Index

Introduction 2

Chapter 1: Preliminary results 4 1.1 Number fields and number rings 4 1.2 Absolute values 7 1.3 Heights in projective and affine space 10 1.4 Heights of polynomials and Mahler’s measure 13

Chapter 2: Dynamical systems 16 2.1 Basic definitions and main properties 16 2.2 Fatou and Julia sets 17 2.3 Periodic points 19 2.4 Canonical heights and preperiodic points 30 2.5 Cyclotomic preperiodic points 33 2.6 Totally v-adic preperiodic points 39 2.7 Length of finite orbits 44 2.8 An explicite example: quadratic number fields 48

Chapter 3: Narkiewicz’s properties 76 3.1 The properties P and SP 76 3.2 The property N 83 3.3 The property V P 87

Bibliography 92

Introduction

The study of complex dynamical system given by polynomial mappings dates back to the birth of holomorphic dynamics and developed particularly from 1920s, when the frenchmen G. Julia and P. Fatou published their articles which gave birth to the homonymous sets. During the century this class of dynamical system was studied by different techniques that came from various parts of mathematichs, from topology to measure theory.

During the ’70s, even the algebraic properties of these systems started to be analysed, giving birth to the so-called “algebraic dynamics” or “arithmetic of dynamical systems”. The first mathematicians to begin those studies were the polish K. K. K. Kubota and W. Narkiewicz, together with the french P. Liardet and the british D. J. Lewis, who obtained the first results about finiteness of invariant sets. See [13],[14] and [15] for references.

In the most recent years we remind, as main results, the determination of finite orbits in number rings, given by the polish school of Wroclaw with W. Narkiewicz([20],[21]), T. Pezda([21],[22]) and R. Marszalek[18], the theory of canonical height, developed by S. Kawaguchi[12] and J. H. Silverman[6], and the finiteness of cyclotomic preperiodic orbits and the resolution of some open problems left by Narkiewicz, given recently by R. Dvornicich and U. Zannier[8].

This thesis wants to collect the main results in term of finiteness of orbits and invariant sets the most exaustively possible. In the same time, we shall also show some conjectures and open problems on the argument.

In the first chapter, we state, without proof, the main results about number fields and number rings, the natural model to work in algebraic dynamics, and the theory of absolute values and local fields. Then we introduce the theory of heights, important instrument to show finiteness results, with their principal properties and conjectures.

In the second chapter, after recalling the fundamental definitions of dynamical systems and Fatou and Julia sets, we deal with finite orbits. We start from periodic orbits and, after proving the famous Baker’s theorem on periodic orbits in algebrically closed fields, we shift to number fields and show how we can reduce the problem to the local case, studying finite extension of the p-adic field ℚ_{p}. Then we study preperiodic orbits and, after showing the most famous results of finiteness, we study the problem of the maximum length of a finite orbit; this time the reduction to the local case proves not to be useful and the main instrument becomes the study of particular unit equations on the group of unity of number rings. Finally we solve an explicit problem: the determination of all finite orbits in quadratic number fields ℚ(\sqrt{m}), with m a squarefree integer; this example applies a lot of the tecniques previously introduced.

In the third final chapter we analyse fully invariant sets for polynomial mappings, focusing on the principal finiteness properties, called “Narkiewicz’s properties”. The two main properties are called P and SP and characterize the fields in which polynomials have not infinite fully invariant sets; in particular we discuss the problem of the inequivalence between the two properties, recentry solved by R. Dvornicich and U. Zannier (see [8]). An important instrument for the inequivalence is another finiteness property, the property N, which generalises the results of finiteness of the height function given in the first chapter. Finally we deal with property V P, which describes the equivalent of fully invariant sets for pairs of polynomials, and discuss about its possible equivalence with SP, which is still an open problem.

Chapter 1 Preliminary results

1.1 Number fields and number rings

A number field is a finite extension of the rational field ℚ. Number fields are the most common and useful extension fields studied and will be the main application fields of the results we are going to show. In this section we want to list without proof the main properties of these fields we are going to explore. For the prooves we refer to [17].

Let K be a number field. We denote by O_{K} the set of the elements α ∈ K that are roots of a monic polynomial over K. The main property of number fields is that the set O_{K} is a ring, called the number ring (or the ring of integers) of K and its elements are the algebraic integers of K.

EXAMPLES 1.1.1 : 1) Let K = ℚ. Then clearly O_{K} = ℤ.

2) Let K = ℚ(\sqrt{m}), where m is squarefree integer. Then we have:

O_{K} = ℤ[\sqrt{m}] if m ≡ 2, 3 mod 4. O_{K} = ℤ[\frac{1+\sqrt{m}}{2}] if m ≡ 1 mod 4.

3) Let K = ℚ(ζ) where ζ is a primitive root of unity. Then O_{K} = ℤ[ζ].

Unluckily the number ring of a number field is not always monogenic, that is, of the form ℤ[α] for some α ∈ ℚ̄. However we have the following result.

Theorem 1.1.1 : Let K be a number field and n = [K : ℚ]. Then O_{K} ≃ ℤ^{n}. In particular any ℤ-basis of O_{K} is also a ℚ-basis of K calles integral basis.

The integral basis enables us to define the discriminant.

Definition 1.1.2 : Let K be a number field with n = [K : ℚ]. Let σ_{1}, ...σ_{n} be the embeddings of K in ℚ̄ and {α_{1}, ..., α_{n}} be an integral basis of K. We define the discriminant of K, denoted as disc(K), as the square of the determinant of the matrix A_{jk} = σ_{j}(α_{k}). It is easy to show that it does not depend on the choice of the basis.

EXAMPLES 1.1.2 : 1) Let K = ℚ(α), with α ∈ ℚ̄. Then |disc(K)| = Norm_{K/ℚ}(f′(α)) where f is the minimal polynomial of α on ℚ.

2) Let K = ℚ(\sqrt{m}) with m a squarefree integer. Then we have:

disc(K) = 4m if m ≡ 2, 3 mod 4; disc(K) = m if m ≡ 1 mod 4.

3) Let K = ℚ(ζ_{p}) with p an odd prime number and ζ a primitive p-th root of unity. Then |disc(K)| = p^{p−2}.

Theorem 1.1.3 (Minkowski) : If K is a proper extension of ℚ, that is K ≠ ℚ, then |disc K| > 1.

Let us now turn back to O_{K} and examine special properties of its ideals

Theorem 1.1.4 : For every number field K, the number ring O_{K} is noetherian, integrally closed and every non-zero prime ideal is maximal. Such a ring is also called a Dedekind domain. In every Dedekind domain every ideal can be uniquely factorized as a product of prime ideals.

The last statement of this theorem expresses a very useful property. In fact not every number ring is a UFD but they all have the unique factorization of ideals. This property enables us to define a relation of divisibility among ideals. We are particularly interested in the study of prime ideals.

Definition 1.1.5 : Let K be a number field and O_{K} be its number ring, p a prime of ℤ and P a prime ideal of O_{K}. We say that P lies over p (or equivalently p lies under P) if P ∩ ℤ = p or equivalently if P |pO_{K}, where pO_{K} is the minimum ideal of O_{K} containing p.

This definition enables us to define two very important quantities in studying prime ideals: the ramification index and the inertial degree.

Definition 1.1.6: If P is a prime ideal of O_{K} lying over p, we define the ramification index e(P/p) of p at P as the maximum power of P dividing pO_{K}. The inertial degree s(P/p) of p at P is the degree of the extension field O_{K}/P over ℤ/pℤ. Both the ramification index and the inertial degree are multiplicative in the extension towers.

Lemma 1.1.7 (degree formula): Let p be a prime number and P_{1}, ...P_{r} be all the prime ideals of O_{K} lying over p. If n = [K : ℚ] we have

∑_{j=1}^{r} e(P_{j}/p)s(P_{j}/p) = n.

We are particularly interested in situations when one among r, e, s is equal to the degree of the extension.

Definition 1.1.8: Let p be a prime number. We say that:

1) p is ramified in K if there exists at least a prime ideal P of O_{K} such that e(P/p) > 1.

2) p is totally ramified in K if e(P/p) = n for some P lying over p.

3) p splits completely in K if r = n (in particular in that case e(P/p) = s(P/p) = 1 for all prime ideals lying over p).

The following theorem enables us to identify more easily the ramified primes Theorem 1.1.9: A prime p is ramified in a number field K ⟺ p|disc(K).

Let us now suppose that K is a Galois extension of ℚ. In that case the Galois group G acts transitively on the set of primes of O_{K} lying over the same prime of ℤ, permuting them; by the unique factorization of ideals, this implies that all the prime ideals lying over p have the same ramification index and inertial degree and the degree formula becomes

re(P/p)s(P/p) = n,

where P is any prime ideal lying over p and r is the number of such prime ideals.

Let G = Gal(K/ℚ); we consider the subgroup

G_{0}(P/p) = {σ ∈ G|σ(α) ≡ α (mod P) ∀α ∈ O_{K}}.

G_{0}(P/p) is called the inertia group of p at P. Generally it depends on P, but if P_{1}, P_{2} are two primes lying over p, we can find an element σ ∈ G such that G_{0}(P_{2}/p) = σG_{0}(P_{1}/p)σ^{−1}, that is, the different inertia groups are conjugate. In particular, if G is abelian, all the prime ideals lying over p have the same inertia group.

Let K_{0} be the subfield of K fixed by G_{0}. K_{0} is called the inertia field of p at P. Again, unless G is abelian, this field depends on P. Moreover G_{0} is not in general a normal subgroup of G and therefore K_{0} is not in general a Galois extension of ℚ. Anyway the extension K/K_{0} is a Galois extension of order e = e(P/p) and therefore the inertia group G_{0}, being its Galois group, has itself order e.

We can also characterize K_{0} has some minimality and maximality property.

Theorem 1.1.10 : In the previous notations, we have that:

1) If K′ is the maximum subfield of K in which p is not ramified, then K′ = K_{0}.

2) If K′ is the minimum subfield of K in which the extension of p is totally ramified, then K′ = K_{0}.

We end this paragraph with a famous theorem describing the structure of the group of invertible elements, or units, of O_{K}.

Theorem 1.1.11 (Dirichlet) : Let K be a number field with [K : ℚ] = n = r + 2s where r is the number of real embeddings of K in ℚ̄ and 2s the number of the complex ones. Let O_{K}^{∗} be the set of the invertible elements of O_{K}. Then we have

O_{K}^{∗} ≃ T × ℤ^{r+s−1},

where T is the multiplicative group of the roots of unity contained n K.

An element η ∈ O_{K} is called a fundamental unit if it belongs to a set of generators of the free part ℤ^{r+s−1}. A unit η such that 1 − η is still a unit is called an exceptional unit.

EXAMPLE 1.1.3: Let K = ℚ(\sqrt{m}), with m a non-zero squarefree integer. Then we have: i) if m < 0, m ≠ −3, −1, then O_{K}^{∗} = {±1} ≃ ℤ/2ℤ.

ii) if m = −1, then O_{K} = \{±1, ±\sqrt{−1}\} ≃ ℤ/4ℤ.
iii) if m = −3 then O_{K} =< ζ_{6} >≃ ℤ/6ℤ.
iv) if m > 0, m ≡ 2, 3 mod 4, then O_{K}^{∗} = \{±1\} × < η >_{ℤ}, with η = 1 + \sqrt{m}.
v) if m > 0, m ≡ 1 mod 4, then O_{K}^{∗} = \{±1\} × < η >_{ℤ} with η = (1 + \sqrt{m})/2.

1.2 Absolute Values

In this section we want to introduce some new topologies on number fields, different from the euclidean one, who will enable us to define the local fields. The proves of these result can be found in [4].

Let K be a field. An absolute value on K is a real-valued function

|.| : K → ℝ^{+}

which satisfies the following conditions:

1) ∀α ∈ K |α| = 0 ⟺ α = 0.
2) ∀α, β ∈ K |αβ| = |α||β|.
3) ∀α, β ∈ K |α + β| ≤ |α| + |β|.

We also define the ultrametric property:

3’) ∀α, β ∈ K |α + β| ≤ max{|α|, |β|}.

We say that an absolute value is non-archimedean if it satisfies the ultrametric property, archimedean if not.

EXAMPLE 1.2.1: 1) Let K be any field; we may define the trivial absolute value on K as |α| = 1 ∀α ∈ K α ≠ 0

2) Take K = ℚ. Then the standard euclidean norm is clearly an archimedean absolute value; we shall indicate is as |.|_{∞} or simply as |.|. In addition, for every prime number p, we can write every α ∈ ℚ as α = p^{r} \frac{a}{b} with a, b ∈ ℤ and gcd(a, b, p) = 1 and define the p-adic absolute value as |α|_{p} = p^{−r}. It is easy to see that they are all non-archimedean absolute values.

Every absolute value is also a norm and then induces a topology on the field on which is defined. We say that two absolute values on the same field are equivalent if they induce the same topology. This is an equivalence relation on the set of all the absolute on a field and an equivalence class of absolute values is called a place.

Theorem 1.2.1: Two non-trivial absolute values |.|_{1} and |.|_{2} are equivalent ⟺ ∃λ > 0 such that ∀α ∈ K we have |α|_{1} = |α|_{2}^{λ}.

The following theorem classifies completely the places on ℚ.

Theorem 1.2.2 (Ostrowski): Every absolute value on ℚ is equivalent either to the standard absolute value or to one of the p-adic absolute values.

Let L an extension of K. Let |.|_{v} an absolute value on K and |.|_{w} an absolute value on L. We say that |.|_{w} extends |.|_{v} and indicate as w|v if ∀α ∈ K |α|_{w} = |α|_{v}.

If K is not complete with respect to the topology induced by the place |.|_{v}, we can define the completion K_{v} as the unique (up to isometric isomorphisms) field such that:

1) K_{v} is complete with respect to the topology induced by an extension of |.|_{v}.

2) K is a dense subset of K_{v}.

EXAMPLE 1.2.2 : If K = ℚ the field ℚ_{p} of p-adic numbers is the completion of ℚ with respect to the p-adic absolute value. The ring ℤ_{p} of p-adic integers is the closure of ℤ in ℚ_{p}. On the other hand ℝ is the completion of ℚ with respect to the archimedean absolute value.

An important fact about the completion K_{v} is that the extensions of v to finite extensions of K_{v} are particularly simple.

Lemma 1.2.3 : Let n = [L : K]. If K = K_{v}, then there is a unique extension |.|_{w} of |.|_{v} to L, given by:

|α|_{w} = |Norm_{L/K}(α)|_{v}^{1/n},

where Norm_{L/K} is the classic norm function. Moreover, we have L = L_{w}, that is L is complete with respect to the extension.

If K is not complete there are at most n extension (not necessarily inequivalent) of |.|_{v}. We say that v splits completely if there are exactly n inequivalent extensions. However we have the following result.

Lemma 1.2.4 : In the preceeding notations, we have

∑_{w|v}[L_{w} : K_{v}] = [L : K].

The quantity [L_{w} : K_{v}] is called local degree of L in w.

Let K be a field with a nonarchimedean place |.|_{v}. The valuation ring of |.|_{v} is the set

R_{v} = {α ∈ K| |α|_{v} ≤ 1}.

It is immediate to see that R_{v} is a ring and that the definition does not depend on the representative of |.|_{v}. R_{v} is a local ring and we denote as m_{v} its unique maximal ideal: obviously

m_{v} = {α ∈ K| |α|_{v} < 1},

while the set

{α ∈ K| |α|_{v} = 1}

is the set of v-units. The quotient field R_{v}/m_{v} is called the residue field of K with respect to v.

Let L be a finite extension of K and |.|_{w} a place on L such that w|v. The inertia degree of w on v is the number s = s(w/v) = [R_{w}/m_{w} : R_{v}/m_{v}]. The set |L^{∗}|_{w} is a multiplicative group and |K^{∗}|_{v} is a subgroup; the index e = e(w/v) = [|L^{∗}|_{w} : |K^{∗}|_{v}] is called the ramification index of w on v. Finally the place v is called discrete if |K^{∗}|_{v} is cyclic.

Proposition 1.2.5 : In the preceeding notations we have

e(w/v)s(w/v) ≤ [L : K]

and the equality vails if v is discrete and K = K_{v}. Moreover, in this case e and s are multiplicative in the extension towers.

If the equality of the proposition vails, we have some particular definitions. We say that the place |.|_{v}:

1) is unramified if e(w/v) = 1 for every extension |.|_{w}.

2) is totally ramified if e(w/v) = [L : K] for some |.|_{w}.

Let K be a number field and O_{K} its number ring, that is, the integral closure of ℤ in K. Let |.|_{v} be a nonarchimedean place on K; then v|p for some p prime. We have that K_{v} is a finite extension of ℚ_{p} and that the closure of O_{K} in K_{v} is the valuation ring of K_{v}. Moreover the classic ramification index and inertia degree of primes, which we have described in the previous paragraph, are obtained as sum of the local ones, that is

e_{p} = ∑_{v|p} e(v/p) s_{p} = ∑_{v|p} s(v/p).

One of the peculiarities of the absolute values on ℚ is that, thanks to unique factorization, ∀α ∈ ℚ^{∗} if we take the product of all the absolute values of x, the result is always 1. This is a very special property and leeds us to the following definition.

Definition 1.2.6 : Let K be a field and M_{K} a set of places on K Then we say that K satisfies the product formula with respect to M_{K} if, ∀α ∈ K^{∗}, we have |α|_{v} = 1 for all but a finite number of v and ∏_{v∈M_{K}} |α|_{v} = 1.

EXAMPLES 1.2.3 : 1) Take K = ℚ, M_{K} = {|.|_{∞}, |.|_{p} p prime}. K clearly satisfies the product formula with respect to M_{K} (it is a consequence of unique factorization).

2) Let K = k(t) where k is an algebrically closed field. Define |f(t)|_{∞} = c^{deg f} with c > 1 and, ∀ξ ∈ K, |f(t)|_{ξ} = c^{−m_{ξ}(f)}, where m_{ξ}(f) is the multiplicity of ξ as root of f. These are all absolute values on K (the first archimedean, the other ones non-archimedean) and it is immediate to check that K satisfies the product formula with respect to them.

3) If K is a field with product formula and L is a finite extension of K, we take M_{L} as the set of absolute values on L which extend the ones on K; then L satisfies the product formula respect to M_{L}, where M_{L} is the set of absolute values on L extending the ones on K. In particular every number field satisfies the product formula.

Lemma 1.2.7 : If K satisfies the product formula with respect to M_{K}, then M_{K} contains only a finite number of archimedean absolute values.

1.3 Heights in projective and affine space

We shall now define the concept of height, a function which “measures the algebraic complexity” of a point and a is basical notion of diophantine geometry. To define this function we need to recall some definitions about the projective space.

Definition 1.3.1 : Let K be a field. The n-dimensional projective space ℙ^{n}(K) is the set of non-zero n + 1-tuple of elements of K with the equivalence relation

(α_{1}, ..., α_{n+1}) ≡ (β_{1}, ..., β_{n+1}) ⟺ ∃λ ≠ 0 ∀j = 1, ..., n + 1 β_{j} = λα_{j}.

The equivalence classes are called the projective points of ℙ^{n}(K). We denote a projective point by P = (P_{1}, ...P_{n+1}) and, if α ∈ K^{n} we denote P_{α} = (α_{1}, ..., α_{n}, 1).

We can now define the concept of height.

Definition 1.3.2 : Let K be a number field and P ∈ ℙ^{n}(K) a projective point, then we define the height of P as

h(P) = ∑_{v∈M_{k}} max_{j}log|P_{j}|_{v}.

Since this is a definition on a projective space, we should check that it does not depend on the choice of projective coordinates for P. We also want to prove that the height does not depend on the choice of the field K, extension of ℚ, in which P is defined.

Lemma 1.3.3 : The height function h does not depend on the choice of K and of the projective coordinates.

Proof: Let L be another field such that P ∈ ℙ^{n}(L). For simplicity we may assume that K ⊆ L. Let us call h_{L} the height computed with the absolute values of L and h_{K} the height computed with the absolute values of K. Then we have

h_{L}(P) = ∑_{w∈M_{L}} max_{j}log|P_{j}|_{w} = ∑_{v∈M_{K}} ∑_{w|v} max_{j}log|P_{j}|_{w} = ∑_{v∈M_{K}} max_{j}log|P_{j}|_{v} = h_{K}(P).

That is what we need. For the coordinates, note that if P′ = (P′_{1}, ..., P′_{n+1}) is another projective representation of P we have P′_{j} = λP_{j} for some λ ∈ K, λ ≠ 0. Then

h(P′) = ∑_{v∈M_{K}} max_{j}log|P′_{j}|_{v} = ∑_{v∈M_{K}} log|λ|_{v} + ∑_{v∈M_{K}} max_{j}log|P_{j}|_{v}

but the first term is zero because of the product formula, so h(P′) = h(P) as we want.

If the coordinates P_{1}, ..., P_{n+1} can be chosen in ℚ, it is easy to see that they can also be chosen in ℤ and we can assume that they have no common factors. In this representation the non-archimedean absolute values give no contribut to the height, so we can computed easily the height using only the archimedean one.

Let us know consider the affine case. If α ∈ K^{n} we can define the height of α as

h(α) = h(P_{α}).

By the equivalence relation defining the projective space we can always choose a system of projective homogenous coordinates in which one of the α_{j} is 1, therefore the height is a non-negative function. The next theorem characterizes the points of height zero in K.

Theorem 1.3.4 (Kronecker): Let α ∈ K. Then h(α) = 0 ⟺ α is a root of unity.

Proof: If α is a root of unity, obviously all of its absolute values are 1. So h(α) = 0. Conversely if h(α) = 0, then |α|_{v} ≤ 1 ∀v ∈ M_{K} and that means that α is an algebraic integer. Let d be the degree of the minimum polynomial of α on ℚ and α = α_{1}, ..., α_{n} its conjugates. Let s_{k} the elementary symmetric functions for k = 0, ..., d. Since α is an algebraic integer, we have that s_{k}(α_{j}^{m}) ∈ ℤ ∀m ∈ ℤ. Since |α_{j}|_{v} ≤ 1 ∀j, v and expanding the elementary symmetric functions we obtain that

∑_{k=0}^{d} |s_{k}(α_{j}^{m})| ≤ ∑_{k=0}^{d}(\frac{d}{k}) = 2^{d}.

Then we have only a finite number of possibilities for the vectors (s_{k}(α_{1}^{m}), ..., s_{k}(α_{d}^{m})). By Dirichlet’s pidgeon-hole principle there must be two integers m and n with m > n with the same vectors. Then it must exist a permutation σ ∈ S_{d} such that (α_{1}^{m}, ..., α_{d}^{m}) = σ(α_{1}^{n}, ..., α_{d}^{n}). If r is the order of σ in S_{d}, then we have α^{m^{r}−n^{r}} = 1 and so α is a root of unity as we want.

Let us now examine some fundamental properties of the height.

Proposition 1.3.5 : If α_{1}, ..., α_{r} ∈ K^{n} then h(α_{1}+...+α_{r}) ≤ h(α_{1}) + ... + h(α_{r}) + log(r).

Proof: Let (α_{1}^{(k)}, ..., α_{n}^{(k)}) be a coordinate vector for α_{k}. We have

h(α_{1} + ... + α_{r}) = ∑_{v∈M_{K}} max_{j}log|α_{j}^{(1)} + ... + α_{j}^{(r)}|_{v}.

Now if v is nonarchimedean, then

|α_{j}^{(1)} + ... + α_{j}^{(r)}|_{v} ≤ max_{k}|α_{j}^{(k)}|_{v}.

Otherwise, If v is archimedean, we have

|α_{j}^{(1)} + ... + α_{j}^{(r)}|_{v} ≤ |r|_{v}max_{k}|α_{j}^{(k)}|_{v}.

Putting all together we have:

h(α_{1} + ... + α_{r}) ≤ log(r) + ∑_{v∈M_{K}} max_{j,k}log|α_{j}^{(k)}|_{v} ≤ log(r) + ∑_{v∈M_{K}} ∑_{k} max_{j}log|α_{j}^{(k)}|_{v}

and permuting the two sums in the last term we have the thesis.

Proposition 1.3.6 : Let K be a Galois extension of ℚ, α ∈ K an algebraic number and σ ∈ Gal(K/ℚ). Then h(α) = h(σ(α)).

Proof: Let |.|_{v} an absolute value of K which extends the p-adic absolute value on ℚ. Then the absolute value |α|_{v′} = |σ(α)|_{v} is another absolute value on K which extends the p-adic one. This shows that the Galois group acts transitively on the absolute values extending |.|_{p}. So we have

∑_{v|p} max_{j}log|α_{j}|_{v} = ∑_{v|p} max_{j}log|σ(α_{j})|_{v}

and the thesis follows immediatly.

Proposition 1.3.7 : Let α ∈ K, α ≠ 0 and λ ∈ ℚ. Then h(α^{λ}) = |λ|h(α). In particular h(α^{−1}) = h(α).

Proof: If λ > 0 the proposition is trivial by the definition of height. So we can reduce to the case λ = −1. For any absolute value |.|_{v} we have

log|α|_{v} = log^{+}|α|_{v} − log^{+}|α^{−1}|_{v},

where

log^{+}(α) = max{0, log(α)}.

Summing both sides on v, the left-hand side is zero by the product formula, while the right-hand side is h(α) − h(α^{−1}), proving the result.

Corollary 1.3.8 (fundamental inequality): If S ⊆ M_{K} is a finite set of places, then

∑_{v∈S} log|α|_{v} ≤ |h(α)|.

Proof: Clearly ∑_{v∈S} log|α|_{v} ≤ h(α) by the definition of height. If we use α^{−1} and the previous proposition we have the other inequality.

Lehmer’s conjecture: For any algebraic number field K there exists a constant C > 0 depending only the field K, such that, ∀α ∈ K̄, we have

h(α) ≥ \frac{C}{[K(α) : K]}.

1.4 Heights of polynomials and Mahler measure

In this section we extend the definition of height also to the set of polynomials. First of all we need to define some norms on polynomials:

Definition 1.4.1 : Let f = ∑_{j=0}^{d} a_{j}x^{j} a polynomial on K[x]. If |.|_{v} is an absolute value on K we define the v-adic norm of f as:

|f|_{v} = max_{j}|a_{j}|_{v}.

∀p ∈ ℝ p ≥ 1 we define the p-adic norm of a polynomial as:

l_{p}(f) = (∑_{j=0}^{d} |a_{j}|^{p})^{1/p}.

If p = +∞, we set

l_{∞}(f) = max_{j}|a_{j}|

Finally, we define the height of f as:

h(f) = ∑_{v∈M_{K}} log|f|_{v}.

Proposition 1.4.2 : Let f ∈ K[x] and g ∈ K[y] be polynomials in different variables. Then h(fg) = h(f) + h(g).

Proof: Trivial by calculation.

Now we shall define a very useful quantity in studying polynomial, which will lead us to some important results in terms of heights.

Definition 1.4.3 : The Mahler measure of a polynomial f ∈ K[x_{1}, ..., x_{n}] is the quantity

M(f) = exp(∫_{T^{n}} log|f(e^{iθ_{1}}, ..., e^{iθ_{n}})| dθ_{1}...dθ_{n}),

where T^{n} is the unitary n-dimensional disc.

The following properties of Mahler measure are immediate by calculation:

1) The Mahler measure is a normalized measure on K[x_{1}, ..., x_{n}].

2) ∀f, g ∈ K[x_{1}, ..., x_{n}] M(fg) = M(f)M(g).

3) ∀α ∈ K M(x − α) = log^{+}|α|.

4) If f = ∑_{j=0}^{d} a_{j}x^{j} and α_{1}, ..., α_{d} are the roots of f, then

log(M(f)) = log|a_{d}| + ∑_{j=1}^{d} log^{+}|α_{j}|.

5) ∀f ∈ K[x_{1}, ..., x_{n}] l_{∞}(f) ≤ 2^{d}M(f).

Proposition 1.4.4 : Let α be an algebraic number and f be its minimal polynomial. Then

log(M(f)) = [ℚ(α) : ℚ]h(α).

Proof: Let d = [ℚ(α) : ℚ] and write f = ∑_{j=0}^{d} a_{j}x^{j}. We choose a field K which contains α and is a Galois extension of ℚ, with Galois group G. Then the set {σ(α)}_{σ∈G} contains every conjugate of α exactly [K : ℚ]/d times. The multiplicativity of the v-adic norm tells us that

|a_{d}|_{v} ∏_{σ∈G} max(1, |σ(α)|_{v})^{d/[K:ℚ]} = 1

for every non-archimedean v ∈ M_{K} Now using this formula and splitting the archimedean and nonarchimedean places we have

[K : ℚ]h(α) = ∑_{v|∞} ∑_{σ∈G} log^{+}|σ(α)|_{v} − \frac{[K : ℚ]}{d} ∑_{v|p} log|a_{d}|_{v} =

= \frac{[K : ℚ]}{d} ∑_{v|∞}(log|a_{d}|_{v} + ∑_{j=1}^{d} log^{+}|α_{j}|_{v})

and the application of property 4 completes the proof.

Theorem 1.4.5 (Northcott): ∀A, B ∈ ℝ B ≥ 0 the set

J_{A,B} = {α ∈ ℚ̄| [ℚ(α) : ℚ] ≤ A, h(α) ≤ B}

is finite.

Proof: Let α ∈ J_{A,B} and A = [ℚ(α) : ℚ]. Let f = ∑_{j=0}^{A} a_{j}x^{j} be the minimal polynomial of α, which can be chosen with coefficients in ℤ. The previous proposition tells us that M(f) ≤ e^{AB} and property 5) shows that max_{j}|a_{j}| = l_{∞}(f) ≤ 2^{A}M(f). Then the coefficients of f are bounded by (2e^{B})^{A}. Since there are A + 1 coefficients in every polynomial, the number of these possible polynomials is bounded by (2(2e^{B})^{A} + 1)^{A+1}. Since each of these polynomials has A roots, the number of elements of J_{A,B} is at most A(2(2e^{B})^{A} + 1)^{A+1} ≤ (5e^{B})^{A^{2}+A}, hence is finite.

Dynamical Systems

2.1 Basic Definitions and main properties

Definition 2.1.1 : A (discrete) dynamical system is a couple (X, f), where X is a topologycal space and f : X → X is a continuous map. For k ∈ ℕ we shall denote by f^{k} the k-th iterate of f, defined inductively by f^{0} = id and f^{k} = f ∘ f^{k−1}. If f is invertible, we set f^{−k} = (f^{−1})^{k}. If α ∈ X, the set {f^{k}(α)}_{k∈ℕ} is called the (forward) orbit of α and, if f is invertible, the set {f^{−k}(α)}_{k∈ℕ} is the backward orbit of α. Finally, the set {f^{k}(α)}_{k∈ℤ} is the full orbit of α.

In our case X = K^{n} where K will be usually a field of algebraic numbers and f will be a polynomial map on it. Since the inverse of a (non-linear) polynomial is not a polynomial, we will consider only the forward orbit of every point α ∈ K^{n} and call it simply the orbit of α.

Our main interest will be the case when the orbit is finite. We have the following definitions:

Definition 2.1.2 : Let (X, f) be a dynamical system. A point α ∈ X is called fixed if f(α) = α; we denote as F ix(f) the set of fixed points of f. If f^{m}(α) = α for some m ∈ ℕ, we say that α is periodic; the minimal such m is called the exact period of α. The set of periodic points of a map is denoted by P er(f), while the set of periodic points of exact period m is denoted by P er_{m}(f). The orbit of a periodic point is also called a cycle. Finally we say that a point is preperiodic if f^{m}(α) is periodic for some m ∈ ℕ; the set of preperiodic points is denoted by Π(f).

The study of periodic and preperiodic points will be one of our main interests. Another interesting situation is the following:

Definition 2.1.3 : Let (X, f) be a dynamical system. A subset Y ⊆ X is called finvariant if f(Y) ⊆ Y, fully f-invariant if f(Y) = Y, completely f-invariant if f(Y) = f^{−1}(Y) = Y.

Clearly every orbit is f-invariant and every cycle is fully f-invariant. The study of fully f-invariant sets will be the main purpose of the next chapter.

The following examples discuss the case deg f = 1, n = 1.

EXAMPLE 2.1.1 : Let f : ℚ̄ → ℚ̄ given by f(x) = ax, with a ∈ ℚ̄^{∗}. Clearly 0 is the only fixed point and f^{k}(x) = a^{k}x so the behaviour of the orbit can be described easily:

1) If |a| < 1, then f^{k}(α) converges to 0 for every α ∈ ℚ̄.

2) If |a| > 1, then f^{k}(α) diverges to infinity for every α ∈ ℚ̄.

3) If |a| = 1, then f^{k}(α) is contained in the set {β ∈ ℂ| |β| = |α|}.

EXAMPLE 2.1.2 : Let f : ℚ̄ → ℚ̄ given by f(x) = x + b, with b ≠ 0. Clearly there are neither fixed nor periodic points and f^{k}(x) = x + kb. Therefore every orbit diverges to infinity.

Combining the results of these examples we have a complete solution of the case deg f = 1 in every dimension. Then from now on we shall consider the case deg f ≥ 2.

The last notion we need is conjugation. This definition characterizes the maps that must be considered identical when referring to their dynamical properties.

Definition 2.1.4 : Two dinamycal systems (X, f) and (Y, g) on topologycal spaces are conjugate if there is a homeomorphism ϕ : X → Y, called conjugation, such that:

ϕ ∘ f ∘ ϕ^{−1} = g.

It is immediate to check that, if f and g are conjugate, the conjugation ϕ sends fixed, periodic and preperiodic points of f to fixed, periodic and preperiodic points of g. The same is true for invariant sets, so all the dynamical properties we shall examine are preserved by conjugation.

According to the type of dynamical system (topologycal, holomorphic,...) the class of homeomorphisms chosen as conjugations changes; in our case we choose the linear polynomials. Since their dynamical is very simple and their inverses are still linear polynomials, they are the right class for our purpose.

2.2 Fatou and Julia sets

In this paragraph we examine two important invariant sets for polynomial dynamical systems and their role in classifying periodic points. In this section K, if not specified, will always be the complex field ℂ.

Before defining and describing these sets, we need a definition:

Definition 2.2.1 : We say that a family {f_{k}} of maps from ℂ to ℂ is equicontinuous at a point α_{0} if ∀ϵ > 0 there exists δ > 0 such that, ∀f ∈ {f_{k}} we have

|α_{0} − α| < δ ⟹ |f(α_{0}) − f(α)| < ϵ.

We say that {f_{k}} is equicontinuous on the set X ⊆ ℂ if it is equicontinuous at every point of X.

Definition 2.2.2 : Let f : ℂ → ℂ a polynomial of deg ≥ 2. The Fatou set of f is the maximal open subset of ℂ on which the family {f^{k}} is equicontinuous. The Julia set of f is its complement in ℂ. The Julia set is denoted by J(f) while the Fatou set is simply denoted as its complement J(f)^{c}.

The following properties are obvious by the definitions:

1) The Julia set is a closed subset of ℂ, while the Fatou set is open.

2) The Julia and Fatou set are completely invariant sets.

3) ∀k ≥ 0 the Julia and Fatou set of f^{k} are the same of f.

Definition 2.2.3 : Let f : ℂ → ℂ be a holomorphic map and α ∈ P er_{m}(f). The multiplier λ of α is defined by

λ = (f^{m})′(α).

(Note that the multiplier is the same for all the periodic orbit). We say that a periodic point α is attracting if |λ| < 1, repelling if |λ| > 1, parabolic if |λ| = 1, but none of the iterates of f is the identity map.

If f is a non-linear polynomial, none of the iterates of f is the identity ever and then every periodic point whose multiplier |λ| = 1 is parabolic. We can then locate every periodic point in the right set by the following theorem

Theorem 2.2.4 : Every attracting periodic point is contained in the Fatou set, while every repelling or parabolic periodic point is contained in the Julia set.

Proof: By the property 3) of Julia set, we can simply consider a fixed point α. If |λ| > 1, then the derivative (f^{k})^{n}(α) = λ^{n} diverges to infinity. Therefore no iterate of f can converge in a neighbourhood of α and then α ∈ J(f). If |λ| < 1 we can apply Taylor’s theorem and choose |λ| < μ < 1 such that

|f(x) − α| ≤ μ|x − α|;

then if we choose a sufficiently small neighbourhood of α we have that f converges uniformly to the constant α on that neighbourhood. Then α ∈ ℂ\J(f). Finally if |λ| = 1 we may suppose for simplicity that α = 0. Writing

f = ∑_{j=0}^{d} a_{j}x^{j},

we have that a_{0} = 0, a_{1} = 1. Let r the minimum index ≥ 2 such that a_{r} ≠ 0. Taking the derivative, we have that

(f^{k})^{r}(0) = r!ka_{r},

which diverges to infinity. Then no iterate of f can converge in a neighbourhood of 0 and we have finished.

The Julia set has also many other important properties. We list here the main ones; their proves may be found in [19]:

1) If deg f ≥ 2 then J(f) is an infinite uncountable set and it has no isolated points.

2) If J(f) contains an interior point, then J(f) = ℂ.

3) If α ∈ P er(f) and U = {β ∈ ℂ|f^{k}(β) → α}, then ∂U ⊆ J(f).

4) If degf ≥ 2 then either J(f) is connected or it has uncountably many connected components.

EXAMPLE 2.2.1 : We take f(x) = x^{d} with d a positive integer. It is easy to check that J(f) = {α ∈ ℂ||α| = 1}

In this example the Julia set has a very easy structure, but generally it has not; usually Julia sets have very complex forms and are the easiest examples of fractals.

Suppose now that K = ℂ_{v} where v is any place, archimedean or non-archimedean. We want to describe some variants of the Julia set for these fields.

Definition 2.2.5 : Let f ∈ ℂ_{v}[x] a polynomial of degree d ≥ 2. The v-adic filled Julia set of f, denoted by F J_{v}(f), is the set of points of ℂ_{v} whose orbits remains bounded, that is,

F J_{v}(f) = {α ∈ ℂ_{v}| |f^{n}(α)|_{v} ↛ +∞}.

If ℂ_{v} = ℂ the Julia set J(f) is always a non-empty closed set and F J_{v}(f) is a compact set containing J(f), but if v is non-archimedean, J(f) can be empty and F J_{v}(f) is in general non-compact. The link between these two sets is in the following theorem, whose proof can be found in [1].

Theorem 2.2.6 : If the Julia set J(f) is non-empty, then it is the topological boundary of the filled Julia set F J_{v}(f).

EXAMPLE 2.2.2 : Let v be a non-archimedean place and f ∈ ℂ_{v}[x] of degree d ≥ 2 such that f(x) = ∑_{j=0}^{d} a_{j}x^{j} with |a_{j}|_{v} ≤ 1 and |a_{d}|_{v} = 1. It is an easy calculation to verify that J(f) is empty and F J_{v}(f) is the closed unitary disc.

2.3 Periodic points

In this paragraph we will examine the main properties of periodic points of polynomials. The main problem will be to classify the integers k for which P er_{k}(f) is empty or not. We start with the case when K is an algebrically closed field and then explore the most interesting case of number fields.

We start with a lemma that assures us that we can write f in a useful form.

Lemma 2.3.1 : Let α ∈ F ix(f). Then ∀n ∈ ℕ we can write

f^{n}(x) = x + (f(x) − x)g_{n}(x),

where g_{n} is a polynomial satisfying the relation

g_{n}(α) = 1 + f′(α) + f′(α)^{2} + ... + f′(α)^{n−1}.

Proof: For all n > 1, applying Lagrange’s theorem and since α is a fixed point, we have that

f^{n}(x) − f^{n−1}(x) = f(f^{n−1}(x)) − f(f^{n−2}(x)) = (f^{n−1}(x) − f^{n−2}(x))f′(ξ),

where ξ is a real number standing between f^{n−1}(x) and f^{n−2}(x). Applying recursively this relation we have

f^{n}(x) − f^{n−1}(x) = (f(x) − x) ∏_{j=1}^{n−1} f′(ξ_{j}) = (f(x) − x)q_{n−1}(x),

where q_{j} is an appropriate polynomial and ξ_{j} is a real number between f^{j}(x) and f^{j−1}(x). Finally we have that

f^{n}(x) − x = (f^{n}(x) − f^{n−1}(x)) + ... + (f(x) − x) = (f(x) − x)(1 + q_{1}(x) + ... + q_{n−1}(x)),

that is, g_{n}(x) = 1 + q_{1}(x) + ... + q_{n−1}(x) and since α is a fixed point f^{j}(α) = α for all j and therefore ξ_{j} = α for all j. Then q_{j}(α) = f′(α)^{j}.

If K is an algebrically closed field every polynomial f ∈ K[x] has at least one fixed point, since the equation f(x) − x is itself a polynomial and then must have at least one root. We want to show that f has also many periodic points.

Theorem 2.3.2 : Let K be an algebrically closed field and f ∈ K[x] with deg f = d ≥ 2. Then the set P er(f) is infinite and the set P er_{p}(f) is non-empty for all but at most d prime numbers.

Proof: Since f has at least a fixed point, we can write

f^{p}(x) = x + (f(x) − x)g_{p}(x).

Let α ∈ K be a root of g_{p}. Then we must have

g_{p}(α) = 1 + f′(α) + ... + f′(α)^{p−1},

therefore f′(α) must be a p-th root of unity. Morevoer f^{p}(α)−α = 0. If P er_{p}(f) is empty, then α must be a fixed point of f; but f has at most d fixed points (they are the roots of the polynomial f(x) − x of degree d) then there are at most d primes for which P er_{p}(f) is empty.

In case K = ℂ there is a more powerful result.

Theorem 2.3.3 (Baker) : If f ∈ ℂ[x] with deg f = d ≥ 2, then P er_{k}(f) is non-empty for all k but at most one exception.

Proof: Suppose that there are two positive integers k_{1} and k_{2} with 1 < k_{1} < k_{2} such that P er_{k_{1}}(f) and P er_{k_{2}}(f) are empty. We define the rational function

g(x) = \frac{f^{k_{2}}(x) − x}{f^{k_{2}−k_{1}}(x) − x} = \frac{P (x)}{Q(x)}

where P and Q are relatively prime polynomials. If q =degQ then we have clearly degP = d^{k_{2}} − d^{k_{2}−k_{1}} + q; it follows easily by calculation that the degree of the numerator of g′ is bounded by

c = d^{k_{2}} − d^{k_{2}−k_{1}} + 2q − 1,

that is, g′ as at most c zeros in the complex field.

Now we need a bound on the number of distinct zeros of g. Let α be a zero of g, then f^{k_{2}}(α) = α. By hypothesis P er_{k_{2}}(f) is empty, then α ∈ P er_{j}(f) and j divides a maximal proper divisor of k_{2}; those divisors are of the form k_{2}/p with p a prime dividing k_{2}, thus α is a zero for the polynomial f^{k_{2}/p}(x) − x for some p. Since this polynomial has degree d^{k_{2}/p}, the number of distinct zeros is bounded by the constant

∑_{p|k_{2}} d^{k_{2}/p}.

We need also a bound on the number of distinct solution of g(x) = 1 If α is such a point, then

f^{k_{2}−k_{1}}(α) = f^{k_{2}}(α) = f^{k_{1}}(f^{k_{2}−k_{1}}(α)),

that is f^{k_{2}−k_{1}}(α) ∈ P er_{k_{1}}(f). By hypothesis P er_{k_{1}}(f) is empty then f^{k_{2}−k_{1}}(α) ∈ P er_{j}(f) where j is a proper divisor of k_{1}; this implies that α is a zero of f^{k_{2}−k_{1}+j} − f^{k_{2}−k_{1}}, which is a polynomial of degree d^{k_{2}−k_{1}+j}. Then the number of solutions of g(x) = 1 is bounded by

∑_{j|k_{1}, j<k_{1}} d^{k_{2}−k_{1}+j} ≤ d^{k_{2}−1}.

Suppose now that α is either a zero of g or a solution of g(x) = 1 of order n. Derivating term by term, we have that α is a zero of g′ of order n − 1. As we have seen above the number of zeros of g′ is bounded by c = d^{k_{2}} − d^{k_{2}−k_{1}} + 2q − 1. Then the number of such α counted with their multiplicities is bounded by

d^{k_{2}} − d^{k_{2}−k_{1}} + d^{k_{2}−1} + 2q − 1 + ∑_{p|k_{2}} d^{k_{2}/p}.

Looking at the degree of the polynomials, we see that the number of such α counted with multiplicity is 2(d^{k_{2}} − d^{k_{2}−k_{1}} + q). This implies

2(d^{k_{2}} − d^{k_{2}−k_{1}} + q) ≤ d^{k_{2}} − d^{k_{2}−k_{1}} + d^{k_{2}−1} + 2q − 1 + ∑_{p|k_{2}} d^{k_{2}/p},

that is,

1 + d^{k_{2}} ≤ ∑_{p|k_{2}} d^{k_{2}/p} + d^{k_{2}−k_{1}} + d^{k_{2}−1}.

Since k_{2} ≥ 3 we have that

∑_{p|k_{2}} d^{k_{2}/p} ≤ d^{k_{2}−2}

and since k_{1} ≥ 2, we finally obtain

d^{k_{2}} < 1 + d^{k_{2}} ≤ d^{k_{2}−2} + d^{k_{2}−k_{1}} + d^{k_{2}−1} ≤ 2d^{k_{2}−2} + d^{k_{2}−1},

but this inequality is impossible for d ≥ 2. Then we have a contradiction and the theorem is proved.

The exception cited in the theorem is effective: for example the polynomial f(x) = x^{2} − x has not periodic points of period 2 (in [19] is eventually prooved that, up to conjugation, this is the only exception). The following corollary shows that the theorem is true also for every algebrically closed field of zero characteristic.

Corollary 2.3.4: If K is an algebrically closed field and f ∈ 𝕂[x] with deg f ≥ 2, then P er_{k}(f) is non-empty for all k but at most one exception.

Proof: Let E be the field generated over ℚ by the coefficients of f. Since it is finitely generated, there is an embedding of E in ℂ; then also the algebraic closure Ē is isomorphic to a subfield of ℂ. We can regard f as a polynomial with coefficients in ℂ via isomorphism. Then, by the theorem, P er_{k}(f) is infinite for all but at most one k, as a subset of ℂ; but P er_{k}(f) ⊆ Ē ⊆ K because K is algebrically closed. This establishes the theorem.

Now we drop the hypothesis of K algebrically closed and pass to study the more interesting case when K is a number field. In this case the previous theorem is no longer true and it can also happen that the set P er(f) is empty.

EXAMPLE 2.3.1 : Let K = ℚ and f(x) = x^{2} + x + 1. Calculating explicitely, it is easy to see that f has no periodic points of every order in K.

Our goal is giving a bound on the maximum integer k such that the set P er_{k}(f) is empty; our main tool will be the use of the number ring O_{K}, which will simplify the discussion. In fact, up to multiplying by a constant, which is a conjugation, every polynomial f ∈ K[x] can be seen as a polynomial in O_{K}[x].

We start with two lemmas extablishing some properties of polynomials on number rings.

Lemma 2.3.5: Let K be a number field and O_{K} be its number ring. Let f ∈ O_{K}[x] and r ∈ ℕ such that P er_{r}(f) is not empty. Then there exists a monic polynomial g ∈ O_{K}[x] having an r-periodic orbit which starts with the elements 0 and 1.

Proof: Let α ∈ P er_{r}(f) and {α = α_{0}, α_{1}, ..., α_{r−1}} its periodic orbit. We consider the polynomial

f_{1}(x) = f(x + α) − α ∈ O_{K}[x].

Putting β_{k} = α_{k} − α we have that {0, β_{1}, ..., β_{r−1}} is an r-periodic orbit for g. We now want to show by induction that every β_{j} is multiple of β_{1}. In fact this is obvious for k = 1 and, if β_{1}, ..., β_{k−1} are multiples of β_{1}, we write

f_{1}(x) = ∑_{j=0}^{d} a_{j}x^{j},

and clearly we have that a_{0} = β_{1} and β_{k} = g(β_{k−1}) ≡ 0 (mod β_{1}). Then all the β_{j}-s are multiples of β_{1}. Now we set f_{2}(x) = \frac{f_{1}(xβ_{1})}{β_{1}}. Then we have clearly f_{2} ∈ O_{K}[x] and has the r-periodic orbit \{0, 1, γ_{2}, ..., γ_{r−1}\}, where γ_{k} = \frac{β_{k}}{β_{1}} . Finally, we put

g(x) = f_{2}(x) + x^{m}(x − 1) ∏_{k=2}^{r−1}(x − γ_{k}),

where m is a sufficiently large integer such that g is monic. Then g is the polynomial we want.

Lemma 2.3.6 : Let f ∈ O_{K}[x] a monic polynomial of degree d ≥ 2 having the r-periodic orbit {0, 1, α_{2}, ..., α_{r−1}}. Then ∀k ∈ ℕ not divisible by r the differences α_{j+k} − α_{j} are associated (that is, they differ by the product of a unit) in O_{K}. In particular all the differences x_{j+1} − x_{j} are units in O_{K}.

Proof: By definitions we have ∀j

α_{j+k} − α_{j} = f(α_{j+k−1}) − f(α_{j−1}).

By Lagrange’s theorem, f(α_{j+k−1}) − f(α_{j−1}) = f(ξ)(α_{j+k−1} − α_{j−1}) and then we have

α_{j+k−1} − α_{j−1}|α_{j+k} − α_{j}.

Since this is true for all j and α_{0} = α_{r} = 0, we have

α_{k} = α_{k} − α_{0}|α_{k+1} − α_{1}|...|α_{k+r} − α_{r} = α_{k},

prooving that the differences are associated. Putting k = 1, since α_{1} = 1 we have that the differences α_{j+1} − α_{j} are units.

Corollary 2.3.7 : Under the same assumptions and notations of the lemma, we have: 1) If k|m then α_{k}|α_{m}.

2) If gcd(k, r) = 1 then α_{k} is a unit in O_{K}. In particular α_{r−1} is a unit.

3) If r is a prime, then all the α_{j} and their differences are units.

Proof: 1) By the lemma, α_{k} divides all the differences α_{mk} − α_{k} and then it divides also α_{mk} ∀m.

2) Let m be the inverse of k modulo r. Since the orbit is r-periodic we have α_{km} = α_{1} = 1; by part 1, α_{k}|α_{km} = 1 and then must be a unit.

3) Obvious from the other two properties.

Now we can give a general theorem which classifies the periodic points for monic polynomials in a large class of number fields.

Theorem 2.3.8 : Let K be a number field such that O_{K}^{∗} = ±1 and f ∈ O_{K}[x] a monic polynomial. Then the set P er_{k}(f) is empty for k ≥ 3.

Proof: Since O_{K} is integrally closed and f is monic, the periodic points of f must lie in O_{K}. Suppose that f has a r-periodic orbit {0, 1, ..., α_{r−1}} with r ≥ 3. By lemma 2.3.6, α_{j+1} − α_{j} ∈ {1, −1} and by part 2 of the corollary α_{r−1} = −1. Then we have α_{1} = 1, α_{r−1} = −1, the intermediate differences are unitary, but none of the intermediate terms is 0. This is a contradiction.

This theorem is valid in particular for ℚ and all the imaginary quadratic fields except ℚ(\sqrt{−1}) and ℚ(\sqrt{−3}). Then the knowledge of the set O_{K}^{∗} can give us some information about periodic points.

If K = ℚ we can say something more.

Theorem 2.3.9 : Let f ∈ ℤ[x] be a monic polynomial. Then one and only one of the following is true:

1) f has neither periodic nor fixed points.

2) f has one fixed point and k ≥ 0 periodic points of period 2.

3) f has k ≥ 1 fixed points and no periodic points of period 2.

4) f has k ≥ 1 periodic points of period 2 and no fixed points.

Proof: Let us suppose that f has at least a 2-periodic orbit {α_{0}, α_{1}} and two fixed points β and γ. By lemma 2.3.5, we may assume α_{0} = 0 and set α_{1} = α. Then we can write

f(x) = (x − β)(x − γ)g(x) + x,

with g ∈ ℤ[x] a suitable polynomial such that g(α) ≠ 0. We have

α = f(0) = βγg(0)

and

0 = f(α) = (α − β)(α − γ)g(α) + α = βγ((γg(0) − 1)(βg(0) − 1)g(α) + g(0)).

Since β, γ ≠ 0 we have that g(α) and g(0) must divide each other then g(α) = ±g(0) ≠ 0. This implies

0 = (γg(0) − 1)(βg(0) − 1) ± 1

and then γg(0) − 1 = ±1 and βg(0) − 1 = ±1 and, since g(0) ≠ 0, we must have

βg(0) = γg(0) = 2,

that is, β = γ and the theorem is prooved.

For quadratic number fields, a complete classification of periodic points is possible, too. We state here the complete result:

Theorem 2.3.10 : Let K = ℚ(\sqrt{m}) with m a squarefree integer and O_{K} its number ring. Let f ∈ O_{K}[x]. Then we have the following:

1) If m ≠ −3, −1, 2, 5, then P er_{k}(f) is empty for k ≥ 3.

2) If m = −1, 2, then P er_{k}(f) is empty for k = 3 and k ≥ 5.
3) If m = −3, then P er_{k}(f) is empty for k = 4, 5 and k ≥ 7.
4) If m = 5, then P er_{k}(f) is empty for k ≥ 5.

We will give a proof of this result, together with a stronger one, in paragraph 2.8

In case of any number field K and a non necessarily monic polynomial, the situation is much more complicated. A complete classification of periodic points is not known for fields of degree larger than 2. We state here the most important result, which we do not prove.

Theorem 2.3.11 : (Narkiewicz) Let K be a number field. There exists a constant B_{K}, depending only on the field K, such that the set P er_{k}(f) is empty for k > B_{K} for every f ∈ K[x].

Now we want to give an explicit bound for the constant B_{K}. Our argument will pass through the use of finite extensions of the p-adic field ℚ_{p}. In this case the number ring O_{K} is a discrete valuation domain and this fact implies a lot of consequence for periodic points.

We need a generalisation of lemma 2.3.6, vailing also for non-monic polynomials.

Lemma 2.3.12 : Let f ∈ O_{K}[x] of degree d and α ∈ P er_{r}(f) and denote α_{j} = f^{j}(α). The following are true:

i) ∃g ∈ O_{K}[x] with deg g ≤ r − 1 such that α ∈ P er_{r}(g). ii) If 0 < k < r then the differences α_{j+k} − α_{j} are associated. iii) If gcd(j − k, r) = 1 then α_{j} − α_{k} is associated to α_{1} − α. Proof: i) Take for g the remainder of the division of f by (x − α)...(x − α_{k−1}). ii) Apply Lagrange’s theorem as in lemma 2.3.6.

iii) Apply lemma 2.3.6 (this part does not require that f is monic).

Let now K be a finite extension of ℚ_{p}, O_{K} its ring of integers, P the only maximal ideal of O_{K} (which lies over p), and π a generator of P, s the inertia degree of P and e the ramification index. Firstly we need a definition:

Definition 2.3.13 : Let f ∈ K[x] and α ∈ P er_{r}(f). We say that the α is a *-periodic point if the differences (α_{j} − α_{k}) are not units in O_{K} for j ≠ k. The set of ∗-periodic point will be denoted as P er^{∗}(f) and B_{K}^{∗} will denote the constant which bounds the maximum length of a ∗-periodic orbit.

Lemma 2.3.14 : If α ∈ P er_{r}(f), then r is a product of primes not exceeding p^{s}.

Proof: We can easily reduce to the case r = q a single prime; in fact if r = q_{1}q_{2} then clearly α ∈ P er_{q_{2}}(f^{q_{1}}) and we can just examine a single prime.

Suppose that q > p^{s}. By part iii of the lemma 2.3.12, we have that α_{j} − α_{k} is associated to α_{1} − α ∀j ≠ k. Since O_{K} is a discrete valuation ring we have the equality of ideals (α_{j} − α_{k}) = (α_{1} − α) = (π)^{m}. If we choose k = 0, we can write, for 1 ≤ j < q, α_{j} = α_{0} + π^{m}η_{j} for some unit η_{j} and obviously η_{j} ≠ η_{k} if j ≠ k and η_{j} − η_{k} is still a unit. Then we have found q − 1 inequivalent non-zero elements of the residue field O_{K}/P, which has p^{s} − 1 non-zero elements. This is absurd. Then q ≤ p^{s}.

We can now enounce the first theorem about length of periodic orbits in local fields.

Theorem 2.3.15 : In the preceeding notations we have

B_{K}^{∗} ≤ (p^{s} − 1)p^{C(p)}

where

C(p) = 1 + \frac{log(e)}{log2}.

Proof: Let r be the length of a ∗-periodic orbit. We shall distinguish three cases:

1) r is equal to a prime q.

2) r is not a multiple of p.

3) r is a power of p.

Let us consider the case 1. Write f(x) = ∑_{j=0}^{d} a_{j}x^{j}. Let α ∈ P er_{q}^{∗}(f), with q prime. Conjugating by x − α, we may suppose that α = 0. Clearly we have

\frac{α_{j+2} − α_{j+1}}{α_{j+1} − α_{j}} = a_{d}(α_{j+1}^{d−1} + ... + α_{j}^{d−1}) + ... + a_{2}(α_{j+1} + α_{j}) + a_{1} ≡ a_{1} (mod P),

therefore

1 = ∏_{j=1}^{q} \frac{α_{j+2} − α_{j+1}}{α_{j+1} − α_{j}} ≡ a_{1}^{q} (mod P).

This implies

a_{1}^{q_{1}} ≡ 1 (mod P),

where q_{1} = gcd(q, p^{s} − 1). Since q is prime, either q|p^{s} − 1 or a_{1} ≡ 1 (mod P). Suppose we are in the second case and write (α) = (π^{m}). Setting j = 0, we have

\frac{α_{2} − α_{1}}{α_{1} − α_{0}} ≡ 1 (mod P),

and α_{2} − α_{1} ≡ α_{1} − α (mod P^{m+1}). Similarly ∀j we have α_{j+2} − α_{j+1} ≡ α_{j+1} − α_{j} ≡ ... ≡ α_{1} − α (mod P^{m+1}). Then

0 = ∑_{j=1}^{q} α_{j+1} − α_{j} ≡ q(α_{1} − α) (mod P^{m+1}).

Then the only possibility is q = p.

Let us now consider the case 2 and let α ∈ P er_{r}^{∗}(f). Up to a conjugation, we may suppose α = 0 and α_{1} = π. Set

c = \frac{α_{2} − α_{1}}{α_{1} − α} ,

then

\frac{α_{j+1} − α_{j}}{α_{j} − α_{j−1}} ≡ c (mod P)

and it is immediate to show by induction that

α_{j} ≡ (1 + c + ... + c^{j−1})π (mod P^{2}).

Let l be the minimum positive integer such that α_{l} ∈ P^{2} and suppose l < r. Then clearly c^{l} ≡ 1 (mod P) and l|r since α_{r} = 0 ∈ P^{2}. If (α_{l}) = (π^{m}) we write

f^{m}(x) = ∑_{j=0}^{t} b_{j}x^{j}.

We have

b_{1} ≡ (f^{l})′(0) ≡ c^{l} ≡ 1 (mod P)

then

α_{(j+2)l} − α_{(j+1)l} ≡ α_{(j+1)l} − α_{jl} (mod P^{m+1})

and

0 = ∑_{j=1}^{r/l}(α_{(j+1)l} − α_{jl}) ≡ \frac{r}{l} (α_{l} − α) (mod P^{m+1})

that is absurd.

Then the only possibility is l = r and α_{1}, ..., α_{r−1} ∉ P^{2}. Suppose now that r does not divide p^{s} − 1, then we have c^{r_{1}} ≡ 1 (mod P), where r_{1} = gcd(r, p^{s} − 1). Using the previous formula and since c ≢ 1 (mod P), we have α_{r_{1}} ∈ P^{2}, which is absurd. Then r|p^{s} − 1.

Consider now the case 3 and suppose r = p^{n} and let α ∈ P er_{r}^{∗}(f). Again we can assume α = 0 and α_{1} = π. For 0 ≤ j ≤ n − 1 we set (α_{p^{j}}) = (π^{m_{j}}) (in particular m_{0} = 1) and l_{j} = (f^{p^{j}})′(0). If j ≤ n − 1 we have

1 = ∏_{k=1}^{p^{n−j}} \frac{α_{(k+1)p^{j}} − α_{kp^{j}}}{α_{kp^{j}} − α_{(k−1)p^{j}}} ≡ l_{j}^{p^{n−j}} (mod P), l_{j} ≡ 1 (mod P).

We can write l_{j} = 1 + η_{j}π^{c_{j}}, where η_{j} is a unit; if l_{j} = 1 we put c_{j} = ∞. We can easily obtain by induction that

α_{p^{j+1}} ≡ (1 + l_{j} + ... + l_{j}^{p−1})α_{p^{j}} (mod P^{2m_{j}}).

Suppose now that m_{j+1} < 2m_{j}; then if l_{j} = 1 then, we have m_{j+1} = m_{j} + e, where e is the ramification index of p. On the other hand, if l_{j} ≠ 1 we have

α_{p^{j+1}} ≡ \frac{(1 + η_{j}π^{c_{j}})^{p} − 1}{η_{j}π^{c_{j}}}α_{p^{j}} (mod P^{2m_{j}}).

Using Newton’s binomial formula for the numerator of this expression we obtain that m_{j+1} ≥ min(m_{j} + e, m_{j} + (p − 1)c_{j}). We arrive at the formula

m_{j+1} ≥ min(2m_{j}, m_{j} + e, m_{j} + (p − 1)c_{j}).

If j = n − 1 we have that, if (p − 1)c_{n−1} ≠ e then

m_{n−1} ≤ e,

otherwise

(p − 1)c_{n−1} = e.

If j ≤ n − 2 we have

l_{j+1} = (f^{p^{j+1}})′(0) = ∏_{k=0}^{p−1}(f^{p^{j}})′(α_{kp^{j}}) ≡ l_{j}^{p} (mod P^{m_{k}})

and this implies

c_{j+1} ≥ min(m_{j}, c_{j} + e, pc_{j}).

In the special case p = 2 we have a stronger inequality. Since

l_{j+1} ≡ l_{j}(l_{j} + (f^{p^{j}})′′(0)α_{p^{j}}) (mod P^{2m_{j}})

and 2|(f^{p^{j}})′′(0) we have

c_{j+1} ≥ min(2m_{j}, c_{j} + e, 2c_{j}, m_{j} + e).

Suppose that for some 0 ≤ j ≤ n − 1 we have min(m_{j}, c_{j}) > e; then the preceding inequalities would imply m_{n−1} > e and c_{n−1} > e. But we have shown that either m_{n−1} ≤ e or c_{n−1}(p − 1) = e, therefore we have a contradiction. Then for every j = 0, ..., n − 1 we have min(m_{j}, c_{j}) ≤ e.

Using the relations we have obtained, it is immediate to show the following facts: i) If m_{j} ≤ e then m_{j} ≥ 2^{j}.

ii) If p is odd then c_{j} ≥ 2^{j−1}.
iii) if p = 2 then c_{j} ≥ 2^{j}.

Finally, combining the all the results, we have that

n ≤ C(p) = 1 + \frac{log(e)}{log2}.

We can now end the proof of the theorem. Combining the cases 1, 2 and 3 we have shown that the length of every ∗-periodic orbit divides the constant

B_{K}^{∗} = (p^{s} − 1)p^{C(p)},

proving the theorem.

The importance of this result is that we can relate the length of all periodic orbits to the ∗-periodic ones.

Lemma 2.3.16 : If α ∈ P er_{r}(f) then r = ab, where a is the length of a ∗-periodic orbit and b ≤ p^{s}.

Proof: Let b be the minimum positive integer such that α_{b} − α is not a unit of O_{K}. Suppose that b does not divide r. Then, by unique factorization r = mb+q with 0 < q < b. By part ii of the lemma 2.3.12, α_{b−q} − α is associated to α_{(m+1)b} − α_{r} = α_{(m+1)b} − α = (α_{(m+1)b} − α_{mb}) + ... + (α_{b} − α) which is not a unit. Then b − q contradicts the minimality of b. Then we have q = 0 and therefore b|r. Suppose that there exist integers 1 ≤ j < k < b such that α_{j} − α ≡ α_{k} − α (mod P), that is, α_{k} − α_{j} is not a unit; but, by lemma 2.3.12, α_{k} − α_{j} is associated to α_{k−j} − α which is a unit by the definition of b, j, k. This is absurd. Then we have that the differences α_{j} − α 1 ≤ j ≤ b − 1 are b − 1 inequivalent non-zero elements of R/P. Therefore b ≤ p^{s}. Finally the set {α, α_{b}, α_{2b}, ..., α_{(a−1)b}} is a ∗-periodic orbit, by definition of b. Then a is the length of a ∗-periodic orbit.

This lemma and the theorem about the ∗-periodic orbits imply the following.

Theorem 2.3.17 : In the previous notations we have

B_{K} ≤ p^{s}B_{K}^{∗} ≤ p^{s}(p^{s} − 1)p^{C(p)}

where C(p) is defined as before.

Let us now come back to the global case. The valor we found for B_{K} in local fields enables us to give a valor in the global case. We consider K a number field with n = [K : ℚ] and O_{K} its number ring. Let P ∈ O_{K} a prime lying over 2 and e and s its ramification index and inertia degree. We have the following result.

Theorem 2.3.18 : If K is a number field, in the previous notations we have

B_{K} = (2^{n} − 1)2^{n+1}.

Proof: We define the localized ring R = (O_{K})_{P}; since R contains an isomorphic copy of O_{K}, the length of periodic orbits in O_{K} is bounded by the length in R. R is a discrete valuation ring so we can apply to R the results for local field having B_{K} ≤ 2^{s}(2^{s} − 1)2^{C(2)}. But in this ring there cannot be a ∗-periodic orbit whose length is a power of 2 and superior to 2^{e}. Then, using the degree formula, we have

B_{K} ≤ 2^{s}(2^{s} − 1)2^{e} ≤ 2^{n/e}(2^{n/e} − 1)2^{e} ≤ 2^{n+1}(2^{n} − 1).

2.4 Canonical heights and preperiodic points

Our next aim is studying preperiodic points, that is, points whose iterates are periodic. Our first step shall be describing a new type of height function, which will be useful to show the dynamical properties of polynomials and to classify some categories of preperiodic points.

We start from the following fact, whose proof can be found in [11].

Lemma 2.4.1 : Let f ∈ ℚ be a polynomial map with degf = d ≥ 2 and h the height function on ℚ̄. Then ∀α ∈ ℚ̄ we have:

h(f(α)) = dh(α) + O(1)

where O(1) depends only on the map f and not on α and is bounded in terms of f only.

Our next step is to define a new height function for which the O(1) term entirely disappears.

Definition 2.4.2 : Let f ∈ ℚ be a polynomial with deg f = d ≥ 2 and h be the height function on ℚ̄. The canonical height function associated to f is a real-valued function

ĥ_{f} : ℚ̄ → ℝ^{+}

which satisfies the following conditions:

1) ∀α ∈ ℚ̄ ĥ_{f}(α) = h(α) + O(1).
2) ∀α ∈ ℚ̄ ĥ_{f}(f(α)) = dĥ_{f}(α).

Theorem 2.4.3 : ∀f ∈ ℚ̄[x] with deg f = d ≥ 2 there exists a unique canonical height function associated to f. This function is defined by:

ĥ_{f}(α) = lim_{n→∞} \frac{h(f^{n}(α))}{d^{n}}.

Proof: Let’s start with the existence. We define ĥ_{f} by the limit formula. First of all we proove that this limit exists, showing that the sequence \frac{h(f^{n}(α))}{d^{n}} is Cauchy. Let C be an upper bound for the O(1) in lemma 2.4.1. For n ≥ m integers we have:

|\frac{h(f^{n}(α))}{d^{n}} − \frac{h(f^{m}(α))}{d^{m}}| = | ∑_{j=m+1}^{n} \frac{h(f^{j}(α))}{d^{j}} − \frac{h(f^{j−1}(α))}{d^{j−1}}|.

Now we set β_{j} = f^{j−1}(α) and we have:

| ∑_{j=m+1}^{n} \frac{h(f^{j}(α))}{d^{j}} − \frac{h(f^{j−1}(α))}{d^{j−1}}| ≤ ∑_{j=m+1}^{n} \frac{|h(f(β_{j})) − dh(β_{j})|}{d^{j}} ≤ ∑_{j=m+1}^{n} \frac{C}{d^{j}} ≤ \frac{C}{d^{m}(d − 1)}.

This inequality shows that the sequence is Cauchy and then the limit exists. Now we have to show that this function satisfies the two properties of the canonical height. For property 1), we put m = 0 in the previous estimate, obtaining:

|\frac{h(f^{n}(α))}{d^{n}} − h(α)| ≤ \frac{C}{d − 1}.

Taking the limit for n going to infinity we have:

|ĥ_{f}(α) − h(α)| ≤ \frac{C}{d − 1}

and property 1) is satisfied. For property 2), we have by definition:

ĥ_{f}(f(α)) = lim_{n→∞} \frac{h(f^{n}(f(α)))}{d^{n}} = d lim_{n→∞} \frac{h(f^{n+1}(α))}{d^{n+1}} = dĥ_{f}(α)

and so both the properties are satisfied and we have proved the existence. For the uniqueness, suppose that ĥf^{′} is another canonical height associated to f. Then the difference g = ĥ_{f} − ĥf^{′} is bounded by property 1) and g(f(α)) = dg(α) by property 2). Hence

∀n ∈ ℕ d^{n}|g(α)| = |g(f^{n}(α))| ≤ O(1).

By hypothesis d > 1; then the only possibility is g(α) = 0, but α is arbitrary so ĥ_{f} = ĥf^{′} and we have proved the uniqueness.

We have also a notion of local canonical height, which decomposes the canonical height in a sum of functions associated to the different places.

Definition 2.4.4 : Let K be a characteristic zero field and |.|_{v} a place on K and f ∈ K[x] with degf = d. The v-adic local canonical height associated to f is the unique function ĥ_{f,v} : ℂ_{v} → ℝ defined as

ĥ_{f,v}(α) = lim_{n→+∞} \frac{1}{d^{n}} log^{+}|f^{n}(α)|_{v}.

The proof of the existence and uniqueness of the local canonical height is identical to the global ones. Now it is an immediate calculation to show that

ĥ_{f}(α) = \frac{1}{r} ∑_{v∈M_{K}} ∑_{j=1}^{r} ĥ_{f,v}(α_{j}),

for all α ∈ K̄, where α_{1}, ..., α_{r} are the conjugate of α over K. Morevoer the local canonical height satisfies itself the formula ĥ_{f,v}(f(α)) = d f̂_{f,v}(α).

The main importance of canonical height in studying polynomial dynamics is given by the following lemma.

Lemma 2.4.5: Let K be a number field, α ∈ K^{∗} and f ∈ K[x] with deg f = d ≥ 2. Then α is preperiodic for f if and only if ĥ_{f}(α) = 0. Moreover Π(f) is a set of bounded height so it contains only finitely many points defined over all extensions of K of bounded degree.

Proof: Let’s first suppose that α is preperiodic. We have:

ĥ_{f}(α) = \frac{ĥ_{f}(f^{n}(α))}{d^{n}} = \frac{h(f^{n}(α)) + O(1)}{d^{n}},

but, since α is preperiodic, the set {h(f^{n}(α))}_{n∈ℕ} takes on only finitely many distinct values so, taking the limit for n going to infinity, we have ĥ_{f}(α) = 0. Conversely, if ĥ_{f}(α) = 0, then

h(f^{n}(α)) = ĥ_{f}(f^{n}(α)) + O(1) = d^{n}ĥ_{f}(α) + O(1) = O(1).

Hence the orbit of α is a set of bounded height, so it is finite. Therefore α is preperiodic. To show the second part of the lemma, we recall that the O(1) in the precedeing formula does not depend on α but only on f. So the height of every preperiodic point is bounded independently of the point. Then Π(f) is a set of bounded height and the final part of the lemma follows from Northcott’s theorem 1.4.5.

This property vails also for all the local canonical heights, but since they are defined on p-adic number fields, they are not useful for calculation. Anyway even the local canonical height has interesting dynamical properties; we will explore them in section 2.6.

2.5 Cyclotomic preperiodic points

In this chapter and in the following ones, we will deal more deeply with the study of preperiodic points. Lemma 2.4.5 assures us that there are only finitely many preperiodic points of bounded degree on K, but nothing assures us that they are globally finite; in fact this is generally not true.

Our first step is to prove that, at least in the case of K algebrically closed and complete, Π(f) is always an infinite set. We need a lemma about some particular points of f.

Lemma 2.5.1 : Let f ∈ K[x] be a polynomial of degree ≥ 2, with K algebrically closed and complete and X ⊆ K a completely invariant finite set. Then X contains at most two elements.

Proof: We may suppose K = ℂ. We recall the Riemann-Hurwitz formula, which says that, if f is non-constant,

∑_{α∈ℂ}(m_{f}(α) − 1) = 2deg(f) − 2,

where m_{f}(α) is the unique integer m such that the limit

lim_{x→α} \frac{f(x) − f(α)}{(x − α)^{m}}

exists, is finite and non-zero. (a proof of this formula may be found in [3]). If k is the number of elements of X, by completely invariance, f must act on X as a permutation, therefore has a finite order q ≥ 1 and degree d ≥ 2. Then, ∀α ∈ X, the equation f^{q}(x) = α has d solutions, all equals to α. By the Riemann-Hurwitz formula applied to f^{q} we have

k(d − 1) ≤ 2d − 2

and since d ≥ 2, we must have k ≤ 2.

Theorem 2.5.2 : Let f ∈ K[x] be a polynomial of degree d ≥ 2 with K algebrically closed and complete. Then f has infinitely many distinct preperiodic points in K.

Proof: The set Π(f) of preperiodic point is clearly completely invariant. If it were finite, there should be at most two preperiodic points. A rapid calculus shows that this is impossible; in fact we may suppose that they are 0 and 1, then f(x) − x = x^{d} and f(x) = x^{d} + x. but the equation f(x) = 1 as at least a nonzero root, which is still preperiodic. This is absurd.

To find some results of finiteness we will look for preperiodic points of a particular form, that is, that lies in some particular extensions of K. The first strong result is obtained in cyclotomic extensions.

Definition 2.5.3 : Let K be a number field. The cyclotomic closure of K, denoted by K^{c}, is the extension obtained by adding to K all the roots of unity.

An interesting result, whose proof can be found in [2], is the following:

Theorem 2.5.4 : If f is a polynomial map such that Π(f) contains infinitely many roots of unity, then f(x) = x^{d} for some d ≥ 2.

In the rest of the section we will proove the main theorem of finiteness about cyclotomic preperiodic points, firstly proved by Dvornicich and Zannier in [8]. We start with a definition

Definition 2.5.5 : Let α ∈ K. We define the quantity

C(α) = max_{σ}|σ(α)|,

where σ ranges over all embeddings of K in ℚ̄, to be the containing of α.

One of the main tools in this proof is the following Loxton’s theorem, whose proof can be found in [16].

Theorem 2.5.6 (Loxton) :Let α ∈ ℚ^{c} be a cyclotomic integer. Then we can write

α = ∑_{j=1}^{b} ζ_{j},

where the ζ_{j} are roots of unity and b is an integer such that b ≤ L(C(α)), where L(x, b) = L(x) is a Loxton’s function, given by

L(x) = x^{2}exp(\frac{log(b)}{loglog(b)}).

We need an extension of this theorem to the case of any number field K.

Theorem 2.5.7 : Let K be a number field and α ∈ K^{c} a cyclotomic integer. Then there exists a constant B and a finite set E ⊆ K with #E ≤ [K : ℚ], both depending only on K, such that we can write

α = ∑_{j=1}^{b} c_{j}ζ_{j},

where c_{j} ∈ E, ζ_{j} are roots of unity and b ≤ #EL(BC(α)) with L a Loxton’s function.

Proof: Let {ω_{1}, ...ω_{r}} an integral basis of K/(K∩ℚ^{c}). Since the extension ℚ^{c}/(K∩ℚ^{c}) is Galois then this is a basis also for Kℚ^{c}/ℚ^{c} and we may write

α = ∑_{j=1}^{r} γ_{j}ω_{j}

with γ_{j} ∈ ℚ^{c}.

Let {ω_{1}^{∗}, ..., ω_{r}^{∗}} be the dual basis with respect to the traceT r_{K/(K∩ℚ^{c})}. We have

T r_{Kℚ^{c}/ℚ^{c}}(ω_{j}ω_{k}^{∗}) = T r_{K/(K∩ℚ^{c})}(ω_{j}ω_{k}^{∗}) = δ_{jk}.

Using the relation for α we have

T r_{Kℚ^{c}/ℚ^{c}}(ω_{j}^{∗}α) = T r_{Kℚ^{c}/ℚ^{c}}(ω_{j}^{∗} ∑_{k=1}^{r} γ_{k}ω_{k}) = ∑_{k=1}^{r} γ_{k}T r_{Kℚ^{c}/ℚ^{c}}(ω_{j}^{∗}ω_{k}) = γ_{j}.

Let m be the minimum integer such that mω_{j}^{∗} is an algebraic integer for all j. Since the basis is fixed, d depends only on the field K. Clearly mγ_{j} = T r_{Kℚ^{c}/ℚ^{c}}(mω_{j}^{∗}α), hence the mγ_{j} are algebraic integers in ℚ^{c}. Then we can apply Loxton’s theorem to these numbers, obtaining

mγ_{j} = ∑_{k=1}^{b_{k}} ζ_{jk},

where the ζ_{jk} are roots of unity and b_{k} positive integers such that

b_{k} ≤ L(C(mγ_{j})).

Then we have

α = ∑_{j=1}^{r} γ_{j}ω_{j} = ∑_{j=1}^{r} ∑_{k=1}^{b_{k}} \frac{ω_{j}}{m} ζ_{jk}.

This is the representation we need if we set E = \{\frac{ω_{1}}{m} , ..., \frac{ω_{r}}{m} \}; we just need to check for the appropriate B. We have

b ≤ ∑_{k=1}^{r} b_{k} ≤ ∑_{k=1}^{r} L(C(mγ_{j}))

and its immediate that C(γ_{j}) ≤ rC(α)max_{j}C(ω_{j}^{∗}) = m′C(α) where m′ is a positive integer depending only on K. We finally have

b ≤ rL(C(mm′α))

prooving our result with B = mm′.

Before stating the theorem, we still need the definition of a particular class of polynomials.

Definition 2.5.8 : The d-th Chebishev polynomial is the unique polynomial T_{d} which, ∀α ∈ ℚ̄, satisfies the realation

T_{d}(α + α^{−1}) = α^{d} + α^{−d}.

Theorem 2.5.9 (Dvornicich-Zannier): Let f be a polynomial of degree d ≥ 2 defined over a number field K. Then the set Π(f)∩K^{c} is finite unless, for some ϵ = ±1, f is conjugated either to (ϵx)^{d} or to T_{d}(ϵx).

Proof: First of all, we recall that we can find a suitable conjugation g such that g ∘ f ∘ g^{−1} is monic and with vanishing second coefficient (Fatou form of f), that is

g ∘ f ∘ g^{−1}(x) = x^{d} + a_{d−2}x^{d−2} + ... + a_{0}.

Since Π(g ∘ f ∘ g^{−1}) = g(Π(f)), we may assume that f is itself in the Fatou form. If α ∈ Π(f), we want to show that the quantity C(α) is bounded independently on α.

Lemma 2.5.10 : There exists a constant C, depending only on f, such that ∀α ∈ Π(f) C(α) ≤ C

Proof: Let C = max_{σ}{2 + |σ(a_{0})| + ... + |σ(a_{d−2})|}. Suppose that, for some σ, |σ(α)| > C. Let f_{σ} be the polynomial f_{σ}(x) = ∑_{j=0}^{d} σ(a_{j})x^{j}; we want to show by induction that the sequence |f_{σ}^{k}(σ(α))| is strictly increasing. The case k = 0 is trivial so we can pass to the inductive step; but

|f_{σ}^{k+1}(σ(α))| = |f_{σ}^{k}(f_{σ}(σ(α)))| > |f_{σ}^{k−1}(f_{σ}(σ(α)))| = |f_{σ}^{k}(σ(α))|.

We have proved that this sequence is increasing. On the other hand σ(α) ∈ Π(f_{σ}) and so the sequence can assume only finitely many values. This is impossible.

Now we want to prove that there exists a positive integer D, depending only on f, such that ∀α ∈ Π(f) Dα is an algebraic integer. In fact, let |.|_{v} be a non-archimedean absolute value on K. If |α|_{v} > max(1, |a_{j}|_{v}), following the argument of the previous lemma, we can show by induction that the sequence |f^{k}(α)|_{v} should be strictly increasing and assume only finitely many values, which is impossible. Then |α|_{v} < max(1, |a_{j}|_{v}). Let D be a positive integer such that Da_{j} is an algebraic integer for every j. Then

|Dα|_{v} ≤ max(|D|_{v}, |Da_{j}|_{v}) ≤ 1

and D is the integer we wanted.

Since now we have no assumption on Π(f)∩K^{c}. From now on we suppose that |Π(f)∩K^{c}| = ∞ and derive that f is of the wanted form. We need a lemma with a particular form of theorem 2.5.7 in our case:

Lemma 2.5.11 : There exists a positive constant b and a finite subset E ⊆ K, both depending only on f and K, such that, ∀α ∈ Π(f) ∩ K^{c} and every k ∈ ℕ we can write

f^{k}(α) = ∑_{j=1}^{b} c_{j}ζ_{j}

where c_{j} ∈ E and ζ_{j} are roots of unity.

Proof: By the preceding observation, there is a positive integer D such that Dα is an algebraic integer and C(Dα) ≤ CD. But Dα ∈ K^{c} then we can apply theorem 2.5.7 to α and obtain the desired representation.

The last lemma suggests to define the set

J_{b,E} = {∑_{j=1}^{b} c_{j}ζ_{j}|c_{j} ∈ E, ζ_{j} ∈ ℚ_{ab}}.

From the two preceding lemmas, it follows that this set satisfies the following properties:

1) There exists a positive integer D′, depending only on f and K, such that ∀ϕ ∈ J_{b,E} D′ϕ is an algebraic integer.

2) There exists a positive constant C′, depending only on f and K, such that ∀ϕ ∈ J_{b,E} C(D′ϕ) ≤ C′.

Now let M be an integer such that d^{M−2} > C^{′2} and α ∈ Π(f) ∩ K^{c}. By the lemma we may write

f^{k}(α) = ∑_{j=1}^{b} c_{jk}ζ_{jk} k = 0, 1, ..., M

where the c_{jk} lie in a finite set E. All these quantities but the c_{jk} and the ζ_{jk} depend only on f and K, not on α.

Let us now fix a subset of b(M +1) elements of E, namely {c_{jk}} for j = 1, ..., b k = 0, ..., M. Since we are assuming that Π(f) ∩ K^{c} is infintite, while the set E is finite, there must be at least one infinite subset Π′ ⊂ Π(f) ∩ K^{c} such that ∀α ∈ Π′ the c_{jk} are the ones we have fixed and do not depend on α.

From the former equation we have

α = f^{0}(α) = ∑_{j=1}^{b} c_{j0}ζ_{j0}

so we can substitute this expression for α in the generic equation. Then we have

f^{k}(∑_{j=1}^{b} c_{j0}x_{j0}) = ∑_{j=1}^{b} c_{jk}x_{jk} ∀k = 1, ..., M;

this is a system of M algebraic equations satisfied by the roots of unity ζ_{jk}. If α ∈ Π′ the coefficients of the system do not depend on α, but the solutions does.

Before going on, we have to recall a well known theorem of algebraic geomerty, whose proof may be found in [4].

Theorem 2.5.12 (Torsion points theorem) : The Zariski closure X′ of the torsion points of a subvariety X is a finite union of torsion cosets, that is, of set βY where Y is a linear irreducible subvariety of X and β a torsion point of X.

Let us know get back to our theorem. We set X ⊆ 𝔾_{m}^{b(M+1)} be the subvariety generated by our system of equations; then the solutions of the system are torsion points in X. Since every α ∈ Π′ gives a different solution and Π′ is infinite, X has infinitely many torsion points. By the torsion points theorem the Zariski closure X′ of these torsion points is a finite union of torsion cosets of 𝔾_{m}^{b(M+1)}.

We consider now the map ψ : X′ → K ψ(ζ_{jk}) = ∑_{j=1}^{b} c_{j0}ζ_{j0}; this map is clearly dominant because the image contains the infinite set Π′. Therefore X′ must contain at least one torsion coset βY not sent to a constant by ψ.

Let x_{jk} = ξ_{jk}t^{e_{jk}} be a parametrization of this coset, with ξ_{jk} roots of unity and e_{jk} ∈ ℤ. Substituting in the system we have

f^{k}(∑_{j=1}^{b} c_{j0}ξ_{j0}t^{e_{j0}}) = ∑_{j=1}^{b} c_{jk}ξ_{jk}t^{e_{jk}} ∀k = 1, ..., M.

Now if we set q(t) = ∑_{j=1}^{b} c_{j0}ξ_{j0}t^{e_{j0}} we have that q(t) ∈ K[t, t^{−1}] and it is not constant because ψ(βY) is not constant. Ordering the terms we can write

q(t) = q_{0}t^{m_{0}} + q_{1}t^{m_{1}} + ... + q_{l}t^{m_{l}},

where q_{j} ∈ J_{b,E} and m_{i} ∈ ℤ |m_{0}| > |m_{1}| > ... > |m_{l}|. We may also suppose that m_{0} ≥ |m_{l}| (the case |m_{l}| ≥ m_{0} is symmetric).

Suppose now that q(t) is not of the shape αt^{m} + βt^{−m} α, β ∈ K^{c}. Then we have l > 0 and m_{0} + m_{1} > 0. We want to show that under our hypothesis this cannot happen.

Calculating explicity, we have that

f(q(t)) = (q(t))^{d} + o.o.t. = q_{0}^{d}t^{dm_{0}} + dq_{0}^{d−1}q_{1}t^{(d−1)m_{0}+m_{1}} + o.o.t.

and that

f^{k}(q(t)) = q_{0}^{d^{k}}t^{d^{k}m_{0}} + d^{k}q_{0}^{d^{k}−1}q_{1}t^{(d^{k}−1)m_{0}+m_{1}}.

Now we denote by ϕ_{0}(k) = q_{0}^{d^{k}} and ϕ_{1}(k) = d^{k}q_{0}^{d^{k}−1}q_{1}. Comparing with the preceeding equations, we note that ϕ_{0}, ϕ_{1} ∈ J_{b,E} and that ϕ_{0}(k)q_{1} = q_{0}ϕ_{1}(k)/d^{k}. By property 1 and 2 of the set J_{b,E}, we have that, ∀k, D′^{2}q_{0}ϕ_{1}(k)/d^{k} is an algebraic integer and then 1 ≤ C(D′^{2}q_{0}ϕ_{1}(k)/d^{k}) ≤ C′^{2}/d^{k}. Setting k = M we have d^{M} ≤ C′^{2}, but this is impossible for our choice of M. Then q(t) is of the desired shape.

The same argument show us also that f(q(t)) and f^{2}(q(t)) are of this shape. Finally we have to prove that these three cases imply the general one. For simplicity we denote by f^{s}(q(t)) = a_{s}t^{m_{s}} + b_{s}t^{−m_{s}} for s = 0, 1, 2.

First of all, if b_{0} = 0, q(t) is a monomial and then f(q(t)) must be a polynomial. So b_{1} = b_{2} = 0 and f(q(t)) and f^{2}(q(t)) are monomials; then the only possibility is that f is a monomial himself and we have the conclusion. For simmetry, the same is true if a_{0} = 0.

Suppose now that a_{0}b_{0} ≠ 0. Then also a_{1}b_{1} ≠ 0 and a_{2}b_{2} ≠ 0. By calculation we have m_{1} = m_{0}d and m_{2} = m_{0}d^{2}. For simplicity we may assume m_{0} = 1.

Let λ = \sqrt{a_{0}b_{0}} and u = a_{0}t/λ. Then we have

q(t) = λ(u + u^{−1})

and, by symmetry,

f^{s}(q(t)) = λ^{d^{s}}(u^{d^{s}} + u^{−d^{s}}) s = 0, 1, 2.

Now we define F (x) = λ^{−d}f(λx). Substituting for f the preceding expression, we find F (u + u^{−1}) = u^{d} + u^{−d} and then F is the d-th Chebishev polynomial. On the other hand, defining G(x) = λ^{−d^{2}}f(λ^{d}x) and substituting for f, we find again that G must be the d-th Chebishev polynomial.

Comparing the equalities for F and G we have

T_{d}(λ^{d−1}x) = λ^{d^{2}−d}T_{d}(x).

Now a simple asyntotic approximation argument shows that λ^{d} = ϵλ and λ^{d^{2}} = (ϵλ)^{d}, with ϵ = ±1. Therefore λ^{d−1} = ±1.

Finally we have f(x) = λ^{d}T_{d}(λ^{−1}x) = ϵλT_{d}(λ^{−1}x). So we put g(x) = ϵλ^{−1}x and we have

g ∘ f ∘ g^{−1}(x) = T_{d}(ϵx)

and then f is conjugated to the d-th Chebishev polynomial and we have finished.

This result is effective; not only it fixes the infiniteness case, but also exhibits the infinite preperiodic points: they are all the roots of unity in the case of a monomial and all the numbers ζ + ζ^{−1}, with ζ a root of unity, in case of a Chebishev polynomial.

2.6 Totally v-adic preperiodic points

In this section we look for preperiodic points contained in particular extensions built by using the non-archimedean absolute values. An important role in this part is played by the canonical height, for which we shall give a version of Lehmer’s conjecture. We start with a definition of a particular extension of K.

Definition 2.6.1: Let K be a field and |.|_{v} a non-archimedean place on K. We say that an extension L of K is totally v-adic of type (e,s) if, for any embedding σ : L → ℂ_{v}, we have that σ(L) is contained in a finite extension of K_{v} whose ramification index and inertial degree are bounded respectively by e and s. In the same way, we say that an element α ∈ K̄ is totally v-adic of type (e,s) if the field K(α) is; if e = s = 1 we briefly say that α is totally v-adic.

The proof of the main theorems of this section makes use of the measure and equidistribution theory. We are not going to explore that theory in detail, then we describe here the main definition and results we are going to use for our proofs. The details can be found in [1].

Definition 2.6.2 : Let A be a subset of ℂ_{v}; the diameter of A is the supremum of the distances between any pair of elements of A, that is

diam(A) = sup_{α,β∈A}|α − β|.

More generally we can define the n-th diameter of A as the supremum over all n-tuples of elements of A of the geometric mean of the pairwise distances between elements of a given n-tuple, that is

diam_{n}(A) = sup_{α_{1},...,α_{n}∈A} ∏_{j≠k} |α_{j} − α_{k}|^{\frac{1}{n(n−1)}} .

Finally, the transfinite diameter of A is the limit

td(A) = lim_{n→+∞}diam_{n}(A).

EXAMPLES 2.6.1 : 1) Let K be a local field and O_{K} its number ring. Then we have

td(O_{K}) = p^{\frac{−1}{e(p^{s}−1)}} ,

where p is the only prime ramified in K and e, s are its ramification index and inertial degree.

2) Let f(x) = ∑_{j=0}^{n} a_{j}x^{j} ∈ ℂ_{v} be a polynomial of degree d ≥ 2. Then

td(F J_{v}(f)) = |a_{d}|v^{\frac{−1}{d−1}} .

3) Let K be a field with product formula. By using the preceeding result and the product formula we have

∏_{v∈M_{K}} td(F J_{v}(f))^{[K_{v}:ℚ_{p}]} = 1

Now we introduce some important measures

Definition 2.6.3 : Let K be a field and |.|_{v} be a place on K (archimedean or not); let A be a compact subset of ℂ_{v} such that td(A) ≠ 0. There exists a unique probability measure μ_{A} supported on A (that is, μ_{A}(B) = 0 if B ∩ A is the empty set) that minimizes the double integral

∫_{A} ∫_{A} −log|α − β|_{v} dμ(α)dμ(β).

The measure μ_{A} is called the equilibrium measure for A and this integral turns out to be equal to −log(td(A)). Moreover, if ℂ_{v} = ℂ and A = F J_{v}(f) with f ∈ ℂ[x], then the equilibrium measure is f-invariant.

Definition 2.6.4 : Let A be a finite subset of ℂ_{v}. The Dirac measure on A is the probability measure

δ_{A} = \frac{1}{#A} ∑_{α∈A} δ_{α},

where δ_{α} is the Dirac measure that is 1 at α and 0 everywhere else.

We also need the definition of a convergence between measures.

Definition 2.6.5 : Let M be a metric space (in our applications M will be ℂ_{v} or ℂ). A sequence {μ_{n}} of measures is said to be weakly convergent to a measure μ if

lim_{n→+∞} ∫_{M} fdμ_{n} = ∫_{M} fdμ

for every bounded continuous function f : M → ℝ. This is called the weak convergence of measures.

We can now define the important notion of equidistribution. A sequence {S_{n}} of subset of ℂ_{v} is said to be equidistributed with respect to the measure μ if the sequence of measures {δ_{S_{n}}} weakly converges to μ.

We have also a weaker variant of this notion, but we need some other definitions to state it.

Definition 2.6.6 : Let A be a subset of ℂ_{v}. A logarithmic distance function for A is a continuous function λ : ℂ_{v} → ℝ such that λ(α) = 0 if α ∈ A, λ(α) > 0 if α ∉ A and that the difference |λ(α) − log^{+}|α|_{v}|_{v} is bounded.

EXAMPLE 2.6.2 : Let f ∈ K[x] with deg f ≥ 2; then the local canonical height function ĥ_{f,v} is a logarithmic distance function for the v-adic filled Julia set F J_{v}(f).

Definition 2.6.7 : Let {S_{n}} be a sequence of finite subsets of ℂ_{v} and A a subset of ℂ_{v}; let λ be a logarithmic distance function for A. Then we say that the sequence S_{n} is λ-taut if

lim_{n→+∞} \frac{1}{#S_{n}} ∑_{α∈S_{n}} λ(α) = 0.

Definition 2.6.8 : Let S_{n}, A be as before. We say that S_{n} is a Fekete sequence for A if

lim_{n→+∞}#S_{n} = +∞

and

lim_{n→+∞} ∏_{α,β∈S_{n}, α≠β} |α − β|v^{\frac{1}{#Sn(#Sn−1)}} = td(A).

We can now state the property we need. In the previous notations, we say that the sequence S_{n} is pseudo-equidistributed with respect to A and λ if S_{n} is λ-taut and S_{n} is a Fekete sequence for A.

All these definitions enable us to state the following theorem.

Theorem 2.6.9 : Let K be a field satisfying the product formula and f ∈ K[x] with deg f ≥ 2 and ∀v ∈ M_{K} let F J_{v}(f) be the v-adic filled Julia set of f. Let {S_{n}} be a sequence of finite distinct Galois-stable subsets of K̄ such that

lim_{n→+∞}ĥ_{f}(S_{n}) = 0.

Then, ∀v ∈ M_{K} the sequence S_{n} is pseudo-equidistributed with respect to F J_{v}(f) and ĥ_{f,v}.

Corollary 2.6.10 : In the same hypothesis, if the set F J_{v}(f) is compact, then {S_{n}} is equidistributed with respect to μ_{F J_{v}(f)}.

We still need the definition of a particular subset

Definition 2.6.11 : Let {S_{n}} and {S′_{n}} two sequence of subsets of ℂ_{v}. We say that {S′_{n}} is a full subsystem of {S_{n}} iff S′_{n} ⊆ S_{n} for all n and

lim_{n→+∞} \frac{#S_{n}}{#S′_{n}} = 1.

We can now turn back to our preperiodic points. The main theorem about totally v-adic extensions is the following

Theorem 2.6.12 : With the preceding notations, let f(x) = ∑_{j=0}^{d} a_{j}x^{j} ∈ K[x] be a polynomial such that |a_{j}|_{v} ≤ 1 for 0 ≤ j ≤ d − 1 and |a_{d}|_{v} = 1. ∀(e, s) ∈ ℕ × ℕ there exists a positive constant C depending on e, s, v such that ĥ_{f}(α) ≥ C for all but finitely many α ∈ K̄ totally v-adic of type (e, s).

Proof: Let L be the compositum of all finite extensions of K_{v} whose ramification index and inertial degree are bounded by e and s. Then L contains all the totally v-adic points of type (e, s) of K̄. Since there are only finitely many extensions of K_{v} with fixed ramification index and inertial degree, we have that L is a finite extension of K_{v}. Denote by e′ and s′ the ramification index and inertial degree of L respectively. Then, to prove the theorem, it suffices to show that there does not exist a sequence {S_{n}} of finite distinct subsets of L such that lim_{n→+∞}ĥ_{f}(S_{n}) = 0.

Suppose that such a sequence {S_{n}} exists. Then this sequence is ĥ_{f,v}-taut. Let us define S′_{n} = {α ∈ S_{n}| |α|_{v} ≤ 1}. We want to show that S′_{n} is a full subsystem of S_{n}; this is equivalent to show that

lim_{n→+∞} \frac{#T_{n}}{#S_{n}} = 0

where T_{n} = S_{n}\S′_{n}.

It is easy to show that, in this particular case, ĥ_{f,v}(α) = log^{+}|α|_{v}, so that

ĥ_{f,v}(S_{n}) = \frac{1}{#S_{n}} ∑_{α∈T_{n}} log^{+}|α|_{v}.

Let p be the characteristic of the residue field of ℂ_{v}. We have that log^{+}|α|_{v} ≥ (1/e′)log(p) for α ∈ T_{n} and it follows that

ĥ_{f,v}(S_{n}) ≥ \frac{#T_{n}}{e′#S_{n}} log(p).

Taking the limit for n going to infinity and recalling that {S_{n}} is ĥ_{f,v}-taut, we have the required condition. Then S′_{n} is a full subsystem of S_{n}.

We have clearly that limsup td(S_{n}) ≤ limsup td(S′_{n}) ≤ td(O_{L}), where O_{L} is the number ring of L. By the results of the examples 2.6.1, we have

td(O_{L}) = p^{−1/e′(p^{s′} −1)} < 1 = td(F J_{v}(f)).

Therefore S_{n} can not be pseudo-equidistributed with respect to F J_{v}(f) and ĥ_{f,v}. This contradicts theorem 2.6.9 and hence the sequence S_{n} can not exist.

Corollary 2.6.13 : There are only finitely many preperiodic points of f that are totally v-adic of type (e, s).

Proof: By Lemma 2.4.5, ĥ_{f}(α) = 0 ⟺ α is preperiodic; in particular ĥ_{f}(α) = 0 for every α preperiodic and totally v-adic of type (e, s). By the theorem, there can be only finitely many of these points.

We are particularly interested in a very special category of totally v-adic points, that is, when |.|_{v} is a p-adic absolute value on ℚ. In that case we have an interesting equivalence.

Lemma 2.6.14 : Let p be a rational prime. Then every α ∈ ℚ̄ is totally p-adic if and only if p splits completely in ℚ(α).

Proof: Trivial by definition.

Thanks to this equivalence we have this specialization of the preceeding theorem for totally p-adic algebraic numbers.

Corollary 2.6.15 : Let f(x) = ∑_{j=0}^{d} a_{j}x^{j} ∈ ℚ[x] be a polynomial and p be a prime such that |a_{j}|_{p} ≤ 1 for 0 ≤ j ≤ d − 1 and |a_{d}|_{p} = 1. Then there exists a positive constant C, depending only on p and f, such that ĥ_{f}(α) ≥ C for all but finitely many totally p-adic α ∈ ℚ̄.

Proof: Immediate by the preceeding theorem.

We end this section giving a proof of a theorem which represents a sort of Lehmer conjecture for the canonical height (but, differently from Lehmer’s conjecture, this is solved).

Theorem 2.6.16 : Let K be a number field and f ∈ K[x] with deg f = d ≥ 2. Assume that J(f) ⊈ ℝ, where J(f) is the Julia set of f. Then there exists a positive constant C, depending only on f, such that ĥ_{f}(α) ≥ C for all but finitely many totally real α ∈ ℚ̄.

Proof: Suppose that there exists a sequence {α_{n}} of totally real algebraic numbers such that ĥ_{f}(α_{n}) → 0. By Corollary 2.6.10, we have that α_{n} is equidistributed with respect to μ_{J}, the equilibrium measure of the Julia set J. Since all the α_{n} are totally real, it follows that the support of μ_{J} must be contained in ℝ, but this support is J, which, by hypothesis, is not a subset of ℝ. This is absurd, then the sequence α_{n} does not exists.

2.7 Length of finite orbits

In this section we want to give a constant for preperiodic points in a field K which turns to be the equivalent of the constant B_{K} defined in 2.3 for periodic points. This results applies essentially to number fields, or in general, to fields of fractions of finitely generated commutative domains of characteristic zero. We start with a defintion.

Definition 2.7.1 : Let f ∈ K[x] be a non-linear polynomial. The length of a point α ∈ K, denoted by m(α), is the minimum positive integer m such that f^{m}(α) = f^{j}(α) for some positive integer j < m. If such an m does not exist, we say that m(α) = ∞.

Clearly m(α) < ∞ ⟺ α is preperiodic. Our aim is to give a bound, depending only on the field K, for the length of any preperiodic point of any non-linear polynomial on K. First of all, we need the following result about unit equations, whose proof may be found in [9].

Theorem 2.7.2 : Let K be a number field and O_{K} its number ring. Then the unit equation

η_{1} + η_{2} + η_{3} = 1

has only a finite number of solutions in O_{K}^{∗} such that η_{j} ≠ 1 for j = 1, 2, 3. We denote by C_{K} the number of these solutions. (The solutions for which one of the η_{j} = 1 are called trivial).

The main result we are going to prove about the length of a finite orbit is the following:

Theorem 2.7.3 : Let K be a number field and D_{K} be an upper bound for the length of finite orbits in K. Then D_{K} is finite and we have

D_{K} ≤ \frac{1}{3}B_{K}(31 + C_{K}) − 1

where B_{K} and C_{K} are the constants defined in 2.3 and in the previous definition.

Proof: Let f ∈ O_{K}[x] be a non-linear polynomial and α ∈ O_{K} a point of finite length m. Clearly the orbit of α must contain a periodic orbit of length r ≤ m. We want to show the we can reduce to the case where r = 1, that is, the orbit of α ends with a fixed point.

Suppose r > 1, then we can write m − r = qr + r_{1} with r_{1} < r. Then the sequence α_{r_{1}}, α_{r_{1}+r}, α_{r_{1}+2r}, ...α_{r_{1}+qr} is a finite orbit for f^{r} ending with a fixed point. Suppose we know that this type of orbit has a number of elements bounded by a constant D; then the orbit of α under f has length bounded by

m = m − r + r = (1 + q)r + r_{1} ≤ Dr + r_{1} ≤ (D + 1)r − 1 ≤ (D + 1)B_{K} − 1,

that is,

D_{K} ≤ (D + 1)B_{K} − 1.

Then we just need to find a bound for finite orbits ending with a fixed point. Up to conjugation we may also assume that the ending fixed point is 0 and that the initial point is α = 1.

Now, by Lemma 2.3.6, we have that α_{j} − α_{k}|α_{j+c} − α_{k+c}. Choosing c = j − k and j ≤ (m + k)/2 we have α_{j} − α_{k}|α_{2j+k} − α_{j} = −α_{j}, that is, α_{j} − α_{k}|α_{j}. By symmetry we may show that in the same hypothesis α_{j} − α_{k}|α_{k}. In particular, choosing k = 0 we have α_{j} − 1|1, that is, α_{j} − 1 is a unit.

Finally, the argument of Lemma 2.3.5 tells us that we can write α_{m−1} = λ_{j}α_{j} with λ_{j} ∈ O_{K} for all j = 1, ..., m − 2. By the preceding relations we have α_{m−1} − α_{j} = λ_{j}α_{j} − α_{j} = (λ_{j} − 1)α_{j}|λ_{j}α_{j} = α_{m−1}, that is, λ_{j} − 1|λ_{j} and this implies that λ_{j} − 1 is a unit in O_{K}.

Now we have all the instruments to end our proof. Obviously we can assume m ≥ 2. The preceding arguments tell us that, if m/2 ≤ j ≤ m − 2, then we have α_{j} = 1 + ϵ_{j}, α_{m−1} = 1 + ϵ_{m−1} and α_{j} = \frac{α_{m−1}}{1−γ_{j}}, where ϵ_{j}, ϵ_{m−1} and γ_{j} are units. The relations among these elements lead to the equation

\frac{ϵ_{j}}{ϵ_{m−1}} − \frac{γ_{j}}{ϵ_{m−1}} − \frac{ϵ_{j}γ_{j}}{ϵ_{m−1}} = 1,

that is, to (m − 3)/2 distinct solutions of the unit equation η_{1} + η_{2} + η_{3} = 1, given by

η_{1}(j) = \frac{ϵ_{j}}{ϵ_{m−1}} ,

η_{2}(j) = −\frac{γ_{j}}{ϵ_{m−1}} ,

η_{3}(j) = −\frac{ϵ_{j}γ_{j}}{ϵ_{m−1}} .

We have to drop the solutions in which one of the η_{k} is 1. But the relation η_{1}(j)+η_{3}(j) = 0 and η_{2}(j)+η_{3}(j) = 0 imply α_{j} = 0, which is a contradiction, while η_{1}(j)+η_{2}(j) = 0 implies ϵ_{j} = γ_{j}, thus α_{j}^{2} − 2α_{j} + α_{m−1} = 0 which can happen for at most two values of j. Then we have (m − 7)/2 distinct non-trivial solutions.

Permuting the η_{k} we obtain 3(m − 7) triples of units solving the equation. We just have to check for some coincidences due to possible equalities among the η_{k}. Now η_{1}(j) = η_{2}(j) implies α_{j}^{2} = α_{m−1}, η_{1}(j) = η_{3}(j) implies α_{j} = 2α_{m−1} and η_{2}(j) = η_{3}(j) implies α_{j} = 2. Then we finally obtain 3(m − 7) − 4 = 3m − 25 distinct triples solving the equations, thus m ≤ (C_{K} + 25)/3 and hence we have

D = 1 + m ≤ (C_{K} + 28)/3.

We finally obtain that

D_{K} ≤ (D + 1)B_{K} − 1 ≤ \frac{1}{3}B_{K}(31 + C_{K}) − 1

and the theorem is proved.

In [21] we have a numerical bound for D_{K}

Theorem 2.7.4 : Let K be a number field with n = [K : ℚ] and let s be the number of archimedean places on K. Then we have

D_{K} ≤ \frac{2}{3}n4^{n}(31 + 2^{1031s}).

The number given in the theorem is enormous and therefore useless for any application. Luckily, we can drastically reduce this number by putting some conditions on the fields we want to consider. We start with the easiest example of the rational field.

Theorem 2.7.5 : Let K = ℚ. Then we have D_{ℚ} = 4.

Proof: We have shown in section 2.3 that B_{ℚ} = 2. Consider first the case of a point α of length m, whose orbit ends with a fixed point, that we can assume to be 0. Then we have f(x) = x(x − α_{m−2})g(x) with g ∈ ℤ[x]. We can assume m ≥ 2. By the arguments used in the theorem, we have that, if j ≥ m/2, then either α_{j} = 2 or α_{j} = 0, but the last case is possible only for j = m − 1, then we must have α_{m−2} = 2 and m − 3 < m/2, that is, m ≤ 5. We want to show that the cases m = 4, 5 are not possible; in fact, in those cases we would have 2 = α_{m−2} = f(α_{m−3}) = α_{m−3}(α_{m−3} − 2)g(α_{m−3}), that is, α_{m−3} and α_{m−3} − 2 are both divisors of 2 and this implies α_{m−3} = 1. Then 1 = α_{m−3} = f(α_{m−4}) = α_{m−4}(α_{m−4} − 2)g(α_{m−4}), implying α_{m−4} = 1, which is absurd. Then m ≤ 3.

Consider now the case of an orbit of length m ending with a 2-periodic orbit and assume α_{m−2} = 0. Then f has the form

f(x) = (x − α_{m−3})(x − α_{m−1})g(x),

with g ∈ ℤ[x]. Moreover, since α_{m−1} = f(0) = α_{m−1}α_{m−3}g(0), we must have α_{m−3} = ±1. Up to multiplying by -1 we may assume α_{m−3} = 1.

Suppose now that m ≥ 5. Then

1 = α_{m−3} = f(α_{m−4}) = (α_{m−4} − 1)(α_{m−4} − α_{m−1})g(α_{m−4}),

that means α_{m−4} − 1 = ±1 and α_{m−4} − α_{m−1} = ±1, that is, α_{m−4} = 2 and α_{m−1} = 3. Substituting in the expression of f, we must have that

2 = α_{m−4} = f(α_{m−5}) = (α_{m−5} − 1)(α_{m−5} − 3)g(α_{m−5}).

As in the previous case this implies that α_{m−5} − 1 and α_{m−5} − 3 are both divisors of 2 and therefore we must have α_{m−5} = 2, which is absurd. Thus m ≤ 4 and the theorem is proved.

Now we show a theorem improving the bound of 2.7.4 for a large class of fields.

Theorem 2.7.6 : Let K be a number field satisfying the following conditions: 1) K does not contain the cyclotomic field ℚ(ζ_{4}) = ℚ(\sqrt{−1}).

2) The unit equation

η_{1} + η_{2} = 1

has no solution for η_{1}, η_{2} ∈ O_{K}^{∗}.

3) If (η_{1}, η_{2}, η_{3}) is a triple of elements of O_{K}^{∗} satisfying the unit equation

η_{1} + η_{2} + η_{3} = 1

then at least one of the η_{j} is equal to 1.

Then B_{K} ≤ 2 and D_{K} ≤ 19.

Proof: We just need to prove that B_{K} ≤ 2, because the hypothesis on the field says that C_{K} = 0 and, in that case, the general bound given in theorem 2.7.3 implies the thesis.

Suppose that there exists a polynomial f ∈ K[x] and α ∈ P er_{k}(f) with k odd. Up to conjugation, we may assume α = 0 and α_{1} = 1. By Lemma 2.3.6, we have that both α_{2} and α_{2} − 1 are units; then, taking η_{1} = α_{2} and η_{2} = α_{2} − 1, we have violated the condition 2. This is absurd. Then every periodic orbit in K has even length.

Suppose that there is a 4-periodic orbit in O_{K}, that we may assume to be {0, 1, α, β}. By Lemma 2.3.6, 1 − α, α − β, β are units and α, β − 1 are associated, that is β − 1 = ϵα with ϵ ∈ O_{K}^{∗}. We have

(1 − α) + (α − β) + β = 1.

Since 1 − α ≠ 1 and β ≠ 1, by condition 3 we must have α − β = 1, that is α − 2 = ϵα. Then we can write

α = \frac{2}{1 − ϵ} β = \frac{1 + ϵ}{1 − ϵ},

leading to

β − ϵβ − ϵ = 1.

Since β ≠ 1 and ϵ ≠ −1 (otherwise we would have α = 1), condition 3 says that ϵβ = −1. This implies ϵ = β and therefore ϵ^{2} = −1, that means ϵ ∈ ℚ(ζ_{4}). This contradicts condition 1 and then we have an absurd. So the theorem is proved.

The conditions on the field given in this theorem may seem too restrictive but they are not. It is easily shown that all but finitely many quadratic number fields and cubic fields with negative discriminant satisfy these conditions; since they are the most common fields used for applications, this bound proves to be a strong result.

In section 2.3 we derived our estimates for B_{K} from the p-adic fields. It is not possible to do the same when evaluating the constant D_{K}. In fact the following theorem shows that, if K is an extension of ℚ_{p}, then D_{K} = ∞.

Theorem 2.7.7 : Let K be an extension of ℚ_{p}. Then, for all n ∈ ℕ, there is a polynomial f_{n} ∈ ℤ_{p}[x] ⊆ O_{K}[x] having a finite orbit of length n + 1. Then D_{K} = ∞.

Proof: Since ℤ ⊆ ℤ_{p} and B_{ℚ} = 2, we may assume n ≥ 2. For any n ≥ 2 we define the polynomial

f_{n}(x) = cx(x − c_{0}p)(x − c_{1}p^{2})(x − c_{2}p^{2^{2}})...(x − c_{n−1}p^{2^{n−1}}) + x^{2} + p^{2^{n}}x ∈ ℤ_{p}[x],

where, for k = 1, ..., n, the c_{k} ∈ ℤ_{p} are defined by induction by

c_{0} = 1 c_{k} = c_{k−1}(c_{k−1} + p^{2^{n}−2^{k−1}})

and c ∈ ℤ_{p} is chosen so that f_{n}(c_{n}p^{2^{n}}) = 0. Clearly 0 is a fixpoint for f_{n} and it is easy to see that the orbit of p under f_{n} is

{p, c_{1}p^{2}, c_{2}p^{2^{2}}, ..., c_{n}p^{2^{n}}, 0}

and all of the elements are distinct. Then f_{n} has a finite orbit of length n + 1.

2.8 An explicite example: quadratic number fields

In this section we want to apply the results we have found to quadratic number fields. The advantage of these fields is that all the calculations are particularly simple and enable us to have the exact valuation of the constants B_{K}, D_{K} and of the orbits realizing them. In all the section we put K = ℚ(\sqrt{m}) with m a squarefree rational integer. In addition, we shall denote

ξ = \frac{1 + \sqrt{5}}{2}.

We start with some definition that reduces the number of orbits we need to study.

Definition 2.8.1 : Let E_{1}, E_{2} two finite sets of O_{K} of the same cardinality. We say that E_{1} and E_{2} are equivalent orbits if there exist two polynomials of the same degree f_{1}, f_{2} ∈ K[x] and a linear polynomial g ∈ K[x] such that E_{j} is a finite polynomial orbit for f_{j} and g is a conjugation between f_{1} and f_{2} such that

g ∘ f_{1} ∘ g^{−1}(E_{1}) = f_{2}(E_{2}).

We say that E_{1} and E_{2} are weakly equivalent orbits if there exists an automorphism σ of O_{K} such that E_{1} and σ(E_{2}) are equivalent orbits. Finally we say that a finite orbit is normalized if it starts with 0 and 1.

Now we give a criterium that enables us to recognize if a finite subset of O_{K} is effectively a finite orbit for a polynomial.

Definition 2.8.2 : Let E ⊆ O_{K} be a finite ordered subset of the form E = {α_{0}, α_{1}, ..., α_{n}}. The Lagrange interpolation polynomial of E is the polynomial f ∈ K[x] with deg f ≤ n − 1 given by

f(x) = ∑_{j=0}^{n−1} α_{j+1}L_{j}(x),

where

L_{j}(x) = ∏_{k=0,k≠j}^{n−1} \frac{x − α_{k}}{α_{j} − α_{k}} .

Theorem 2.8.3 : E is a finite orbit for O_{K} ⟺ the Lagrange interpolation polynomial of E has coefficients in O_{K}.

Proof: If the Lagrangian polynomial has coefficients in O_{K}, then it is itself a polynomial realising E as an orbit. Conversely, if E is a finite orbit and f is any polynomial realising it, we take

g = ∏_{j=0}^{n−1}(x − α_{j})

and, writing f(x) = p(x)g(x) + q(x), with p, q ∈ O_{K}[x], it is easy to note that q must be the Lagrangian polynomial, which therefore has coefficients in O_{K}.

We need some lemmas about the particular periodic orbits we shall find.

Lemma 2.8.4 : If there exists an exceptional unit η ∈ O_{K}^{∗}, then {0, 1, η, 0} is a 3-periodic orbit in K and every normalized 3-periodic orbit is of this form. Moreover, if α ∈ O_{K} is such that α|η^{2} − η + 1 then {0, α, αη, 0} is a 3-periodic orbit and every 3-periodic orbit of K is of this form.

Proof: To prove the first part it suffices to note that {0, 1, η, 0} is a 3-periodic orbit for the polynomial

f(x) = \frac{1 − η(1 − η)}{η(1 − η)}x^{2} + \frac{η^{2}(1 − η) − 1}{η(1 − η)}x + 1.

For the second part, suppose {0, α, β, 0} is a 3-periodic orbit. Then, by Lemma 2.3.6, α|β and if we set η = β/α, then {0, 1, η, 0} is a 3-periodic orbit. Then by the first part η and 1 − η are units. The polynomial realizing the 3-periodic orbit {0, α, β, 0} is of the form f(x) = (x−β)(ax+ b), with a, b ∈ O_{K} and we have α = f(0) = −bβ, while β = f(α) = (α − β)(aα − β). Then we get

b = −η^{−1} η(1 − η)aα = η^{2} − η + 1,

thus α|η^{2} − η + 1 and we have finished.

Lemma 2.8.5 : If K is a quadratic number field, the length of a periodic orbit in K is of the form 2^{a}3^{b} with a, b non-negative integers.

Proof: By Corollary 2.3.7 we deduce that, if p is a prime dividing the length of a cycle, then p < m, where m is the infimum of the cardinalities of the sets O_{K}/I for I a non-zero ideal of O_{K}. Since in every quadratic field we have #O_{K}/2 = 4, we have the result.

Lemma 2.8.6 : Let {0, 1, α_{2}, ...α_{n−1}, 0} a n-periodic orbit in O_{K} with n ≥ 3 and I an ideal of O_{K} such that #O_{K}/I < n. Then I contains at least one of the α_{j}.

Proof: The residue classes mod I of the α_{j} are n elements of O_{K}/I therefore they cannot be all distinct. Then α_{j} − α_{k} ∈ I for some j, k. By Corollary 2.3.7, this implies that α_{j−k} ∈ I.

Now we need a lemma about solutions of unit equations in quadratic fields.

Lemma 2.8.7 : Let K = ℚ(\sqrt{m}). The following are true:

1) The solutions of

η_{1} + η_{2} = 1

with η_{1}, η_{2} ∈ O_{K}^{∗} are

(ζ_{6}, −ζ_{6}^{2}), (1 + ξ, −ξ), (ξ, 1 − ξ), (ξ − 1, 2 − ξ),

where ζ_{6} is a primitive sixth root of unity. Therefore they exist only if m = −3, 5.

2) The solutions of

η_{1} + η_{2} = 2,

with η_{1}, η_{2} ∈ O_{K}^{∗}, are

(1, 1), (1 + \sqrt{2}, 1 − \sqrt{2}), (1 + ξ, 1 − ξ), (2 − ξ, ξ).

Therefore if m ≠ 2, 5 there is only one such solution.

3)The equation

η_{1} + η_{2} + η_{3} = 1,

with η_{1}, η_{2}, η_{3} ∈ O_{K}^{∗}, has non-trivial solutions ⟺ m = 2, 5. If m = 2 the solutions, up to conjugation, are

(3 + 2\sqrt{2}, −1 − \sqrt{2}, −1 − \sqrt{2}), (1 + \sqrt{2}, −1, 1 − \sqrt{2}).

If m = 5 they are

(1+ξ, 1+ξ, −2ξ−1), (−ξ, 3ξ+2, −2ξ−1), (−ξ, −ξ, 2ξ+1), (2−ξ, ξ, −1), (1−ξ, −1−ξ, 2ξ+1)

4) If m = 2, the solutions of

η_{1} + η_{2} = \sqrt{2},

with η_{1}, η_{2} ∈ O_{K}^{∗}, are

(1, −1 + \sqrt{2}), (−1, 1 + \sqrt{2}).

5) If m = 5, the solutions of

η_{1} + η_{2} = \sqrt{5}.

with η_{1}, η_{2} ∈ O_{K}^{∗}, are

(ξ − 2, ξ + 1), (ξ − 1, ξ).

6) If m = 3, the only solution of

η_{1} + η_{2} = 1 + \sqrt{3},

with η_{1}, η_{2} ∈ O_{K}^{∗}, is

(−1, 2 + \sqrt{3}).

7) Let η ∈ O_{K}^{∗} such that #O_{K}/(η − 1) = 2. If m = −1 then η = ±\sqrt{−1}. If m = 2 then η = ±1 ± \sqrt{2}. If m = 3 then η = 2 ± \sqrt{3}. If m ≠ −1, 2, 3 such an η does not exists.

Proof: 1) The proof can be found in [18].

2) If m < 0 the result is obvious, then we can assume m > 0. Let (η_{1}, η_{2}) be a solution. If m ≡ 2, 3 (mod 4), we can write η_{1} = (α_{1} + β_{1}\sqrt{m})/2, η_{2} = (α_{2} + β_{2}\sqrt{m})/2 with α_{1} ≡ β_{1} (mod 2) and α_{2} ≡ β_{2} (mod 2) and 2|α_{1}, β_{1}, α_{2}, β_{2}. Then α_{2} = 4 − α_{1} and β_{2} = −β_{1}. Therefore

Norm_{K/ℚ}(η_{1}) = \frac{α_{1}^{2} − β_{1}^{2}m}{4} = 4 − 2α_{2} + Norm_{K/ℚ}(η_{2}),

that is

α_{2} = \frac{Norm_{K/ℚ}(η_{2}) − Norm_{K/ℚ}(η_{1}) + 4}{2}.

Since η_{1}, η_{2} are units we have Norm_{K/ℚ}(η_{j}) = ±1. Then, considering all the possibilities, we have

α_{2} ∈ {1, 2, 3}.

Then we have also

β_{2}^{2}m = α_{2}^{2} − 4Norm_{K/ℚ}(η_{2}) ∈ {0, 5, 8}.

Now if β_{2}^{2}m = 0, then β_{2} = 0 and we have the solution (1, 1), which occurs in all quadratic fields. If β_{2}^{2}m = 5, the only possibility is m = 5 and β_{2} = ±1; this relation gives the solutions (1 + ξ, 1 − ξ) and (2 − ξ, ξ). Finally, if β_{2}^{2}m = 8, since m is squarefree, we have m = 2 and β_{2} = ±2; this relation furnishes the solution (1 + \sqrt{2}, 1 − \sqrt{2}).

3) If m < 0 the result is obvious, then we can assume m > 0. Suppose that (η_{1}, η_{2}, η_{3}) is a non-trivial solution and let η ∈ O_{K}^{∗} be a fundamental unit of O_{K}. Suppose that η_{1} = −1, then we have η_{2}+η_{3} = 2 and we can use part 2 to build the solutions (−1, 1+\sqrt{2}, 1−\sqrt{2}) for m = 2 and (−1, 2 − ξ, ξ), (−1, 1 + ξ, 1 − ξ) (which is conjugate to (1 − ξ, −1 − ξ, 2ξ + 1)) for m = 5. We can then assume that none of the η_{j} = −1 and, for some non-zero integers a ≥ b ≥ c, we can rewrite the equation as

±η^{a} ± η^{b} ± η^{c} = 1.

If c < 0, that is, c = −t, with t > 0, and we write r = a + t and s = b + t, the equation becomes

η^{r} ± η^{s} ± η^{t} = ±1,

with r ≥ s ≥ t ≥ 1. Suppose that r = s; then 2η^{r} = ±1∓η^{t} ≤ 1 + η^{r} and therefore 1 < η^{r} ≤ 1. This is absurd, then we must have r > s ≥ 1 and

η^{r} ≤ |1 + η^{s} + η^{t}| ≤ 1 + 2η^{r−1},

thus

η ≤ 2 + η^{1−r} ≤ 3.

This can happen only if m = 2, η = 1 + \sqrt{2} or m = 5, η = ξ. Let us examine the case m = 2, where η = 1 + \sqrt{2}; then η^{−1} = η − 2 and 2 + η^{−1} = η ≤ 2 + η^{1−r}, which implies 1 ≤ r ≤ 2. Since r = 1 is impossible, we must have r = 2, s = 1, t = 1 and the equation becomes

η^{2} − η − η = 1,

which is the solution we want. Finally, let us consider the case m = 5, where η = ξ. It is immediate to show by induction that

ξ^{j} = F_{j−1} + F_{j}ξ

where F_{j} is the j-th Fibonacci number. Then our equation is equivalent to the system

±F_{r−1} ± F_{s−1} ± F_{t−1} = 1
±F_{r} ± F_{s} ± F_{t} = 0.

The triples of the same signs are impossible, then we must have either

F_{r−1} + F_{s−1} − F_{t−1} = 1
F_{r} + F_{s} − F_{t} = 0

or

F_{r−1} + F_{s−1} − F_{t−1} = −1
F_{r} + F_{s} − F_{t} = 0.

Now it is easy to see that the only possibilities for (r, s, t) are (3, 2, 2), (3, 1, 1) and (4, 3, 1) and they give the last solutions we were looking for.

4) Proceed to the calculations as in part 2.

5) Proceed to the calculations as in part 2.

6) Proceed to the calculations as in part 2.

7) Write η = (α + β\sqrt{m})/2, then the condition |Norm_{K/ℚ}(η − 1)| = 2 implies that either α = Norm_{K/ℚ}(η) + 3 or α = Norm_{K/ℚ}(η) − 1. If Norm_{K/ℚ}(η) = 1 then either α = 0, which implies m = −1 and η = ±\sqrt{−1}, or α = 4, which implies m = 3 and η = 2 ± \sqrt{3}. Finally, if Norm_{K/ℚ}(η) = −1 then we have α = ±2, which implies m = 2 and η = ±1 ± \sqrt{2}.

Corollary 2.8.8 : If m ≠ −3, 5 and there is an r-periodic orbit in O_{K}, then r is a power of two. If m ≠ −1, 2, 5, then r ≠ 4. In particular, if m ≠ −3, −1, 2, 5, then B_{K} = 2.

Proof: The result follows immediatly from the previous lemma and Theorem 2.7.6.

Lemma 2.8.9 : Let K be a quadratic field and let C′ the cardinality of the set of the non-associated non-zero α ∈ O_{K} such that the unit equation η_{1} + η_{2} = α has more than one solution. Then C′ ≤ 3C_{K}, where C_{K} is the constant defined in 2.7

Proof: If α = η_{1} + η_{2} = η_{3} + η_{4}, with all the η_{j} distinct units, then

\frac{η_{1}}{η_{3}} + \frac{η_{2}}{η_{3}} + \frac{−η_{4}}{η_{3}} = 1

and none of the summands is equal to 1. Permuting the η_{j} we see that every such α gives origin to at most three different solutions, proving the result.

Corollary 2.8.10 : If C_{K} = 0 and there exists an r-periodic orbit in O_{K}, then r is not divisible by 8 or by the square of an odd prime.

Proof: Let q = p^{k}, with k ≥ 2, be a prime power dividing r and {0, 1, α_{2}, ..., α_{q−1}, 0} a normalized q-periodic orbit. By Lemma 2.3.6, if j < q does not divide q, then α_{j}, α_{q−p−j} and α_{q−p} − α_{j} are units; therefore

α_{q−p} = α_{j} + (α_{q−p} − α_{j}).

Then the unit equation η_{1} + η_{2} = α_{p−q} has at least ϕ(q)/2 solutions, where ϕ is the Euler function. The previous lemma implies ϕ(q) ≤ 2. Hence we have q = p or q = 4.

Corollary 2.8.11 : If m ≠ 2, 5 then r ∈ {1, 2, 3, 4, 6}.

Proof: The result follows from the preceding corollary and Corollary 2.8.8.

Now we have all the instruments to prove the theorem about polynomial cycles in quadratic fields. Our proof will also describe the inequivalent periodic orbits and the polynomials realizing them.

Theorem 2.8.12 : Let K = ℚ(\sqrt{m}) with m a squarefree integer and let O_{K} be its number ring. Let f ∈ O_{K}[x]. Then we have the following results:

1) If m ≠ −3, −1, 2, 5, then P er_{k}(f) is empty for k ≥ 3.

2) If m = −1, then P er_{k}(f) is empty for k = 3 and k ≥ 5.

3) If m = −3, then P er_{k}(f) is empty for k = 4, 5 and k ≥ 7.

4) If m = 2, then P er_{k}(f) is empty for k = 3 and k ≥ 5.

5) If m = 5, then P er_{k}(f) is empty for k ≥ 5.

Proof: 1) The result follows from Corollaries 2.8.8 and 2.8.10. Moreover, from Lemma 2.3.5, we obtain that the only normalized 2-periodic orbit is {0, 1, 0}, realized by f(x) = 1 − x and that every non-normalized 2-periodic orbit is equivalent to an orbit of the form {0, α, 0}, with α ≠ 0, realized by f(x) = α − x.

2) The result follows from Corollaries 2.8.8 and 2.8.10. Moreover, we see immediatly that the set \{0, 1, 1+\sqrt{−1}, \sqrt{−1}, 0\} is a normalized 4-periodic orbit realized by the polynomial f(x) = \sqrt{−1}x + 1; we want to show that every normalized 4-periodic orbit is weakly equivalent to this one.

Let {0, 1, α_{2}, α_{3}, 0} be such an orbit, then, by Lemma 2.3.6, α_{2}−1, α_{3}−α_{2} and α_{3} are units and α_{3} − 1|α. Moreover (1 − α_{2}) + (α_{2} − α_{3}) + α_{3} = 1, then, by Lemma 2.8.7, we have that the only possibility is α_{2} = 1 ± \sqrt{−1} and α_{3} = ±\sqrt{−1}, proving the result. This also shows that every non-normalized 4-periodic orbit is weakly equivalent to an orbit of the form \{0, α, α(1 + \sqrt{−1}), α\sqrt{−1}, 0\}, with α ≠ 0, realized by the polynomial f(x) = \sqrt{−1}x + α.

3) The result follows from Corollaries 2.8.8 and 2.8.10. The same argument of part 2) shows that every normalized 3-periodic orbit is equivalent to {0, 1, ζ_{6}, 0} and every 6periodic orbit is equivalent to {0, 1, 1+ζ_{6}, 2ζ_{6}, −1+2ζ_{6}, −1+ζ_{6}, 0}. Moreover, multiplying every element of a normalized periodic orbit by pairwise non-associated non-zero elements of O_{K}, we see that there are infinitely many weakly inequivalent 3 and 6-periodic orbits.

4) Lemmas 2.8.5 and 2.8.6 show that P er_{k}(f) is empty if k is not a power of 2. Then we just have to show that P er_{4}(f) can be not empty, but P er_{8}(f) is always empty.

Let us explicitly build the possible normalized 4-periodic orbits. If there exists such an orbit, it must have the form {0, 1, α_{2}, α_{3}, 0} and, by Lemma 2.3.6, α_{3}, α_{2} − α_{3}, 1 − α_{2} must be units and satisfy the relation (1 − α_{2}) + (α_{2} − α_{3}) + α_{3} = 1. If none of the summands is 1, Lemma 2.8.7 ensures that there are only two possible orbits of this form. On the other hand, if one of the summands equals 1, the only possibility is α_{2} −α_{3} = 1; by Lemma 2.3.6, η = (α_{3} − 1)/α_{2} is a unit, then η = 1 − 2/α_{2}, which means that α_{2}|2. Then we have only three possibilities: either α_{2} = 2η_{1} or α_{2} = η_{1} or α_{2} = \sqrt{2}η_{1} with η_{1} ∈ O_{K}. In the first case we have η + η_{1}^{−1} = 1, which contradicts Lemma 2.8.7; then this case is impossible. In the second case we have α_{2} − ηα_{2} = 2; by Lemma 2.8.7, we must have α = 1 ± \sqrt{2} and this is absurd because it contradicts the fact that α − 1 must be a unit, then this case is impossible, too. Finally in the third case we have η_{1} − ηη_{1} = \sqrt{2} and Lemma 2.8.7 implies that η_{1} ∈ \{1, −1, 1 + \sqrt{2}, −1 + \sqrt{2}\}. Then we have found four possible weakly inequivalent normalized 4-periodic orbits, given by

\{0, 1, 2 + \sqrt{2}, −1 − \sqrt{2}, 0\}, \{0, 1, \sqrt{2}, 1 + \sqrt{2}, 0\},
\{0, 1, \sqrt{2}, −1 + \sqrt{2}, 0\}, \{0, 1, 2 + \sqrt{2}, 1 + \sqrt{2}, 0\}.

They are realized respectively by the polynomials

(−6 + 4\sqrt{2})x^{3} + (10 − 7\sqrt{2})x^{2} + (−3 + 4\sqrt{2})x + 1,
−2x^{3} + (2 + 3\sqrt{2})x^{2} − (1 + 2\sqrt{2})x + 1,
−2(2 + \sqrt{2})x^{3} + 6(1 + \sqrt{2})x^{2} − 3(1 + \sqrt{2})x + 1,
2(2 − \sqrt{2})x^{3} − 6x^{2} + 3(1 + \sqrt{2})x + 1.

In the same way we see that there are four weakly inequivalent 4-periodic orbits, not equivalent to a normalized one, given by

\{0, \sqrt{2}, 2 + 2\sqrt{2}, −2 − \sqrt{2}, 0\}, \{0, \sqrt{2}, 2, 2 + \sqrt{2}, 0\},
\{0, \sqrt{2}, 2, 2 − \sqrt{2}, 0\}, \{0, \sqrt{2}, 2 + 2\sqrt{2}, 2 + \sqrt{2}, 0\}.

They are realized by the polynomials

(−3 + 2\sqrt{2})x^{3} + (−7 + 5\sqrt{2})x^{2} + (−3 + 4\sqrt{2})x + \sqrt{2},
−x^{3} + (3 + \sqrt{2})x^{2} − (1 + 2\sqrt{2})x\sqrt{2},
−(2 + \sqrt{2})x^{3} + 3(2 + \sqrt{2})x^{2} − 3(1 + \sqrt{2})x + \sqrt{2},
(2 − 2\sqrt{2})x^{3} − 3\sqrt{2}x^{2} + 3(1 + \sqrt{2})x + \sqrt{2}.

We are just left to show that P er_{8}(f) is empty for all f ∈ K[x]. Suppose that there exists such a normalized orbit, of the form {0, 1, α_{2}, α_{3}, ..., α_{7}, 0}. By Lemma 2.3.6, α_{3}, α_{5}, α_{7} must be units and {0, α_{2}, α_{4}, α_{6}, 0} is a 4-periodic orbit, then it must be weakly equivalent to one of the listed above. None of the elements of this 4-orbit is a multiple of 3 + \sqrt{2}, whose norm is 7, then none of the α_{j} is. This is absurd, because contradicts Lemma 2.8.6. Then there can not be such an orbit and therefore P er_{8}(f) is empty.

5) First of all, we deal with P er_{3}(f). By Lemma 2.8.4, we have that all the normalized 3-periodic orbits are equivalent to an orbit of the form {0, 1, η, 0}, where η is an exceptional unit. The exceptional units of ℚ(\sqrt{5}) are ξ, −ξ, 1 + ξ, 1 − ξ and they give us three inequivalent normalized 3-periodic orbits

{0, 1, ξ, 0}, {0, 1, −ξ, 0}, {0, 1, 1 + ξ, 0},

realized by the polynomials

−2x^{2} + (1 + ξ)x + 1,
2(1 − ξ)x^{2} + (−3 + ξ)x + 1,
2(1 − ξ)x^{2} + (−2 + 3ξ)x + 1.

About non-normalized 3-periodic orbits, Lemma 2.8.4 tells us that they have the form {0, α, αη, 0} where α|η^{2} − η + 1; but the only such α in O_{K} is 2. Then there are three weakly inequivalent non-normalized 3-periodic orbits in O_{K}, given by

{0, 2, 2ξ, 0}, {0, 2, −2ξ, 0}, {0, 2, 2(1 + ξ), 0},

realized by the polynomials

−x^{2} + (1 + ξ)x + 2,
(1 − ξ)x^{2} + (−3 + ξ)x + 2,
(1 − ξ)x^{2} + (−2 + 3ξ)x + 2.

Now we want to show that P er_{6}(f) is empty. Suppose that there is a normalized 6-periodic orbit {0, 1, α_{2}, α_{3}, α_{4}, α_{5}, 0} in O_{K}. By Lemma 2.3.6, α_{5} is a unit and, since {0, α_{2}, α_{4}, 0} and {0, α_{3} − 1, α_{5} − 1, 0} are 3-periodic orbits, either α_{2}, α_{4}, α_{3} − 1, α_{5} − 1 are all units or they are all associated with 2. Since #O_{K}/(\sqrt{5}) = 5 and #O_{K}/(2) = 4, Lemma 2.8.6 tells us that there exist indices 2 ≤ j, k ≤ 5 such that 2|α_{j} and \sqrt{5}|α_{k}.

Suppose that all the mentioned numbers are units. Then in particular α_{2} and α_{5} are exceptional units. Then the only non-zero non-unit element in the orbit is α_{3}, therefore 2\sqrt{5}|α_{3}, but α_{3} is associated to α_{5} − α_{2}, the difference of two exceptional units. Then, up to associated, α_{3} ∈ \{±1, ±2, ±\sqrt{5}\}. Since none of these elements is divisible by 2\sqrt{5}, we have an absurd.

Then we must have that α_{2}, α_{4}, α_{3} − 1, α_{5} − 1 are associated with 2; in particular α_{2} = 2η_{2}, α_{3} − 1 = 2η_{3} with η_{2}, η_{3} ∈ O_{K}^{∗}. Moreover, by the results about the 3-periodic orbits, there exist exceptional units ϵ_{4}, ϵ_{5} such that α_{4} = ϵ_{4}α_{2}, α_{5} − 1 = ϵ_{5}(α_{3} − 1). Then α_{4} is associated with 2 and it follows that α_{3} must be divisible by \sqrt{5}. By Lemma 2.3.6, η_{1} = α_{2} − 1 = 2η_{2} − 1 is a unit and we have the unit equation

\frac{1}{η_{2}} + \frac{η_{1}}{η_{2}} = 2.

But then lemma 2.8.7 says that η_{2} ∈ {1, 2 − ξ, −ξ, ±1 + ξ}, implying that α_{2} ∈ {2, 4 − 2ξ, −2ξ, ±2 + 2ξ}. Moreover, we can write η_{4} = α_{5} − α_{4} = α_{5} − 2ϵ_{4}η_{2} ∈ O_{K}^{∗} and we get

\frac{α_{5}}{ϵ_{4}η_{2}} + \frac{−η_{4}}{ϵ_{4}η_{2}} = 2.

Up to conjugation we may restrict to the cases α_{2} ∈ {2, 4 − 2ξ, −2ξ}.

Suppose α_{2} = 2, which means η_{2} = 1. The preceding equation and Lemma 2.8.7 tell us that α_{5} ∈ \{ϵ_{4}, ϵ_{4}(1 + ξ), ϵ_{4}(1 − ξ)\}. The relation \sqrt{5}|α_{3}|α_{5} − 2 implies α_{5} ≡ 2 (mod \sqrt{5}). If α_{5} = ϵ_{4} then α_{4} = 2α_{5}, which implies α_{4} ≡ 4 (mod \sqrt{5}) and this is absurd because \sqrt{5}|α_{3}|α_{4} − 1. Then the only possibility is that α_{5} is a product of two exceptional units, that is,

α_{5} ∈ {±ξ, ±(1 + 2ξ), 2 + 3ξ, ±(2 − ξ), 3 − 2ξ}.

The condition α_{5} ≡ 2 (mod \sqrt{5}) enables us to restrict to the cases α_{5} ∈ \{1+2ξ, −ξ, 3−2ξ\}. From α_{3}|α_{4} − 1, we infer that 2ϵ_{4} = α_{4} ≡ 1 (mod \sqrt{5}), that is, ϵ_{4} ≡ 3 (mod \sqrt{5}). Since ϵ_{4} is an exceptional unit, we have ϵ_{4} ∈ {ξ, 1 − ξ} and α_{4} ∈ {2ξ, 2 − 2ξ}. The condition α_{5} − α_{4} ∈ O_{K}^{∗} lets us with only three possibilities: α_{4} = 2 − 2ξ, α_{5} = −ξ, 3 − 2ξ and α_{4} = 2ξ, α_{5} = 1 + 2ξ. In all of these cases α_{4} − 1 is associated with \sqrt{5}, which means \sqrt{5}|α_{3}|α_{4} − 1|\sqrt{5}, that is α_{3} = ϵ\sqrt{5}, with ϵ ∈ O_{K}^{∗}. The condition that α_{3} − 1 is associated with α_{2} implies that ϵ\sqrt{5} − 1 = α_{3} − 1 = 2η with η ∈ O_{K}^{∗}, leading to the unit equation

ϵ\sqrt{5} − 2η = 1.

By Lemma 2.8.7, the only solutions are ϵ = 1, η = ξ − 1 and ϵ = −1, η = ξ. Thus α_{3} = ±\sqrt{5}. The condition α_{4} − α_{3} ∈ O_{K}^{∗} excludes the cases α_{3} = \sqrt{5}, α_{4} = 1 − \sqrt{5} and α_{3} = −\sqrt{5}, α_{4} = 1 + \sqrt{5} and the condition that α_{5} − α_{3} is associated with 2 excludes the case α_{3} = −\sqrt{5}, α_{5} = −ξ. Then the two only possible orbits are

\{0, 1, 2, −\sqrt{5}, 1 − \sqrt{5}, 2 − \sqrt{5}, 0\} \{0, 1, 2, \sqrt{5}, 1 + \sqrt{5}, 2 + \sqrt{5}, 0\};

in both of these cases the Lagange interpolation polynomial does not have coefficients in O_{K} and therefore they can not be periodic orbits. Hence the case α_{2} = 2 is not possible.

Suppose α_{2} = 4 − 2ξ. As in the previous case we obtain that

\frac{α_{5}}{η_{2}ϵ_{4}} ∈ \{1 + ξ, 1 − ξ, 2 − ξ, ξ, 1\}.

Since α_{2} ≡ 3 (mod \sqrt{5}) and α_{4} ≡ 1 (mod \sqrt{5}), we have ϵ_{4} ∈ \{−ξ, ξ − 1\}. Since α_{5} ≡ α_{2} ≡ 3 (mod \sqrt{5}), we have α_{5} = η_{2}ϵ_{4} = (2 − ξ)ϵ_{4}, thus α_{5} ∈ \{1 − ξ, −3 + 2ξ\} and α_{4} ∈ \{2 − 2ξ, −6 + 4ξ\} and since \sqrt{5}|α_{3}|α_{4} − 1, we obtain α_{3} = ϵ\sqrt{5} with ϵ ∈ O_{K}^{∗}. As in the previous case, this leads to the unit equation ϵ\sqrt{5} − 2η = 1, that implies ϵ = ±1. The examination of the four arising cases shows that there are no possibles 6-periodic orbits in this case, too.

Suppose finally that α_{2} = −2ξ. This time we get

\frac{α_{5}}{ϵ_{4}} ∈ \{−1 − 2ξ, 1, 1 − ξ, −1 − ξ, −ξ\}

and since α_{2} ≡ 4 (mod \sqrt{5}), we get ϵ_{4} ∈ \{1 + ξ, 2 − ξ\} and thus α_{4} ∈ \{−2 − 4ξ, 4\}; the condition α_{4} ≡ 1 (mod \sqrt{5}) takes us to the conclusion α_{4} = −2 − 4ξ and ϵ_{4} = 1 + ξ, therefore α_{5} ∈ \{1 + ξ, −2 − 3ξ\}. As in the previous case we obtain again that α_{3} = ±\sqrt{5}. Examining the Lagrange interpolation polynomials, we see that all the four arising cases are not possible. Then we have shown that there are no possible 6-periodic orbits in O_{K}.

Let us now show that P er_{9}(f) is always empty. If {0, 1, α_{2}, ..., α_{8}, 0} would be a normalized 9-periodic orbit, Lemma 2.3.6 implies that α_{2}, α_{4}, α_{5}, α_{7}α_{8} ∈ O_{K}^{∗}. Moreover {0, α_{3}, α_{6}, 0} is a 3-periodic orbit and therefore, by the structure of these orbits, α_{3}, α_{6} are units or associated with 2. Then none of the elements of the orbit can be divisible by \sqrt{5}, contradicting Lemma 2.8.6. Then P er_{9}(f) is empty.

Now we want to deal with P er_{4}(f). Let {0, 1, α_{2}, α_{3}, 0} be a normalized 4-periodic orbit in O_{K}. According to Lemma 2.3.6, α_{3}, α_{3} − α_{2}, α_{2} − 1, α_{2}/(1 − α_{3}) ∈ O_{K} and they satisfy the equation (1 − α_{2}) + (α_{2} − α_{3}) + α_{3} = 1. If one of the summands equals 1, the only possibility is α_{2} = 1 + α_{3} and (1 + α_{3})/(1 − α_{3}) is a unit. Hence 1 − α_{3}|2 and therefore either 1 − α_{3} is a unit or 1 − α_{3} = 2η with η ∈ O_{K}^{∗}. In the first case we have

α_{3} ∈ {±ξ, 1 + ξ}

and we obtain two inequivalent orbits. In the second case we get the unit equation

\frac{1}{η} + \frac{−α_{3}}{η} = 2

and, using Lemma 2.8.7, we have

α_{3} ∈ {1 + 2ξ, −3 + 2ξ, 3 − 2ξ, −1 − 2ξ},

which gives us two other inequivalent orbits. Otherwise, if none of the summands equals 1, Lemma 2.8.7 gives us 24 possible pairs of values for (α_{2}, α_{3}) and, cheking the Lagrange interpolation polynomials, we find four other inequivalent 4-periodic orbits. Totally there are eight non-equivalent 4-periodic orbits in O_{K}, given by

{0, 1, 1 + ξ, ξ, 0}, {0, 1, 1 − ξ, −ξ, 0}, {0, 1, 2 + 2ξ, 1 + 2ξ, 0}, {0, 1, −2 + 2ξ, −3 + 2ξ, 0},

{0, 1, −ξ, 1 + ξ, 0}, {0, 1, 1 + ξ, −ξ, 0}, {0, 1, 1 − ξ, 2 − ξ, 0}, {0, 1, ξ − 1, ξ, 0}.

These orbits are realized, respectively, by the polynomials

2(3 − ξ)x^{3} − 3(2 + ξ)x^{2} + 6ξx + 1, 2(−3 + ξ)x^{3} + 3(4 − 3ξ)^{2} + 6(−1 + ξ)x + 1,

(7 − 4ξ)x^{3} + 3(−3 + ξ)x^{2} + 3(1 + ξ)x + 1, −(3 + 4ξ)x^{3} + 3(1 + 3ξ)x^{2} − 3(1 + ξ)x + 1, 2(−3 + 2ξ)x^{3} + (11 − 7ξ)x^{2} + 2(−3 + ξ)x + 1, 2(3 − 2ξ)x^{3} + (−8 + 5ξ)x^{2} + 2x + 1,

2ξx^{3} + (2 − 3ξ)x^{2} − 2x + 1, 2ξx^{3} − (4 + 3ξ)x^{2} + 2(1 + ξ)x + 1.

Finally, we need to show that P er_{8}(f) is empty. Let {0, 1, α_{2}, ..., α_{7}, 0} be a normalized 8-periodic orbit. By Lemma 2.3.6, α_{3}, α_{5}, α_{7} ∈ O_{K}^{∗}. On the other hand, {0, α_{2}, α_{4}, α_{6}, 0} is a 4-periodic orbit and therefore it must be equivalent to one of the previous list. In all of these cases none of the elements of the orbit is divisible by \sqrt{5}, contradicting Lemma 2.8.6 and the theorem is proved.

Now we want to prove a stronger result. Not only we shall calculate the constant D_{K} for all quadratic number fields, but we shall also exstablish the number of finite orbits of a given length containing a periodic orbit of given length. Before starting, we state a result about finite orbits of ℤ, which is obtained from Theorem 2.7.5 with some elementary calculation.

Lemma 2.8.13 : A finite maximal orbit in ℤ is equivalent to one of the following orbits: {1, 2, 0, 0}, {2, 1, 0, 3, 0}, {2, 4, 0, 0}, {±1, 0, n, 0} with n ≠ 3 or {n, 0, 0} with n ≠ 2, 4.

Now we can state the main result. In the following, we denote by R(m, n, k) the number of weakly inequivalent finite orbits in ℚ(\sqrt{m}) of length n containing a periodic orbit of length k.

Theorem 2.8.14 : The following are true:

1) i) D_{K} = 4 if m ≠ −3, 2, 5.

ii) D_{K} = 6 if m = 2, 5. iii) D_{K} = 7 if m = −3.

2) i) R(m, 3, 1) = ∞ if m ≥ 2.

ii) R(m, 3, 1) = 2 if m < 0, m ≠ −3, −1 and 2 does not have a proper divisor in O_{K}. iii) R(m, 3, 1) = 3 if m < 0, m ≠ −3, −1 and 2 has a proper divisor in O_{K}. iv) R(m, 3, 1) = 5 if m = −1.

v) R(m, 3, 1) = 3 if m = −3.

3) i) R(m, 4, 1) = 5 if m = 2.

ii) R(m, 4, 1) = 1 if m = −3, −1, 3.

iii) R(m, 4, 1) = 16 if m = 5.

iv) R(m, 4, 1) = 0 in all the other cases.

4) i) R(m, 5, 1) = 4 if m = 5.

ii) R(m, 5, 1) = 0 if m ≠ 5.

5) R(m, 3, 2) = ∞ for every squarefree integer m.

6) i) R(m, 4, 2) = ∞ if m > 0.

ii) R(m, 4, 2) = 2 if m = −3, −1.

iii) R(m, 4, 2) = 1 if m < 0, m ≠ −3, −1.

7) i) R(m, 5, 2) = 1 if m = −3.

ii) R(m, 5, 2) = 6 if m = 2. iii) R(m, 5, 2) = 39 if m = 5. iv) R(m, 5, 2) = 0 if m ≠ −3, 2, 5.

8) i) R(m, 6, 2) = 4 if m = 2. ii) R(m, 6, 2) = 6 if d = 5. iii) R(m, 6, 2) = 0 if m ≠ 2, 5.

9) i) R(m, 4, 3) = 4 if m = −3. ii) R(m, 4, 3) = 30 if m = 5. iii) R(m, 4, 3) = if m ≠ −3, 5.

10) i) R(m, 5, 3) = 2 if m = −3. ii) R(m, 5, 3) = 28 if m = 5. iii) R(m, 5, 3) = 0 if m ≠ 2, 5.

11) i) R(m, 6, 3) = 10 if m = 5. ii) R(m, 6, 3) = 0 if m ≠ 5.

12) i) R(m, 5, 4) = 4 if m = 2. ii) R(m, 5, 4) = 10 if m = 5. iii) R(m, 5, 4) = 0 if m ≠ 2, 5.

13) i) R(m, 6, 4) = 2 if m = 5. ii) R(m, 6, 4) = 0 if m ≠ 5.

14) i) R(m, 7, 6) = 1 if m = −3. ii) R(m, 7, 6) = 0 if m ≠ −3. Proof: 2) Let {α, β, γ, γ} be a 3-orbit ending with a fixed point. Up to conjugation,

we may assume that γ = 0. By Lemma 2.3.6, α|β and α − β|β, that implies β − α = ηα, with η ∈ O_{K}^{∗}, η ≠ −1. Thus we have shown that every 4-orbit containing a fixed point is of the form {α, (1 + η)α, 0, 0} with η ∈ O_{K}^{∗}, η ≠ −1 and α|1 + η. Now part i) is immediately established from the fact that the unit group is infinite and therefore we have infinite possibilities for η. If m < 0, m ≠ −3, −1, the only possibility is η = 1 and α|2; this implies parts ii) and iii). Parts iv) and v) can be easily obtained by calculation and the resulting orbits are respectively

\{1, 2, 0, 0\}, \{1 + \sqrt{−1}, 2 + 2\sqrt{−1}, 0, 0\}, \{2 + 2\sqrt{−1}, 4\sqrt{−1}, 0, 0\},
\{1, 1 + \sqrt{−1}, 0, 0\}, \{1 + \sqrt{−1}, 2, 0, 0\},

and

{1, 1 + ζ_{6}, 0, 0}, {1 + 2ζ_{6}, 3 − ζ_{6}, 0, 0}, {1, 2 − ζ_{6}, 0, 0}.

3) Again we may assume that the orbit ends with the fixed point 0 and has the form {α_{0}, α_{1}, α_{2}, 0, 0}. Then it is realized by the polynomial

f(x) = x(x − α_{2})(ax + b),

with a, b ∈ O_{K}. In particular

α_{1} = f(α_{0}) = α_{0}(α_{0} − α_{2})(aα_{0} + b)

and

α_{2} = f(α_{1}) = α_{1}(α_{1} − α_{2})(aα_{1} + b).

These relations imply that α_{1}|α_{2}, that is, α_{2} = λα_{1} with λ ∈ O_{K}, and that α_{1}−α_{2}|α_{2}, that is, 1 − λ|λ, therefore 1 − λ = η_{1} ∈ O_{K}^{∗}. Moreover α_{0}(α_{0} − α_{2})|α_{1}|α_{2}, thus α_{2} = (1 − η_{2})α_{0}, with η_{2} ∈ O_{K}^{∗}. We also have that α_{1} − α_{0}|α_{2} − α_{1} = −η_{1}α_{1}|α_{1}, hence α_{1} = μα_{0} and μ−1|μ, therefore μ−1 = η_{0} ∈ O_{K}^{∗}. We have obtained that all the elements of the orbit are divisible by α_{0} and then, up to conjugation, we may assume α_{0} = 1 and the orbit becomes {1, 1 − η_{0}, 1 − η_{2}, 0, 0}. The relation α_{2} = (1 − η_{2})α_{0} = (1 − η_{1})(1 + η_{0})α_{0} leads us to the unit equation η_{1} = η_{2} + η_{0} − η_{0}η_{1}, that is

\frac{η_{2}}{η_{1}} + \frac{η_{0}}{η_{1}} + (−η_{0}) = 1.

If we assume that none of the summands equals 1, then, by Lemma 2.8.7, the only possibilities are m = 2, 5. Let us first examine the case m = 2. By Lemma 2.8.7, we have the following possibilities: η_{0} = η_{2} = −3 − 2\sqrt{2}, η_{0} = 1, η_{2} = −3 + \sqrt{2}, and η_{0} = ±(1 + \sqrt{2}), η_{2} = −1, 3 + 2\sqrt{2}, but the first two violate the condition α_{1}(α_{1} − α_{2})|α_{2}. Then this case gives us four possible orbits. In the case m = 5, Lemma 2.8.7 gives us eleven possible pairs of values for (η_{0}, η_{2}) and therefore we have found eleven different orbits for this case. Assume now that none of the summands equals 1. In that case the only possibility is η_{0} = η_{1}, beacuse η_{0} = −1 implies α_{1} = 0 and η_{1} = η_{2} implies α_{2} = 0 and these cases are not possible. Then the orbit becomes {1, 1 + η_{0}, 1 − η_{0}^{2}, 0, 0} and it is realized by the polynomial

f(x) = x(x − 1 + η_{0}^{2})(ax + b).

In particular,

1 + η_{0} = f(1) = η_{0}^{2}(a + b)

and

1 − η_{0}^{2} = f(1 + η_{0}) = η_{0}(1 + η_{0})^{2}(a(1 + η_{0}) + b).

Thus we have 1 − η_{0} = η_{0}(1 + η_{0})(a(1 + η_{0}) + b), and we deduce that

a + b = \frac{1 + η_{0}}{η_{0}^{2}} , a(1 + η_{0}) + b = \frac{1 − η_{0}}{η_{0}(1 + η_{0})}.

This shows that 1 + η_{0}|2. Therefore we have three possibilities: either 1 + η_{0} is a unit or it is associated with 2 or it is a proper divisor of 2. In the case that 1 + η_{0} is a unit, we have the unit equation (1 + η_{0}) + (−η_{0}) = 1 and Lemma 2.8.7 implies that either m = −3, η_{0} = −ζ_{6} or m = 5, η_{0} ∈ {ξ, −ξ, −1 − ξ} and we can realize the respective orbits. In the case that 1 + η_{0} is associated with 2, that is, 1 + η_{0} = 2ϵ, with ϵ ∈ O_{K}^{∗}, we have the unit equation

\frac{1}{ϵ} + \frac{η_{0}}{ϵ} = 2.

Then, by lemma 2.8.7, being the case η_{0} = ϵ = 1 impossible, the possibilities are m = 2, η_{0} = −3 + 2\sqrt{2} and m = 5, η_{0} = 1 + 2ξ, −3 + 2ξ and we have the corresponding orbits. In the case that 1 + η_{0} is a proper divisor of 2, then Norm_{K/ℚ}(1 + η_{0}) = ±2. Then we can apply lemma 2.8.7, obtaining the possibilities m = −1, η_{0} = \sqrt{−1}, m = 2, η_{0} = ±1 + \sqrt{2} and m = 3, η_{0} = −2 + \sqrt{3}. Anyway the case m = 2 gives orbits equivalent to the ones found before, so we do not need to consider them. We finally list our results: in case m = 2, we have found five orbits given by

\{1, 2 + \sqrt{2}, 2, 0, 0\}, \{1, 2 + \sqrt{2}, −2 − 2\sqrt{2}, 0, 0\}, \{1, −\sqrt{2}, −2 − 2\sqrt{2}, 0, 0\},
\{1, −\sqrt{2}, 2, 0, 0\}, \{1, −2 + 2\sqrt{2}, −16 + 2\sqrt{2}, 0, 0\},

proving part i). We have found, for m = −3, the orbit {1, 1 − ζ_{6}, 1 − ζ_{6}^{2}, 0, 0}, for m = −1, the orbit \{1, 1 + \sqrt{1}, 2, 0, 0\} and. for m = 3, the orbit \{1, −1 + \sqrt{3}, −6 + 4\sqrt{3}, 0, 0\}, proving part ii). In case m = 5, we have found the sixteen weakly inequivalent orbits

{1, −ξ, 1 − ξ, 0, 0}, {1, −ξ, −2ξ, 0, 0},
{1, 1 + ξ, 2 + ξ, 0, 0}, {1, 1 + ξ, ξ, 0, 0},
{1, 1 + ξ, 2 + 2ξ, 0, 0}, {1, ξ − 1, 2 − ξ, 0, 0},
{1, ξ − 1, 4 − 2ξ, 0, 0}, {1, 1 − ξ, 2 − ξ, 0, 0},
{1, 1 − ξ, −2ξ, 0, 0}, {1, ξ, 3 − ξ, 0, 0},
{1, ξ, 2, 0, 0}, {1, 1 + ξ, −ξ, 0, 0},
{1, 1 − ξ, −ξ, 0, 0}, {1, −ξ, −1 − 3ξ, 0, 0},
{1, 2 + 2ξ, −4 − 8ξ, 0, 0}, {1, −2 + 2ξ, −4 − 8ξ, 0, 0},

proving part iii). In all the other cases we have found no orbits, as part iv states. A final check on the Lagrange interpolation polynomials of the orbits shows that they are all the possible 4orbits ending with a fixed point and 3) is proved.

4) Again assume that 0 is the ending fixed point and let {α_{0}, α_{1}, α_{2}, α_{3}, 0, 0} be a desired 5-orbit. Then the 4-orbit {α_{1}, α_{2}, α_{3}, 0, 0} must be weakly equivalent to one of the orbits listed in 3). Therefore we may assume that α_{1} = 1 and thus α_{0} must be a unit. Moreover we have α_{0} − 1|f(α_{0}) − f(1) = 1 − α_{2}. Looking at all the orbits listed in 3, we see that 1 − α_{2} is always a unit and then so does α_{0} − 1; hence α_{0} is an exceptional unit. By Lemma 2.8.7, O_{K} contains an exceptional unit if and only if m = −3, 5. Suppose m = −3, then α_{0} = ζ_{6} and the orbit becomes {ζ_{6}, 1, 1 − ζ_{6}, 2 − ζ_{6}, 0, 0}; this orbit can not exist, otherwise we would have

\sqrt{−3}|(1 − ζ_{6}) − ζ_{6}|(2 − ζ_{6}) − 1|1,

which is a contradiction. This proves part ii). We are left to prove part i). Take m = 5. By Lemma 2.8.7, we have

α_{0} ∈ {ξ, −ξ, 1 + ξ, −1 + ξ, 1 − ξ, 2 − ξ}.

Attaching these possible values to the sixteen orbits of 3.iii, we obtain all the possible orbits. In all of them 1 − α_{3} is a unit. Since α_{0} − α_{3}|α_{1} = 1, α_{0} − α_{3} must also be a unit. Let us consider the last two orbits; in those cases the condition that α_{0} − α_{2} must be a unit implies that α_{0} ∈ {−ξ, 1 + ξ} and α_{0} ∈ {ξ, −1 + ξ} respectively. In all of those cases the Lagrange interpolation polynomial has not coefficient in O_{K} and then they can not be 5-orbits in O_{K}. Let us look at the other orbits. We see that α_{2}, η_{1} = α_{0} −1, η_{2} = α_{0} −α_{2} ∈ O_{K}^{∗} and satisfy the unit equation

α_{2} + η_{2} + (−η_{1}) = 1.

If some of the summands is equal to 1, the only possibility is η_{2} = 1, that is, α_{0} = 1 + α_{2}. This fact excludes the first, the third and the eighth orbit of the list, because they already contain the element 1 + α_{2}. Moreover, we have to exclude the fourth, the fifth, the sixth, the seventh, the ninth, the tenth, the twelfth, the thirteenth and the fourteenth orbit, in which α_{0} − α_{3} = 1 + α_{2} − α_{3} is not a unit. We are left with the second and the eleventh orbit and a simple check shows that they can be prolonged, giving us two orbits of the required form. If none of the summands is equal to 1, Lemma 2.8.7 gives five different possibilities for α_{2}, that is α_{2} ∈ {−ξ, 1 + ξ, 1 − ξ, ξ, ξ − 1}. Let us examine all of these cases. If α_{2} = −ξ, we have to analyse the first, the second and the fourteenth orbits of the list. The condition that α_{0} = α_{2} + η_{2} ∈ O_{K}^{∗} implies η_{2} = 2ξ + 1, α_{0} = ξ + 1 and α_{3} ∈ {1−ξ, −2ξ, −1−3ξ}. Since α_{0} −α_{3} must be a unit, the only possibility is α_{3} = 1−ξ. Anyway the resulting orbit is not possible beacuse its Lagrangian interpolation polynomial has not coefficients in O_{K}. Therefore this case is not possible. If α_{2} = 1+ξ, we consider the third, the fourth, the fifth and the twelfth orbits. In order to satisfy all the conditions we must have η_{2} = −1−2ξ, α_{0} = −ξ and α_{3} ∈ {2+ξ, ξ, 2+2ξ, −ξ}. But since α_{0} −α_{3}, the only possibility is α_{3} = 2+2ξ and we have another acceptable orbit. If α_{2} = 1 − ξ, we consider the eighth, the ninth and the thirteenth orbit. We obtain that η_{2} = 2ξ + 1 and α_{3} ∈ {2 − ξ, −2ξ, −ξ} and the condition α_{0} − α_{3} ∈ O_{K}^{∗} gives α_{2} = −2ξ and α_{0} = 2 − ξ. Anyway the Langrange interpolation polynomial of this orbit has not coefficients in O_{K}, therefore this case is not possible. If α_{2} = ξ, we consider the tenth and the eleventh orbit and we obtain η_{2} = −1, α_{0} = ξ − 1 and α_{3} ∈ {3 − ξ, 2}. However in all of these cases α_{0} − α_{3} is not a unit and therefore this case is not possible. If α_{2} = ξ − 1, we consider the sixth and seventh orbit. We get η_{2} ∈ {−1 + ξ, 3 − 2ξ} and cheking the four arising cases, we obtain another acceptable orbit. We have finally collected four weakly inequivalent orbits, given by

{1 − ξ, 1, −ξ, −2ξ, 0, 0}, {1 + ξ, 1, ξ, 2, 0, 0},

{−ξ, 1, 1 + ξ, 2 + 2ξ, 0, 0}, {2 − ξ, 1, ξ − 1, 4 − 2ξ, 0, 0},

proving part i). Then 4) is proved.

5) We can assume that the 2-periodic point is 0, hence the orbit is {α_{0}, 0, α_{1}, 0}. The polynomial realizing this orbit has the form a(x − α_{0})(x − α_{1}), with a ∈ O_{K}, a ≠ 0. Then α_{1} = f(0) = aα_{0}α_{1} and, since α_{1} ≠ 0, we have 1 = aα_{0}, that is, α_{0} ∈ O_{K}^{∗}. This shows that for any unit α_{0}, we can build infinitely many 3-orbits containing a 2-periodic orbit, varying α_{1} ∈ O_{K}, α_{1} ≠ 0, α_{0}. This proves 5).

6) Again assume that the periodic point is 0. A 4-orbit containing a 2-periodic orbit has the form {α_{0}, α_{1}, 0, α_{2}, 0}. And the polynomial realizing it is

f(x) = (x − α_{1})(x − α_{2})(ax + b)

with a, b ∈ O_{K}. We have

α_{2} = f(0) = bα_{1}α_{2}

and since α_{2} ≠ 0, we have that α_{1} ∈ O_{K}^{∗}. Therefore we may assume α_{1} = 1 and b = 1. It follows that

1 = f(α_{0}) = (α_{0} − 1)(α_{0} − α_{2})(aα_{0} − 1),

therefore η_{1} = α_{0} − 1 and η_{2} = α_{0} − α_{2} are units. Moreover, since a = (1 + η_{1}η_{2})/η_{1}η_{2}(1 + η_{1}) ∈ O_{K}, we have that α_{0} = 1 + η_{1}|1 + η_{1}η_{2} ≡ 1 + η_{2} mod α_{0}. Hence any 4-orbit ending with a 2-periodic orbit has the form {1+η_{1}, 1, 0, 1+η_{1} −η_{2}, 0}, with η_{1}, η_{2} ∈ O_{K}^{∗} such that 1 + η_{1}|1 + η_{2}. Then we have just to check the number of such units in quadratic fields. If m < 0, m ≠ −3, −1, the only pair of units that satisfies the required condition is clearly (1, −1), then there is only one possible such orbit; this proves part iii). If m = −1, we have additionally the Pair (\sqrt{−1}, −\sqrt{−1}) and, if m = −3, the pair (ζ_{6}, 1 − ζ_{6}); this proves ii). Finally, for positive m, let θ_{m} be a fundamental unit of ℚ(\sqrt{m}) and consider the relations

\frac{1 + η_{2}}{1 + η_{1}} = \frac{1 − θ_{m}^{rs}}{1 − θ_{m}^{r}} , \frac{1 + η_{2}}{1 + η_{1}} = \frac{1 + θ_{m}^{rs}}{1 + θ_{m}^{r}} , \frac{1 + η_{2}}{1 + η_{1}} = \frac{1 − θ_{m}^{rs}}{1 + θ_{m}^{r}} ,

with r, s ∈ ℤ. It is easy to check that there are infinitely many pairs of units which satisfy at least one of these realtions, giving us infinitely many orbits of the required form.

7) From 6), we deduce that every 5-orbit containing a 2-periodic orbit has the form {α_{0}, 1 + η_{1}, 1, 0, 1 + η_{1} − η_{2}, 0} with η_{1}, η_{2} ∈ O_{K}^{∗} such that 1 + η_{1}|1 + η_{2} and the orbit is generated by the polynomial f(x) = (x − 1)(x − 1 − η_{1} + η_{2})g(x). By Lemma 2.3.6, we have that ϵ_{1} = 1 + η_{1} − α_{0} and α_{0} must be units. Then we obtain the unit equation

α_{0} + ϵ_{1} + (−η_{1}) = 1.

If some of the summands equals 1, the only possibility is ϵ_{1} = 1, hence α_{0} = η_{1}. Now η_{1} − 1 = α_{0} − 1|(1 + η_{1}) − (1 + η_{1} − η_{2}) = η_{2} and it follows that η_{1} is an exceptional unit. Therefore we must have m = −3, 5.

Suppose m = −3. Then, by Lemma 2.8.7, η_{1} = ζ_{6} and it follows that

η_{2} ∈ {ζ_{6}, 1, 1 − ζ_{6}, −1, −ζ_{6}, −1 + 3ζ_{6}}.

Anyway, an immediate calculation shows that the only possible case is η_{2} = 1 − ζ_{6}, since all the other ones lead to contradictions. Then we have found one possible orbit. Examine now the case m = 5. By Lemma 2.8.7, we have

α_{0} = η_{1} ∈ {ξ, −ξ, 1 + ξ}.

In the first two cases 1 + η_{1} is a unit and, since η_{2} − 1|1 + η_{1}, we infer that η_{2} is an exceptional unit, too. This enables us to obtain eight possible orbits. If η_{1} = 1 + ξ, then 1 − η_{2}|2 + ξ|\sqrt{5}, therefore either η_{2} is an exceptional unit or 1 − η_{2} is associated with \sqrt{5}. In the first case, the condition \sqrt{5}|2 + ξ|η_{1} − η_{2} implies that the only possiblity is η_{2} = 2 − ξ and we have another possible orbit. In the second case, writing 1 − η_{2} = ϵ_{2}\sqrt{5} with ϵ_{2} ∈ O_{K}^{∗}, we obtain the unit equation

\frac{1}{ϵ_{2}} + \frac{−η_{2}}{ϵ_{2}} = \sqrt{5},

and Lemma 2.8.7 tells us that η_{2} ∈ {2 + 3ξ, 5 − 3ξ, −1 − ξ, −2 + ξ}. However all of these cases violate the condition \sqrt{5}|η_{1} −η_{2}. Then we have obtained in total nine possible orbits. Suppose now that, in the equation α_{0} + ϵ_{1} + (−η_{1}) = 1, none of the summands is equal to 1. By Lemma 2.8.7, this can happen only if m = 2, 5. We shall examine these two cases. If m = 2, Lemma 2.8.7 gives us the following possibilities:

α_{0} = 3 + 2\sqrt{2}, ϵ_{1} = −1 − \sqrt{2}, η_{1} = 1 + \sqrt{2},
α_{0} = −1, ϵ_{1} = 1 + \sqrt{2}, η_{1} = −1 + \sqrt{2},
α_{0} = −1 − \sqrt{2}, ϵ_{1} = −1 − \sqrt{2}, 3 + 2\sqrt{2}, η_{1} = −3 − 2\sqrt{2}, 1 + 2\sqrt{2},
α_{0} = 1 ± \sqrt{2}, ϵ_{1} = −1, η_{1} = −1 + \sqrt{2}, 1.

The first two cases are not possible, because they violate the condition α_{0} − 1|1 + η_{1}. Since α_{0} − 1 − η_{1} + η_{2} = η_{2} − ϵ_{1}, Lemma 2.8.7 implies that α_{0} − 1 − η_{1} + η_{2} is not a unit. Since 1 + η_{1} = f(α_{0}) = (α_{0} − 1)(α_{0} − 1 − η_{1} + η_{2})g(α_{0}), we get that (1 + η_{1})/(α_{0} − 1) is not a unit, too. Then we are left with the cases α_{0} = −1 − \sqrt{2}, η_{1} = −3 − 2\sqrt{2} and α_{0} = 1 ± \sqrt{2}, η_{1} = 1. If α_{0} = −1 − \sqrt{2}, we set β = 1 + η_{1} − η_{2} − α_{0} and ϵ_{3} = β − α_{0}. We have that β|2 and

β = −\sqrt{2} − 1 − η_{2} ≡ \sqrt{2} mod 2,

therefore β is associated with \sqrt{2}. Write β = η_{3}\sqrt{2} with η_{3} ∈ O_{K}^{∗} and we have the unit equation

\frac{ϵ_{3}}{η_{3}} + \frac{α_{0}}{η_{3}} = \sqrt{2}.

By Lemma 2.8.7, we have η_{3} ∈ \{±ϵ_{3}, ±(1 + \sqrt{2})\}; the case η_{3} = 1 + \sqrt{2} leads to η_{1} = η_{2}, which is impossible, while the other cases give us three possible orbits. If α_{0} = 1 + \sqrt{2} then ϵ_{3} = η_{1} − η_{2} − 1 and β = −η_{2} − 1 − \sqrt{2}|2. As in the previous case, we obtain that β = η_{3}\sqrt{2}, getting the unit equation

\frac{−η_{2}}{η_{3}} + \frac{1 − \sqrt{2}}{η_{3}} = \sqrt{2}

and Lemma 2.8.7 implies that η_{3} ∈ \{±η_{2}, ±(1 + \sqrt{2})\}. The first two cases imply η_{1} = η_{2} and ϵ_{3} = 5 − 2\sqrt{2} ∉ O_{K}^{∗} respectively and therefore are impossbile. The other cases give us other two new orbits. Finally if α_{0} = 1 − \sqrt{2}, we have η_{1} = 1 and again β = η_{3}\sqrt{2}, obtaininig

\frac{1 + \sqrt{2}}{η_{3}} + \frac{−η_{2}}{η_{3}} = \sqrt{2}

and Lemma 2.8.7 gives us the cases η_{3} ∈ \{±η_{2}, ±(1 + \sqrt{2})\}. The cases η_{3} = ±(1 + \sqrt{2}) give orbits conjugated to the previous found ones and the case η_{3} = η_{2} implies η_{1} = η_{2}, which is absurd. Therefore we are left with the case η_{3} = −η_{2}, which gives us the sixth last possible orbit. If m = 5, Lemma 2.8.7 gives us a list of 24 possible pairs of values for α_{0}, η_{1}.

(1 + ξ, 1 + 2ξ), (1 + ξ, −1 − ξ), (−1 − 2ξ, −1 − ξ), (−ξ, 1 + 2ξ),
(−ξ, −2 − 3ξ), (2 + 3ξ, 1 + 2ξ), (2 + 3ξ, ξ), (−1 − 2ξ, −2 − 3ξ),
(−1 − 2ξ, ξ), (−ξ, −1 − 2ξ), (−ξ, ξ), (1 + 2ξ, ξ),
(2 − ξ, 1), (2 − ξ, −ξ), (ξ, −2 + ξ), (ξ, 1),
(−1, −ξ), (−1, −2 + ξ), (1 − ξ, −1 − 2ξ), (1 − ξ, 1 + ξ),
(−1 − ξ, −1 − 2ξ), (−1 − ξ, −1 − ξ), (1 + 2ξ, −1 + ξ), (1 + 2ξ, 1 + ξ).

We see that 1 + η_{1} is either a unit or associated with 2 or with \sqrt{5}. If 1 + η_{1} is a unit, which occurs in twelve cases, the equality 1 + η_{1} = (α_{0} − 1)(α_{0} − 1 − η_{1} + η_{2})g(α_{0}), implies that α_{0} − 1 and γ = α_{0} − 1 − η_{1} + η_{2} are units. Therefore α_{0} is an exceptional unit and we have only four pairs with this property. Moreover we have

\frac{η_{2}}{ϵ_{1}} + \frac{−γ}{ϵ_{1}} = 1,

which implies that η_{2} = ϵ_{1}η_{4}, with η_{4} an exceptional unit. Removing the orbits with duplications, this method produces sixteen new acceptable orbits. If 1+η_{1} is associated with 2, there are eight pairs with this property. In this case α_{0} −1|1+ η_{1}|2, which eliminates two pairs, and γ|2; therefore γ must be a either a unit or associated with 2. But the second case would imply α_{0} ≡ 1 mod 2, which is absurd, therefore γ ∈ O_{K}^{∗}. We obtain the equation

\frac{η_{2}}{ϵ_{1}} + \frac{−γ}{ϵ_{1}} = 1,

which implies again that η_{2} = ϵ_{1}η_{4} with η_{4} an exceptional unit. Cheking this condition, we find other eleven possible orbits. Finally, if 1 + η_{1} is associated with \sqrt{5}, we have four possible pairs of values. Again we have that α_{0} − 1|\sqrt{5} and that γ must be a unit or associated with \sqrt{5}, excluding two of the pairs. In the first case we have

\frac{γ}{η_{2}} + \frac{ϵ_{1}}{η_{2}} = 1

and this implies again that η_{2} = ϵ_{1}η_{4} for some exceptional unit η_{4}. This gives us other four orbits. In the second case, writing γ = \sqrt{5}ϵ_{4}, with ϵ_{4} ∈ O_{K}^{∗}, we have

\frac{η_{2}}{ϵ_{4}} + \frac{ϵ_{1}}{ϵ_{4}} = \sqrt{5}

and Lemma 2.8.7 gives us the possible values of η_{2}. It is easy to check that no orbits arise in this way. Now we group together our results. If m ≠ −3, 2, 5, there are no possible orbits of this type; this proves part iv). If m = −3, we have found the only orbit

{ζ_{6}, 1 + ζ_{6}, 1, 0, 2ζ_{6}, 0},

proving part i). If m = 2, we have found six orbits given by:

\{−1 − \sqrt{2}, −2 − 2\sqrt{2}, 1, 0, −5 − 4\sqrt{2}, 0\}, \{−1 − \sqrt{2}, −2 − 2\sqrt{2}, 1, 0, −1 − 2\sqrt{2}, 0\},

\{−1 − \sqrt{2}, −2 − 2\sqrt{2}, 1, 0, −3 − 2\sqrt{2}, 0\}, \{1 + \sqrt{2}, 2, 1, 0, −1 + 2\sqrt{2}, 0\},
\{1 + \sqrt{2}, 2, 1, 0, 3, 0\}, \{1 − \sqrt{2}, 2, 1, 0, 5 + 2\sqrt{2}, 0\},

proving part ii). Finally, if m = 5, we have found thirty-nine weakly inequivalent orbits, proving part iii). These orbits are given by

{ξ, 1 + ξ, 1, 0, 2ξ, 0}, {ξ, 1 + ξ, 1, 0, −1 + 2ξ, 0}, {ξ, 1 + ξ, 1, 0, 1 + 2ξ, 0},
{ξ, 1 + ξ, 1, 0, 2, 0}, {−ξ, 1 − ξ, 1, 0, −2ξ, 0}, {−ξ, 1 − ξ, 1, 0, −1, 0},

{−ξ, 1 − ξ, 1, 0, 2 − 2ξ, 0}, {−ξ, 1 − ξ, 1, 0, 1 − 2ξ, 0}, {1 + ξ, 2 + ξ, 1, 0, 2ξ, 0},

{1 + ξ, −ξ, 1, 0, 3 + 4ξ, 0}, {1 + ξ, −ξ, 1, 0, −2 − 4ξ, 0}, {1 + ξ, −ξ, 1, 0, 2 + 2ξ, 0},

{1 + ξ, −ξ, 1, 0, −1 − 2ξ, 0}, {−ξ, 1 + ξ, 1, 0, −2 − 4ξ, 0}, {−ξ, 1 + ξ, 1, 0, 3 + 4ξ, 0},

{−ξ, 1 + ξ, 1, 0, −1 − 2ξ, 0}, {−ξ, 1 + ξ, 1, 0, 2 + 2ξ, 0}, {2 − ξ, 1 − ξ, 1, 0, 2, 0},

{2 − ξ, 1 − ξ, 1, 0, 1 − 2ξ, 0}, {2 − ξ, 1 − ξ, 1, 0, 2 − 2ξ, 0}, {2 − ξ, 1 − ξ, 1, 0, 3 − 2ξ, 0},

{ξ, −1 + ξ, 1, 0, 2ξ, 0}, {ξ, −1 + ξ, 1, 0, −1, 0}, {ξ, −1 + ξ, 1, 0, −1 + 2ξ, 0},

{ξ, −1 + ξ, 1, 0, −2 + 2ξ, 0}, {1 + ξ, 2 + 2ξ, 1, 0, 3 + 4ξ, 0}, {1 + ξ, 2 + 2ξ, 1, 0, 1 + 2ξ, 0},

{−ξ, 2 + 2ξ, 1, 0, −3 − 6ξ, 0}, {−ξ, 2 + 2ξ, 1, 0, 3 + 4ξ, 0}, {−ξ, −2ξ, 1, 0, −1 − 2ξ, 0},

{−ξ, −2ξ, 1, 0, 1 − 2ξ, 0}, {2 − ξ, 2, 1, 0, 1 − 2ξ, 0}, {2 − ξ, 2, 1, 0, 3, 0},
{ξ, 2, 1, 0, −1 + 2ξ, 0}, {ξ, 2, 1, 0, 5 − 2ξ, 0}, {1 − ξ, −2ξ, 1, 0, −1 − 4ξ, 0},

{1 − ξ, −2ξ, 1, 0, 1 − 2ξ, 0}, {−ξ, −1 − 3ξ, 1, 0, −2 − 4ξ, 0}, {1 − ξ, 2 + ξ, 1, 0, 4 + 4ξ, 0}.

Therefore 7 is proved

8) If m ≠ −3, 2, 5 then R(m, 6, 2) = 0, otherwise, cutting away the first term of the orbit, we would obtain a 5-orbit ending with a 2-periodic orbit, which does not exist in that case. We shall examine those three cases, attempting to prolong the orbits. Condider the case m = −3. A 6-orbit ending with a 2-periodic orbit would have the form {α, ζ_{6}, 1 + ζ_{6}, 1, 0, 2ζ_{6}, 0}. By Lemma 2.3.6, η_{1} = α + ζ_{6} and η_{2} = α − ζ_{6} − 1 are units and, since η_{1} − η_{2} = 1, Lemma 2.8.7 implies that η_{1} = ζ_{6}, 1 − ζ_{6}, that is, α = 2ζ_{6}, 1 and both of these cases are not possible. Examine the case m = 2. Consider an orbit of the form \{β, −1 − \sqrt{2}, −2 − 2\sqrt{2}, 1, 0, α, 0\}, where α ∈ \{−5 − 4\sqrt{2}, −1 − 2\sqrt{2}, −3 − 2\sqrt{2}\}. Then η_{1} = β + 1 + \sqrt{2} and η_{2} = β − 1 are units and we have

\frac{η_{1}}{1 + \sqrt{2}} + \frac{−η_{2}}{1 + \sqrt{2}} = \sqrt{2}.

By Lemma 2.8.7, we can determine the values of η_{1}, η_{2} and β, obtaining two possible orbits. If we consider an orbit of the form \{β, 1 + \sqrt{2}, 2, 1, 0, α, 0\}, then η_{1} = β − 1 − \sqrt{2} and η_{2} = β − 1 are units satisfying η_{2} − η_{1} = 1 and again this determines the possible values of β; we obtain other two weakly inequivalent orbits in this way. Finally, if we have an orbit of the form \{β, 1 − \sqrt{2}, 2, 1, 0, α, 0\}, then η_{1} = β − 1 − \sqrt{2} and η_{2} = β − 2 are units satisfying η_{1} − η_{2} = 1 + \sqrt{2} ∈ O_{K}^{∗} and this is impossible. Then we have totally four weakly inequivalent orbits given by:

\{2+\sqrt{2}, −1−\sqrt{2}, −2−2\sqrt{2}, 1, 0, −5−4\sqrt{2}, 0\}, \{−\sqrt{2}, −1−\sqrt{2}, −2−2\sqrt{2}, 1, 0, −1−2\sqrt{2}, 0\},

\{\sqrt{2}, 1 + \sqrt{2}, 2, 1, 0, 3, 0\}, \{\sqrt{2}, 1 + \sqrt{2}, 2, 1, 0, −1 + 2\sqrt{2}, 0\}.

Let now m = 5 and let {β, α_{0}, α_{1}, 1, 0, α_{4}, 0} be a 6-orbit containing a 2-periodic orbit. Dropping β, we must obtain one of the orbits listed in 7). If α_{1} is a unit, then even β, β−α_{1} are units. Moreover α_{0}, β − α_{0}, β − 1 are units, too, and therefore β, α_{0}/β and α_{1}/β must be exceptional units. A rapid check shows that this is impossible. Suppose then that α_{1} is associated with \sqrt{5}, that is, α_{1} = ϵ\sqrt{5} with ϵ ∈ O_{K}^{∗}; it follows that β is an exceptional unit and η = α_{1} − β is a unit, too. Then we get

\frac{β}{ϵ} + \frac{η}{ϵ} = \sqrt{5}

and Lemma 2.8.7 gives us the possibilities for β. A quick check shows that neither in that case there are possible orbits. Finally, let α_{1} be associated with 2. Then η_{1} = β − 1 and η_{2} = β − α_{0} are units satisfying

α_{0} + η_{2} + (−η_{1}) = 1.

If one of the summands equals 1, we must have η_{2} = 1 and therefore β = 1 + α_{0}. This gives origin to two possible orbits. If none of the summands is 1, Lemma 2.8.7 gives us the possible values of α_{0} and β. A check of all resulting orbits shows that there are other four possible orbits in this case. In total we have found the six weakly inequivalent orbits

{1 + ξ, ξ, 2, 1, 0, −1 + 2ξ, 0}, {1 − ξ, −ξ, −2ξ, 1, 0, 1 − 2ξ, 0},
{−ξ, 1 + ξ, 2 + 2ξ, 1, 0, 3 + 4ξ, 0}, {1 + ξ, −ξ, 2 + 2ξ, 1, 0, 3 + 4ξ, 0},
{1 − ξ, 2 − ξ, 2, 1, 0, 1 − 2ξ, 0}, {−ξ, 1 − ξ, −2ξ, 1, 0, 1 − 2ξ, 0}.

Therefore 8) is proved.

9) By Theorem 2.8.12, we have that R(m, n, 3) = 0 for m ≠ −3, 5, then we can consider only these two cases. Assume that the periodic point is 0. A 4-orbit containing a 3-periodic orbit has the form {α_{0}, 0, α_{2}, α_{3}, 0} and it is realized by the polynomial f(x) = (x − α_{0})(x − α_{3})(ax + b) ∈ O_{K}[x]. By Lemma 2.3.6, we have that α_{0}|α_{2}|α_{3} − α_{2}|α_{3} and α_{3}|α_{2}, therefore α_{2} = λα_{0} with λ ∈ O_{K} and α_{3} = η_{1}α_{2} with η_{1} ∈ O_{K}^{∗} and (η_{1}−1)α_{2} = η_{2}α_{2} with η_{2} ∈ O_{K}^{∗}. Hence η_{1} is an exceptional unit. Since

η_{1}α_{2} = α_{3} = f(α_{2}) = (α_{2} − α_{0})(α_{2} − α_{3})(aα_{2} + b) = α_{2}(α_{2} − α_{0})(1 − η_{1})(aα_{2} + b),

we obtain that α_{0}α_{2}(λ − 1)(1 − η_{1})|α_{2}|α_{3}, therefore α_{0} is a unit and we may assume α_{0} = 1. Renaming α_{2} = α, our orbit becomes {1, 0, α, η_{1}α, 0} with η_{1} an exceptional unit and 1 − α ∈ O_{K}^{∗}. Now we have that α = f(0) = bη_{1}α, therefore b = η_{1}^{−1}. Moreover we have

aα = \frac{η_{1}}{(α − 1)(1 − η_{1})} − b = \frac{η_{1}^{2} − (α − 1)(1 − η_{1})}{η_{1}(α − 1)(1 − η_{1})}

and since α − 1 and 1 − η_{1} are units, this implies that α|η_{1}^{2} − η_{1} + 1. Let us now examine the case m = −3. In this case η_{1}^{2} − η_{1} + 1 = 0 for every exceptional unit and then the condition α|η_{1}^{2} − η_{1} + 1 = 0 is trivial. We only have to ensure that α − 1 ∈ O_{K}^{∗}. Excluding repetitions, we obtain four inequivalent orbits, given by:

{1, 0, 2, 2ζ_{6}, 0}, {1, 0, 1 + ζ_{6}, −1 + 2ζ_{6}, 0},
{1, 0, ζ_{6}, −1 + ζ_{6}, 0}, {1, 0, 2 − ζ_{6}, 1 − ζ_{6}, 0}.

If m = 5, the conditon that η_{1} is an exceptional unit implies that η_{1}^{2} − η_{1} + 1 is associated with 2 and hence α|2. Up to conjugation, we may assume that η_{1} ∈ {±ξ, 1 + ξ}. Now, if α is a unit, then it is an exceptional unit and, excluding duplications, we have fifteen orbits of this type. If α is associated with 2, α = 2η_{2}, with η_{2} ∈ O_{K}^{∗}. Then we have

\frac{α − 1}{η_{2}} + \frac{1}{η_{2}} = 2

and Lemma 2.8.7 gives us all the possibilities for α, obtaining fifteen other orbits. Then there are thirty orbits of this type, given by:

{1, 0, ξ, 1 + ξ, 0}, {1, 0, ξ, −1 − ξ, 0}, {1, 0, ξ, 1 + 2ξ, 0},
{1, 0, −ξ, 1 + ξ, 0}, {1, 0, −ξ, −1 − ξ, 0}, {1, 0, −ξ, −1 − 2ξ, 0},
{1, 0, 1 − ξ, −1, 0}, {1, 0, 1 − ξ, −ξ}, {1, 0, −1 + ξ, −1, 0},
{1, 0, −1 + ξ, ξ, 0}, {1, 0, 1 + ξ, 1 + 2ξ, 0}, {1, 0, 1 + ξ, −1 − 2ξ, 0},
{1, 0, 1 + ξ, 2 + 3ξ, 0}, {1, 0, 2 − ξ, −1 + ξ, 0}, {1, 0, 2 − ξ, 1 − ξ, 0},
{1, 0, 2, 2ξ, 0}, {1, 0, 2, −2ξ, 0}, {1, 0, 2, 2 + 2ξ, 0},
{1, 0, 4 − 2ξ, −2 + 2ξ, 0}, {1, 0, 4 − 2ξ, 2 − 2ξ, 0}, {1, 0, 4 − 2ξ, 2, 0},
{1, 0, −2ξ, −2 − 2ξ, 0}, {1, 0, −2ξ, 2 + 2ξ, 0}, {1, 0, −2ξ, −2 − 4ξ, 0},
{1, 0, 2 + 2ξ, 2 + 4ξ, 0}, {1, 0, 2 + 2ξ, −2 − 4ξ, 0}, {1, 0, 2 + 2ξ, 4 + 6ξ, 0},
{1, 0, −2 + 2ξ, 2, 0}, {1, 0, −2 + 2ξ, −2, 0}, {1, 0, −2 + 2ξ, 2ξ}.

Thus we have proved 9).

10) Again, by Theorem 2.8.12, R(m, 5, 3) = 0 if m ≠ −3, 5. Then we need just to consider these two cases. A 5-orbit ending with a 3-periodic orbit has the form {α_{0}, 1, 0, α_{3}, α_{4}, 0} and cutting the initial term α_{0} we must obtain one of the orbits listed in 9). It follows that 1 − α_{3} is always a unit and therefore α_{0} and 1 − α_{0} are units, too. Then α_{0} must be an exceptional unit. All we have to check is that the orbits obtained in this way are realized, looking at their Langrangian interpolation polynomial. The result is that, if m = −3, we have only two weakly inequivalent orbits, given by

{ζ_{6}, 1, 0, 2, 2ζ_{6}, 0}, {ζ_{6}, 1, 0, 1 + ζ_{6}, −1 + 2ζ_{6}, 0},

while if m = 5 we have twenty-eight different weakly inequivalent orbits, given by:

{1 + ξ, 1, 0, ξ, 1 + 2ξ, 0}, {1 + ξ, 1, 0, −ξ, −1 − 2ξ, 0}, {−ξ, 1, 0, 1 − ξ, −1, 0},
{−1 + ξ, 1, 0, 1 − ξ, −1, 0}, {ξ, 1, 0, −1 + ξ, −1, 0}, {1 − ξ, 1, 0, −1 + ξ, −1, 0},
{−ξ, 1, 0, 1 + ξ, −1 − 2ξ, 0}, {ξ, 1, 0, 1 + ξ, 1 + 2ξ, 0}, {1 + ξ, 1, 0, 2, 2ξ, 0},
{1 + ξ, 1, 0, 2, 2 + 2ξ, 0}, {ξ, 1, 0, 2, 2ξ, 0}, {1 − ξ, 1, 0, 2, −2ξ, 0},

{−1 + ξ, 1, 0, 2, 2ξ, 0}, {1 − ξ, 1, 0, 4 − 2ξ, 2 − 2ξ, 0}, {−1 + ξ, 1, 0, 4 − 2ξ, −2 + 2ξ, 0}, {2 − ξ, 1, 0, 4 − 2ξ, 2 − 2ξ, 0}, {2 − ξ, 1, 0, 4 − 2ξ, 2, 0}, {1 + ξ, 1, 0, −2ξ, −2 − 4ξ, 0},

{−ξ, 1, 0, −2ξ, 2 + 2ξ, 0}, {−ξ, 1, 0, −2ξ, −2 − 4ξ, 0}, {1 + ξ, 1, 0, 2 + 2ξ, −2 − 4ξ, 0},

{1 + ξ, 1, 0, 2 + 2ξ, 4 + 6ξ, 0}, {−ξ, 1, 0, 2 + 2ξ, −2 − 4ξ, 0}, {1 + ξ, 1, 0, −2 + 2ξ, 2ξ, 0},

{ξ, 1, 0, −2 + 2ξ, 2, 0}, {ξ, 1, 0, −2 + 2ξ, 2ξ, 0}, {−1 + ξ, 1, 0, −2 + 2ξ, −2, 0},
{−1 + ξ, 1, 0, −2 + 2ξ, 2ξ, 0},

and 10) is proved.

11) Again we need to consider only the cases m = −3, 5. If m = −3, by the results of 10), such an orbit must have the form {α_{0}, ζ_{6}, 1, 0, α_{4}, α_{5}, 0}, therefore α_{0} and 1 − α_{0} must be units, so α_{0} ∈ {ζ_{6}, −ζ_{6}^{2}}. Anyway in both of these cases the Lagrangian interpolation polynomial has not coefficients in O_{K} and hence the orbit is not realizable. If m = 5, the results of 9) and 10) show that the desired orbits must have the form {α_{0}, ϵ_{1}, 1, 0, α_{4}, η_{1}α_{4}, 0} with ϵ_{1}, η_{1} exceptional units and α_{4} is a unit or associated with 2. Suppose that α_{0} is a unit, too. By Lemma 2.3.6, also 1 − α_{0} is a unit, and therefore α_{0} is an exceptional unit. Cheking the Lagrangian interpolation polynomials of the orbits obtained in such a way, we find ten weakly inequivalent orbits. If α_{0} is not a unit, we use the unit equation

(1 − α_{0}) + (α_{0} − η_{1}α_{4}) + η_{1}α_{4} = 1.

The Lemma 2.8.7 gives the possible values of α_{0}. Anyway a rapid check of the Lagrangian polynomials shows that none of those values leads to a possible orbit. Then we have ten acceptable orbits, given by:

{ξ, 1 + ξ, 1, 0, 2, 2ξ, 0}, {1 + ξ, ξ, 1, 0, 2, 2ξ, 0},
{ξ, −1 + ξ, 1, 0, 2, 2ξ, 0}, {2 − ξ, 1 − ξ, 0, 4 − 2ξ, 2 − 2ξ, 0},
{−ξ, 1 + ξ, 1, 0, −2ξ, −2 − 4ξ, , 0}, {−ξ, 1 + ξ, 1, 0, 2 + 2ξ, −2 − 4ξ, 0},
{1 + ξ, −ξ, 1, 0, 2 + 2ξ, −2 − 4ξ, 0}, {ξ, 1 + ξ, 1, 0, −2 + 2ξ, 2ξ, 0},
{−1 + ξ, ξ, 1, 0, −2 + 2ξ, 2ξ, 0}, {ξ, −1 + ξ, 1, 0, −2 + 2ξ, 2ξ, 0}.

Therefore 11) is proved.

12) Theorem 2.8.12 tells us that R(m, 5, 4) = 0 if m ≠ −1, 2, 5. Then we havejust to examine these cases. If m = −1, the results on the 4-periodic orbits of ℚ(\sqrt{−1}) show that a desired orbit must have the form \{α_{0}, 0, α_{2}, (1 + \sqrt{−1})α_{2}, \sqrt{−1}α_{2}, 0\}, with α_{2} ≠ 0. This orbit is generated by the polynomial f(x) = (x − α_{0})(x − \sqrt{−1}α_{2})(ax^{2} + bx + c), with a, b, c ∈ O_{K}. Since α_{2} = f(0) = cα_{0}\sqrt{−1}α_{2}, we obtain that α_{0} is a unit. Then we may assume α_{0} = 1 and rename α_{2} = α. Then

(1 + \sqrt{−1})α = f(α) = α(α − 1)(1 − \sqrt{−1})(aα^{2} + bα + c),

which implies that η = α − 1 is a unit. Then we have that β = (1 + \sqrt{−1})α − 1 = (1 + \sqrt{−1})(1 + η) − 1|\sqrt{−1}α = \sqrt{−1}(1 + η)|1 + η, but this is impossible, because 1 + η is either 2 or associated with 1 + \sqrt{−1}. Then there are no such orbits.

If m = 2, let \{α_{0}, 0, α_{2}, α_{3}, α_{4}, 0\} be our orbit. The results on 4-periodic orbits in ℚ(\sqrt{2}) show that either α_{2}, α_{4} ∈ O_{K}^{∗} and α_{3} is associated with \sqrt{2}, or α_{2}, α_{4} are associated with \sqrt{2} and α_{3} is associated with 2. The polynomial realizing the orbit is f(x) = (x − α_{0})(x−α_{4})g(x) with g(x) ∈ O_{K}[x]. Since α_{3} = f(α_{2}) = (α_{2} −α_{0})(α_{2} −α_{4})g(α_{2}) and α_{3} is associated with α_{2} −α_{4}, we have that η_{1} = α_{2} −α_{0} ∈ O_{K}^{∗}. Moreover α_{2} = f(0) = α_{0}α_{4}g(0) and it follows that α_{2} can not be a unit, otherwise α_{0} would be a unit, too, and the relation

\frac{α_{2}}{η_{1}} + \frac{α_{0}}{η_{1}} = 1

would contradict Lemma 2.8.7. Therefore we may assume α_{2} = \sqrt{2}. By the structure of 4-periodic orbits, we have that either α_{3} = 2 + 2\sqrt{2} and α_{4} = ±(2 + \sqrt{2}) or α_{3} = 2 and α_{4} = 2 + \sqrt{2}. Now the condition \sqrt{2} = f(0) = ±α_{0}(2 + \sqrt{2})g(0) implies that α_{0} is a unit. Moreover α_{3} = f(\sqrt{2}) = (\sqrt{2} − α_{0})(\sqrt{2} − α_{4})g(\sqrt{2}), ±(2 + \sqrt{2}) = f(α_{3}) = (α_{3} − α_{0})(α_{3} − α_{4})g(α_{3}) and the fact that α_{3} − α_{4} is associated with \sqrt{2} imply that η_{2} = α_{3} − α_{0} is a unit. Then either

\frac{η_{2}}{−1 + \sqrt{2}} + \frac{−η_{1}}{−1 + \sqrt{2}} = \sqrt{2}

or

\frac{η_{2}}{1 + \sqrt{2}} + \frac{−η_{1}}{1 + \sqrt{2}} = \sqrt{2}

and Lemma 2.8.7 shows that either η_{1} or η_{2} equals ±1 ± \sqrt{2}. The examination of the resulting values of α_{0} gives origin to four acceptable weakly inequivalent orbits, given by:

\{1 + \sqrt{2}, 0, \sqrt{2}, 2 + 2\sqrt{2}, 2 + \sqrt{2}, 0\}, \{1 + \sqrt{2}, 0, \sqrt{2}, 2, 2 + \sqrt{2}, 0\},
\{1 + \sqrt{2}, 0, \sqrt{2}, 2 + 2\sqrt{2}, −2 − \sqrt{2}, 0\}, \{−1, 0, \sqrt{2}, 2 + 2\sqrt{2}, 2 + \sqrt{2}, 0\}.

Finally, let m = 5; by the structure of 4-periodic orbits in ℚ(\sqrt{5}), our orbit must be weakly equivalent to {α_{0}, 0, 1, α_{3}, α_{4}, 0} with α_{3} an exceptional unit or associated with 2. By Lemma 2.3.6, α_{0}, α_{4}, η = α_{0} − α_{3} are units and α_{0} − 1|α_{3}. If α_{3} is a unit, then α_{0} is an exceptional unit. Cheking all the possibilities for α_{3}, α_{4} in this case, we obtain six different orbits. If α_{3} is associated with 2, a rapid check shows that we must have either α_{3} = 2 + 2ξ, α_{4} = 1 + 2ξ or α_{3} = −2 + 2ξ, α_{4} = −3 + 2ξ. In the first case, we have

\frac{α_{0}}{1 + ξ} + \frac{−η}{1 + ξ} = 2

and Lemma 2.8.7 gives only two possibilities for α_{0}. In the second case, we obtain

\frac{α_{0}}{−1 + ξ} + \frac{−η}{−1 + ξ} = 2

and again Lemma 2.8.7 gives us two orbits. A check on their interpolation polynomials shows that these orbits are effectively realized. We have in total ten orbits given by:

{−ξ, 0, 1, 1 + ξ, ξ, 0}, {2 − ξ, 0, 1, 1 − ξ, −ξ, 0}, {1 − ξ, 0, 1, −ξ, 1 + ξ, 0},

{ξ, 0, 1, 1 + ξ, −ξ, 0}, {−ξ, 0, 1, 1 − ξ, 2 − ξ, 0}, {2 − ξ, 0, 1, −1 + ξ, ξ, 0},

{−ξ, 0, 1, 2 + 2ξ, 1 + 2ξ, 0}, {ξ, 0, 1, −2 + 2ξ, −3 + 2ξ, 0}, {1 + ξ, 0, 1, 2 + 2ξ, 1 + 2ξ, 0},

{−1 + ξ, 0, 1, −2 + 2ξ, −3 + 2ξ, 0}.

Therefore 12) is proved.

13) By the results of 12) and of Theorem 2.8.12, we only need to consider the cases m = 2, 5. If m = 2, cutting the initial term α_{0}, we must obtain one of the four orbits listed in 12). In all of those cases, we get that α_{0} and β = α_{0} − α_{1} divide α_{1} ∈ O_{K}^{∗} and therefore are units themselves. This implies that

\frac{α_{0}}{β} + \frac{α_{1}}{β} = 1;

by Lemma 2.8.7, this can not happen and therefore there are no such orbits. Let us consider then the case m = 5. By 12), the desired orbit must have the form {α_{0}, α_{1}, 0, 1, α_{4}, α_{5}, 0} with α_{1} an exceptional unit. Therefore α_{0} must be an exceptional unit, too, and α_{0} − α_{1} ∈ O_{K}^{∗}. Cheking all the possible cases, we obtain two acceptable orbits:

{1 + ξ, −ξ, 0, 1, 2 + 2ξ, 1 + 2ξ, 0}, {−1 + ξ, ξ, 0, 1, −2 + 2ξ, −3 + 2ξ, 0}.

Then 13) is proved.

14) By Theorem 2.8.12, R(m, n, 6) = 0 if m ≠ −3, then we just need to consider this case. By the result on 6-peiodic orbits in ℚ(\sqrt{−3}), our orbit must have the form {α_{0}, 0, α_{2}, (1+ζ_{6})α_{2}, 2ζ_{6}α_{2}, (−1+2ζ_{6})α_{2}, (−1+ζ_{6})α_{2}, 0} with α_{0}, α_{2} ≠ 0. The polynomial realizing this orbit is f(x) = (x − α_{0})(x − ζ_{6}α_{2} + α_{2})g(x). Therefore we have α_{2} = f(0) = α_{0}(−1 + ζ_{6})α_{2}g(0), hence α_{0} must be a unit. Moreover (1 + ζ_{6})α_{2} = f(α_{2}) = (α_{2} − α_{0})(2 − ζ_{6})α_{2}g(α_{2}), hence η_{1} = α_{2} − α_{0}|(1 + ζ_{6})/(2 − ζ_{6}) and therefore is a unit. Finally 2ζ_{6}α_{2} = f(α_{2} + ζ_{6}α_{2}) = 2α_{2}(α_{2} + ζ_{6}α_{2} − α_{0})g(2ζ_{6}α_{2}) and this implies that η_{2} = α_{2} + ζ_{6}α_{2} − α_{0} is also a unit. Then we have

\frac{η_{1}}{η_{2}} + \frac{ζ_{6}η_{1}}{η_{2}} + \frac{ζ_{6}α_{0}}{η_{2}} = 1.

By Lemma 2.8.7, one of the summands must be equal to 1 and the only possibility is ζ_{6}η_{1} = η_{2}, which implies α_{0} = α_{2}/(1 − ζ_{6}) = ζ_{6}α_{2}. Then all the elements of the orbit are divisible by α_{2} and the condition on the lagrangian polynomial implies that α_{2} ∈ O_{K}^{∗}, therefore we may assume α_{2} = 1, obtaining the orbit

{ζ_{6}, 0, 1, 1 + ζ_{6}, 2ζ_{6}, −1 + 2ζ_{6}, −1 + ζ_{6}, 0},

realized by the polynomial

f(x) = ζ_{6}x^{6} + 6(1 − ζ_{6})x^{5} − 15x^{4} + 20ζ_{6}x^{3} + 15(1 − ζ_{6})x^{2} + (−6 + ζ_{6})x + 1

and all the 7-orbits containing a 6-periodic orbit are weakly equivalent to this one. Therefore 14) is proved.

1) By the results of all the other parts, we only need to show that there are no orbits longer than the previous ones. In particular we just need to show that R(m, 6, 1) = R(m, 7, 2) = R(m, 7, 3) = R(m, 7, 4) = R(m, 8, 6) = 0 for every squarefree m. Let us start with R(m, 6, 1). By the results of 4, we can reduce to the case m = 5 and the orbit having the form {α_{0}, α_{1}, 1, α_{3}, α_{4}, 0, 0} with α_{1} ∈ O_{K}^{∗}. Then also α_{0} is a unit and α_{0} − 1|α_{1} − α_{3} ∈ O_{K}^{∗}, therefore α_{0} is an exceptional unit. Cheking all the possible cases, we have three possibilities: either α_{0} − 1 + ξ and α_{0} + ξ are units or α_{0} − 1 − ξ and α_{0} − ξ are units or α_{0} − 1 − ξ and α_{0} + ξ are units. Anyway, Lemma 2.8.7 shows that none of these cases is possible. Examine now R(m, 7, 2); by the results of 8), we only need to deal with m = 2, 5 and try to prolong the orbits listed in 8). If m = 2 and α is the initial element prolonging the orbit, in the first two cases we obtain that η_{1} = α − 2\sqrt{2} and η_{2} = α + 2 + 2\sqrt{2} are units satisfying

\frac{η_{2}}{3 + 2\sqrt{2}} + \frac{−η_{1}}{3 + 2\sqrt{2}} = \sqrt{2}

and Lemma 2.8.7 implies that α ∈ \{1, −5 + 4\sqrt{2}, 5 + 3\sqrt{2}, −1 + 3\sqrt{2}\}. However the calculation of the lagrangian polynomial shows that none of these orbits is possible. In the second case we have that η_{1} = α + \sqrt{2} and η_{2} = α − 1 are units satisfying

\frac{η_{1}}{1 + \sqrt{2}} + \frac{−η_{2}}{\sqrt{2}} = 1,

which can not happen by Lemma 2.8.7. Finally for the last two orbits we obtain that η_{1} = α − \sqrt{2} and η_{2} = α − 2 are units satisfying

\frac{η_{1}}{−1 + \sqrt{2}} + \frac{−η_{2}}{−1 + \sqrt{2}} = \sqrt{2},

that is α ∈ \{1, 1 + \sqrt{2}, −1 + 2\sqrt{2}, 3 − \sqrt{2}\} and again the calculation of the lagrangian polynomial shows that there are no possible orbits of this type. Therefore R(2, 7, 2) = 0. In the case m = 5, Lemma 2.3.6 shows that, in order to prolong a 6-orbit with an element α, α must be an exceptional unit. However a rapid check shows that, in any case, one of the divisibility conditions of lemma 2.3.6 is violated. Therefore R(5, 7, 2) = 0. Now we need to show that R(m, 7, 3) = 0. By 11), we can reduce to m = 5. If we try to prolong one of the orbits listed in 11) with an element α, we see that α must be an exceptional unit. Again this shows that in any case one of the divisibility conditions of Lemma 2.3.6 is violated. Therefore R(5, 7, 3) = 0. Let us now examine R(m, 7, 4). Again we can reduce to the case m = 5 and look for an α ∈ O_{K} prolonging the existing 6-orbits. A rapid calculation shows that α must be an exceptional units and this fact violates the divisibility conditions. Finally, dealing with R(m, 8, 6), we have just to examine the case m = −3. The only possible orbit is {α, ζ_{6}, 0, 1, 1 + ζ_{6}, 2ζ_{6}, −1 + 2ζ_{6}, −1 + ζ_{6}, 0}. In particular Lemma 2.3.6 implies that α, α−ζ_{6}, α−1 ∈ O_{K}^{∗}, which is clearly impossible. Then our proof is complete.

The arguments we have used to prove Theorems 2.8.12 and 2.8.14 can be used also in case of fields of larger degree over ℚ. Anyway, already in the case of cubic fields, the number of solutions of the equations given in Lemma 2.8.7 is extremely large and therefore this task seems to be unsurmountable. In [18], [20] and [21] some results about cubic fields can be found, but the classification of finite orbits in every number field remains an open problem.

Narkiewicz’s Properties

In this chapter we continue the study of polynomials as dynamical systems. We focuse on fully invariant sets and analyse the case of finiteness; this will lead us to consider some particular finiteness properties collected and studied by Narkiewicz in [20] and hence called Narkiewicz’s properties. If not differently said, we denote by K a generic extesion of ℚ (finite or infinite) and by f a generic polynomial of K[x] with deg f ≥ 2.

3.1 The properties P and SP

The properties P and SP describe the fields in which polynomial maps have only finite fully invariant sets. We start with a lemma that describes the behaviour of polynomials in a very simple case.

Lemma 3.1.1 : Let f be a non-constant polynomial which maps the rational field ℚ onto itself. Then f is linear.

Proof: Suppose not. Up to multiplying by a constant, we may assume that f has the form

f(x) = ∑_{j=0}^{d} a_{j}x^{j} a_{j} ∈ ℤ.

We may also assume that a_{d} > 0; then (looking at f as a real function) there exists a real number α such that f is increasing in [α, +∞). Since f is onto, ∀n ∈ ℤ we may choose a rational number α_{n} = \frac{p_{n}}{q_{n}} ∈ ℚ with gcd(p_{n}, q_{n}) = 1 such that f(α_{n}) = n. Since f is increasing, we can choose an n_{0} such that α_{n+1} > α_{n} for n ≥ n_{0}. In particular, since q_{n} must divide a_{d}, we have α_{n+1} − α_{n} ≥ 1/a_{d} for n ≥ n_{0}. Hence, for sufficiently large n,

1 = f(α_{n+1}) − f(α_{n}) = f′(ξ_{n})(α_{n+1} − α_{n}) ≥ \frac{f′(ξ_{n})}{a_{d}}.

If d ≥ 2, then the last term diverges to infinity and this is a contradiction. Then d = 1 and f is linear.

The lemma tells us that neither ℚ nor an infinite subset of itself can be a fully invariant set for a non-linear polynomial. The property P is a generalisation of this fact for a field K.

Definition 3.1.2: Let K be a field. We say that K satisfies the property P if every non-linear polynomial f ∈ K[x] does not have any infinite fully invariant set.

Before defining the property SP, we need another definition for polynomial maps of many variables.

Definition 3.1.3 : Let f_{1}, ..., f_{n} ∈ K[x_{1}, ..., x_{n}] and let F : K^{n} → K^{n} be the map whose j-th component is f_{j}. We say that F is an admissible polynomial map if none of the f_{j} is linear and the leading forms of all the f_{j} do not have any trivial common zero in ℚ̄.

Definition 3.1.4 : Let K be a field. We say that K satisfies the property SP if, for every n ≥ 1 and every admissible polynomial map F : K^{n} → K^{n}, we have that F does not have any infinite fully invariant set.

The requirement of the map being admissible is necessary. For example, if we take K = ℚ, n = 2 and set

f_{1}(x_{1}, x_{2}) = x_{1}^{2} − x_{1}x_{2} + x_{1}, f_{2}(x_{1}, x_{2}) = x_{1}x^{2} − x_{2}^{2} + x_{2}

and

F = (f_{1}, f_{2})

we have that the set

Δ = {(α, α)|α ∈ ℚ} ⊆ ℚ^{2}

is an infinite fully invariant set for F (in fact it is contained in F ix(F)). This happens becuase F is not admissible; the leading form of F is identically zero on this set.

Clearly, if a field satisfies the property SP, then it satisfies P trivially, too. The converse is not true, but it is not easy to prove. We shall discuss it in the next section, constructing explicitly a counterexample.

The property SP is generally difficult to test. To simplify it, we give a theorem which ensures us that we need to test it just on some particular polynomial maps.

Theorem 3.1.5 : Let K be a field such that, ∀n ≥ 1 and every homogeneous admissible polynomial map F (that is, f_{j} is homogeneous ∀j = 1, ..., n), F does not have any infinite fully invariant set. Then K has the property SP.

Proof: Let F : K^{n} → K^{n} be any admissible polynomial map, with F = (f_{1}, ..., f_{n}) and degf_{j} = d_{j}. We can write

f_{j} = ∑_{k=0}^{d_{j}} g_{j,k}

where g_{j,k} is the homogeneous part of degree k of f_{j}. Then we define

fj^{′} = ∑_{k=0}^{d_{j}} x_{n+1}^{k}g_{j,d_{j}−k}

and

fn+1^{′} = x_{n+1}^{2}.

We can now define a homogeneous admissible polynomial map

F′ = (f1^{′}, ..., fn+1^{′}) : K^{n+1} → K^{n+1}.

Let X ⊆ K^{n} be a fully invariant set for F and define

Y = {(α_{1}, ..., α_{n}, 1) ∈ K^{n+1}|(α_{1}, ..., α_{n}) ∈ X}.

It is immediate to check that Y is a fully invariant set for F′ and hence is finite. Then X is also finite and this implies that K has the property SP.

Now we shall prove a technical lemma (in [8] it is called “The main lemma”), which will be usefull to construct fields which satisfy or do not satisfy the properties P and SP.

Lemma 3.1.6 (Narkiewicz) : Let X be a non-empty set and F : X → X be a surjective map. Let f, g : X → ℝ^{+} be two real valued functions satisfying the following conditions:

1) ∀a > 0 the set {α ∈ X|f(α) + g(α) ≤ a} is finite;

2) There is a constant A > 0 such that f(α) > A implies f(F (α)) > f(α);

3) ∀b > 0 there exists B_{b} > 0 such that f(α) ≤ b and g(α) > B_{b} imply g(F (α)) > g(α). Then X is finite.

Proof: Since F is surjective, we can define by induction a sequence of elements of X such that α_{0} ∈ X is any element of X and F (α_{j+1}) = α_{j}. Let j_{0} be the minimum index j such that f(α_{j}) > A. Then by condition 2) f(F (α_{j_{0}})) = f(α_{j_{0}−1}) > f(α_{j_{0}}). This is absurd because it contradicts the minimality of j_{0}. Then the only possibility is f(α_{j}) ≤ A ∀j. Put b = A in condition 3) and let j_{1} be the smallest index j such that g(α_{j}) > B_{A}. Applying condition 3), the same argument as before shows that we have a contradiction and then g(α_{j}) ≤ B_{A} ∀j. Then we have f(α_{0}) + g(α_{0}) ≤ A + B_{A}. But α_{0} is arbitrary and applying condition 1) we have the final thesis.

Corollary 3.1.7 : Let X be a non-empty set and F : X → X be a surjective function. Assume that there exists a function f : X → ℝ^{+} which satisfies the conditions:

1) ∀a ≥ 0 the set {α ∈ X|f(α) ≤ a} is finite;

2) ∃A ≥ 0 such that f(α) > A implies f(F (α)) > f(α).

Then X is finite.

Proof: Apply the lemma with g = 1.

We are now able to prove whether or not some fields satisfy the properties P and SP.

Theorem 3.1.8 : The rational field ℚ satisfies properties P and SP.

Proof: Since SP implies P we need only to prove SP. Let F : ℚ^{n} → ℚ^{n} be an admissible polynomial map F = (f_{1}, ..., f_{n}). By Theorem 3.1.5, we can assume that all the f_{j}’s are homogeneous. If deg f_{j} = d_{j} we put

d = min_{j}{d_{j}} D = max_{j}{d_{j}}

and, up to multiplying by a constant, we can also assume that f_{j}(x_{1}, ..., x_{n}) ∈ ℤ[x_{1}, ..., x_{n}]. Let now X be a fully invariant set for F; we will prove that X is finite. We can write every element α ∈ ℚ^{n} in the form

α = (\frac{p_{1}}{q} , ..., \frac{p_{n}}{q}),

where p_{1}, ..., p_{n}, q ∈ ℤ, q > 0 and (p_{1}, ..., p_{n}, q) = 1. We set

f(α) = q g(α) = max_{j}|p_{j}|.

We have to prove that f and g satisfy the three conditions of Lemma 3.1.6. The first one is obvious, we have to check the other two. We observe that

F (α) = (q^{D−d_{1}} \frac{f_{1}(p_{1}, ..., p_{n})}{q^{D}}, ..., q^{D−d_{n}} \frac{f_{n}(p_{1}, ..., p_{n})}{q^{D}})

and so if we put M = (q^{D}, q^{D−d_{1}}f_{1}(p_{1}, ..., p_{n}), ..., q^{D−d_{n}}f_{n}(p_{1}, ..., p_{n})), we have

f(F (α)) ≥ \frac{q^{D}}{M} .

Now we invoke Hilbert’s Nullstellensatz which gives us the existence of λ_{k} ≥ 0, c_{k} ∈ ℤ and G_{jk} ∈ ℤ[x] such that

q^{D−d} ∑_{j=1}^{n} G_{jk}(x_{1}, ..., x_{n})f_{j}(x_{1}, ..., x_{n}) = c_{k}x_{k}^{λ_{k}}q^{D−d} ∀k = 1, ..., n.

Evaluating this expression in (p_{1}, ..., p_{n}) we obtain that M|q^{D−d}c_{j}p_{j}^{λ_{j}} ∀j = 1, ..., n and this leads to M ≤ B_{0}q^{D−d}, with B_{0} not depending on α. We finally have

f(F (α)) ≥ \frac{q^{d}}{B_{0}} = \frac{1}{B_{0}} f(α)^{d}

and since d ≥ 2, if f(α) is sufficiently large we have

f(F (α)) > f(α)

and condition 2) is proved. Let us shift to condition 3) and choose a B > 0. If the condition were not true, we would find a sequence {α_{j}} of elements of ℚ^{n} such that f(α_{j}) ≤ B ∀j and

lim_{j→+∞} \frac{g(F (α_{j}))}{g(α_{j})^{d}} = 0.

As before we can write

α_{j} = (\frac{p_{1j}}{q_{j}} , ..., \frac{p_{nj}}{q_{j}}).

By choosing a suitable subsequence, we can assume that q_{j} = q ≤ B ∀j and that there exists an index k_{0} such that

g(α_{j}) = |p_{k_{0}j}| ∀j

and the limits

λ_{k} = lim_{j→+∞} \frac{p_{kj}}{p_{k_{0}j}}

exist ∀k = 1, ..., n.

Since q ≤ B we can find a constant c > 0 such that

max_{k}{|f_{k}(p_{1j}, ..., p_{nj})|} ≤ cg(F (α_{j})).

Substituting in the limit relation, we have that

lim_{j→+∞} \frac{|f_{k}(p_{1}, ..., p_{n})|}{p_{k_{0}j}^{d}} = 0,

hence

lim_{j→+∞} \frac{|f_{k}(p_{1}, ..., p_{n})|}{p_{k_{0}j}^{d_{k}}} = 0.

This implies that F_{k}(λ_{1}, ..., λ_{n}) = 0 ∀k and then λ_{k} = 0 ∀k. This is absurd, because |λ_{k_{0}}| = 1, then condition 3) is also satisfied.

We can now apply the main lemma that ensures us that X is finite. Then ℚ satisfies the property SP (and therefore also P).

We have shown that ℚ satisfies the two properties. On the other hand, the cyclotomic closure ℚ^{c} is an example of a field which does not satisfy any of these properties; in fact, in the previous chapter we have proved that every monomial and every Chebishev polynomial has infinitely many cyclotomic preperiodic points in ℚ̄, all of them lying in ℚ^{c}. Then the set of cyclotomic preperiodic points is an example of infinite fully invariant set and so ℚ^{c} does not satisfy the properties P and SP. This statement is general: if K is any characteristic zero field, then K^{c} does not satisfy any of the two properties.

Now we want to show that SP behaves well under finite extensions.

Theorem 3.1.9 : If K is a field satisfying property SP and L is a finite extension of K, then L has the property SP.

Proof: Let n = [L : K]; we can choose a basis ω_{1}, ..., ω_{n} of this extension and a positive integer m.

∀α ∈ L^{m} we can write

α = (α_{1}, ..., α_{m})

and ∀k = 1, ..., m

α_{k} = ∑_{j=1}^{n} α_{kj}ω_{j},

with α_{kj} ∈ K. Then we can define

α̂ = (α_{11}, ..., α_{1n}, α_{21}, ..., α_{mn}) ∈ K^{mn}

and the map ψ : L^{m} → K^{mn} ψ(α) = α̂ is clearly one-to-one because {ω_{j}} is a basis.

Now let f ∈ L[x_{1}, ..., x_{m}] be a polynomial. We can define a n-tuple of polynomials f̂_{1}, ..., f̂_{n} ∈ K[x_{11}, ..., x_{mn}] by

f(∑_{j=1}^{n} x_{1j}ω_{j}, ..., ∑_{j=1}^{n} x_{mj}ω_{j}) = ∑_{j=1}^{n} f̂_{j}(x_{11}, x_{12}, ..., x_{mn})ω_{j}.

Let F = (f_{1}, ..., f_{m}) : L^{m} → L^{m} be an admissible polynomial map. By Theorem 3.1.5, we may assume that all the f_{j} are homogeneous. Using the map ψ, we may associate with every f_{j} a n-tuple of polynomials f̂_{jk} and obtain a polynomial map

F̂ = (f̂_{11}, ..., f̂_{mn}) : K^{mn} → K^{mn}.

Clearly all of the f̂_{jk} are homogeneous and non-linear and an easy calculation shows that, if F is admissible, then also F̂ is. Now let X ⊆ L^{m} be a fully invariant set for F. Then the set

X̂ = ψ(X) ⊆ K^{mn}

is clearly a fully invariant set for F̂ and #X ≤ #X̂, since ψ is one-to-one; but K satisfies the property SP so X̂ must be finite. Then X is also finite and then L satisfies the property SP, too.

One could ask if the same proof (or a similar one) works also for the property P. Unluckily this is not true; if we try to adapt the preceeding proof (that is, we put m = 1), we see that it is necessary that K satisfies the property SP and so this does not prove what we want. We will see in the next section that P does not behave well under finite extensions and this fact will lead us to conclude that the two properties are inequivalent.

Since now we have only built examples of finite extensions of ℚ satisfying the two properties. We want to end this section building an infinite one.

The first observation we make is that, since ℚ̄ is countable, then also the set of all non-linear polynomials with coefficients lying in ℚ̄ is countable. Then we can fix an enumeration of them, that is, we can order them in a sequence f_{1}, f_{2}, .... In the same way, the set of all rational primes is also countable and we can enumerate them respecting the order p_{1} < p_{2} < p_{3}.... Finally we fix the notation

E_{n} = {α ∈ ℚ̄| ∃j ≤ n α ∈ P er(f_{j})}.

We need a lemma about finiteness properties of these sets

Lemma 3.1.10 : For any positive integer d and any n ∈ ℕ, the set

{α ∈ E_{n}| [ℚ(α) : ℚ] ≤ d}

is finite.

Proof: Each such set is a finite union of sets of type

{α ∈ E_{n}| ∃j ≤ n α ∈ P er(f_{j}), [ℚ(α) : ℚ] ≤ d};

but all of these sets are of bounded height by Theorem 2.4.5, because periodic points are in particular preperiodic. Then they are all finite set by Northcott’s Theorem 1.4.5.

We can turn back to our construction. We define by induction a sequence of ascending extensions of the rationals in this way: put K_{0} = ℚ and if K_{n−1} has already been defined, we choose K_{n} as an extension of K_{n−1} such that [K_{n} : K_{n−1}] = p_{n} and

K_{n} ∩ E_{n} = K_{n−1} ∩ E_{n};

such an extension always exists beacuse any number field L has an extension of bounded degree not containing a fixed finite set of algebraic numbers disjoint from L. In our case this set is {α ∈ E_{n}| [K_{n−1}(α) : K_{n−1}] ≤ p_{n}}, which is finite by the preceeding lemma.

Finally we put

K = ∪_{n=0}^{+∞}K_{n}.

K is an infinite extension of ℚ and cannot be generated by a set of algebraic numbers of bounded degree either; in fact it contains elements of degree dividing p for any prime number p.

We now have to show that K satisfies the property P. Let f ∈ K[x] be a non-linear polynomial of degree d and X ⊆ K be a fully invariant set for f. By the preceding observation, since K is countable, so is K[x]; then we can enumerate the non-linear polynomials in such a way that f = f_{s}. Denote by r the smallest integer such that K_{r} contains all the coefficients of f, that is, f ∈ K_{r}[x]. We can choose α ∈ X and n ∈ ℕ such that α ∈ K_{n} and p_{n} > d.

We define by induction a sequence of elements of X in this way: fix α_{0} = α, α_{m+1} = f(α_{m}); by the same formula we may define also α_{−1}, α_{−2}... Let d_{1} = [K_{r}(α_{−j−1}) : K_{r}(α_{−j})]; since f(x) − α_{−j} vanishes at α_{−j−1} and has degree d, we have that d_{1}|d < p_{n}. In particular, ∀j, we have that d_{2} = [K_{r}(α_{−j}) : K_{r}] can not be divisible by any prime ≥ p_{n}. But α_{−j} ∈ K so there must be a suitable b_{j} ≥ r such that α_{−j} ∈ K_{b_{j}}; however, since [K_{b_{j}} : K_{r}] = p_{r+1}...p_{b_{j}}, we obtain that [K_{b_{j}} : K_{r}]/[K_{b_{j}} : K_{r}(α_{−j})] is a divisor of p_{r+1}...p_{b_{j}}. Thus we must have α_{−j} ∈ K_{r} for every positive j, but K_{r} is a finite extensione of ℚ and therefore satisfies the property P; hence α must lies in a periodic orbit in K_{r}. This implies

X ⊆ P er(f) ∩ K = P er(f_{s}) ∩ K = P er(f_{s}) ∩ K_{s} ⊆ K_{s}.

Then, if we put m = r + s, then X ⊆ K_{m} and f ∈ K_{m}; since K_{m} satisfies the property P, then we have the finiteness of X and this shows that K satisfies the property P.

3.2 The property N

We now define a new property which represents a generalisation of Northcott’s Theorem 1.4.5. This property is weaker than P and SP and will help us to construct a field that has P but not SP.

Definition 3.2.1 : Let E ⊆ ℚ̄. We say that E satisfies the property N if, ∀B > 0, the set

E_{B} = {α ∈ E|h(α) ≤ B},

where h is the usual height function, is finite.

Norhtcott’s Theorem tells us that every number field satisfies this property and so does every union of number fields of degree ≤ d ∀d ∈ ℕ. We now state an important result, that ensures us the N is satisfied also by some classes of algebraic number fields of infinite degree over ℚ.

Let K be a number field. We denote by K^{(d)} the compositum of all extensions L of K of degree [L : K] ≤ d. It is easy to see that K^{(d)} is a Galois extension of K. We also denote by K_{ab}^{(d)} the compositum of all abelian extension of K of degree ≤ d; clearly K_{ab}^{(d)} is the maximal subfield of K^{(d)} abelian over K. It is an open problem to determine whether K^{(d)} satisfies N or not. Anyway we have the following result:

Theorem 3.2.2 : ∀K, d, the field K_{ab}^{(d)} satisfies the property N.

Proof: We may assume for simplicity that K contains the field ℚ(ζ_{d!}). Let α ∈ K_{ab}^{(d)} with h(α) ≤ B and denote L = K(α); clearly L is an abelian extension of K whose exponent divides d!. Let p be a prime unramified in K, |.|_{v} be a place of K extending the p-adic absolute value, |.|_{w} be a place of L extending |.|_{v} and e = e(w/v) be the ramification index. If p > d then is tamely ramified in L because any prime dividing |Gal(L/K)| does not exceed d; since p is tamely ramified, we have that G_{0}(L/K) ≃ ℤ/eℤ, where G_{0} is the inertia group.

Let β = p^{1/e} and consider the field L(β); clearly the extension L(β)/L is unramified at all the places extending |.|_{v} and then we have e(w′/v) = e for every |.|_{w′} extending |.|_{v}. Then G_{0}(L(β)/K) ≃ ℤ/eℤ. Let L_{0} be the subfield of L(β) fixed by G_{0}(L(β)/K); clearly |.|_{v} is unramified in L_{0}. Moreover [L(β) : L_{0}] = |G_{0}(L(β)/K) = e and L_{0} ∩ K(β) = K because |.|_{v} is unramified in L_{0} and totally ramified in K(β). Hence [L_{0}(β) : L_{0}] = e and then L_{0}(β) = L(β). Then α ∈ L_{0}(β) and we may write

α = ∑_{j=0}^{e−1} a_{j}β^{j}, a_{j} ∈ L_{0}.

Clearly the conjugates of β over L_{0} are ζ_{e}^{j}β for j = 0, ..., e − 1. Therefore we have that T r_{L_{0}(β)/L_{0}}(β^{j}) vanishes if j is not a multiple of e and equals e if j = 0. Hence

a_{j} = \frac{1}{e}T r_{L_{0}(β)/L_{0}}(\frac{α}{β^{j}}) = \frac{1}{ep^{j/e}} ∑_{k=0}^{e−1} \frac{α_{k}}{ζ_{e}^{jk}} ,

where α_{k} are the conjugates of α in L_{0}(β). Since h(α_{k}) = h(α) ≤ B and using Proposition 1.3.5, we have

h(a_{j}p^{j/e}) ≤ 2log(e) + ∑_{k=0}^{e−1} h(α_{k}) ≤ 2log(d!) + Bd!.

Set γ = a_{j}p^{j/e} and let |.|_{w} be an absolute value of L_{0}(β) extending |.|_{v}. By definition |γ|_{w} = |Norm_{L_{0}(β)_{w}/ℚ_{p}}(γ)|_{p}^{1/m} where m = [L_{0}(β) : ℚ]. Set m_{w} = [L_{0}(β)_{w} : ℚ_{p}]: a simple calculation shows that

|log|γ|_{w}| ≥ \frac{m_{w}}{em} log(p).

Thus we have

2h(γ) = h(γ) + h(γ^{−1}) ≥ ∑_{w|v} |log|γ|_{w}| ≥ \frac{log(p)}{em} ∑_{w|v} m_{w}.

But clearly ∑_{w|v} m_{w} = [L_{0}(β) : K] and [L_{0}(β) : ℚ] ≤ [L_{0}(β) : K][K : ℚ]. Then if γ ≠ 0 we have

2h(γ) ≥ \frac{log(p)}{e[K : ℚ]}.

Comparing with the preceeding estimate we have that either a_{j} = 0 or

log(p) ≤ 2e[K : ℚ](2log(d!) + Bd!).

Let S be the set of primes that do not satisfy this condition. If p ∈ S then we must have a_{j} = 0 for 1 ≤ j ≤ e − 1, that is, α ∈ L_{0}, which is an abelian extension of K unramified above p. Then the field K(α) is unramified above any p ∈ S. By the structure theorem of abelian finite groups applied to Gal(K(α)/K), we may see K(α) as a compositum of cyclic extensions of degree at most d! unramified above any prime in S. Since the degree of these cyclic extensions is bounded, then also their discriminants must be bounded and this implies there can be only a finite number of these extensions. Then there are only finitely many distincts fields K(α) and consequently only a finite number of possibilities for α. Then we have proved that K_{ab}^{(d)} satisfies the property N.

Now we want to prove the relations between the property N and the properties P and SP.

Theorem 3.2.3 : Let K ⊆ ℚ̄. If K satisfies the property N, it also satisfies the properties P and SP.

Proof: Apply corollary 3.2.7, taking F an admissible polynomial map in K, X a fully invariant set for F and f the restriction of the height function to X.

Corollary 3.2.4 : The field K = ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5}, ...) satisfies the properties P, SP and N.

Proof: K is a subfield of ℚ_{ab}^{(2)} and therefore satisfies N. By the previous theorem, it satisfies also P and SP.

We are ready to prove the inequivalence of P and SP. Let p, q be two odd prime numbers such that q|p − 1. Let K = ℚ(ζ_{p}, ζ_{p^{2}}, ..., ζ_{p^{n}}, ...). K is a cyclotomic Galois extension of ℚ with Galois group

G ≃ ℤ_{p}^{∗} ≃ ℤ_{p} × (ℤ/pℤ)^{∗} ≃ ℤ_{p} × ℤ/(p − 1)ℤ;

then, by Galois correspondence between subfields of K and subgroups of G, there exists a unique subfield K′ ⊆ K such that [K : K′] = q and H = Gal(K/K′) ≃ ℤ/qℤ. It is clear that K does not satisfy the property P. In fact it contains infinitely many roots of unity, then their set is an infinite fully invariant set for every monomial x^{d}, d ≥ 2. What we need to prove is the following:

Theorem 3.2.5 (Dvornicich-Zannier) : K′ satisfies the property P but does not satisfy either property N or property SP.

Proof: First of all we note that the only roots of unity in K′ are ±1. In fact if ζ ≠ ±1 is a root of unity in K′, then its order divides 2p^{r} for some r. If the order of ζ is > 2 then K′ must contain a primitive p-th root of unity ξ. But the Galois group H has order q and cannot fix ξ. This is absurd.

Suppose now that K′ has not the property P and let f ∈ K[x] be a polynomial with an infinite fully invariant set X. Then Theorem 2.5.9 tells us that there exists a linear polynomial g(x) = ax + b such that g^{−1} ∘ f ∘ g is either a monomial or a Chebishev polynomial. Then we have two possibilities: g(X) is contained either in the set of roots of unity or in the set {ζ + ζ^{−1}} with ζ root of unity.

Suppose we are in the first case; then we can find α, β ∈ ℚ̄ such that αζ + β ∈ K′ for infinitely many ζ roots of unity. Let σ ∈ H, then we must have σ(αζ + β) = αζ + β, that is

αζ + β − σ(αζ + β) = αζ − σ(α)σ(ζ) + (β − σ(β)) = 0.

To conclude we need the following lemma, whose proof may be found in [7].

Lemma 3.2.6 : Let

a_{0} + ∑_{j=1}^{k} a_{j}ζ^{n_{j}} = 0

be a linear relation between roots of unity. If the relation is indecomposable, that is a_{0} ≠ 0 and no proper subsum vanishes, then only a finite number of roots of unity can satisfy the relation.

Now if β ∉ K′ then β − σ(β) ≠ 0 and therefore we have an indecomposable linear relation and so only finitely many roots of unity can satisfy it; this fact contradicts our hypothesis and then it is absurd. Then we must have β ∈ K′ and so the relation becomes

αζ = σ(α)σ(ζ)

for every σ ∈ H.

Let ζ_{0} be a root of unity satisfying the relation; dividing term by term we obtain ζ/ζ_{0} = σ(ζ/ζ_{0}). Then ζ/ζ_{0} ∈ K′ and is a root of unity. But K′ contains only finitely many roots of unity and then we have a contradiction.

Let us now consider the case g(X) ⊆ {ζ + ζ^{−1}}. Again we can find α, β ∈ ℚ̄ such that α(ζ + ζ^{−1}) + β ∈ K′ for infinitely many roots of unity ζ. For every σ ∈ H we have

αζ + αζ^{−1} − σ(α)σ(ζ) − σ(α)σ(ζ)^{−1} + β − σ(β) = 0.

Again, supposing that δ ∉ K′, we can find a σ ∈ H such that σ(β) ≠ β. Again our relation must have some vanishing subsum, otherwise only a finite number of roots of unity can satisfy it; one of these subsums must contain σ(β) − β, then this subsum has only a finite number of solutions and then also the global relation does. This is absurd, then we must have β − σ(β) = 0. Multiplying by ζ we have

αζ^{2} + α − σ(α)σ(ζ)ζ − σ(α)σ(ζ)^{−1}ζ = 0.

The coefficients of this relation dipend only on α, so we have only a finite number of possible coefficients for σ ∈ H. Again we have to suppose that ∀σ ∈ H, our relation has some vanishing subsum. There are three possibilities:

i) 1 + ζ^{2} = 0.

ii) ζσ(ζ) = ασ(α)^{−1} and ζσ(ζ) = σ(α)α^{−1}, which implies α^{2} = σ(α)^{2}.

iii) ζσ(ζ)^{−1} = ασ(α)^{−1} and ζσ(ζ)^{−1} = σ(α)α^{−1}, which implies again α^{2} = σ(α)^{2}.

Now case i) implies there are only a finite number of solutions and this is absurd. Suppose then we are in case ii) or iii). We can rewrite our relation in the form α(ζ + ζ^{−1}) = σ(α)σ(ζ + ζ^{−1}); squaring both sides we have ζ^{2} + ζ^{−2} ∈ K′ for infinitely many roots of unity ζ. We have obtained that [K′(ζ^{2}) : K′] ≤ 2 and this implies that ξ = ζ^{24} is a p-power root of unity and then it lies in K. Moreover

T_{12}(ζ^{2} + ζ^{−2}) = ξ + ξ^{−1} ∈ K′.

This fact tells us that K′ ⊆ K′(ξ) ⊆ K and [K′(ξ) : K′] ≤ 2; but [K : K′] = l is odd, then the only possibility is ξ ∈ K′. But the only roots of unity in K′ are ±1 and then we have only a finite number of possibilities for ξ. This is absurd.

Then we have shown that every fully invariant set X cannot be of the special infinite forms and then must be finite. Then K′ satisfies the property P.

To show that K′ does not satisfy the property SP we observe that, if it does, then K, being a finite extension of K′, would also satisfy SP; but then K would satisfy P as well and this is false. Then K′ does not satisfy SP and we have shown that the two properties are inequivalent.

Finally, if K′ satisfies N, it has also to satisfy SP and this is false. Then K′ does not satisfy N either.

3.3 The property VP

The property V P (also called property K since it was introduced by polish mathematician K.K.Kubota) is different from the other properties we have defined, since it involves not only a single polynomial or polynomial map, but two. Anyway it has many relations with properties P and SP. Before introducing it, we need some results and some terminology about pairs of polynomials.

Theorem 3.3.1 : Let f_{1} ∈ ℚ[x] be a polynomial of degree d_{1} not injective on ℚ, that is, there exists α, β ∈ ℚ α < β such that f_{1}(α) = f_{1}(β). Then ∃f_{2} ∈ ℚ[x] with deg f_{2} = d_{2} = 2d_{1} and an infinite subset X ⊆ ℚ such that f_{1}(X) = f_{2}(X).

Proof: Consider the polynomial

f(x) = (x − α)^{2} + x + β − α = x^{2} + (1 − 2α)x + (α^{2} − α + β)

and set f_{2}(x) = f_{1}(f(x)). Clearly d_{2} = 2d_{1}. We define by induction the sequence α_{0} = α, α_{n+1} = f(α_{n}) (note that in particular α_{1} = f(α) = β)

We note that f is increasing in [α, +∞); in fact if γ > α and since β > α we have

f(γ) = (γ − α)^{2} + γ + β − α > (γ − α)^{2} + γ > γ.

Then the set X = {α_{n}|n ∈ ℕ} is infinite and clearly f(X) ⊆ X; applying f_{1} to both sides we obtain f_{2}(X) = f_{1}(f(X)) ⊆ f_{1}(X).

To show the other inclusion, we note that

f_{1}(α_{0}) = f_{1}(α) = f_{1}(β) = f_{1}(f(α)) = f_{2}(α) ⊆ f_{2}(X).

On the other hand if j ≥ 1 we have

f_{1}(α_{j}) = f_{1}(f(α_{j−1})) = f_{2}(α_{j−1}) ⊆ f_{2}(X).

Then f_{1}(X) = f_{2}(X).

The preceeding theorem resembles a sort of “negation” of the property P for a pair of polynomials, where X plays the role of the fully invariant set. The fundamental point is the non-injectivity of f_{1}. The theorem fails to be true if f_{1} is injective (we will prove it later). Then we can infer that injectivity is what we need to define a property similar to P for polynomial pairs. To go further we need the definition of the analogue of periodic and preperiodic points for a polynomial pair.

Definition 3.3.2 : Let A = {α_{1}, .., α_{r}} be a finite subset of ℚ. We say that A is a (f_{1}, f_{2})-periodic orbit of period r if ∀j = 2, ..., r we have

f_{2}(α_{j}) = f_{1}(α_{j−1}), f_{2}(α_{1}) = f(α_{r}).

The points α_{1}, ..., α_{r} are called (f_{1}, f_{2})-periodic points of period r and the set of all (f_{1}, f_{2})- periodic points is denoted by P er(f_{1}, f_{2}).

Definition 3.3.3 : Let f_{1}, f_{2} ∈ ℚ[x] such that degf_{1} < degf_{2}. Let B = {α_{1}, α_{2}, ...} be an infinite subset of ℚ such that

f_{2}(x_{j}) = f_{1}(x_{j−1}) ∀j ∈ ℕ;

then we say that B is a (f_{1}, f_{2})-preperiodic orbit and the points α_{j} are called the (f_{1}, f_{2})- preperiodic points. We denote the set of all (f_{1}, f_{2})-preperiodic points by Π(f_{1}, f_{2}).

Since now we have taken f_{1}, f_{2} ∈ ℚ[x], but we may also assume that f_{1}, f_{2} ∈ ℤ[x]; in fact multiplying for the least common multiple of the denominators does not change the periodic and preperiodic points. In the following lemmas we will see that (f_{1}, f_{2})-periodic and preperiodic points can be only of certain forms.

Lemma 3.3.4 : Let f_{1}, f_{2} ∈ ℤ[x] as before. If f_{2} is monic then P er(f_{1}, f_{2}) ⊆ ℤ and Π(f_{1}, f_{2}) contains at most finitely many non-integral points.

Proof: We can write

f_{1}(x) = ∑_{j=0}^{d_{1}} a_{j}x^{j}, f_{2}(x) = ∑_{j=0}^{d_{2}} b_{j}x^{j} a_{j}, b_{j} ∈ ℤ.

Let {α_{k}} be a (f_{1}, f_{2})-periodic orbit and ∀k write α_{k} = p_{k}/q_{k} with gcd(p_{k}, q_{k}) = 1 and q_{k} > 0. We have

f_{1}(α_{k}) = \frac{b_{d_{1}}p_{k}^{d_{1}} + ...}{q_{k}^{d_{1}}} = f_{2}(α_{k+1}) = \frac{p_{k+1}^{d_{2}}}{q_{k+1}^{d_{2}}}

and the second fraction is irreducible. Then ∀k there must exist c_{k} ∈ ℤ^{+} such that

q_{k}^{d_{1}} = c_{k}q_{k+1}^{d_{2}}.

Multiplying by k and setting C = ∏ c_{k} and Q = ∏ q_{k} we have

Q^{d_{1}} = CQ^{d_{2}},

that is, CQ^{d_{2}−d_{1}} = 1. Since they are integers, the only possibility is C = 1 and Q = 1. This implies that all the q_{k} are equal to 1.

If now {α_{k}} is a (f_{1}, f_{2})-preperiodic orbit we can still write the preceding relations. Now let us suppose that there exists an index k_{0} such that q_{k_{0}+1} ≥ q_{k_{0}}. Then we have

q_{k_{0}}^{d_{1}} = c_{k_{0}}q_{k_{0}+1}^{d_{2}} ≥ c_{k_{0}}q_{k_{0}}^{d_{2}}.

Multyplying by q_{k_{0}}^{−d_{1}} we have

1 ≥ c_{k_{0}}q_{k_{0}}^{d_{2}−d_{1}}.

Since all the quantities are integers, q_{k_{0}} = 1 and we have q_{k} = 1 ∀k ≥ k_{0}. Hence all but finitely many α_{k} are integers.

If q_{k+1} < q_{k} ∀k then {q_{k}} is a strictly decreasing sequence of positive integers and then must be eventually equal to 1.

Removing the hypothesis of f_{2} monic, the following lemma shows that much more is true.

Lemma 3.3.5 : Let f_{1}, f_{2} ∈ ℤ[x] with d_{1} = degf_{1} < degf_{2} = d_{2}. Then the set P er(f_{1}, f_{2}) is finite and the set Π(f_{1}, f_{2}) is the empty set.

Proof: Writing f_{1}, f_{2} as in the previous lemma, we put

g_{i}(x) = b_{d_{2}}^{d_{2}−1}f_{i}(x/b_{d_{2}}).

It is clear that g_{i} ∈ ℤ[x] and that g_{2} is monic. Moreover if {α_{k}} is a (f_{1}, f_{2})-periodic or preperiodic orbit, then {b_{d_{2}}α_{k}} is a (g_{1}, g_{2})-periodic or preperiodic orbit. So we can assume that f_{2} is monic. By the previous lemma we have P er(f_{1}, f_{2}) ⊆ ℤ. We now want to show that ∃C > 0 depending only on f_{1}, f_{2} such that ∀α ∈ P er(f_{1}, f_{2}) we have |α| ≤ C; this will imply the finiteness of P er(f_{1}, f_{2}). First of all we choose a periodic orbit {α_{k}} of period r such that max(|α_{k}|) is sufficiently large; then we must have |α_{j−1}| < |α_{j}| (since d_{1} < d_{2}). Repeating this argument we have

|α_{0}| = |α_{r}| < |α_{r−1}| < ... < |α_{1}| < |α_{0}|

which is absurd. Then the only possibility is that max(|α_{k}|) is bounded. Since this argument does not depend on the choice of the periodic orbit, we conclude that all of them are bounded by a constant depending only on f_{1}, f_{2} and then there is only a finite number of them. Now let {α_{k}} be a preperiodic orbit. By the previous lemma, removing at most a finite number of terms, we can assume that {α_{k}} ⊆ ℤ. The same argument as before tells us that, if |α_{j}| is sufficiently large, we have

0 ≤ |α_{j}| < |α_{j−1}| < ... < |α_{1}| < |α_{0}|.

Then |α_{0}| > j − 1; this is absurd since j can be arbitrarily large.

We can now state the main theorem about polynomial pairs in ℚ[x].

Theorem 3.3.6 : Let f_{1}, f_{2} ∈ ℚ[x] with d_{1} =deg f_{1} <deg f_{2} = d_{2} and X ⊆ ℚ such that f_{1}(x) ⊆ f_{2}(X) and f_{1}|_{X} is injective. Then X is finite.

Proof: First of all, up to multilying by the least common multiple of the denominators, we may assume that f_{1}, f_{2} ∈ ℤ[x]. We construct a sequence {α_{k}} of elements of X in the following way: choose α_{0} any element of X and, given α_{j}, let α_{j+1} be any element that satisfies f_{2}(α_{j+1}) = f_{1}(α_{j}) (by hypothesis at least one of these elements exists). Letting α_{0} vary over all elements of X, the union of these sequences covers the set X. Let us now suppose that ∃r, s ≥ 0 r < s such that α_{r} = α_{s} and that r is the smallest index with that property. If r > 0 we have:

f_{1}(α_{r−1}) = f_{2}(α_{r}) = f_{2}(α_{s}) = f_{1}(α_{s−1}).

Since f_{1} is injective, this relation contradicts the minimality of r. Then the only possibilities are r = 0 or r does not exist and therefore {α_{k}} must be a (f_{1}, f_{2})-periodic or preperiodic orbit. By the previous lemma, there are no (f_{1}, f_{2})-preperiodic orbits and there are only a finite number of periodic ones. Then we have shown that the set X is a union of periodic orbits and therefore must be finite.

What we have shown is just a sort of “dimension one”-case of the property V P; now we have all the elements to state the complete property.

Definition 3.3.7 : Let K be a field. We say that K satisfies the property VP if ∀n > 0, for every pair of admissible polynomial maps F_{1}, F_{2} such that, if D_{1} is the maximum of the degrees of polynomials defining F_{1} and d_{2} is the minimum of the degrees of polynomials defining F_{2}, then D_{1} < d_{2} and that ∀X ⊆ K^{n} such that F_{1}(X) ⊆ F_{2}(X) and F_{1}|_{X} is injective, then X is finite.

It is easy to generalise the preceeding proof to every dimension and to show that ℚ satisfies the property V P. We now want to show that V P behaves well under finite extensions.

Theorem 3.3.8 : Let K be a field satisfying the property V P and L be a finite extension of K of degree n. Then also L satisfies the property V P.

Proof: Let F = (f_{1}, ..., f_{m}), G = (g_{1}, ..., g_{m}) : L^{m} → L^{m} be two admissible polynomial maps such that

min{deg f_{j}} > max{deg g_{j}}.

and X ⊆ L^{m} such that G(X) ⊆ F (X) and G|_{X} is injective. Fixing a K-basis of L ω_{1}, ..., ω_{n}, we can apply the map ψ and construct the polynomial maps F̂, Ĝ, as in the proof of Theorem 3.1.9. By the property of ψ, F̂ and Ĝ are also admissible polynomial maps. We just have to check that they satisfy the right conditions on the set X̂ = ψ(X). Let α ∈ X; there exists β ∈ X such that F (β) = G(α) and then ∀j

∑_{k=1}^{n} f_{jk}(β̂)ω_{k} = ∑_{k=1}^{n} g_{jk}(α̂)ω_{k}.

Since {ω_{k}} is a basis, we have f_{jk}(β̂) = g_{jk}(α̂) ∀j, k and hence F̂(β̂) = Ĝ(α̂). Then Ĝ(X̂) ⊆ F̂(X̂). Finally we have to show that Ĝ|X̂ is injective; if ∃α, β ∈ X such that Ĝ(α̂) = Ĝ(β̂), then g_{jk}(α̂) = g_{jk}(β̂) ∀j, k. Since ψ in injective this implies G(α) = G(β), but G is injective on X and then α = β and α̂ = β̂. Then Ĝ|X̂ is injective. Since K satisfies the property V P it follows that X̂ is finite and so is X. Then also L satisfies V P.

Obviously we have that

V P ⟹ SP ⟹ P

and in section 3.2 we have shown that the second arrow can not be inverted. It is conjectured that even the first one can not, but this is still an open problem.

Bibliography

[1] M. H. Baker and L. C. Hsia, Canonical heights, transfinite diameters and polynomial dinamics, 2004, available at arxiv:math.NT/0305181 v2.

[2] M. H. Baker and R. Rumely, Equidistribution of small points, rational dynamics and potential theory, 2005, available at arxiv:math.NT/0407426.

[3] A. F. Beardon, Iteration of rational functions, Springer-Verlag 1991.

[4] E. Bombieri and W. Gubler, Heights in diophantine geometry, Cambridge University Press, 2006.

[5] E. Bombieri and U. Zannier, A note on heights in certain finite extensions of ℚ, Rend. Mat. Acc. Lincei, 12 (2001), 5-14.

[6] G. S. Call and J. H. Silverman, Canonical heights on varieties with morphisms, Comp. Math., 89 (1993), 163-205.

[7] R. Dvornicich and U. Zannier, On sums or roots of unity, Monatsch. Math., 129 (2000), 97-108.

[8] R. Dvornicich and U. Zannier, Cyclotomic diophantine problems, to appear in Duke Math. J.

[9] J. H. Evertse, The number of solutions of decomposable form equations, Invent. Math., 122 (1995), 559-601.

[10] F. Halter-Koch and W. Narkiewicz, Finiteness properties of polynomial mappings, Math. Nachr., 159 (1992), 7-18.

[11] M. Hindry and J. H. Silverman, Diophantine geometry: an introduction, Springer-Verlag 1999.

[12] S. Kawaguchi, Canonical heights for random iterations in certain variables, 2005, available at arxiv:math.AG/0510633 v1.

[13] K. K. Kubota, Note on a conjecture of W. Narkiewicz, J. Number Th., 4 (1972), 181-190.

[14] D. J. Lewis, Invariant sets of morphisms in projective and affine number spaces, J. Algebra, 20 (1972), 419-434.

[15] P. Liardet, Sur une conjecture de Narkiewicz, Comptes Rendus Acad. Sci. Paris, 274 (1972), 1836-1838.

[16] A. Loxton, On the maximum modulus of cyclotomic integers, Acta Arit. 22 (1972), 69-85.

[17] D. Marcus, Number fields, Springer-Verlag 1977.

[18] R. Marszalek and W. Narkiewicz, Finite polynomial orbits in quadratic rings, Ramanujan J., 12 (2006), 91-130.

[19] J. Milnor, Dynamics in one complex variable, Princeton University Press 2006.

[20] W. Narkiewicz, Polynomial mappings, Springer-Verlag 1995.

[21] W. Narkiewicz and T. Pezda, Finite polynomial orbits in finitely generated domains, Monatsch. Math., 124 (1997), 309-316.

[22] T. Pezda, Polynomial cycles in certain local domains, Acta Arith., 66 (1994), 11-22.