorem 2.1] showsthat { Rational power series } ⊆ {
Algebraic power series } ⊆ {
Differentiably finite power series } . Algebraic power series comprise the power series F ( z ) ∈ K [[ z ]] which satisfy a nonzero polynomialequation P (cid:0) z, F ( z ) (cid:1) = 0 with P ( z, y ) ∈ K [ z, y ]. The set of differentiably finite power series F ( z ) Mathematics Subject Classification.
Primary: 11B37; Secondary: 11B25, 34A30.
Key words and phrases.
Holonomic sequences, Skolem-Mahler-Lech theorem, zero sets, recurrences, differentialequations, algebraic functions.The authors thank NSERC for supporting this project. are those that satisfy a non-trivial homogeneous linear differential equation with rational functioncoefficients:(1.1) d X i =0 Q i ( z ) d i dz i F ( z ) = 0 . Definition 1.2.
Let K be a field and let f : N → K be a K -valued sequence. We say that f ( n )is holonomic if the power series ∞ X n =0 f ( n ) z n ∈ K [[ z ]]is differentiably finite.It is straightforward to show that a K -valued sequence f ( n ) is holonomic if it satis-fies a polynomial-linear recurrence ; that is, there exist a natural number d and polynomials P ( z ) , . . . , P d ( z ) ∈ K [ z ], not all zero, such that(1.2) d X i =0 P i ( n ) f ( n − i ) = 0 , for all sufficiently large n (see Stanley [33, Theorem 1.5]). It will be convenient to use thischaracterization of holonomic sequences throughout this paper.Differentiably finite power series and holonomic sequences have been extensively studied byseveral authors [10, 11, 15, 16, 33, 35], and many important sequences arising naturally in com-binatorics, number theory, and algebra have been shown to be holonomic [9, 16, 21, 27, 33]. Anumber of significant results, including the following, have employed holonomic methods: (i)Ap´ery’s proof of the irrationality of ζ (3) used a sequence satisfying a polynomial-linear recur-rence (see van der Poorten’s survey [27]); (ii) Wilf and Zeilberger [35] showed how the theoryof holonomic sequences could be applied to the “automatic” proving of combinatorial identities;and (iii) Bousquet-M´elou [9] applied the theory to enumeration of walks in the quarter plane.In light of the well-behaved nature of the zero set of a sequence satisfying a linear recurrenceover a field of characteristic zero, it is natural to ask whether a similar result holds for holonomicsequences. Indeed, Rubel [31, Problem 16] asked: does the conclusion of the Skolem-Mahler-Lechtheorem hold for all holonomic sequences? The strongest result in this direction is due to B´ezivin[7] and Methfezzel [26]: the set of zeros of a holonomic sequence can be expressed as a union offinitely many arithmetic progressions and a set of zero density. (B´ezivin assumed that 0 and ∞ are not irregular singular points of the corresponding linear differential equation.) Laohakosol[19] and B´ezivin and Laohakosol [8] showed that the answer to Rubel’s question is ‘yes’, providedadditional technical conditions on the associated differential equation hold. (There is an error inthe first paper of Laohakosol which is repaired in the subsequent paper with B´ezivin.)We are able to give an affirmative answer to Rubel’s question in the case that the recurrencehas a reciprocal property (see Stanley [33, § Theorem 1.3.
Let K be a field of characteristic zero, let d be a positive integer, and let P ( z ) , . . . , P d ( z ) ∈ K [ z ] be polynomials with P d ( z ) a nonzero constant. Suppose that f : N → K is a sequence satisfying the polynomial-linear recurrence (1.3) f ( n ) = d X i =1 P i ( n ) f ( n − i ) for all n sufficiently large. Then (cid:8) n ∈ N : f ( n ) = 0 (cid:9) is a union of finitely many arithmeticprogressions and a finite set. This theorem generalizes the Skolem-Mahler-Lech theorem, which asserts that the conclusionholds provided P ( z ) , . . . , P d ( z ) are constant polynomials. We note that the reciprocal prop-erty of Stanley mentioned above is given by the assumption that P d ( z ) is a nonzero constant ERO SETS OF HOLONOMIC SEQUENCES 3 polynomial—this allows the recurrence to be rewritten as f ( n ) = d − X i =1 λP d − i ( n + d ) f ( n + i ) − λf ( n + d ) , where λ = − /P d ( z ), a nonzero element of K . Thus the definition of f ( n ) extends to all n ∈ Z .The proof of Theorem 1.3 uses p -adic methods and consists of three steps. The first step usesan argument of Lech to show that one can assume the base field K is Q p , for some prime p , andthe nonzero coefficients of the polynomials P i ( z ) in (1.3) are units in the p -adic integers. Thenext step involves showing that there exists a positive integer b such that for c ∈ { , . . . , b − } ,the sequence f ( bn + c ) can be embedded in an ‘analytic arc’, in the sense that there exists a p -adicanalytic map g : Z p → Z p such that g ( n ) = f ( bn + c ). We then use a theorem of Strassman (see §
2) to show that (cid:8) n ∈ N : f ( bn + c ) = 0 (cid:9) is either finite or N . The difficult step is the secondstep, in which one must find a p -adic analytic map that agrees with a subsequence on the naturalnumbers. In the case that f ( n ) satisfies a linear recurrence over a field K , it is well-known (see,for example, Everest, van der Poorten, Shparlinski, and Ward [13, Section 1.1.6]) that thereexist constants γ i,j ∈ K and α , . . . , α m ∈ K such that, for some positive integer N and all n sufficiently large, we have f ( n ) = m X i =1 N X j =0 γ i,j n j α ni . Lech used this fact to show that if K = C p and | α i − | p < p − / ( p − then the map g ( z ) := m X i =1 N X j =0 γ i,j z j α zi will be an analytic map that converges absolutely on the closed unit ball of C p . Such an expressionfor f ( n ) is not available when one works with holonomic sequences, and additional work is neededto obtain the embedding into a p -adic analytic arc.2. Preliminaries
This section has some of the p -adic results necessary to prove Theorem 1.3. We start withan embedding theorem due to Lech [20]—this result can be regarded as a p -adic analogue of theLefschetz principle. Lemma 2.1.
Let K be a finitely generated extension of Q and let S be a finite subset of K .Then there exist infinitely many primes p such that K embeds in Q p ; moreover, for all butfinitely many of these primes, every nonzero element of S is sent to a unit in Z p .Proof. The first-named author [3, Lemma 3.1] used the Chebotarev density theorem (see Lang[18, Theorem 10, p. 169]) to show that the set of primes p having the required property haspositive density. (cid:3) The other main result from p -adic analysis required is Strassman’s theorem [34], which assertsthat if a power series f ( z ) ∈ Q p [[ z ]] converges in the closed p -adic unit disc B Q p (0; 1) := (cid:8) α ∈ Q p : | α | p ≤ (cid:9) (= Z p ) , and has infinitely many zeros in this disk, then it is identically zero. A power series f ( z ) ∈ Q p [[ z ]]that converges in the closed p -adic unit disc will be called a rigid p -adic power series .To prove Theorem 1.3, we use the following ring MP p [ z ] of polynomials. Definition 2.2.
Given a prime p , let MP p [ z ] be the subring of the polynomial ring Q p [ z ]consisting of the polynomials which map Z p into itself.Clearly Z p [ z ] ⊆ MP p [ z ] ⊆ Q p [ z ]. A theorem of Mahler [25, p. 49-50] shows that for p a prime,(2.4) MP p [ z ] = ( m X i =0 α i (cid:18) zi (cid:19) : m ≥ α i ∈ Z p ) . JASON P. BELL, STANLEY N. BURRIS, AND KAREN YEATS Proof of Theorem 1.3
In this section we prove our main result. We begin with a simple lemma.
Lemma 3.1.
Let p be prime and suppose that Q ( z ) , . . . , Q d ( z ) ∈ MP p [ z ] are such that each iscongruent modulo p MP p [ z ] to a polynomial in MP p [ z ] of degree at most N . Then there exist H ( z ) , . . . , H d ( z ) ∈ MP p [ z ] , each of degree at most N + 1 , with H (0) = · · · = H d (0) = 0 , suchthat (3.5) H ( z + 1) ... H d ( z + 1) ≡ H ( z ) ... H d ( z ) − Q ( z ) ... Q d ( z ) (cid:0) mod p MP p [ z ] (cid:1) . Proof.
It is no loss of generality to assume that Q ( z ) , . . . , Q d ( z ) each have degree at most N .Thus Q i ( z ) = N X k =0 α i,k (cid:18) zk (cid:19) with each α i,k ∈ Z p . Let(3.6) H i ( z ) := − N X k =0 α i,k (cid:18) zk + 1 (cid:19) . Then H i ( z ) ∈ MP p [ z ], and it is of degree at most N + 1. It is easy to check that this gives asolution to equation (3.5), using the identity (cid:18) z + 1 k + 1 (cid:19) − (cid:18) zk + 1 (cid:19) = (cid:18) zk (cid:19) . Furthermore, H i (0) = 0 for 1 ≤ i ≤ d . (cid:3) To create analytic maps for the modified version of the p -adic analytic arc lemma, we will usethe following lemma about a subalgebra of MP p [ z ]. Lemma 3.2.
Given a prime p and a positive integer m , let S m := ( γ + m X i =1 p i H i ( z ) (cid:12)(cid:12)(cid:12) γ ∈ Z p , H i ( z ) ∈ MP p [ z ] , deg (cid:0) H i ( z ) (cid:1) ≤ i − ) and T m := S m + ( γ + n X i =1 p i H i ( z ) (cid:12)(cid:12)(cid:12) n ≥ , γ ∈ Z p , H i ( z ) ∈ MP p [ z ] , deg (cid:0) H i ( z ) (cid:1) ≤ i − ) . Then T m is a Z p -subalgebra of MP p [ z ] . Moreover, if α ∈ p Z p and β ∈ Z p , then for anypolynomial P ( z ) ∈ Z p [ z ] , one has P ( β + αz ) ∈ T m , for m ≥ . In particular, Z p [ pz ] ⊆ T m for m ≥ .Proof. Since T m is closed under addition and multiplication by p -adic integers, to show T m is a Z p -algebra, it is sufficient to show that T m is closed under multiplication. Since T m is containedin the Z p -span of the constant function 1 and elements of the form p i H ( z ) for some i ≥ H ( z ) ∈ MP p [ z ] a polynomial of degree at most 2 i −
1, it suffices to check that if i and j are positive integers and H ( z ) , G ( z ) ∈ MP p [ z ] are polynomials of degrees at most 2 i − j − p i H ( z ) · p j G ( z ) lies in T m . Note that this follows since H ( z ) G ( z ) ∈ MP p [ z ] has degree at most 2( i + j ) − p i + j L ( z )for some L ( z ) ∈ MP p [ z ] of degree at most 2( i + j ) − m ≥ , α ∈ p Z p , β ∈ Z p , and P ( z ) ∈ Z p [ z ]. Then α in p Z p implies β + αz ∈ T m ;and since T m is a Z p -algebra, it follows that P ( β + αz ) ∈ T m , and so, in particular, Z p [ pz ] ⊆ T m . (cid:3) ERO SETS OF HOLONOMIC SEQUENCES 5
Given a positive integer d and a ring R with identity element, we let M d ( R ) denote the set of d × d matrices with entries from R . When working with a matrix ring M d ( R ), we take I to bethe d × d identity matrix. Lemma 3.3. (Analytic arc lemma) Let d be a natural number, let p ≥ be prime, let v =[ v · · · v d ] T ∈ Z dp , and let A ( z ) := (cid:0) a ij ( z ) (cid:1) ∈ M d (cid:0) Z p [ pz ] (cid:1) with A (0) ≡ I (cid:0) mod p M d (cid:0) Z p (cid:1) . Thenthere exist rigid p -adic power series f ( z ) , . . . , f d ( z ) such that f i (0) = v i for i ∈ { , . . . , d } and f ( z + 1) ... f d ( z + 1) = A ( z ) · f ( z ) ... f d ( z ) . Proof.
Let S m and T m be as in the statement of Lemma 3.2. The desired tuple (cid:0) f ( z ) , . . . , f d ( z ) (cid:1) of power series will be successively approximated by tuples of polynomials (cid:0) G ,j ( z ) , . . . , G d,j ( z ) (cid:1) ,that is, f i ( z ) − G i,j ( z ) ∈ p j MP p [ z ], for 1 ≤ i ≤ d and j ≥
0. Let G i, ( z ) := v i for 1 ≤ i ≤ d. It will be proved, by induction on m , that one can recursively find H i,m ( z ) ∈ MP p [ z ], for1 ≤ i ≤ d , such that by setting(3.7) G i,m ( z ) := v i + m X k =1 p k H i,k ( z ) , one has the following three conditions holding:(i) H i,m (0) = 0 for 1 ≤ i ≤ d ;(ii) H i,m ( z ) ∈ S m , for 1 ≤ i ≤ d ;(iii) and G i,m ( z + 1) − d X j =1 a i,j ( z ) G j,m ( z ) ∈ p m +1 MP p [ z ] . The base case of the induction is m = 0. In this case, conditions (i) and (ii) are vacuous, and(iii) holds since A ( z ) is congruent modulo p to the identity matrix.Let m ≥ H i,k ( z ) have been found, for 1 ≤ i ≤ d and 0 ≤ k ≤ m , suchthat conditions (i)–(iii) hold for 0 ≤ k ≤ m . The method is to find polynomials H i,m +1 ( z ) ∈MP p [ z ] such that, with G i,m +1 ( z ) defined as in (3.7), conditions (i)–(iii) hold.By (iii), there are polynomials Q i,m ( z ) ∈ MP p [ z ], for 1 ≤ i ≤ d , such that(3.8) p m +1 Q i,m ( z ) = G i,m ( z + 1) − d X j =1 a i,j ( z ) G j,m ( z ) . Then Definition (3.7) and condition (ii) show that G ,m ( z ) , . . . , G d,m ( z ) as well as G ,m ( z + 1) , . . . , G d,m ( z + 1)are in S m . Thus, by Lemma 3.2 and the fact that the a ij ( z ) are in Z p [ pz ], p m +1 Q i,m ( z ) is in the Z p -algebra T m . It follows that p m +1 Q i,m ( z ) = γ i,m + n X k =1 p k Q i,m,k ( z )for some γ i,m ∈ Z p , and for some polynomials Q i,m,k ( z ) ∈ MP p [ z ] such that deg (cid:0) Q i,m,k ( z ) (cid:1) ≤ k − k ≤ m and deg (cid:0) Q i,m,k ( z ) (cid:1) ≤ k − k > m .Consequently, p m +1 Q i,m ( z ) is equivalent modulo p m +2 MP p [ z ] to the polynomial γ i,m + m +1 X k =1 p k Q i,m,k ( z ) , a polynomial in MP p [ z ] of degree at most 2 m . Hence Q i,m ( z ) is congruent modulo p MP p [ z ] toa polynomial in MP p [ z ] of degree at most 2 m . JASON P. BELL, STANLEY N. BURRIS, AND KAREN YEATS
Note that the definition (3.7) of G i,m +1 ( z ) can be replaced by G i,m +1 ( z ) := G i,m ( z ) + p m +1 H i,m +1 ( z ) . To satisfy property (iii), it is sufficient to find H i,m +1 ( z ) ∈ MP p [ z ], 1 ≤ i ≤ d , such that G i,m ( z + 1) + p m +1 H i,m +1 ( z + 1) − d X j =1 a i,j ( z ) (cid:0) G j,m ( z ) + p m +1 H j,m +1 ( z ) (cid:1) is in p m +2 MP p [ z ], for i ∈ { , . . . , d } . This expression is congruent modulo p m +2 MP p [ z ] to p m +1 Q i,m ( z ) + p m +1 H i,m +1 ( z + 1) − p m +1 d X j =1 a i,j ( z ) H j,m +1 ( z ) . However, since each a i,j ( z ) ∈ δ i,j + p MP p [ z ], we see that this simplifies as p m +1 Q i,m ( z ) + p m +1 H i,m +1 ( z + 1) − p m +1 H i,m +1 ( z )modulo p m +2 MP p [ z ]. It therefore suffices to solve the system(3.9) Q i,m ( z ) + H i,m +1 ( z + 1) − H i,m +1 ( z ) ≡ p MP p [ z ]) , for i ∈ { , . . . , d } , where each Q i,m is congruent modulo p MP p [ z ] to a polynomial of degree atmost 2 m .The hypotheses of the statement of Lemma 3.1 are satisfied by the system (3.9), so we concludethat there exists a solution (cid:2) H ,m +1 ( z ) , . . . , H d,m +1 ( z ) (cid:3) ∈ MP p [ z ] d with H i,m +1 (0) = 0 for1 ≤ i ≤ d and with H i,m +1 ( z ) of degree at most 2( m + 1) −
1. Thus conditions (i)–(iii) aresatisfied, completing the induction step.We set f i ( z ) := v i + ∞ X j =1 p j H i,j ( z ) . Then each H i,j ( z ) ∈ MP p [ z ] is of degree at most 2 j − H i,j ( z ) = j − X k =0 γ i,j,k (cid:18) zk (cid:19) , with γ i,j,k ∈ Z p . (Let γ i,j,k = 0 for k > j − f i ( z ) = v i + ∞ X j =1 p j j − X k =0 γ i,j,k (cid:18) zk (cid:19)! = v i + ∞ X k =0 β i,k (cid:18) zk (cid:19) , (3.10)in which β i,k := ∞ X j =1 p j γ i,j,k is absolutely convergent p -adically, since each γ i,j,k ∈ Z p . To show that the series (3.10) definesan analytic function on Z p , we must establish that | β i,k | p / | k ! | p → k → ∞ (see Robert [30,Theorem 4.7, p. 354]); that is, for any j > β i,k /k ! ∈ p j Z p for all sufficiently large k .To do this, we note that γ i,j,k = 0 if j < ( k + 1) /
2. Hence β i,k = X j ≥ ( k +1) / p j γ i,j,k . It follows that | β i,k | p ≤ p − ( k +1) / . Since 1 / | k ! | p < p k/ ( p − , we see that β i,k /k ! → p > f , . . . , f d are rigid analytic maps on Z p .Finally, observe that the argument above showed that f i ( z ) ≡ G i,j ( z ) (cid:0) mod p j MP p [ z ] (cid:1) . ERO SETS OF HOLONOMIC SEQUENCES 7
It then follows from property (iii) above that f i ( z + 1) ≡ d X ℓ =1 a i,ℓ ( z ) f ℓ ( z ) (cid:0) mod p j MP p [ z ] (cid:1) for i ∈ { , . . . , d } .Since this holds for all j ≥
1, we conclude that f ( z + 1)... f d ( z + 1) = A ( z ) · f ( z )... f d ( z ) . Finally, we have f i (0) = v i + ∞ X j =1 p j H i,j (0) = v i , which concludes the proof. (cid:3) We are almost ready to prove our main result. The one remaining thing we need is to showhow the analytic arc lemma applies to our situation. We accomplish this with the followinglemma.
Lemma 3.4.
Let p be a prime number, let d be a natural number, and let P ( z ) , . . . , P d ( z ) ∈ Z p [ z ] . Suppose that f : N → Z p is a sequence satisfying the polynomial-linear recurrence f ( n ) = d X i =1 P i ( n ) f ( n − i ) for n ≥ d . Let B ( z ) := · · ·
00 0 1 · · · ... ... . . . ... ... · · · P d ( z − d ) P d − ( z − d ) · · · P ( z − d ) P ( z − d ) , and let v := [ f (0) , . . . , f ( d − T . Then f ( n ) ... f ( n + d − = B ( n ) B ( n − · · · B (1) v , for every positive integer n . Furthermore, if P d ( z ) is a constant that is a unit in Z p then thedeterminant of B ( z ) is a constant that is a unit in Z p .Proof. For each positive integer n let v n := f ( n )... f ( n + d − . Then v n = B ( n ) v n − , for n ≥
1, leads to v n = B ( n ) B ( n − · · · B (1) v . The determinant of B ( z ) is P d ( z − d ); hence if P d ( z ) = α , a unit in Z p , then det (cid:0) B ( z ) (cid:1) equalsthat unit. (cid:3) JASON P. BELL, STANLEY N. BURRIS, AND KAREN YEATS
Proof of Theorem 1.3.
It is no loss of generality to assume that the recurrence in the statementof Theorem 1.3 holds for all n ≥ d . Let S denote the finite subset of K consisting of the nonzero f ( n ), where 0 ≤ n < d , along with all nonzero coefficients of the polynomials P ( z ) , . . . , P d ( z ).Then we let K = Q ( S ), the subfield of K generated by the elements of S . By Lemma 2.1,there exists a prime p ≥ K into Q p such that the image of S iscontained in the units of Z p . By identifying K with its image in Q p , we see that there is no lossof generality in assuming that P ( z ) , . . . , P d ( z ) ∈ Z p [ z ], with the nonzero coefficients of the P i ( z )being units of Z p ; in particular, P d ( z ) is a unit in Z p . Furthermore, after this identification, asimple induction shows that f : N → Z p .Letting B ( z ) and v be as defined in the statement of Lemma 3.4, one has, for n ≥ d ,(3.11) f ( n )... f ( n + d − = B ( n ) B ( n − · · · B (1) v , and the determinant of B ( z ) is a unit in Z p . In addition, observe that ( z + p ) i − z i ∈ p Z p [ z ] forall nonnegative integers i and hence, by linearity, we have that Q ( z + p ) − Q ( z ) ∈ p Z p [ z ] for all Q ( z ) ∈ Z p [ z ]. Thus we have B ( z + p ) − B ( z ) ∈ p M d (cid:0) Z p [ z ] (cid:1) , since the entries of B ( z ) are in Z p [ z ].Let ϕ : M d ( Z p ) → M d ( Z /p Z ) be the canonical surjection. For n ∈ Z , B ( n ) ∈ M d ( Z p ), andthus ϕ (cid:0) B ( n ) (cid:1) ∈ M d ( Z /p Z ). Since M d ( Z /p Z ) is a finite ring, there exist natural numbers m and m with m < m such that ϕ (cid:0) B ( pm ) · · · B (1) (cid:1) = ϕ (cid:0) B ( pm ) · · · B (1) (cid:1) . Since the determinant of B ( z ) is a unit of Z p , the determinant of ϕ (cid:0) B ( i ) (cid:1) is nonzero, for i ∈ Z .Thus ϕ (cid:0) B ( i ) (cid:1) is in GL d ( Z /p Z ) and hence ϕ (cid:0) B ( pm ) · · · B ( pm +1) (cid:1) is the identity. Furthermore,since ϕ (cid:0) B ( n + p ) (cid:1) = ϕ (cid:0) B ( n ) (cid:1) for n ∈ Z , it follows that for i ∈ Z , ϕ (cid:0) B ( pm + i ) · · · B ( pm +1+ i ) (cid:1) is similar to ϕ (cid:0) B ( pm ) · · · B ( pm + 1) (cid:1) and hence(3.12) ϕ (cid:0) B ( pm + i ) · · · B ( pm + 1 + i ) (cid:1) = I . Let b := p ( m − m ) . Suppose that c ∈ { , . . . , b − } . Then Equation (3.12) gives(3.13) ϕ (cid:0) B ( c + b ( n + 1)) · · · B ( c + bn + 1) (cid:1) = I , for n ∈ Z . Define the matrix A ( z ) ∈ M d (cid:0) Z p [ z ] (cid:1) by A ( z ) := B ( c + bz ) B ( c + bz − · · · B ( c + bz − b + 1) . Since p divides a , we see that for each integer i , the matrix-valued function B ( i + bz ) ∈ M d (cid:0) Z p [ pz ] (cid:1) and hence A ( z ) ∈ M d (cid:0) Z p [ pz ] (cid:1) . Moreover, Equation (3.13) shows that for n ∈ Z we have ϕ (cid:0) A ( n ) (cid:1) = I and in particular A (0) ≡ I (mod p M d ( Z p )). Let v := B ( c ) B ( c − · · · B (1) v . Then from (3.11), for n ≥ f ( c + bn )... f ( c + bn + d − = A ( n ) A ( n − · · · A (1) v . Since the hypotheses in the statement of Lemma 3.3 are satisfied by A ( z ) and v , we also havethat there exist rigid p -adic power series f ( z ) , . . . , f d ( z ) such that, for n ≥ f ( n )... f d ( n ) = A ( n ) A ( n − · · · A (1) v . Thus for n ≥ f ( n ) = f ( c + bn ) . ERO SETS OF HOLONOMIC SEQUENCES 9
By Strassman’s theorem, { n ∈ N : f ( n ) = 0 } is either finite or equal to N . Hence { n ∈ N : f ( c + bn ) = 0 } is either finite or equal to N . Let X := (cid:8) c ∈ { , . . . , b − } : f ( c + bn ) = 0 for n ∈ N (cid:9) . Then (cid:8) n ∈ N : f ( n ) = 0 (cid:9) = Z ∪ [ c ∈ X (cid:0) b N + c (cid:1) , where Z is a finite set consisting of all n such that f ( n ) = 0 and n ≡ c ′ (mod b ) for some c ′ ∈ { , . . . , b − } \ X . The result follows. (cid:3) Concluding remarks and examples
We make some general remarks about Theorem 1.3 in this section.There are some difficulties that arise with trying to extend this argument to the collection ofall holonomic sequences. For example, if we take f ( n ) = (cid:18) nn (cid:19) , then f ( n ) is holonomic as it satisfies the polynomial linear recurrence( n + 1) f ( n + 1) − n + 1) f ( n ) = 0for n ≥
0. Observe that there is no way to select a prime p ≥ a such thatfor all c ∈ { , . . . , b − } we have a p -adic rigid analytic map G c ( z ) such that G c ( n ) = f ( bn + c )for all n ∈ N . To see this, suppose that we have such a prime p and a natural number b and let c = 0 and let g ( z ) be a p -adic rigid analytic map with g ( n ) = f ( bn ). Since g is continuous, thereis some j ≥ (cid:12)(cid:12) g ( z + p j ) − g ( z ) (cid:12)(cid:12) p < /p for z ∈ Z p . This then gives that f ( bp j k ) ≡ f (0) = 1 (mod p )for all natural numbers k . But f ( pk ) ≡ f ( k ) (mod p ) for all natural numbers k and thus wemust have f ( bk ) ≡ p )for all natural numbers k . But if we choose ℓ such that p ℓ ≤ b < p ℓ +1 and choose k to be thesmallest natural number such that 2 bk > p ℓ +1 then we see that p divides f ( bk ) using the formula (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) nn (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) p = p P j ≥ ⌊ n/p j ⌋−⌊ n/p j ⌋ , and noting that each term appearing in the sum on the right-hand side is non-positive and thatthe term is strictly negative when j = ℓ + 1. This is a contradiction.In an earlier paper, the authors [5] considered the set of zero coefficients to functions arisingfrom solutions to a system of equations with certain prescribed properties. In many cases, thesolution sets were algebraic functions and hence differentiably finite. The hypotheses guaranteedthat the zero set was a finite union of arithmetic progressions along with a finite set, althoughthe methods used were very different and relied on studying the behaviour of power series definedby systems of equations.It is also interesting to ask what happens in positive characteristic. Lech [20] gave exam-ples that showed that the conclusion of the Skolem-Mahler-Lech theorem does not hold if oneeliminates the hypothesis that the field have characteristic 0. For example if p is a prime and K = F p ( t ), then f ( n ) = (1 + t ) n − t n − K but the zero set of f is { , p, p , . . . } .Derksen [12] showed that the zero set S of a linear recurrence over a field of characteristic p > p expansionof a number n and accepts the number if and only if n ∈ S . We call such sets p - automatic sets (see the book of Allouche and Shallit [2] for a more precise definition). Recently, Adamczewskiand the first-named author [1] have shown that the set of zero coefficients of an algebraic powerseries over a field of characteristic p > p -automatic. It thus seems reasonable to conjecturethat if f ( n ) is a holonomic sequence taking integer values and p is a prime number, then the zeroset of the reduction of f ( n ) (mod p ) is a p -automatic set.We also observe that f ( n ) is a holonomic Z -valued sequence that satisfies a polynomial-linearrecurrence of the form f ( n ) = d X i =1 P i ( n ) f ( n − i ) , then the argument we employed to prove that the zero set of f ( n ) is a finite union of infinitearithmetic progressions along with a finite set can be proved under more general conditions. Inparticular, it is sufficient that there exist infinitely many primes p for which P d ( x ) does not haveany roots modulo p . The Chebotarev density theorem (see Lang [18, Theorem 10, p. 169]) givesthat this will occur precisely when there is some automorphism of the splitting field of P d ( x )over Q whose natural action on the roots of P d ( x ) does not have any fixed points.Finally, we note that Theorem 1.3 is ineffective in the sense that if we know that our recurrencehas only finitely many zeros, we cannot effectively bound the size of the largest zero in termsof data coming from the recurrence and the initial terms of the recurrence. Indeed, this is anotoriously difficult problem for linear recurrences and much work has been done on this problemby Evertse, Schlickewei, and Schmidt [14], who showed how one can obtain a quantitative versionof the Skolem-Mahler-Lech theorem that bounds the number of exceptional zeros and the lengthsof the arithmetic progressions in terms of such data. References [1] B. Adamczewski and J. P. Bell,
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Department of Mathematics, Simon Fraser University, 8888 University Dr., Burnaby, BC, V5A 1S6,Canada
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