Cyclotomic Diophantine Problems

Transcription

1 Cyclotomic Diophantine Problems (Hilbert Irreducibility and Invariant Sets for Polynomial Maps) R. Dvornicich and U. Zannier Abstract. In the context which arose from an old problem of Lang regarding the torsion points on subvarieties of G d m, we describe the points that: lie in a given variety, are defined over the cyclotomic closure k c of a number field k and map to a torsion point under a finite projection to G d m. We apply this result to obtain a sharp and explicit version of Hilbert s Irreducibility Theorem over k c. Concerning the arithmetic of dynamics in one variable, we obtain by related methods a complete description of the polynomials having an infinite invariant set contained in k c. In particular, we answer a number of longstanding open problems by Narkiewicz, which he eventually collected explicitly in the book [N]. AMS2000 Classification: 11G10 (11R18 12E25 37F10) 1. Introduction and statements Equations in roots of unity have a long tradition. We may recall for instance an old problem of Lang regarding (the finiteness of the set of) the solutions of algebraic equations f(x, y) = 0 with x, y roots of unity: this was given simple but ingenious solutions by Ihara, Serre, Tate,..., who proved that if there are infinitely many such solutions, then f must have a factor of the shape ax m y n + b or ax m + by n. This result was later interpreted in terms of torsion points of G 2 m lying on the curve defined there by f and was generalized (by Liardet, Laurent and others) to arbitrary subvarieties of G d m. We recall here a final result, also for later reference, where we use the terminology torsion coset (in G d m) to denote a translate of a connected algebraic subgroup (i.e. a torus) by a torsion point: Torsion Points Theorem. Let V/Q be an algebraic subvariety of G d m. Then the Zariski closure of the set of torsion points in V is a finite union of torsion cosets in G d m. There are several proofs of this theorem; for instance it may be recovered as a particular case of a theorem by S. Zhang on small points on subvarieties of G n m; see e.g. [BG, Thm ] for an elementary approach. Now, torsion points in G d m are special cases of cyclotomic points, namely, points defined over the maximal cyclotomic extension k c of a given number field k. In this paper we modify Lang s question as follows: Describe the solutions to f(x, y) = 0 with x a root of unity and y k c. In other words, we weaken the condition and just ask for the cyclotomic points which have a projection which is torsion. Very special cases of this go back to a question of J. Robinson, solved by Mordell, about the equation 1 + ζ = α n, for a root of unity ζ and an α Q c (see [Mo]). Our Theorem 1 deals with this situation for the general case of an algebraic variety with a finite map to G d m: it is a structure theorem which is in a way a complete description, analogous to the Torsion Points Theorem. Theorem 1 will be applied to a sharp Hilbert s Irreducibility Theorem for cyclotomic extensions (see Corollaries 1,2,3 below). The main principles of our proofs admit other applications. 1

2 First, we investigate cyclotomic preperiodic points for polynomial maps in one variable (see below for definitions). The preperiodic points (for rational and polynomial maps) have been widely studied in the context of One variable dynamics (see [M]), from the topological and analytical viewpoint. More recently, also their arithmetic has been the object of attention (see below for some references); we note that very interesting examples come from torsion points in (commutative) algebraic groups. In the simplest example of G m these are the roots of unity, which are just the preperiodic points for the monomials x d for any d 2 (see also [KS, Theorem 32] for a characterization in several variables). One could ask whether the roots of unity are preperiodic also for other polynomial maps. Here we answer a much more general question, namely: To characterize the polynomial maps having infinitely many preperiodic points in the cyclotomic closure of Q. In Theorem 2 below we give a complete classification for polynomial maps in one variable. (It would be interesting to investigate the case of several variables; for other questions in this context this has been done in [KS].) Secondly, in a cyclotomic context, we discuss more generally invariant sets for a polynomial map f, i.e. sets Γ satisfying f(γ) = Γ. Extending Theorem 2 (see Theorem 2 ) we classify the polynomials of degree 2 having an infinite invariant set in the cyclotomic closure of a number field; actually, we give a full list of the possible polynomials and associated invariant sets. As applications, we provide complete answers to a number of longstanding open questions by Narkiewicz (see Theorem 3 below), which he eventually collected explicitly in [N]. We proceed to give some notation and our precise statements, grouping them according to the described scheme. In the subsequent sections we shall present the proofs, which strongly depend on the Torsion Points Theorem and on an extension of a theorem of Loxton, obtained in 2. Some notation. In the sequel we shall use the following symbols and notation: k: a number field, with ring of integers O = O k. k = Q: the field of algebraic numbers. C(α): maximum of the absolute values of the conjugates over Q of the algebraic number α. H(α), h(α) = log H(α): the absolute Weil height of the algebraic number α (see [BG]). G m : the variety A 1 \ {0} endowed with the multiplicative group law. U d : set of torsion points in G d m; we shall identify U 1 with the set of roots of unity. k c : the cyclotomic closure k(u 1 ) of k, i.e., the field generated over k by all roots of unity. ζ n : a primitive n-th root of unity. Cyclotomic points on varieties and Hilbert Irreducibility Theorem We immediately state our main result here, adding another bit of notation: For an affine variety V with a regular map π : V G d m we let J = J V = J V,π = {η V (k c ) : π(η) U d }. Theorem 1. Let V/k be an affine variety irreducible over k c and let π : V π(v ) G d m be a morphism of finite degree defined over k. Also, let J = J V,π be as above and assume that it is Zariski dense in V. Then: (i) the Zariski closure of π(v ) in G d m is a torsion coset βt (β a torsion point, T a subtorus); (ii) there exist an isogeny µ : T T and a birational map ρ : T V defined over k c such that π ρ = β µ. Remarks. (i) The result has a converse, specified after the proof. (ii) Naturally, if J is not Zariski dense in V, one may apply the result to its Zariski closure, provided the restriction of π to this closure has still finite degree. (iii) Similarly, we could consider the points P V (k) whose projection π(p ) lies in a fixed subgroup G G d m of finite rank (the above case being the case of rank 0). One expects an analogous conclusion. The case of a finitely generated group of rank 1, in dimension 1, may be treated with the methods of [CZ] or [CZ2] (in this last paper a statement similar to Thm. 1 is 2

3 conjectured). However, we do not know of any methods suitable to cover a greater generality. For instance, already the case when V is defined by the equation (y 2 = 1 + u + v), where y k and u, v belong to a group of S-units in k, seems quite out of the present known methods. As simple applications of Theorem 1, we have some corollaries, which may be seen as (strong) versions of Hilbert s Irreducibility Theorem over k c. For simplicity, we restrict to the case of one polynomial in two variables (but see Remark (iv) below). Corollary 1. Let f k c [X, Y ]. (a) If f(x m, Y ) is irreducible over k c for all positive integers m deg Y f, then f(ζ, Y ) is irreducible in k c [Y ] for all but finitely many roots of unity ζ. (b) If f(x m, Y ) is reducible over k c for some m > 0, then f(ζ, Y ) is reducible in k c [Y ] for all but finitely many roots of unity ζ. Remarks. (i) One can effectively check whether f(x m, Y ) is reducible over k c, for any given m > 0. In fact, one can first find the factorization of f over Q, which is a classical question. Then one can test all the proper products of irreducible factors and check whether their coefficients are defined over k c. To deal with this last point, note that an algebraic number lies in k c if and only if it lies in a finite cyclotomic extension of k whose degree and discriminant may be bounded in terms of its minimal polynomial. (ii) When f(x m, Y ) is irreducible over k c for every positive integer m, the finite set of roots of unity ζ such that f(ζ, Y ) is reducible can be effectively found. This follows from the present proof, taking into account that: (A) The results of 2 below are effective; (B) The Zariski closure of the set of torsion points in a given algebraic variety can be effectively found: see remarks in [BZ] or [BG, Remark 4.2.5]. On the contrary, we do not know whether the number of exceptional roots of unity may be bounded in terms only of the degree of f. (iii) Since we have a finiteness statement, a similar result follows directly for any finite set of polynomials. (On the contrary, the Hilbert s Irreducibility Theorem over number fields needs further arguments to treat several polynomials instead of a single one.) (iv) Theorem 1 may be applied in a similar way to obtain a version of Corollary 1 for polynomials f(x 1,..., X n, Y ) in several variables, such that f(x m1 1,..., Xn mn, Y ) is irreducible for a suitably large set of positive integers m 1,..., m n ; the conclusion will be that f(ξ 1,..., ξ n, Y ) is irreducible in k c [Y ] for torsion points (ξ 1,..., ξ n ) lying outside a certain finite union of algebraic subgroups. (See also the proof of Cor. 3 below for an example and see [Sch3] for a kind of function-field analogue.) This kind of extension leads to a substantial simplification to some proofs in [Z2] and also to the proof of conjectures by van der Poorten which extend the so-called Pisot s d-th root conjecture. These applications are the object of recent work by A. Ferretti and U. Zannier. (v) This Corollary may be compared with the results of Dèbes [D] on the irreducibility of f(at m, Y ) over k[y ], where a, t k are fixed and m N. See also [CZ] for the case f(u, Y ), with a variable S-unit u k. Unlike the present situation (see previous Remark (iv) to Thm. 1), the methods used to obtain such results do not generalize in a direct way to higher dimension. (See [CZ2] for some results.) In the case when we only know that f(x, Y ) is irreducible, we can still get the irreducibility of many specializations by changing their shape. For instance, we have: Corollary 2. For every irreducible polynomial f k c [X, Y ], there exists a finite subset U f k c such that for all u k c \ U f, f(ζ + u, Y ) is irreducible in k c [Y ] for all but finitely many roots of unity ζ. In particular, this implies that k c is a hilbertian field. This last fact is known (being a corollary of a rather general result by Weissauer, see [W] or also [V, Cor. 1.28]) but the present statement gives explicit Hilbert sets. (The proof will show that a suitable U f may be effectively determined.) 3

4 Actually, it is easy to get Universal (cyclotomic) Hilbert Sequences, in the sense of [FJ]. For instance we have the following corollary. Corollary 3. The sequence {a n := ζ n + ζ n+1 } n N is a Universal Cyclotomic Hilbert Sequence. Namely, for every irreducible polynomial f k c [X, Y ], the polynomial f(a n, Y ) is irreducible over k c for all but finitely many n N. Preperiodic points We start by recalling the definition of preperiodic points. Definition 1. Let φ k(x) be a rational map. We say that P P 1 ( k) is a periodic point for φ if for some positive integer k we have φ (k) (P ) = P, where φ (k) denotes the k-th iterate. We say that P is preperiodic if φ (m) (P ) is periodic for some positive integer m. We let Π φ be the set of preperiodic points for φ. Clearly, if deg φ = 1 the preperiodic points are very easily studied (e.g. if φ has finite order, then every point is preperiodic). Therefore we shall always assume that deg φ 2 in what follows. Also, we shall deal only with the case when φ = f k[x] is a polynomial. The arithmetic of preperiodic points has recently been studied in several papers by M. Baker, A. Chambert-Loir, A. Mimar, R. Rumely, L. Szpiro, T. Tucker, especially from the point of view of equidistribution under Galois action (see for instance [ST], [BR]; for further recent work, also in several variables, see [KS]). Here we are concerned with the field of definition of such points; for instance it is well-known (and easily proved) that, given a rational map φ k(x) of degree 2 there can be only finitely many preperiodic points for φ of bounded degree over Q. (See [N] Ch. 9 and also below in this paper; see also [Ba], Cor. 1.9, for a remarkable lower bound for the degree of the fields of definition). However, for a field of algebraic numbers of infinite degree not much seems to be known. In this paper, as mentioned above, we apply our methods to investigate the preperiodic points lying in k c, for a polynomial map. An interesting example comes from the roots of unity, which clearly are preperiodic for all monomial maps x ζx d, where ζ is any root of unity and d 2. Are there other polynomial maps f such that Π f contains infinitely many roots of unity? Using the above mentioned equidistribution results (e.g. [BR], but see also the unpublished Mimar s thesis, Columbia University) one can prove that this is not the case: in fact, by the quoted works it follows that the set of Galois conjugates of a preperiodic point tends to be uniformly distributed in C around the whole Julia set of f, as the degree of the point tends to. Then, if there were infinitely many preperiodic roots of unity, the unit circle would be f-invariant. But it is an easy exercise to show that a polynomial leaving the unit circle invariant is necessarily a monomial. Our methods not only give an alternative proof of this, but enable us to answer a much more general question: What are the polynomials f for which Π f k c is infinite? We shall show that there are essentially two cases: Theorem 2. Let f k[x] be a polynomial of degree d 2 and let Π = Π f be as above the set of preperiodic points for f. Then Π k c is finite unless for some polynomial L Q[x] of degree 1 and for some ɛ = ±1, (L f L 1 )(x) is either (ɛx) d or T d (ɛx). Here T d (x) denotes, as usual, the d-th Chebishev polynomial: it is the unique polynomial satisfying the identity T d (x + x 1 ) = x d + x d. We recall that both x d and T d (x) have the same parity of d; this fact allows to remove the ɛ except when d is odd and we are in the Chebishev case. The linear polynomial L need not be defined over k; however it can be proved that it may be chosen over a certain radical extension of k (actually a quadratic one in the Chebishev case ). 4

5 We also note that the exceptional cases of the theorem are genuine exceptions to finiteness. In fact, the finiteness property is preserved under conjugation by polynomials of degree 1, so it suffices to check that (ɛx) d and T d (ɛx) have infinitely many preperiodic points in Q c, which is straightforward. (The nonzero preperiodic points are just roots of unity in the first case and numbers of the form ζ + ζ 1, ζ a root of unity, in the second case.) We finally observe that in the cases when the set Π k c is finite, it may be effectively computed in terms of f. This will appear from our proofs. It would be interesting to compute explicit bounds for #(Π k c ). The property (P) of Narkiewicz We recall from [N], p. 69, the following definition, introduced already in [N2]: Definition 2. We say that a field K has the property (P) if no polynomial f K[x] of degree d > 1 admits an infinite set Γ K such that f(γ) = Γ. Often a set Γ such that f(γ) = Γ is called invariant (for f). To appreciate the relevance of this concept in the context of preperiodic points, note that any union Γ of full periods of periodic points for a polynomial f is invariant. Hence the property (P) for K easily implies that every polynomial over K, of degree > 1, has only finitely many preperiodic points in K. In particular, the cyclotomic extension K = k c does not have the property (P). The first purpose of this section is to provide a solution (in the negative) to Open Problem XV in [N], stated as follows: Is the property (P) preserved under finite extensions? In Theorem 3 below we show in particular that this is not the case, giving an explicit counterexample. In [N] certain sufficient criteria are given for the property (P) to hold; in particular, (P) holds for number fields. These criteria implicitly relate property (P) to another arithmetical property, called property (N) in [BZ2]; more precisely, we have: Definition 3. We say that a subset K of Q has the Northcott property, or the property (N), if for every positive B there are only finitely many elements x K such that h(x) B, where h( ) is the absolute logarithmic Weil height on Q. For each integer d, the union of number fields of degree d over Q has this property, as was proved by Northcott himself (see [BG, Th ]), so in particular every number field has the property (N). However the property holds as well for certain fields of algebraic numbers of infinite degree over Q. For instance in [BZ2] the following is proved (see also [BG, Thm ]): Theorem BZ. For every number field k and every positive integer d, the composite field inside Q of the abelian extensions of k of degree d has the property (N). Taking into account the well-known properties (see [BG]) of the Weil height, Lemma 9.2(b) of [N] immediately implies that: For every field K of algebraic numbers, the property (N) for K implies the property (P) for K. (Actually, by the arguments in [N, Theorem 9.3], the conclusion holds as well for the property (SP) there defined for several polynomials.) Hence, Theorem BZ implies in particular that the field Q( 2, 3, 5,... ) has the property (P), providing a positive answer to Open Problem XVII in [N]. With the methods of proof of Theorem 2 above, we could give a completely different solution to this question as well. However, in a sense we can prove more than that: namely, we can produce a field K of algebraic numbers, having the property (P), which not only has infinite degree over Q, but has not the property (N), contrary to the previously known examples. It turns out that this field K has a finite extension without the property (P). (We also note that the property (N) is preserved by taking a finite extension, as follows from the usual simple argument in Northcott s Theorem; so the existence of such a field K itself proves that (N) and (P) are not equivalent.) 5

6 Therefore we collect all such results in the following single statement: Theorem 3. Let p be a prime such that p 1 has an odd prime factor l. Let K be the field generated over Q by the roots of unity of p-power order and let K be the unique subfield of K such that [K : K] = l. Then: (i) K has the property (P); (ii) K has not the property (P); (iii) K has not the property (N). We remark that Gal(K /Q) = Z p = Z p F p, so that this Galois group has in fact a unique subgroup of order l, corresponding to K. In particular, as announced above, (i) and (ii) imply a negative answer to Open Problem XV in the list of [N, p. 110]. In turn, this automatically implies a negative answer also to Open Problem XIV, asking whether the properties (P) and (SP) (similarly defined for several variables, see [N, p. 68]), are equivalent: now, Thm. 9.4 in [N] says that (SP) is preserved by finite extensions, which excludes the equivalence. Actually, the field K in this theorem has not the property (SP). (Our counterexample also answers, in the negative, to part of Open Problem XIX ). Another feature of the field K is that it cannot be generated over Q by elements of bounded degree. This follows from the structure of the Galois group, recalled a few lines above. (The existence of a field of algebraic numbers, with the property (P), but not generated by elements of bounded degree was proved by K. Kubota and P. Liardet in 1976, (see [N, Thm. 10.8]). However their construction is rather indirect and it seems difficult to detect other properties of the field so obtained.) Theorem 3 is just an instance of what can be said and it could be easily generalized or varied in many ways. In fact, Theorem 3 results from a complete classification of polynomials f k c [x] of degree d 2 and infinite invariant sets Γ for f, Γ k c. This classification in turn appears through Theorem 2 and Proposition 1 of 5 below, which in practice give the precise obstructions for k c to have the property (P). Acknowledgements. We thank P. Corvaja for careful reading of the paper and for suggesting some simplifications. We also thank M. Baker, W. Narkiewicz, J. Pila, J. Silverman, L. Szpiro and T. Tucker for several other helpful remarks and references. 2. Extension of a theorem of Loxton For any algebraic number α, let C(α) be the maximum of the absolute values of the conjugates of α over Q. Suppose now that α is a cyclotomic integer, namely, an algebraic integer contained in some cyclotomic field. Then certainly α is a sum of (not necessarily distinct) roots of unity, α = b ξ i. A theorem of Loxton (see [L]) shows that one can choose the roots of unity ξ i so that b L(C(α)); here L : R + R + is a suitable function, specified in Loxton s paper. In particular, one can take L(x) ɛ x 2+ɛ (see [L] for a more precise statement). We shall refer to any function L(x) such that the above statement is true, as a Loxton function. Roughly speaking, the proof-principle of Loxton is as follows: a cyclotomic integer α may be written (not uniquely!) as the value f(ζ) of a polynomial f Z[x] at a root of unity ζ. A bound for the conjugates of α amounts to a bound for the absolute values f(ζ t ), for certain integers t. For a suitable choice of f one can combine these bounds with Parseval formula, to obtain a bound for the L 2 -norm of the coefficient-vector of f. Since the coefficients are integers, this leads to the sought conclusions. In this paper we shall need an extension of this result to algebraic integers α contained in a cyclotomic extension of a given number field k. In this case, α cannot be written, in general, as a 6

7 sum of roots of unity, but it remains true that α is a sum α = b η iξ i where η i are elements in a finite subset of k independent of α. Our problem is to bound the number b of summands; we shall see that it is indeed possible to bound b in terms of C(α) and k only. More precisely, using Loxton s theorem, we shall prove the following: Theorem L. There exists a number B = B k and there exists a finite set E = E k k with #E [k : Q] such that any algebraic integer α k c can be written as α = b η iξ i, where η i E, ξ i are roots of unity and b #E L(BC(α)), where L : R R is any Loxton function. The proof will show that B and E can be computed in terms of a basis for k/q. Proof. We choose once and for all a basis Ω = {ω 1,..., ω r } for k/(k Q c ), made up of algebraic integers. Note that, since Q c /(k Q c ) is Galois, this is a basis also for kq c /Q c, so we may write α = r γ i ω i, (2.1) where γ i Q c. Let Ω = {ω1,..., ωr} be the dual basis of Ω with respect to the trace map T = T rk Q k c. Then we have T r kqc Q (ω iω c j ) = T (ω i ωj ) = δ ij. Multiplying (2.1) by ω j and taking T rkqc Q c we get T r kqc Q c (ω j α) = γ j. (2.2) Let now d be the minimum positive integer such that dωj is an algebraic integer for all j. Since we have chosen the basis Ω once and for all, depending only on k, this integer d depends only on k. Now, we have dγ j = T r kqc Q c (dω j α), hence dγ j are algebraic integers in Q c. By the theorem of Loxton, for each j = 1,..., r we may write dγ j = b j ξ ij, j = 1,..., r, where the ξ ij are roots of unity and the b j are integers such that In view of (2.1) we have 0 b j L(C(dγ j )). α = r b j j=1 ω j d ξ ij. (2.3) Let then E = { ω1 d,..., ωr d } and observe that (2.3) gives a representation α = b l=1 η lξ l where the η l E, the ξ l are roots of unity and 0 b r j=1 b j r j=1 L(C(dγ j)). Further, from (2.2) we immediately obtain that C(γ j ) rc(α) max C(ω j ) =: d C(α), where d is a positive number depending only on k. Finally, we get 0 b rc(dd α), proving the result with B k := dd. Remark. We note that a conclusion like in Theorem L is not valid if instead of C(α) we consider the maximum of the conjugates of α over a fixed number field K Q (which would amount to the omission of some places of Q). For instance it suffices to consider the numbers (1 2) n and the field Q( 2) to obtain a counterexample. 7

8 3. Proof of Theorem 1 and Corollaries Proof of Theorem 1. Let x 1,..., x d be the restrictions of the coordinates on G d m to π(v ) in G d m. Since V is affine and defined over k we may pick y k[v ] to be a primitive element for k(v )/k(x 1,..., x d ), satisfying a nontrivial irreducible (over k c ) equation f(x 1,..., x d, y) = 0, where f k[x 1,..., x d, Y ] is monic in Y. Multiplying y by a positive integer if necessary we may assume that f has coefficients in the ring of integers of k and is still monic. Now let Q J, so ζ = π(q) π(j) is a torsion point, y(q) lies in k c and f(ζ, y(q)) = 0. In view of our normalization for f, y(q) will be automatically an algebraic integer. Moreover C(y(Q)) C f where C f is a number depending only on f; this holds because the coordinates of ζ are roots of unity. By Theorem L we may write b y(q) = η i ξ i, (3.1) where η i k may depend on ζ π(j) and on Q but vary in a certain prescribed finite set E depending only on k, and where b is bounded only in terms of C f and k; by adding the number 0 to the set E we may assume that b is in fact a fixed positive integer (dependent only on f and k), for all ζ π(j). The vector (η 1,..., η b ) E b has only finitely many possibilities (their number depending on f and k), so we may assume that this vector is fixed for all Q in a Zariski dense subset J of J. Let us consider the set of torsion points Z := {(ζ, ξ) := (ζ, ξ 1,..., ξ b ) G d+b m Q J, ζ = π(q), y(q) = η 1 ξ η b ξ b }, and its Zariski closure Z in G d+b m. Since J is Zariski dense in V, the projection p : G d+b m G d m on the first d coordinates sends Z to a Zariski-dense subset of π(v ), and hence p( Z) is Zariski-dense in π(v ). The set Z is contained in the variety W G d+b m defined by W : f(x 1,..., x d, η 1 t η b t b ) = 0. (3.2) By the Torsion Points Theorem Z is a finite union of torsion cosets, denoted τ H, where τ G d+b m is a torsion point and H is a connected algebraic subgroup of G d+b m. Therefore p( Z) is also a finite union of torsion-cosets, because p is a group homomorphism; in particular, p( Z) is Zariski-closed in G d m, and hence p( Z) = π(v ). Since π(v ) is irreducible over k c, there exists a torsion coset τ H, which is a component of Z, such that p(τ H) = π(v ); this proves part (i) of the conclusion, on putting β := p(τ ), T := p(h). To go on, we show that we may find a connected algebraic subgroup H 0 of H such that p : H 0 p(h) is an isogeny, so in particular dim H 0 = dim p(h) = dim π(v ) = dim V. We have an exact sequence 1 K H p(h) 1 (3.3) where K := ker p. We shall use a few elementary facts on duality on tori; namely, for an algebraic subgroup G of some G n m, we set Λ G := Hom(G, G m ), the set of algebraic-group homomorphisms from G to G m. (All the groups and homomorphisms appearing here are defined over Q c.) It is well-known that Λ H is free abelian of rank = dim H and that Λ p (H) is a lattice in Λ H, of rank = dim p(h). Pick then a lattice Γ in Λ H such that Q Γ is a complement of Q Λ p(h) in Q Λ H. Then we define H 0 := γ Γ ker γ. 8

9 It is an easy exercise to show that H 0 K = H, H 0 K <. (3.4) Note now that we can choose such a Γ to be a primitive lattice, i.e. such that Λ H /Γ is torsionfree; for instance, choose any Γ and then consider its saturation Γ := {σ Λ H : m 0, mσ Γ}. If this condition is satisfied, then H 0 turns out to be a connected algebraic subgroup. By conditions (3.4), dim H 0 = dim H dim K = dim p(h) and ker p H0 is finite. Thus p(h 0 ) is an irreducible subvariety of p(h) of the same dimension, and therefore equal to p(h) = T, proving the above claim. Now, we have τ H 0 τ H W and there is a map ψ : W V, where V is the hypersurface f(x 1,..., x d, y) = 0: we just set ψ(x 1,..., x d, t 1,..., t b ) := (x 1,..., x d, η 1 t η b t b ). Note that ψ is defined over k, because the η i lie in k. However, there is a birational map ψ : V V, also defined over k, such that π ψ is the projection on the first d coordinates. Taking all of this into account, we obtain a rational map ϕ : τ H 0 V, defined over k c (because τ is a torsion point), such that π ϕ = p τ H0. This proves part (ii) because p : H 0 p(h 0 ) = T is an isogeny. To conclude, we show that we can choose H 0 so that ϕ is birational. Let h be a generic point of τ H 0 (over Q). We define K 0 := ker p H 0 (a finite group), K 1 := {x H 0 : ϕ(xh) = ϕ(h)}. Observe that K 1 K 0 because π ϕ = p. Actually, K 1 is a subgroup, because if h is generic, also xh is generic, for any x K 0. Observe also that #ϕ(k 0 h) deg π. Hence [K 0 : K 1 ] deg π. On the other hand, both p and ϕ factor through the canonical projection p 0 : τ H 0 τ H 0 /K 1, so we may write ϕ = φ p 0, p = p 1 p 0, where deg p 1 = [K 0 : K 1 ] deg π. Finally, H 0 /K 1 and T are subtori of some G N m, of the same dimension, hence there exists an isomorphism θ : T H 0 /K 1. Composing this isomorphism with p 1 and φ we obtain maps µ = p 1 θ and ρ = φ (τ θ) such that π ρ = β µ and deg µ deg π. On the other hand, the first equality implies deg µ = deg π deg ρ deg π, so deg ρ = 1 and the theorem follows. Remark. We note that the conclusion is in a way best-possible, since the theorem has a kind of converse, as we are going to explain. Let π : V π(v ) G d m be a morphism of finite degree, where π(v ) = βt is a torsion coset and V/k is an absolutely irreducible affine variety. Suppose further that there exists an isogeny µ : T T and a rational map ρ : T V defined over k c, such that π ρ = β µ. Then we claim that J = J V,π is Zariski-dense in V. In fact, let Z be the set of torsion points in T, which is obviously Zariski dense in T, and observe that µ sends Z in itself. Define Z as the intersection of Z with the domain of regularity of ρ; Z is again Zariski dense in T and ρ(z ) is Zariski dense in V, because ρ is dominant (in fact, π ρ : T βt is dominant and dim T = dim V ). Note that ρ(z ) is contained in V (k c ), because Z T (k c ) and ρ is defined over k c. Moreover, π sends ρ(z ) into the set of torsion points of G d m, because π ρ equals βµ and β is a torsion point. Therefore ρ(z ) J, proving the assertion. Proof of Corollary 1. (a) Preliminary to the proof, we recall (in our notation) a statement, familiar in the theory of Hilbert s Irreducibility Theorem, from Schinzel s book [Sch, 1.9, Lemma 2]: There exists a polynomial g in k c [X, Y ] and a nonzero polynomial G k c [X] with the following property: For a τ in an extension L of k c such that G(τ) 0, f(τ, Y ) is reducible in L[Y ] if and only if g(τ, Y ), has a root Y = y L. Let us argue by contradiction, supposing that there exists an infinite set Z of roots of unity such that f(ζ, Y ) is reducible over k c for all ζ Z. Throwing away a finite subset of Z, we may assume that G(ζ) 0 for all ζ Z. By the above statement (applied with L = k c ) we may also assume that g(ζ, Y ) has a root in k c for each ζ Z, where g is as in the statement. Replacing g 9

10 with a suitable one of its irreducible factors in k c [X, Y ], we may assume, going to an infinite subset of Z, that g itself is irreducible over k c. We now apply Theorem 1 with V the subvariety of G 2 m defined by g = 0 and with π equal to the projection to G m given by the X-coordinate. By definition J V,π is Zariski-dense in V, since V is a curve and since π(v ) is infinite because it contains Z. Let T, β, ρ, µ be as in the conclusion of the theorem. Then T must be equal to G m (and so we can take β = 1) because T contains π(v ) which is dense. Therefore µ must be the multiplication (in G m ) by an integer m 0 and such that m deg π = deg Y g. Note that by composing with x x 1 on T we may assume that m > 0. The rational map ρ is expressed by (R(x), S(x)), where R, S are rational functions in k c (x), if x is a coordinate on G m. Finally, the relation π ρ = β µ means that R(x) = x m, so we obtain that g(x m, S(x)) = 0 identically. Recall that we had changed the original polynomial g with one of its factors, but plainly this conclusion remains true for the original g itself. By choosing τ = x m, L = k c (x) we obtain that g(τ, Y ) = 0 has a root Y = S(x) in L, whence, by the same criterion, f(x m, Y ) is reducible over k c (x). We now prove that if this happens for some m > 0, we may take m deg Y f. Suppose that f(x m, Y ) has r distinct irreducible factors, monic in Y, over k c (x). The substitutions x θx, where θ m = 1, act on the set of factors. The stabilizer of a given factor q(x, Y ) has index r r in the group of m-th roots of unity. Therefore q(x, Y ) is of the shape q 1 (x m/r, Y ). But then f(x r, Y ) is also reducible over k c (x). Since r r deg Y f, this proves the claim. (See also the proof of Corollary 2, which gives implicitly another argument.) (b) Conversely, suppose that f(x m, Y ) is reducible over k c, so we may write f(x m, Y ) = p(x, Y )q(x, Y ) (by Gauss Lemma) for some p, q k c [x][y ] of positive degree in Y. For all but finitely many roots of unity ζ, the polynomials p(ζ, Y ), q(ζ, Y ) will be of positive degree in Y, whence f(ζ, Y ) will be reducible in k c (ζ)[y ] = k c [Y ]. Proof of Corollary 2. We start with the following general remark. Let g k c [X, Y ] be irreducible but such that g(x m, Y ) is reducible for some positive integer m. Let Ω be an algebraic closure of k c (X). We contend that the extension L = k c (X, y) of k c (X) determined by a root y Ω of g(x, Y ) = 0 contains k c (X 1/d ) for some divisor d > 1 of m. (In particular, L/k c (X) is ramified above X = 0.) In fact, write X = T m (for T Ω), so the extension k c (T )/k c (X) is Galois (and totally ramified above X = 0). Hence [k c (T, y) : k c (T )] = [L : L k c (T )]. Now, by the above assumption, g(x, Y ) = g(t m, Y ) is reducible over k c [T, Y ], whence [L : L k c (T )] < [L : k c (X)]. Therefore L k c (T ) is a nontrivial extension of k c (X), included in k c (T ). But such an extension is necessarily of type k c (T m/d ) for some divisor d of m, d > 1. Suppose now that for a u k c the polynomial f(ζ + u, Y ) is reducible in k c [Y ] for an infinity of roots of unity ζ. Then, by Corollary 1 applied to f(x +u, Y ), and by the previous claim applied with g(x, Y ) = f(x +u, Y ), the extension of k c (X) determined by a root z of f(x, Z) = 0 contains some field k c ((X + u) 1/du ), for some d u > 1 (and necessarily d u deg Y f). However, this may hold only for finitely many u k c, since the compositum of k c (X +u s ) 1/dus, s = 1,..., S, is linearly disjoint from k c ((X +u) 1/du ) over k c (X), for u u 1,..., u S ; this concludes the argument. (In fact, the argument says that we must have u d u deg Y f, where the product is extended over the relevant u s.) Proof of Corollary 3. With the same arguments of Corollary 1 we reduce to prove the following: For every irreducible polynomial f k c [X, Y ] of degree > 1 in Y, there exist only finitely many n N such that the polynomial f(a n, Y ) has a root in k c. To prove this statement, we apply Theorem 1 to the variety V defined in affine 3-space by f(x 1 + X 2, Y ) = 0, letting π be the projection on the plane (X 1, X 2 ). Assume that f(a n, Y ) has a root in k c for all n lying in an infinite set M N. Then π(j) Z M := {(ζ n, ζ n+1 ) : n M}. 10

11 Now, by the Torsion Points Theorem, the Zariski-closure of Z M in G 2 m is a finite union of torsion cosets. However, every proper torsion coset of G 2 m may contain only finitely many elements of Z M : in fact, otherwise some nontrivial relation ζnζ a n+1 b = γ would hold for infinitely many n M, where a, b are fixed integers (not both zero) and γ is a fixed root of unity. In turn, this implies that (da/n) + (db/(n + 1)) is an integer, for some fixed nonzero integer d. Plainly, this may hold only for finitely many n if a, b are not both 0, a contradiction. In conclusion, Z M is Zariski-dense in G 2 m and a fortiori J is Zariski-dense in V. Hence we can apply the conclusion of Theorem 1; in practice, after a monoidal transformation of G 2 m (see [BG, 3], especially Def ) we obtain an identity f(x a 1 + X b 2, r(x 1, X 2 )) = 0 for suitable nonzero integers a, b and a rational function r k c (X 1, X 2 ). Specializing X 2 u k we see that f(x a 1 + u b, Y ) is reducible over k c. As in the proof of Cor. 2 this implies that u b must lie in a certain finite set related to f, a contradiction which concludes the proof. 4. Cyclotomic preperiodic points: proof of Theorem 2 Proof of Theorem 2. We start by normalizing f(x) in the so-called Fatou normal form (see [M], p. 242). Namely, it is an easy matter to check that, since d 2, we may by replace f(x) with L f L 1, where L(x) = µx + ν is a suitable polynomial of degree 1, to assume that f is monic and with vanishing second coefficient, that is f(x) = x d + a 2 x d a d. (4.1) Two remarks are in order: first, note that it is sufficient to prove the conclusion of the Theorem after enlarging k; hence we may suppose that L has coefficients in k. Next, note that the set of preperiodic points for L f L 1 is just the image L(Π f ). Hence it will suffice to prove the result for a normalized polynomial. We go on by observing that if α Π, then C(α) C f (see 2 for this notation), for some positive C f depending only on f. In fact, let C f be the maximum of 2 + a σ a σ d, where σ ranges over all embeddings of k in C. Suppose now that α has some conjugate α (over Q) with α > C f. Then, on the one hand α is preperiodic for some polynomial f σ ; on the other hand it is immediately verified by induction (note that d 2) that the sequence { (f σ ) (r) (α ) } r N is strictly increasing, a contradiction. (In the terminology of polynomial dynamics, the set {α : C(α) > C f } is contained in the intersection of the attractive basins of the polynomials f σ around the fixed point, which is superattracting; see [M]). Next, we prove that D f α is an algebraic integer for some positive integer D f depending only on f. In fact, let v be a finite place of k, normalized in some way and extended to k = Q. Then if α v > max(1, a j v ) we have a contradiction again, because { f (r) (α) v } r N is strictly increasing. Hence we deduce that α v max(1, a j v ) for every finite place v. Let now D f be a positive integer such that D f a j is an algebraic integer for j = 2,..., d. Then D f α v max( D f v, D f a j v ) 1 for all finite places v, proving the claim. From now on we shall assume that Π k c is infinite and we shall eventually derive the stated conclusion. We shall need the following consequence of Theorem L of 2, which we state as a lemma. Lemma 1. Let f k[x] be as above. There exist a positive b = b f,k and a finite set E = E f,k k with the following properties. For every α Π k c and every integer r 0, we can write f (r) (α) in the form c 1 ξ c b ξ b, where c i E and where the ξ i are roots of unity. Proof of lemma. First observe that it is sufficient to prove that every α Π k c may be written in the stated shape, for some b, E independent of α. In fact, it will then suffice to apply this conclusion to the f-iterates of α, which are also f-preperiodic and also lie in k c. 11

12 Let then α Π k c. By the observations preceding the lemma, there exist a positive integer D f and a positive C f such that D f α is an algebraic integer with C(D f α) D f C f. Since α k c we may apply Theorem L of 2; the conclusion therein gives the existence of the sought representation, with a b and an E depending only on f and k, not on α. This concludes the proof. In the sequel it will be convenient to denote Φ := {c 1 ξ c b ξ b : c i E {0} and ξ i are roots of unity}. We note two important properties of this set: Property 1: There exists a positive integer D = D f,k dependent only on f, k, such that D φ is an algebraic integer for all φ Φ. Property 2: There exists a positive C = C f,k dependent only on f, k, such that C(D φ) C for all φ Φ. These properties follow immediately from the fact that the set E and the integer b (i.e. the maximum number of summands) of Lemma 1 depend only on f, k. Let us now fix an integer M, sufficiently large in terms of f and k; specifically, it suffices that d M 2 > C 2, where C = C f,k appears in Property 2 above, of the set Φ. Let α Π k c. By the lemma we may write f (r) (α) = b c ir ξ ir, r = 0, 1,..., M, (4.2) where b is a positive integer as in the lemma, the c ir lie in the set E of the lemma and where the ξ ir are roots of unity. Note that these quantities depend on α but b, E, M depend only on f, k. Since we are assuming that Π k c is infinite and since E is finite, we may pick an infinite subset Π of Π k c such that for α Π the c ir are fixed for i = 1,..., b, r = 0,..., M, namely do not depend on α. In those of the equations (4.2) in which r is positive we may use the first equation (with r = 0) to substitute for α on the left side. In this way we obtain the following system of M algebraic equations in the unknowns x ir, satisfied by the roots of unity ξ ir : f (r) ( b c i0 x i0 ) = b c ir x ir, r = 1,..., M. (4.3) Also, note that for α Π, the system (4.3) does not depend on α, since the coefficients c ir are fixed. (Of course, the solutions (ξ ir ) depend on α.) We view the corresponding points (ξ ir ), i = 1,..., b, r = 0,..., M, as torsion points in the subvariety X of G b(m+1) m defined by the system (4.3); hence the subvariety X has infinitely many torsion points. By the Torsion Points Theorem, the Zariski closure X of the torsion points in X is a finite union of torsion cosets of G b(m+1) m. Moreover the map which sends the point (ξ ir ) of X to b c i0ξ i0 P 1 is generically surjective, because the image of the set of torsion points contains the infinite subset Π of the cyclotomic preperiodic points α. Therefore X must contain at least one torsion coset of G b(m+1) m of dimension 1 which is not sent to a constant by the mentioned map. 12

13 Say that this torsion coset is parametrized by x ir = ζ ir t eir, where the ζ ir are roots of unity and the e ir integers, not all zero. This means that we have identically: f (r) ( b c i0 ζ i0 t ei0 ) = b c ir ζ ir t eir, r = 1,..., M. (4.4) Putting q(t) := b c i0ζ i0 t ei0, we have that q(t) k c [t, t 1 ] is not constant, by the above choice of the torsion coset. After rearranging the terms we may also write, q(t) = q 0 t m0 + q 1 t m q l t m l, (4.5) where the q i k c are nonzero, the m i are distinct integers, not all zero and m 0 > m 1 >... > m l. Moreover, changing if necessary t into t 1, we may assume that m 0 m l. We finally note that the coefficients q i lie in the set Φ defined after the proof of Lemma 1. Suppose now that q(t) is not of the shape at m + bt m, a, b k c. Equivalently, in (4.5) we have l > 0 and m 0 + m 1 > 0. We are going to show that d M C 2, which is a contradiction with the present choice of M. In the sequel by O(t h ) we shall mean a Laurent polynomial containing only powers of t with exponent h. Note that f(q(t)) = (q 0 t m0 + q 1 t m ) d + O(t (d 2)m0 ) = q d 0t dm0 + dq d 1 0 q 1 t (d 1)m0+m1 + O(t (d 2)m0 ). More generally, we obtain f (r) (q(t)) = q dr 0 t dr m 0 + d r q dr 1 0 q 1 t (dr 1)m 0+m 1 + O(t (dr 2)m 0 ). Note that from the fact that (d r 1)m 0 + m 1 > (d r 2)m 0 it follows that the error term in these equations is of lower order than the first two terms. Comparing with the equations (4.4) for r = 1,..., M, we deduce that φ 0 (r) := q dr 0 Φ, φ 1 (r) := d r q dr 1 0 q 1 Φ, and also q 0, q 1 Φ. Hence φ 0 (r)q 1 = q 0 φ 1 (r)/d r, whence ψ := D 2 q 0 φ 1 (r)/d r is a nonzero algebraic integer, by Property 1 above of the set Φ. On the other hand, by Property 2, C(ψ) = C(D 2 q 0 φ 1 (r)/d r ) C 2 /d r. For r = M we have thus obtained the required contradiction, since C(ψ) 1. To go on, observe that, for s = 0, 1, 2, equations (4.4) imply f (r) ( b c is ζ is t eis ) = b c i,r+s ζ i,r+s t ei,r+s, r = 1,..., M s. (4.6) The same argument shows that, if one of q(t), f(q(t)), f(f(q(t))) is not of the shape at m +bt m, then d M 2 C 2, which again is a contradiction with the present choice of M. We thus conclude that each of q(t), f(q(t)), f(f(q(t))) is of the mentioned special shape. Then the proof will be finished after the following lemma: Lemma 2. Let f C[t] be a polynomial of the shape (4.1). Suppose that q C[t, t 1 ] is nonconstant and such that, for s = 0, 1, 2, f (s) (q(t)) = a s t ds + b s t ds, for some a s, b s C and integers d s. Then there exist λ C, ɛ = ±1 such that λf(λ 1 ɛx) is either x d or T d (x). 13

14 Proof. A first case occurs when q(t) is a monomial, i.e. a 0 b 0 = 0. By changing possibly t into t 1 we may assume that b 0 = 0. Since f(q(t)) is then a polynomial, we deduce b 1 = b 2 = 0, hence f(x) = x d, falling into the first case of the conclusion. Let us then consider from now on the case a 0 b 0 0; then necessarily a s b s 0 for s = 0, 1, 2 and d 1 = d 0 d, d 2 = d 0 d 2. By writing t in place of t d0 we may assume d 0 = 1. Let λ C be such that λ 2 = b 0 /a 0 and put u := t/λ, so t = λu. Then we find q(t) = c(u+u 1 ), where c = a 0 λ. Therefore, by symmetry in u, u 1, f (s) (c(u + u 1 )) = c s (u ds + u ds ), s = 1, 2 for suitable c 1, c 2 C, where actually, since f is monic, we have c s = c ds. Put F (x) := c 1 1 f(cx). The identity with s = 1 yields F (u + u 1 ) = u d + u d, hence F (x) = T d (x). Also, the first identity f(c(u+u 1 )) = c 1 (u d +u d ) yields f(f(c(u+u 1 ))) = f(c 1 (u d +u d )). Substituting the left term by means of the second identity we find f(c 1 (u d +u d )) = c 2 (u d2 +u d2 ), whence, setting G(x) = c 1 2 f(c 1x), we again find G(x) = T d (x). Comparing the equalities found so far, we get T d (c 1 c 1 x) = c 2 c 1 1 T d(x). But now a simple argument (see for instance Lemma 5, Ch. 1.4 of [Sch]) implies c 1 = ɛc, c 2 = ɛ d c 1, where ɛ = ±1. Taking into account the above formulas for c 1, c 2, all of this amounts to c d 1 = ɛ = ±1. Using again that F (x) = T d (x) we find f(x) = c 1 T d (c 1 x) = c d T d (c 1 x) = ɛct d (c 1 x). Putting H(x) = ɛc 1 x this gives (H f H 1 )(x) = T d (ɛx), as required. Example. In special cases one may obtain a different and simpler proof of Theorem 2, as we show in the following example. Namely, we let f(x) = x d ± 1 where d 4. Observe first that the polynomial g(x) = x d x 1 has a unique real root ξ > 1. A calculation shows that ξ < (1 + 5)/2 for d 4. Now, if x > ξ then it is immediate to check that f(x) > x, so x cannot be preperiodic. Hence, if u C is a preperiodic point for f, we must have u σ ξ for all conjugates u σ of u over Q. Also, u must be an algebraic integer, since f is monic. In particular, the Weil absolute height of u must satisfy H(u) ξ. Suppose now that u Q c. Then, by a theorem of Schinzel [Sch2], either u is 0 or a root of unity or H(u) (1 + 5)/2. By the above inequalities this entails that u is 0 or a root of unity. But this conclusion must hold also for v = f(u), which is preperiodic as well. A simple analysis of the remaining possibilities gives the following. Case f(x) = x d 1 : For any d, 1 and 0 are the only cyclotomic periodic points. Other cyclotomic preperiodic points exist if d is even (the d-th roots of unity). These are the only ones, unless 3 d. In this case, write d = 6kd with (d, 6) = 1: then the only other cyclotomic preperiodic points are (i, 6) = 1}. {ζ i 36k Case f(x) = x d + 1 : There are cyclotomic preperiodic points if and only if d 2, 4 (mod 6): these are the primitive 6-th roots of unity (periodic) and, for any divisor k of d, the primitive 3k-th roots of unity. It is perhaps worth noticing that, for d = 2 and f(x) = x 2 1, there are in fact cyclotomic preperiodic points other than the roots of unity, namely, (±1 ± 5)/2. The example can be generalized to other polynomials; when f is not monic Schinzel s result may not be sufficient, but one can sometimes replace it by a more general (but weaker) lower bound for the height in cyclotomic fields by Amoroso and Dvornicich [AD]. Remarks. (i) Concerning the proof of Theorem 2, equations (4.4) say that the polynomials f (r) (q(t)), r = 1,..., M, have few terms. Now, it is possible to bound from below the number of 14

15 terms in certain composite polynomials, which would allow a different approach to a proof. This kind of bound is carried out in [Z]. (ii) We note that the same method of proof should work if f(x) is a rational function, with the (strong) condition that it has a superattracting fixed point in P 1, being in the polynomial case. (See [M]: a superattracting fixed point is such that f is ramified above it; when f is a polynomial, it is totally ramified above the fixed point.) The conclusion should be that if f is not a polynomial there are only finitely many preperiodic points in k c. It would be interesting to obtain a full generalization of our result to all rational functions. This would apply in particular to the Lattès maps (see [M]), which come from duplication in elliptic curves E, and would imply the finiteness of cyclotomic torsion points on E. This last result however has been proved by Ribet for general abelian varieties and further sharpened by M. Baker and C. Petsche [BP]. 5. The property (P) and the proof of Theorem 3 We start with a slight extension of Theorem 2; for the sake of brevity we shall only outline the arguments of the proof, which follow exactly the same pattern as for Theorem 2. Theorem 2. Let f k[x] be a polynomial of degree d 2. Suppose that there exists an infinite set Γ k c such that f(γ) = Γ. Then for some polynomial L of degree 1 and for some ɛ = ±1, (L f L 1 )(x) is either (ɛx) d or T d (ɛx). Proof. Observe that it is plainly sufficient to prove the statement for a finite extension of k in place of k. Hence, as in the proof of Theorem 2, we may normalize f and suppose that it is given by (4.1). Now, suppose by contradiction that f is not of the mentioned special type but that there exists an infinite Γ k c with f(γ) = Γ. Then f has only finitely many preperiodic points in k c, by Theorem 2. Pick then a non-preperiodic element γ 0 Γ. Since Γ f(γ) we may construct a sequence {γ n } n N of elements of Γ k c such that f(γ n+1 ) = γ n for all n N. The γ n are pairwise distinct since otherwise γ 0 would be preperiodic (and in fact periodic). Now, let C f be as in the proof of Theorem 2. Then, following the simple relevant argument of the proof of Theorem 2, it is easy to see that for all n N we have C(γ n ) C := max(c f, C(γ 0 )). Similarly, if D f is as in the proof of Theorem 2 and if D 0 is a positive integer such that γ 0 is an algebraic integer, then, putting D := D f D 0, we have that D γ n is an algebraic integer for all n N. At this point we have, in analogy with Lemma 1, the following immediate consequence of Theorem L of 2: Lemma 1. Let f k[x], {γ n } k c be as above. There exist a positive b and a finite set E k with the following properties. For every n N we can write γ n in the form c 1 ξ c b ξ b, where c i E and where the ξ i are roots of unity. Following again the steps of the proof of Theorem 2, we denote by Φ the set of all sums of the shape c 1 ξ c b ξ b where where c i E {0} and where the ξ i are roots of unity. We note that γ n Φ for all n N. Also, arguing once more as in Theorem 2, we fix an integer M, sufficiently large in terms of f, k, C, D, but independent of n, and we consider the equations f (r) (γ n ) = γ n r, for r = 0,..., M, and for n M. Since γ n Φ, we may write, for n M and for r = 0, 1,..., M, γ n r = b 15 c irn ξ irn,