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Galois Theory and Polynomial Orbits
by
Kalyani K. Madhu
Submitted in Partial Fulfillment
of the
Requirements for the Degree
Doctor of Philosophy
Supervised by
Professor Thomas J. Tucker
Department of MathematicsArts, Sciences and Engineering
School of Arts and Sciences
University of RochesterRochester, New York
2011
ii
Curriculum Vitae
Kalyani Madhu was born in Seattle, Washington in 1962. She earned her Bachelor
of Arts in English with Distinction at Cornell University in 1984. She received
a Masters of Arts in Mathematics at the State University of New York College
at Brockport in 2005. She received a Master of Arts in Mathematics at the
University of Rochester in 2008. She began her graduate studies in Mathematics
at the University of Rochester in 2006. There she pursued her interest in Algebraic
Number Theory and Arithmetic Dynamics under the guidance of her doctoral
advisor, Thomas J. Tucker.
iii
Acknowledgments
I would like to thank all the good people of the Math department at the Uni-
versity of Rochester for creating a warm, supportive environment. In particular I
would like to thank Jonathan Pakianathan, Nick Rogers, and Shannon Starr for
their help. I would like to thank Adam Towsley for working with me, helping me,
and for his friendship. I would like to thank Joan Robinson for her knowledge
and friendly interest. Without her our department would be a much colder place.
And I would like to thank my advisor, Tom Tucker, a true mensch.
I would like to thank my parents for their endless childcare help, for their
enthusiastic support of my studies, and for forgetting to mention to me when I
was little that girls are not analytical thinkers. I would like to thank my children,
Noah, Hannah, Jo and Susi, for their patience and love, and my daughters for
their proofreading. I would like to thank my husband, Ashwani, for his enduring,
unequivocal support.
iv
Abstract
We address two questions arising from the iteration of the polynomial f(x) =
xm + c. The first question concerns orbits of points in finite fields. Let Em be the
set of rational primes congruent to 1 modulo m. Let p ∈ Em and f(x) ∈ Fp[x].
We show that, with some restrictions on c, the proportion of points in Fp that are
periodic points of f goes to zero as p ∈ Em goes to infinity. The second question
concerns orbits of rational numbers. Let f(x) ∈ Q[x], and α ∈ Q. We show that,
with some restrictions on c, the subset of primes of Em that divide the orbit of α
has natural density zero.
We use Galois theoretic methods developed by R.W.K. Odoni for both ques-
tions. Odoni showed that, in general, prime divisors of the orbits of polynomials
in Z[x] are a sparse set in spec Z [18, Theorem III(Colloquial version)]. Rafe
Jones [12] showed that sets of primes dividing the orbits of certain quadratic
maps have density zero. We continue this process of uncovering specific maps for
which Odoni’s generalization holds.
v
Table of Contents
Curriculum Vitae ii
Acknowledgments iii
Abstract iv
1 Introduction 1
1.1 Galois Groups and Wreath Products . . . . . . . . . . . . . . . . 3
1.2 Stability and Reducibility of Iterates . . . . . . . . . . . . . . . . 5
1.3 Frobenius Automorphisms and Chebotarev’s Density Theorem . . 7
2 Periodic Points in Finite Fields 9
2.1 Periodicity and Roots . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Maximality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 Proof of Theorem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . 18
3 Primes Dividing the Orbits of Rational Numbers 22
3.1 Roots and Galois Groups . . . . . . . . . . . . . . . . . . . . . . . 23
3.2 Proof of Theorem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.3 Quadratics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
vi
4 Conclusion 49
Appendices 52
Bibliography 59
1
1 Introduction
A dynamical system is a pair (S, φ) where S is a set and φ is a map from S
to S. As φ is a self-map, we may compose φ with itself and consider iterates
of φ. By φn we mean n such compositions φ ◦ φ ◦ · · · ◦ φ. The notion of an
orbit naturally arises. Suppose s ∈ S; the forward orbit of s under φ is the set
Oφ(s) := {s, φ(s), . . . , φn(s), . . . }. Arithmetic dynamics concerns the study of
such orbits in sets that are of arithmetic interest.
Definition 1.1. The orbit of a point α is periodic if, for some n > 0, fn(α) = α.
A point is preperiodic if it has finite orbit. Hence, periodic points are always
preperiodic. A point is strictly preperiodic if it is preperiodic, but not periodic. A
point is called wandering if it is not preperiodic.
We will fix our map to be always f(x) = xm + c, a polynomial map, and
address two questions.
Let p be a rational prime not dividing m or c, and let f(x) = xm + c ∈ Fp[x].
As Fp is a finite set, all orbits are also finite, and so the interesting distinction
is between periodic and strictly preperiodic points. Several natural questions
regarding periodicity arise. For example, one might ask whether or not a particular
point in Fp is f -periodic, or for what kinds of maps a particular point is periodic.
We are able to answer a similar question in the finite field setting. We show that
2
for the map f and large p in the class of primes congruent to 1 modulo m, we
may expect very few points in Fp to be periodic for p. That is, if we take the limit
as p goes to infinity of primes p ≡ 1 mod m, we will find that the proportion of
periodic points in Fp goes to zero. This is Theorem 2.3, which we prove in chapter
2.
For the second question we regard f(x) = xm + c as a polynomial over the
rational numbers. Let α ∈ Q and consider Of (α). This is a sequence of rational
numbers. We will consider the set of rational primes that divide the numerator
of some element in Of (α).
Definition 1.2. Let K be a field. Let an = fn(α) for some α ∈ K. Let p be a
prime of OK. We say p divides the orbit of α if vp(an) > 0 for some n ∈ Z≥0.
Note that our definition does not consider primes at which elements of the
orbit have negative valuation. This is because our map is a polynomial, and so
there are only finitely-many such primes.
Definition 1.3. The natural density of a set P of primes of Z, denoted D(P ), is
given by
D(P ) = limn→∞
#{primes in P ∩ Z≤n}#{primes in Z≤n}
.
Let Em = {p ∈ Z | p is prime and p ≡ 1 mod m}. We consider the set of
primes dividing the orbit under f of a rational number α. We will show that,
with some restrictions on m and c, the set of primes in Em that divide Of (α) has
natural density zero. This is Theorem 3.2, which we prove in Chapter 3.
The link between these two questions has largely to do with the method of
proof. We consider the successive Galois groups of the iterated polynomials, and
make use of the correspondence between primes dividing orbits and the cycle type
of the elements in the Frobenius conjugacy classes of those primes in the Galois
group of a polynomial. This technique was introduced by R.K.S. Odoni [18].
3
Odoni considered generic polynomial maps F over fields of characteristic zero. He
shows that the Galois group of the nth iterate of F [18, Theorem 1] is the n-fold
wreath product of the symmetric group Sk with itself, where k is the degree of F.
He also shows that the proportion of elements in this group whose action on the
roots of Fn fixes at least one root goes to zero as n increases.
Rafe Jones showed via properties of Galois groups that the set of primes divid-
ing the orbit under certain quadratic polynomials of any fixed integer point has
natural density zero [12, Theorem 1.2]. Our second result shows that Jones’ result
may be extended, with some restrictions, to maps of the form f := x 7→ xm+c. As
in the quadratic case, it is more difficult to determine the Galois groups here than
it is in the generic case, as one must deal with the issue of stability of iterates,
which we define in Section 1.2. We find that if there exists a prime q ∈ Z such
that vq(c) > 0, then f is eventually stable. This is Proposition 3.7.
1.1 Galois Groups and Wreath Products
We would like to establish some properties of iterated polynomials and their Ga-
lois groups. We begin with notation that will be used with minor modifications
throughout this paper. Let K be a field and f(x) ∈ K[x] a polynomial. Let
x = f 0(x) and fn(x) = f ◦ fn−1(x). Let Kn be the splitting field of fn over K.
Let Gn =Gal(Kn/K) and Hn ⊂ Gn the Galois group of Kn over Kn−1.
If α is a root of fn in K, then α ∈ Kn, and so f(α) ∈ Kn as well. As α is a
root of fn, f(α) is a root of fn−1. Hence, we have containments Kn−1 ⊂ Kn, and
restriction maps ψn−j,n from Gn to Gn−j, where j ≥ 1. That is, let σ ∈ Gn. Then
ψn−j,n(σ) is the restriction of σ to Kn−j. If σ fixes α, then σ fixes f(α), because
σ ◦ f = f ◦ σ. Contrapositively, if f(α) = β, and σ(β) 6= β, then σ(α) 6= α.
We may create a tree graph whose vertices are the roots of successive iterates
of f and whose edges connect roots and their images and inverse images under
4
f . As σ ◦ f = f ◦ σ, adjacency in the graph is preserved under the action of
the Galois groups of successive iterates. Hence Gn is a subgroup of the group of
automorphisms of the tree (see [2]).
Suppose that fn−1 is separable of degree r and that {β1, . . . , βr} is the set of
roots of fn−1. A useful fact to note is that
fn(x) =r∏i=1
(f(x)− βi).
Let Li be the splitting field of f(x)− βi over Kn−1. Then Kn is the compositum
of the Li.
Definition 1.4. Let G,H be two permutation groups and r the degree of G. Then
the wreath product of G by H, which we denote G[H] is the semidirect product of
the Cartesian product of r copies of H by G in which G acts on Hr by permuting
the indices.
Remark 1.5. The wreath product can be defined much more generally, and enjoys
a wide variety of notation in the literature. This definition works well in the
context of iterated Galois groups.
The wreath product is associative in the sense that G[H[K]] ∼= G[H][K], so we
may refer to wreath powers [G]n, where [G]1 = G, and [G]n = [G]n−1[G]. We will
make extensive use of a Lemma of Odoni [18, Lemma 4.1].
Lemma 1.6. Let K be any field and f, g monic polynomials in K[x]. Let f ◦ g(x)
be separable over K, and let the degree of f be k and the degree of g be `, k, ` ≥ 1.
Then f(x) is also separable over K. Let F be the Galois group of f over K.
Then the Galois group of f ◦ g over K can be embedded in F [S`], where S` is the
symmetric group on ` letters.
This tells us that, if fn is separable for all n and the degree of f = m, then
Gn ⊆ Gn−1[Sm].
5
We can write an element of the wreath product F [G] as (τ ; π1, · · · , πr) where
τ ∈ F , r is the degree of F considered as a permutation group, and (π1, · · · , πr)
is an element of Gr. Consider the iteration of a polynomial f of degree m in K[x],
and suppose that iterates of f never factor, so that fn−1 is irreducible of degree
r := mn−1. We may write the roots of fn as {αi,j} where f(αi,j) = βi, a root of
fn−1. Then the i range between 1 and r, and the j range between 1 and m. If
σ = (τ ; π1, · · · , πd) ∈ Gn ⊆ Gn−1[Sm], then σ(αi,j) = ατ(i),πi(j).
Example 1.7. Let f(x) = x2 − 3. Then Gal(f 2/Q) ∼= D4∼= [S2]2 .
0 0
√3 −
√3
√3 +√3 −
√3 +√3
√3−√3 −
√3−√3
β1 β2
α1,1 = 1 α1,2 = 3 α2,1 = 2 α2,2 = 4
We compare permutation and wreath notation for the same group.
Permutation↔Wreath
e↔ (e; e, e) (1432)↔ ((12); (12), e) (13)(24)↔ (e; (12), (12))
(14)(23)↔ ((12); (12), (12)) (13)↔ (e; (12), e) (24)↔ (e; e, (12))
(12)(34)↔ ((12); e, e) (1234)↔ ((12); e, (12))
Lemma 1.8. Let Pn be the proportion of elements of Gn that fix at least one root.
If Gn∼= [G]n, then
limn→∞
Pn = 0.
Proof. This is Odoni’s [18, Lemma 4.3], which we prove in Section 2.2.
1.2 Stability and Reducibility of Iterates
Iterates of irreducible polynomials are not necessarily irreducible. This issue be-
comes important in the identification of the Galois groups of iterates. We will
6
discuss irreducibility of iterates in Chapter III, but we give the basic definitions
here.
Definition 1.9. A polynomial over a field K is called stable if all of its iterates
are irreducible.
A polynomial is called eventually stable if there are integers k, ` such that
f `(x) =k∏i=1
gi ◦ fni(x)
and gi ◦ fn(x) is irreducible for all n ≥ 0.
Example 1.10. Let f(x) = x2 − 169
.
f(x) =
(x− 4
3
)(x+
4
3
)= g1 (x) g2 (x)
f 2(x) =
(x2 − 16
9− 4
3
)(x2 − 16
9+
4
3
)= g1(f(x))g2(f(x))
= g1(f(x))
(x− 2
3
)(x+
2
3
)= g1(f(x))g2,1(x)g2,2(x)
f 3(x) = g1(f 2(x))
(x2 − 16
9− 2
3
)(x2 − 16
9+
2
3
)=
(x4 − 32
9x2 +
4
81
)(x2 − 22
9
)(x2 − 10
9
)=
(x2 − 2x+
2
9
)(x2 + 2x+
2
9
)(x2 − 22
9
)(x2 − 10
9
)= g1,1(x)g1,2(x)g2,1(f(x))g2,2 (f (x))
In Chapter III we will show that f factors no further and so is eventually
stable. By contrast f(x) = x2 − 1 is not eventually stable.
7
1.3 Frobenius Automorphisms and Chebotarev’s
Density Theorem
Definition 1.11. Let K be a field and p a prime of OK, and suppose that OK/p
is the finite field of q elements. Suppose that L is a finite Galois extension of K
with Galois group G := Gal (L/K), and that p is unramified in L. Let P be a
prime of OL, with P ∩ OK = p. Then the extension OL/P over OK/p is Galois
with generator σP called the Frobenius automorphism at P. For x ∈ OL/P,
σP(x) = xq.
We have defined the Frobenius automorphism as generating the Galois group
of OL/P over OK/p, a definition that specifies P. However, we can also define
it from below. The decomposition groups of each of the Pi extending p contain
(as long as p is unramified) a unique element σ that satisfies σ(x) ≡ xq mod Pi.
The unique elements for each i coming from primes that extend a single p are
conjugate, and so we may refer to Frob(p), the conjugacy class of σPi for each i
(see, for example,[10, III.2,p.125]).
Suppose that f is a polynomial in K[x] and E a simple extension obtained
by adjoining to K a root of f . Let L be the Galois closure of E over K. Let
G =Gal(L/K). Suppose that a prime p of OK is unramified in L. The factoring
of a prime p of OK in OE corresponds exactly to the cycle type of the elements
in Frob(p) ⊂ G. That is, if
pOE =s∏i=1
Pi,
then the elements of Frob(p) will be composed of s disjoint cycles each of length
fi, where fi is the residue degree of Pi. Hence, if the reduction of f mod p has
a linear factor, elements of Frob(p) will have a cycle of length one. In particular,
if f has a root modulo p, then the elements of Frob(p) will fix a root of f .
8
We are able to relate the density of primes with certain cycle types to the size
of their Frobenius conjugacy classes via the Chebotarev Density Theorem. (We
use the statement given by Lenstra in [23, p.18].)
Theorem 1.12. Let K be a number field and L a finite Galois extension of K
with Galois group G. Then, for any conjugacy class C of G, the density of the set
of primes p of K for which Frob(p) = C exists and equals #C/#G.
There are analogous theorems in the function field setting, (see for example
[19, Theorems 9.13a,b]), but we will be using an effective version of the Theorem
instead, which we will state in context in Chapter 2.
9
2 Periodic Points in Finite
Fields
Let f := x 7→ xm + c be a map over Fp. Let Em := {primes p | p ≡ 1 mod m}.
We show that, if p ∈ Em, then, with some restriction on c, the proportion of
periodic points of Fp under f goes to zero as p goes to infinity if p ∈ Em.
2.1 Periodicity and Roots
For some prime p ∈ Z and m ≥ 2, let f(x) = xm + c ∈ Fp[x]. Let α ∈ Fp. As
Fp is finite, so is the orbit of α under f . Thus α is at least preperiodic, so the
important distinction is between whether α is periodic or strictly preperiodic.
We begin with two examples of orbits. In both examples the map is the same,
but the congruence classes of p modulo m are different. We see that x 7→ xm + c
will be a permutation on Fp for primes p that are congruent to 2 modulo 3, and
not for primes congruent to 1 modulo 3, because if p ≡ 1 mod 3, then Fp contains
the primitive cube roots of unity.
10
Example 2.1. Orbit graph of f(x) = x3 + 2 over F43
27 34 4 23 0 2 10 3 1 37 41 8
13 6
29
26
33
32
20 15
24
2819 39
9
11
12
1617 35 38
18 22
4025 21
42
7
31 14
5 3036
In Example 2.1, f has five periodic points out of forty-three.
Example 2.2. Orbit graph of f(x) = x3 + 2 over F17
0
2
1016
1
3
12
13
6 14
9
415
117
5
8
In Example 2.2, all points are periodic.
We wish to consider the primes modulo which f is not a permutation, so
that we have a combination of periodic and strictly preperiodic points. We limit
ourselves to primes p such that Fp contains a primitive mth root of unity. This
will guarantee that our map is m-to-one except over c, where the map ramifies.
In the example of F43, the number of periodic points is fairly small. This agrees
with our intuition about the matter. Picking the points in the orbit of α, where
11
α ∈ Fp and p ≡ 1 mod m, is like picking with replacement from a set of p items.
Just as in the birthday problem, if we pick about√p items, the probability that
we have picked some item twice approaches 0.50. Thus we would expect orbits in
Fp to be about√p in size. Each of the
√p possibilities for a duplicate point is
equally likely, so we would not expect to return to the beginning and re-pick the
initial point. Thus we would expect the orbit of a periodic point to be shorter
than√p, and we should expect to see preperiodic orbits.
Further, as our maps are m-to-one, we experience branching as we move back-
ward through the orbit. That is, if a point has a pre-image under f , then it has
m pre-images. All the pre-images of strictly preperiodic points are themselves
strictly preperiodic, and exactly one of the pre-images of a periodic point is pe-
riodic. So heuristically we would expect to have few periodic points, and many
strictly preperiodic ones.
In the case f(x) = xm + c and p ≡ 1 mod m, we are able to prove that this
heuristic is correct.
Theorem 2.3. Let f(x) = xm + c, with m, c ∈ Z and m ≥ 2. Suppose that m, c
are such that 0 is not a preperiodic point of f over Z. Let Em := {primes p | p ≡ 1
mod m}. Let Pf,p refer to the proportion of points in Fp that are periodic points
of f . Then,
limp→∞p∈Em
Pf,p = 0.
The proof of Theorem 2.3 follows from a Proposition and its Corollary given
in the next two Sections. Here we will reformulate the question of periodicity in
finite fields as one about primes in function fields so that we may apply techniques
from Galois theory.
We begin by noticing that a point α ∈ Fp is periodic if and only if α appears
in the set of images of fn for every n. This is not hard to see, as, if α is periodic,
12
then we can find its nth preimage for every n by moving backward through the
periodic cycle of which α is a member.
Definition 2.4. A point α has exact period k under f if fk(α) = α, but f j(α) 6= α
if j < k.
Suppose that α has exact period k. For any n ∈ N, let j be the least integer
such that jk ≥ n. Then fn(f jk−nα)) = α.
Consider the function field Fp(t). For every n, the polynomial fn := fn(x)− t
is an element of the ring Fp(t)[x]. We will be working with the Galois group of
fn over Fp(t). Let pα := (t − α), a principal prime ideal of Fp[t]. We see that
fn has a root modulo pα for every n if and only if α is periodic under f , so we
may productively rephrase the question about periodic points in finite fields as a
question about prime ideals in function fields.
Suppose that α is periodic under f . When we factor fn modulo pα for suc-
cessive n, we will find that the reduced polynomial has a linear factor for each n.
If F ∈ K[x], and F mod p =∏s
i=1 gi(x), where the gi are distinct, irreducible
over OK/p, and each gi has degree fi, then the elements in Frob(p), acting as
permutations on the roots of fn, will be composed of s disjoint cycles, each of
length fi.
If K = Fp(t) and F = fn, and α is periodic under f , then elements in Frob(pα)
in Gn := Gal (fn/K) will have at least one cycle of length one corresponding to
the linear factor coming from the root of fn mod pα, and so each will fix at least
one root of fn.
Murty and Scherk [13] provide an effective Chebotarev Density Theorem in
the function field case. We have modified it somewhat to fit our circumstances.
Let Kn be the splitting field of fn over Fp(t) and Gn := Gal (Kn/Fp(t)). Let gn
be the genus of the curve given by fn, which is equal to the genus of the function
field Kn. We think of P1(Fp) as Fp ∪∞.
13
Theorem 2.5. Let ψ be the points of P1(Fp) that are unramified in Kn, and
ψC = {β ∈ ψ | Frob(β) = C} where C is a conjugacy class in Gn. Let D be the
set of points of P1(Fp) ramifying in Kn. Then∣∣∣∣#ψC − #C
#Gn
#ψ
∣∣∣∣ < 2gn#C
#Gn
√p+ #D.
We wish to consider the union of all conjugacy classes of Gn that have fixed
points. Suppose that there are N conjugacy classes Ci ⊂ Gn whose elements have
at least one fixed point. Let C = ∪Ni=1Ci. By the triangle inequality∣∣∣∣#ψC − #C
#Gn
#ψ
∣∣∣∣ < N∑i=1
∣∣∣∣#ψCi − #Ci#Gn
#ψ
∣∣∣∣ < 2gn#C
#Gn
√p+N#D.
We require that zero is not a preperiodic point of f if considered as a map on
Z. This allows us to choose, for sufficiently large p, an M such that f i(0) and
f j(0) are distinct if i, j < M and i 6= j. We will show in Section 2.2 that, given
such an M and p, Gn∼= [Cm]n, as in Definition 1.4. By Lemma 2.6, we see that
for every ε > 0 there is an n such that #C#Gn
< ε. We choose such an n and note
that there is a sufficiently large p0 to ensure that there exists M ∈ Fp0 , M > n,
such that, if i 6= j, i, j < M , then f i(0) 6≡ f j(0) mod p0. Then for any p > p0,
we will still have that the Galois group of fn over Fp(t) is [Cm]n. In Section 2.3
we will show that, for sufficiently large p, N,D, and gn depend on fn, and not
on p. Then, for fixed n, dividing by #ψ in the inequality allows us to make the
bound arbitrarily small. Thus the proportion of points in Fp that are periodic for
f approximates the proportion of elements in the Galois group of fn that fix at
least one root, and we may take this approximation to be as close as we like.
Then we need only consider the Galois groups that arise from successive it-
erations of f and identify the proportion of elements of each Galois group that
have fixed points. In the following Section, we will show in Proposition 2.9 and
its Corollary that Gn∼= [Cm]n. We stated in Chapter 1 that the proportion of
elements of [Cm]n that fix one or more points approaches zero as n approaches
14
infinity. We prove it here for completeness, and because the indicatrix function
defined in the proof is of intrinsic interest. The proof is due to Odoni [18, Lemma
4.3].
Lemma 2.6. Let K be a field, and let f(x) ∈ K[x], with deg(f(x)) ≥ 2. Let
G = Gal(f/K) and suppose Gn := Gal (fn/K) ∼= [G]n. Let Fn be the proportion
of elements of Gn that fix at least one root of fn. Then limn→∞ Fn = 0.
Proof. Let ΦG(X) = 1#G
∑σ∈GX
rσ where rσ is the number of points fixed by σ.
This polynomial in Q[X] is called the indicatrix and is attributed by Odoni to
Polya. It has the property that, if F and G are two finite groups, then ΦF [G] =
ΦF ◦ ΦG. Clearly Fn = 1 − ΦGn(0) = 1 − ΦnG(0). Note that ΦG(1) = 1. From
Burnside’s Lemma ( [7, Theorem 17.1]) we have that Φ′G(1) = 1, because G acts
transitively. So we have a polynomial on the interval (0,1] whose first and second
derivatives are positive, and whose graph is tangent to the line x = y at x = 1.
Thus the graph of ΦG lies entirely above the line x = y on [0, 1), so ΦG(x) > x.
Then for all points a ∈ [0, 1) and all n ∈ N, ΦnG(a) > Φn−1
G (a). As ΦnG(a) ≤ 1 for
all n, it must be that the the sequences {ΦnG(a)}n≥1 converge for every a ∈ [0, 1].
Since ΦG is continuous, each sequence, including {ΦnG(0)}n≥1, converges to a fixed
point of ΦG in [0, 1], that is, to 1.
2.2 Maximality
We recall the notation of Section 2.1. That is, let f,m, and c be as in Theorem
2.3. Consider fn := fn(x) − t as a polynomial in Fp(t)[x] where p ∈ Em. Let
Kn refer to the splitting field of fn over Fp(t). Let Gn = Gal (Kn/Fp(t)), and
Hn = Gal (Kn/Kn−1). Let r = deg(fn−1). Let β1, β2, . . . , βr be the roots of
fn−1(x)− t. We use this notation throughout the following section.
Claim 2.7. The polynomial fn(x)− t is separable over Fp(t) for each n.
15
Proof. As fn generates a prime ideal of Fp(t)[x], it is irreducible. The roots
m√t− c, ζm m
√t− c, . . . , ζm−1
mm√t− c
of f1 are distinct, as t 6= c. Suppose that fn−1 is separable with distinct roots
β1, . . . , βr. Then each root of fn has the form ζjmm√βi − c. Each is distinct, as the
βi are distinct, c ∈ Fp, and βi is transcendental over Fp, so βi − c is non-zero.
We may write
fn(x)− t =r∏i=1
(f(x)− βi).
For each i, we choose one of the m elements of the set f−1(βi) and name it αi.
Then f(x)− βi splits in Kn−1(αi) := Li. Thus Kn is the compositum of the Li.
Definition 2.8. Let Hn = Gal (Kn/Kn−1). Let Ji = Gal (Li/Kn−1). Note Ji ∼= Jk
for all i, k. Then we say Hn is maximal if #Hn = #Jri .
Proposition 2.9 and its Corollary show that, given our conditions, Hn is max-
imal for each n.
Proposition 2.9. Let N ∈ N such that for i, j < N , i 6= j, we have f i(0) 6≡ f j(0)
mod p. Let n < N . Then:
(i) There is a prime p of Fp[t] that is unramified in Kn−1 and ramifies in Kn.
(ii) Let P ⊆ OKn−1 such that P ∩ Fp[t] = p. Then P ramifies in OLi to degree
m for exactly one i ∈ {1, . . . , r}, and is unramified in OLj if j 6= i.
Proof. We will begin by showing that pn := (t− fn(0)) ramifies in Kn and is un-
ramified in Kn−1. This will prove (i). Let f(αi) = βi. Let Cf,n = Fp[t, x]/(fn(x)−
t). Then Fp[t][αi] ∼= Cf,n. As Li contains all of Kn−1, it is not necessarily true
that OLi ∼= Fp[t][αi]. However, Kn is also the compositum of the Fp(t)(αi), so
a prime ramifies in Fp(t)(αi) for some i if and only if it ramifies in Kn. As
16
Fp[t, x]/(fn(x)− t) ∼= Fp[x], Fp[t][αi] is a principal ideal domain, thus a Dedekind
domain, and so we may consider the unique factorization of its ideals.
We may write fn(x)− t = fn(0)− t+∏d
j=1 gj(x)ej , where gj(x) is irreducible
over Fp(t). Thus the maximal ideals of Cf,n containing pn are generated by gj(x)
for some j ∈ {1, . . . , d}, and pn ramifies in Cf,n if and only if ej > 1 for some j.
Let Mj = (gj(αi)), the image of one of the maximal ideals extending pn under the
isomorphism between Cf,n and Fp[t][αi].
We will show that pn ramifies in Fp[t][αi] if and only if ddx
(fn)(αi) ∈ Mj,
because (fn)′(αi) =∑d
j=1 ejgj(αi)ej−1g′j(αi)
∏k 6=j gk(αi)
ek . If j 6= k, gk(αi) /∈ Mj,
because gk(αi) generates the ideal Mk. As Fp[t][αi]/Mj is a finite extension of the
perfect field Fp[t]/pn, and gj(x) is irreducible in Fp[t]/pn, we have that g′j(αi) /∈Mj.
Hence (fn)′(αi) ∈Mj for some j if and only if ej > 1 if and only if Mj is ramified
over Kn−1.
We have ddx
(fn(x)− t) = mn∏n
i=1(fn−i(x))m−1, so (fn)′(αi) = mn∏n
j=1 βm−1j ,
where f j(βj) = t. Mj ramifies if and only if it contains βk for some k, so if and
only if it contains the product of all the roots of fk(x)− t, which is fk(0)− t.
Suppose a maximal ideal M of Cf,n contains βj. Then M ∩Fp[t] = pj. Then if
M is ramified over Fp[t], M ∩ Fp[t] is one of the pi for i ∈ {1, · · · , n}. As n < N ,
these primes are distinct. In particular, pn = (t − fn(0)) ramifies in OKn and is
unramified in OKn−1 . This concludes the proof of part (i).
For part (ii), let Pi be a prime of OKn−1 extending pn and containing f(0)−βi.
Let f(αi) = βi. We will show that Pi ramifies in Li = Kn−1(αi) and not in Lj
if j 6= i. Recall that fn(0) − t =∏r
j=1(f(0) − βj), so any prime extending pn
contains f(0)− βi for some i.
As Pi is unramified over Fp[t], vPi(f(0) − βi) = 1. Thus Pi
(OKn−1
)Pi
can
be generated by f(0) − βi. Then any element of Pi
(OKn−1
)Pi
has the form
u(f(0) − βi)k for some k ≥ 1 and unit u. Any prime ideal M of OLi extending
17
Pi must contain αi. Hence any element of Pi (OLi)M has the form uαkmi . Then
Pi (OLi)M ⊂Mm (OLi)M , so Pi ramifies to degree m in Li.
We must show that Pi is unramified in OLj if j 6= i. We write f(x) − βj =
h(x) +∏s
k=1 gk(x)ek where h(x), hence h′(x), is in PiOKn−1 [x]. Then
f ′(αj) = h′(αj) +s∑
k=1
ek(gk(αj))ek−1g′k(αj)
∏k 6=`
g`(αj).
So a prime M of OLj extending Pi ramifies over OKn−1 if and only if it contains
f ′(αj). If M ramifies over OKn−1 , then αj ∈ M , so f(0) − βj ∈ M ∩ OKn−1 . As
Pi contains f(0) − βi and is unramified in Kn−1, f(0) − βj /∈ Pi. Hence Pi is
unramified in OLj .
Corollary 2.10. With the suppositions of Proposition 2.9, Hn is maximal for
each n.
Proof. By Proposition 2.9, we have that Kn is the compositum of extensions
of Kn−1 none of which is unramified, and in each of which a prime ramifies to
degree m that doesn’t ramify in any other. It is clear that Gal(Li/Kn−1) ∼= Cm.
So [Li : Kn−1] = m for each i. Fix Lr arbitrarily. Let L =∏r−1
k=1 Lk. Then
[Kn : L] = m, because any prime of OKn−1 containing f(0) − βr is unramified
in L, but ramifies to degree m in Kn. As r was arbitrarily chosen, we have that
[Kn : Kn−1] = mr.
The Galois Group of Gn
We will show that, if Hn is maximal, then Gn∼= Gn−1[Cm].
Even if Hn is maximal, it does not follow immediately that Gn∼= Gn−1[Cm].
For example, let f(x) = x3 + 2 over Fp(t) where p ≡ 2 mod 3 and p > 11.
Then #G2 = #S3[C3], but G2 6∼= S3[C3]. This is because the field K1 contains
an extension of the constant field Fp, so ζ3 is not fixed by G1. Thus we can get
18
automorphisms of the form ((12); (12), (12), (12)) /∈ S3[C3], but not, for example
((12); e, e, e) ∈ S3[C3].
Lemma 2.11. With the notation and conditions of Theorem 2.3 and Proposition
2.9, if Hn is maximal, then Gn∼= Gn−1[Cm].
Proof. Let ζm be a fixed primitive mth root of unity, and let {β1, . . . , βr} be
the roots of fn−1. We will refer to m√βi − c as αi,1. Then αi,2 = ζmαi,1, αi,3 =
ζ2mαi,1, . . . , αi,m = ζm−1
m αi,1.
We know that Gn ⊆ Gn−1[Sm]. We will show that an arbitrary element σ of
Gn has the form σ = (τ ; π1, · · · , πd), where each πi ∈ Cm. This would imply that
Gn ⊆ Gn−1[Cm], and by counting our proof would be complete.
As ζm must be fixed by σ, we know
σ (αi,j) = σ(ζj−1m αi,1
)= ζj−1
m σ (αi,1.)
Suppose σ(αi,1) = ατ(i),k+1 = ζkmατ(i),1. Then σ(αi,j) = ζj−1+km ατ(i),1 for each j.
That is, we could write πi in permutation notation as
(1, k + 1, 2k + 1, · · · , (m− 1)k + 1)
where each term is reduced modulo m. So πi = (123 · · ·m)k ∈ Cm.
2.3 Proof of Theorem 2.3
Proof. From Proposition 2.9 and Corollary 2.11, we know that for each n, Gn∼=
Gn−1[Cm]. As G1∼= Cm, we may conclude by induction that Gn
∼= [Cm]n. From
Lemma 2.6, we know that, if Fn refers to the proportion of elements of Gn that
fix at least one root of fn, then limn→∞ Fn = 0. For any fixed p, we can not take
this limit, as we must have that f i(0) 6= f j(0) in Fp if i, j < n, and for sufficiently
large n in a fixed Fp, this condition will not hold. However, as we are taking p in
19
the infinite class Em, then, for very large n we will always find a p such that the
condition does hold.
Let Ci =Frob(pi) where pi = (t − ci), pi is unramified, and ci ∈ Fp is f -
periodic. Let C = ∪Ni=1Ci. Then, with the notation of Theorem 2.5, ψC is the set
of unramified periodic points in Fp. We see that Pf,p ≤ #ψC+n+1p+1
.∣∣∣∣#ψC + n+ 1
p+ 1− #C
#Gn
∣∣∣∣ < ∣∣∣∣#ψC + n+ 1
p+ 1− #ψCp− n
∣∣∣∣+
∣∣∣∣#ψCp− n
− #C
#Gn
∣∣∣∣<
∣∣∣∣#ψC + n+ 1
p+ 1− #ψCp− n
∣∣∣∣+ 2gn#C
#Gn
√p
p− n+N#D
p− n.
Let ε > 0 and fix n large enough so that #C#Gn
< ε4. We know that there exists a
prime p0 ∈ Em such that
(i) p0 > m, and
(ii) if i, j < n, i 6= j, then f i(0) 6≡ f j(0) mod p0.
Let Pm = {p ∈ Em | p ≥ p0}. Any prime of Pm will also have both properties.
It is clear that N depends only on Gn. Let us consider #ψ and D. The only
critical number of f is zero, and so the only primes of Fp[t] ramifying in OKn are
the pfk(0) for each k ∈ {1, . . . , n}, and the point at infinity. (This fact is shown in
the proof of Proposition 2.9.) Thus #ψ = p− n and #D = n+ 1.
The genus gn also depends only on fn and can be bounded for each n inde-
pendent of choice of p as long as p ∈ Pm. (The following theorem is [24, III.5.6
Corollary], and Abhyankar’s lemma is [24, III.8.9].)
Let F/K be an algebraic function field over K. That is, F/K is a finite,
algebraic extension of K(t) where t ∈ F is transcendental over K. Let P ∈ PFdenote a maximal ideal of some valuation ring OP of F/K.
20
Theorem 2.12. Suppose that F ′/F is a finite separable extension of function
fields having the same constant field K. Let g (resp. g’) denote the genus of F/K
(resp. F ′/K). Then
2g′ − 2 ≥ [F ′ : F ] · (2g − 2) +∑P∈PF
∑P ′|P
(e(P ′ | P )− 1) · degP ′
Equality holds if and only if F ′/F is tame.
Lemma 2.13 (Abhyankar’s Lemma). Let F ′/F be a finite separable extension of
functions fields. Suppose that F ′ = F1F2 is the compositum of two intermediate
fields F ⊆ F1, F2 ⊆ F ′. Let P ′ ∈ PF ′ be an extension of P ∈ PF , and set
Pi := P ′ ∩ Fi. Assume that at least one of the extensions Pi | P is tame. Then
e(P ′ | P ) = lcm{e(P1 | P ), e(P2 | P )}.
We have that Kn/Fp(t) is a finite separable extension of function fields with
constant field Fp, and we assume p ∈ Pm. Then gcd(p,m) = 1, so the the extension
is tamely ramified.
We have that Kn is the compositum of the fields Li. From Proposition 2.9
we know that the finite primes of Fp(t) either ramify to degree m in Li or are
unramified, and that there are n such primes. By Abhyankar’s lemma we see
that pi ramifies to degree m in Ki, as it ramifies to degree m in each subfield Li.
By Propostion 2.9, no further ramification occurs, and each ramifies to degree m
in Kn. Hence there are #Gnm·degP ′
primes of Kn extending each one. The prime at
infinity ramifies to the degree of fn and have degree 1, so there are #Gn/mn of
them. The genus of Fp(t) = 0. This gives us
2gn − 2 = −2#Gn + n#Gn
m(m− 1) +
#Gn
mn(mn − 1),
which is independent of p.
Then we may select p ∈ Pm so that both∣∣∣∣#ψC + n+ 1
p+ 1− #ψCp− n
∣∣∣∣
21
and
2gn#C
#Gn
√p
p− n+N#D
p− nare less than ε
4.
Then∣∣∣#ψC+n+1
p+1− #C
#Gn
∣∣∣ < ε2, so #ψC+n+1
p+1< ε.
Work in this Section was done in collaboration with T. Tucker and P. Kurlberg.
22
3 Primes Dividing the Orbits of
Rational Numbers
In this Chapter we will consider the orbits of rational numbers under the polyno-
mial map f(x) = xm + c ∈ Q[x],m ≥ 2. Let a ∈ Q. We will show that the set of
primes that both divide Of (α) and are elements of Em has natural density zero.
This is Theorem 3.2.
This question can be related to periodic points in finite fields. For example,
suppose a prime p ∈ Z divides the orbit of a exactly once. Suppose that vp(an0) >
0. Then if we considered f as a map over Fp, we would see that an = 0 in Fp,
where an is the reduction of an modulo p. If we know that p does not divide the
orbit of a again, we know that 0 ∈ Fp is strictly preperiodic under f .
The linkage between the two questions of periodicity of points in finite fields
and the density of primes dividing rational orbits extends to the manner of proof.
That is, we can relate the question of primes dividing rational orbits to fixed
points in Galois groups. We will show that the factorization in the splitting field
of fn of some primes coming from the critical orbit (that is, Of (0),) forces the
Galois group of fn to be large, just as in the function field case. The Galois groups
that arise can be shown to have few fixed points; this in turn controls the density
of primes dividing any orbit.
Iteration of f in Q[x] gives rise to some complications that are not present
23
in the function field case. The polynomial fn(x) − t is always irreducible over
Fp(t), whereas fn(x) is not necessarily irreducible over Q. So we must address the
stability of f . Further, (fn(0)− t) is always a prime of Fp[t], and for large enough
p, the primes (f i(0)− t) and (f j(0)− t) are distinct primes. In the rational case,
the numerator of fn(0) may be composite, so it may be divisible by a power of
a prime. Further, it is possible that fn(0) has positive valuation only at primes
that also divided f j(0) for some j < n. Both of these possibilities will affect
the ramification of primes in the splitting field Kn, and so the maximality of the
Galois group Gn.
In Section 3.1 we present the theorem and set up notation and the basic idea
behind the proof. In Section 3.1.2 we show that, under certain very general
circumstances, the factoring of iterates of f is controlled. That is, we show that
f is eventually stable. In Section 3.1.3 we discuss the Galois groups of successive
iterates of f and show that they are maximal for infinitely many n. In Section
3.1.4 we make a counting argument concerning the proportion of elements of the
Galois groups that fix at least one root, and, in Section 3.2, we assemble this
information and prove the Theorem.
To obtain the result concerning eventual stability, we are required to restrict
our choice of constant c. We require that there be a prime q of Z such that
vq(c) > 0. In the case that m = 2, we are able to relax this condition somewhat,
which we do in Section 3.3.
3.1 Roots and Galois Groups
Let f(x) = xm + c ∈ Q[x], m ≥ 2, and suppose that c = rs
with r, s relatively
prime integers. Let α ∈ Q, and consider Of (α). Although this orbit is a set of
rational numbers, we see that the only denominators in the orbit are products of
s and primes dividing the denominator of α.
24
Definition 3.1. Let K be a number field and R its ring of integers. Let S be a
finite set of primes of K including the infinite primes. Then the set
RS := {a ∈ K | vp(a) ≥ 0 for all primes p /∈ S}
is called the set of S-integers of K.
We may consider Of (α) as a set of S-integers with
S = {p ∈ Z | p is prime and vp(c) < 0 or vp(α) < 0} ∪M∞.
Thus our orbit is contained in a ring, rather than a field, and we can discuss prime
ideals containing elements in the orbit without ambiguity.
Theorem 3.2. Let f(x) = xm + c ∈ Q[x], with m ≥ 2, the numerator of c is not
±1, and Of (0) is infinite. Let Em = {p ∈ Z | p is prime, and p ≡ 1 mod m}.
Let α ∈ Q, and Of (α) the forward orbit of α under f . Let an = fn(α). Let
S = {p ∈ MQ | vp(c) < 0 or vp(α) < 0} ∪M∞, and RS the ring of S-integers of
Q. Let Pf,m,α = {p ∈ Z | p ∈ Em and an ∈ (p)RS for some n}. Then the natural
density of Pf,m,α equals 0.
The proof of the theorem follows from the correspondence between primes
dividing orbits and their Frobenius classes.
Remark 3.3. Suppose an ∈ Of (α) is contained in (p)RS, where p ∈ Em. Suppose
fn(x) is irreducible for all n ≥ 1. Then fn(x) has a root modulo (p)RS. Unless p
ramifies in Kn, automorphisms in Frob(p) ⊂ Gn will fix at least one root of fn.
In Section 3.1.2, we will address the possibility that there is a k ∈ N such that
fk is not irreducible. Then we will consider polynomials of the form g ◦ fn(x),
where g ◦ f i(x) is irreducible for all i ≥ 0. We note that if Kn is the splitting
field of g ◦ fn(x) over Q, and Gn = Gal(Kn/Q), then, unless p ramifies in Kn,
automorphisms in Frob(p) ⊂ Gn will fix at least one root of g ◦ fn.
25
Therefore we may relate the density of primes dividing an orbit via Cheb-
otarev’s Density Theorem to the proportion of automorphisms in Gn that have
fixed points. Let Fn be the proportion of automorphisms in Gn that fix at least
one root of g ◦ fn. We will show in Section 3.1.4
limn→∞
Fn = 0.
We will make some adjustments to our notation. We will show that f is
eventually stable as in definition 1.9. That is, we show that there is a fixed
integer N , depending only on f , such that further iteration of f after fN causes
no further factorization. We will then consider maps of the form g◦fn where g is a
polynomial divisor of fk for some k, and g◦fn is irreducible for all n ≥ 0. If a prime
p ∈ Em, then Fp contains a primitive mth root of unity. The factorization of g◦fn
modulo p, and thus the permutation group cycle type of the elements in Frob(p),
will reflect the presence of the mth roots of unity in Fp. Then automorphisms of
the Galois group of g ◦ fn that are in the Frobenius class of a prime belonging to
Em are contained in the subgroup that fixes Q(ζm), so we will be considering only
that subgroup. We will adjust our notation accordingly. Let K = Q(ζm) where
ζm is a primitive mth root of unity. Let Kn be the splitting field of g ◦ fn over K.
Let Gn = Gal (Kn/K). Let Hn := Gal (Kn/Kn−1).
3.1.1 Rigid Divisibility
Definition 3.4. Rigid Divisibility Sequence
Let A = {ai}i∈N ⊂ K be a sequence in a number field K. Then A is a rigid
divisibility sequence if it has two properties:
(i) For any prime p of OK and natural numbers k, n, if vp(an) = en > 0, then
vp(an) = vp(akn).
(ii) If vp(an) = en > 0 and vp(ak) = ek > 0, then vp(agcd(k,n)) > 0.
26
A consequence of the second condition is that if vp(an) = en > 0 and vp(ak) =
ek > 0, then ek = en.
Lemma 3.5. Let K be a number field, and let f(x) = xm+c ∈ K[x], and suppose
that Of (0) is infinite. Let an = fn(0). Then the sequence a1, a2, a3, . . . is a rigid
divisibility sequence.
Proof. Proof of property (i):
Suppose that for some n and some prime p, we have that vp(an) = e > 0. We
wish to show that for every k ∈ N, vp(akn) = e. We proceed by induction.
If k = 1, then an = akn. Suppose the condition holds for t ≤ k − 1. Then
fkn(0) = fn ◦ f (k−1)n(0). We may write fn(x) = fn(0) +∑mn−1
i=1 aixmi. Since
vp(∑mn−1
i=1 ai(f (k−1)n(0)
)mi) ≥ em, then it must be that
vp(fkn(0)
)= vp
(fn(0) +
mn−1∑i=1
ai(f (k−1)n(0)
)mi)= vp(f
n(0)) = e.
Proof of property (ii)
By property(i), if vp(ad) > 0 for some d, and the dth is the first term of the
sequence contained in p, then 0 has exact period d under f in Fp, where p = Q∩p.
For any k = `d+ r, ak mod p will be zero if and only if r = 0. So if both ak and
an are in p, then d | gcd(k, n).
We wish to allow the possibility the f is not stable. Then we will be considering
properties of the translation of iterates of f by a polynomial divisor g of fk for
some k, where g ◦ fn is irreducible for all n. Thus we wish to extend the rigid
divisibility property to a translated orbit of zero. While {g ◦ fn(0)}n≥1 is not a
rigid divisibility sequence, {g ◦ fkn(0)}n≥0 is. (See Example 1.10.)
Lemma 3.6. Let {fn(0)}n≥1 be a rigid divisibility sequence. Suppose that g is
a polynomial divisor of fk. Then the sequence a1, a2, . . . , an, . . . , where an =
g ◦ fk(n−1)(0), is a rigid divisibility sequence.
27
Proof. Proof of property (i)
Suppose that for some prime p, vp(an) = e > 0. As g divides fk, vp(fk ◦
fk(n−1)(0) = vp(fkn(0)) ≥ e.As {fn(0)}n≥1 is a rigid divisibility sequence, vp(f
knr(0)) ≥
e for all r ∈ N. Consider vp(an`) for any natural number `. Suppose that the degree
of g ◦ fk(n−1) = md. Write
g ◦ fk(n−1)(x) = g ◦ fk(n−1)(0) +d∑i=1
bixmi.
Then
an` = g◦fk(n`−1)(0) = g◦fk(n−1)(fk(n`−n)(0)
)= g◦fk(n−1)(0)+
md−1∑i=1
bi(fkn(`−1)(0)
)mi.
Then, by the ultrametric property, vp(an`) = vp(g ◦ fk(n−1)(0)) = e. Property(ii)
follows immediately.
3.1.2 Eventual Stability
Proposition 3.7. Let c ∈ Q, and suppose that there exists a prime q of Z such
that vq(c) > 0. Then, for any m ∈ Z≥2, f(x) = xm + c is eventually stable over
Q.
Recall from Chapter 1 that a polynomial is eventually stable if there is a finite
number of iterates with new factorization. That is, for some N ∈ N, if n > N ,
then
fn(x) =s∏i=1
gi ◦ fni(x)
where gi ◦ fk(x) is irreducible for all k ≥ 0. Hence, the maximum number of
factors of fn for any n is s.
We will devote Section 3.1.2 to the proof of Proposition 3.7. We take the
quadratic case separately, because its proof provides a simpler example of the
proof for m > 2. The proof for m > 2 follows.
We begin with a Lemma given by Stoll [25].
28
Lemma 3.8. Let f(x) = x2 +c where c ∈ Q. Suppose f is irreducible and n is the
least integer greater than one such that fn is reducible. Then fn(x) = g(x)g(−x)
where g(x) is irreducible over Q.
Proof. Let
fn(x) = (x− α1)(x+ α1) · . . . · (x− α2n−1)(x+ α2n−1)
with αi ∈ Q. If fn has a proper, even divisor, g, then we may write it as follows,
renumbering as necessary:
g(x) = (x− α1)(x+ α1) · . . . · (x− αd)(x+ αd) = (x2 − α21) · . . . · (x2 − α2
d)
Here d < 2n−1. We know that if α is a root of fn, then f(α) is a root of fn−1. So
we could write
fn−1(x+ c) = (x+ c− f(α1)) · . . . · (x+ c− f(α2n−1)) =2n−1∏i=1
(x− α2i ).
Then fn−1(x + c) has proper divisor g(x) =∏d
i=1(x − α2i ). We know that g is a
polynomial in Q[x], because it has the same coefficients as g. Thus fn−1 is not
irreducible, contrary to our supposition. So fn can have no proper, even divisor.
Suppose that g(x) is a non-constant, proper divisor of fn. Then so is g(−x), as
fn is even. Let h(x) be a polynomial divisor of both g(x) and g(−x) of minimal
positive degree. Then h(x) 6= h(−x), as h(x) divides fn, so it can’t be even.
Hence, h(x) and h(−x) are relatively prime, as any common divisor would be a
divisor of g of degree less than that of h. As h(x) divides both g(x) and g(−x),
it must be that h(−x) divides both g(−x) and g(x). Then h(x)h(−x) divides
g(x), meaning that fn has a proper even divisor. So no such h exists. Then g(x)
and g(−x) have no common polynomial divisor, so g(x)g(−x) divides fn(x). As
g(x)g(−x) is even, it must be that g(x)g(−x) = fn(x).
29
Corollary 3.9. Let f be quadratic and g an irreducible polynomial divisor of
fk for some k. If n ≥ 1 is the least integer such that g ◦ fn is reducible, then
g ◦ fn(x) = h(x)h(−x) where h is irreducible.
Proof. As in the proof of the preceding Lemma, noting that g ◦ fn is even for all
n ≥ 1, and letting r := deg(g ◦ fn), we write g ◦ fn as
g ◦ fn(x) = (x− α1)(x+ α1) · . . . · (x− α r2)(x+ α r
2)
with αi ∈ Q. If α is a root of g ◦ fn, then f(α) is a root of g ◦ fn−1. Hence
g ◦ fn−1(x+ c) =
r/2∏i=1
(x+ c− f(αi)) =
r/2∏i=1
(x− α2i ).
If g ◦fn has a proper even divisor, then, as above, g ◦fn−1 factors. The remainder
of the proof is the same as that of 3.8.
We prove Proposition 3.7 for quadratics.
Proof. We may build a sequence of polynomials g1, g2, . . . with the following prop-
erties. Let g1 be a divisor of fk where k ≥ 1 and fk−1 is irreducible. (Recall that
f 0(x) = x.) For each subsequent i, let gi be a divisor of gi−1 ◦ fk where k ≥ 1
and gi−1 ◦ fk−1 is irreducible. Such a sequence will be finite if and only if f is
eventually stable. We show that this sequence is necessarily finite.
Let q be a prime of Z such that vq(c) = e. We begin by showing that vq(g1(0)) =
e2. Suppose first that f is reducible. Then f(x) = −g(x)g(−x), where g(x) =
x −√−c. Then f(0) = −g(0)2, so vq(g(0)) = 1
2vq(f(0)). If f is irreducible, we
apply Lemma 3.8. If k is the least integer so that fk factors, then g1(0)2 = fk(0),
so vq(g1(0)) = 12fk(0) = e
2, by the rigid divisibility property. By Corollary 3.9,
vq(gi(0)) = 12vq(gi−1(0)). Thus vq(gk(0)) = 2−ke. There is a k0 such that 2−k0e
will not be divisible by 2, and gk0 ◦ fn(0) can not be a square for n ≥ 0. Then
gk0 ◦ fn will be irreducible for n ≥ 0, so the sequence will terminate with gk0 .
30
In the proof of the quadratic case, the set of roots of g ◦ fn occurred in pairs
±α. The idea was that if α and −α are Galois conjugates, then a previous iterate
must have factored. So our sets of Galois conjugates consisted of exactly one root
from each ± pair. In the case m > 2, the same idea applies. We outline it here.
Suppose that g ◦ fn−1 is irreducible and g ◦ fn factors. Roots of g ◦ fn are
naturally partitioned into sets of the form Ri := {f−1(βi)} where βi is a root
of g ◦ fn−1. They are also partitioned into sets whose members are all Galois
conjugates. That is, let g ◦ fn =∏d
j=1 hj where the hj are irreducible. Let
Cj = {α ∈ Q | hj(α) = 0}. The sets Cj form another partition of the roots of
g ◦ fn.
If m is prime, we will pick exactly one element from each Ri to assemble any
particular Cj. This is because if m is prime and α, ζjmα are two distinct roots of
hj then the element σ in the Galois group of hj that maps α to ζjmα will generate
Cm. So, if two elements of Ri are Galois conjugates, then Ri is a subset of some
Cj. Then by the transitivity of Gn−1, there is only one set Cj, so g ◦ fn is not
reducible.
If m is not prime, different elements of one Ri could be contained in the same
Cj, but the way we partition Ri into various Cj has to match a possible cycle type
of an element of Cm that is not a generator of Cm..
We prove Proposition 3.7 for m ≥ 3.
Proof. The following lemma is an analogue of Lemma 3.8 for m ≥ 3.
Lemma 3.10. Let f(x) = xm + c for m ≥ 3, and let g(x) be an irreducible
polynomial divisor of fk for some k. We allow the possibility that g(x) = x.
Let g ◦ fn−1 be irreducible over Q(ζm) for some n ∈ N, and suppose that g ◦ fn
factors. Let β be a root of g ◦ fn−1 in Q. Then f(x) − β factors over Q(ζm, β),
and f(x) − β =∏d−1
i=0 h(ζ imx) where d is a divisor of m, and h(x) is irreducible
over Q(ζm, β).
31
Proof. We make use of a classical result, Capelli’s Lemma (see [6] for a proof).
Lemma 3.11. Let u(x), v(x) be polynomials over a field K, and suppose β is any
root of u(x) in K. Then (u ◦ v)(x) is irreducible over K if and only if u(x) is
irreducible over K and v(x)− β is irreducible over K(β).
Since we assume that g ◦ fn factors and g ◦ fn−1 does not, Capelli’s Lemma
tells us that we may consider the factorization of f(x) − β in Q(ζm, β). Let
α = m√β − c. Let k be the least positive integer such that αk ∈ Q(ζm, β).
We claim that k is a divisor of m. Let d ∈ Z be such that 0 ≤ m − dk < k.
Then, as αm = αdkαm−dk, it must be that m−dk = 0, because αm−dk ∈ Q(ζm, β),
but m− dk < k.
Let d = mk
. We see that h(x) = xk − αk is the minimal polynomial of α over
Q(ζm, β). Note that
f(x)− β = (xk − αk)(xk − ζdαk)(xk − ζ2dα
k) · · · · · (xk − ζd−1d αk),
and
xk − ζjdαk = ζjd
(ζd−jd xk − αk
)= ζjd
(ζk(d−j)m xk − αk
)= ζjdh(ζd−jm x).
Then
f(x)− β =d−1∏j=0
(xk − ζjdα
k)
=d−1∏j=0
ζjd
d−1∏j=0
h(ζd−jm x) =d−1∏j=0
h(ζjmx).
This concludes the proof of Lemma 3.10
Corollary 3.12. Let f(x) = xm + c where m ≥ 3. Let g(x) be an irreducible
polynomial divisor of fn for some n. Let g ◦ fn−1 be irreducible over Q(ζm) and
suppose that g ◦ fn factors. Then g ◦ fn(x) =∏d−1
j=0 h(ζjmx) where d is a divisor of
m.
32
Proof. Let the degree of g ◦ fn−1 = r. If we consider the factorization given in
Lemma 3.10, as β was arbitrarily chosen, it must be that f(x)− βi has the above
factorization over Q(ζm, βi) for some divisor ki of m for each i ∈ {1, . . . , r}. For
each i, we fix αi, where f(αi) = βi, and let di = mki
. Then we have f(x) − βi =∏di−1j=0 hi(ζ
jmx), where hi(x) = xki − αkii . Let Ri,j refer to the set of roots of
hi(ζj−1m x). Then α1 ∈ R1,1. There are di such sets of order ki for each i. (It turns
out that di = dk,∀i, k, but we haven’t used this fact.)
As we assume that g ◦fn is not irreducible, we let g ◦fn(x) =∏t
s=1 hs(x) with
hs ∈ K for each s. Let Hs refer to the Galois group of hs over K. Let Gn−1 be the
Galois group of g ◦ fn−1 over K. Then Gn−1 acts transitively on the βi. Further
Gn−1 is a subgroup of Hs for each s, so Hs also acts transitively on the βi.
We may select a set of r − 1 automorphisms {τi | 2 ≤ i ≤ r} from H1
in such a way that τi maps R1,1 into {f−1(βi)}, exactly one τi for each i. Let
C1 := R1,1 ∪ τ2(R1,1) ∪ · · · ∪ τr(R1,1).
Without loss of generality, we may number the hs so that h1(α1) = 0. We
claim that C1 is the complete set of Galois conjugates of α1. Then we would have
that h1(x) = h(x)∏r
i=2(xk−τi(α1)k). Clearly, for each i, every element of τi(R1,1)
is conjugate to α1. If there were γ ∈ {f−1(βi)} − τi(R1,1) such that γ is a root
of h1(x), then there must be an automorphism of H1 mapping τi(R1,1) ∪ {γ} to
{f−1(β1)} such that the image includes some γ ∈ {f−1(β1)} − (R1,1). But then
γ is conjugate to α1 by transitivity. This is not possible, as the conjugates of α1
that are in {f−1(β1)} must be in R1,1. Then we must have that C1 is the complete
set of Galois conjugates of α1.
We let Cj := ζj−1m C1. There are d1 such sets, each of order rk1. The Cj
partition the set of roots of g ◦ f into sets of Galois conjugates, giving us the
desired factorization: g ◦ fn =∏d−1
j=0 h(ζjmx) where d := d1 and h := h1.
Proof. Proof of Proposition 3.7
33
Consider the sequence of polynomials in Q(ζm)[x], {g1, g2, . . . } with the following
properties. First, g1 is a divisor of fn for some n ≥ 1, where fn−1 is irreducible.
For subsequent terms in the sequence, gi is a divisor of gi−1 ◦ fk for some k ≥ 1
where gi−1 ◦ fk−1 is irreducible. We will show that any such sequence must be
finite. This will be equivalent to the eventual stability of f over Q(ζm), for if f
were not eventually stable, we could build an infinite sequence of this form. If f
is eventually stable over Q(ζm), then it is eventually stable over Q.
We have assumed that there is a rational prime q such that vq(c) > 0. We
select a prime q, of Q(ζm) extending q, and note that vq(c) = e > 0. Let D :=
{d1, d2, . . . , dk} be all the positive divisors of m including m, but excluding 1. We
claim that vq(g1(0)) = 1dj1vq(f(0)), for some dj1 ∈ D, and, for i ≥ 2, vq(gi(0)) =
1dj2vq(gi−1(0)) for some dj2 ∈ D. We do not claim that dj1 6= dj2 . We proceed by
induction.
Suppose that the nth is the least iterate of f that is reducible. By Corol-
lary 3.12, fn(x) =∏dj0−1
`=0 g1(ζ`mx), where dj0 ∈ D. Then fn(0) = (g1(0))dj0 ,
so vq(g1(0)) = 1dj0vq(f
n(0)). Now consider vq(gi(0)) and vq(gi−1(0)). We write
gi(x) = gi(0) +∑t
k=1 akxk. Now
∑dk=1 ak(f
n(0))k = fn(0)∑d
k=1 akfn(0)k−1.
In Section 3.1.1 we showed that the orbit of 0 under f forms a rigid divisibil-
ity sequence. Hence vq(fn(0)) = vq(c) = e. Then vq(
∑tk=1 ak(f
n(0))k) ≥ e,
whereas vq(gi(0)) ≤ vq(g1(0)) = 1dj1vq(f
n(0)) < e. By the ultrametric property,
vq(gi ◦ fn(0)) = vq(gi(0)).
From 3.12 we know that if k is the least integer such that gi−1 ◦ fk factors,
then gi−1 ◦ fk(x) =∏dji
`=0 gi(ζ`mx). Thus gi−1(0) = (gi(0))dji , and vq(gi(0)) =
1djivq(gi−1(0)) = es−1
i , where si is a product of elements of D. We see that si =
djisi−1. By the pigeonhole principle, there will eventually be an i0 such that
es−i0 = vq(gi0 ◦ fn(0)) will fail to be divisible by any member of D.
From 3.12, gi0 ◦fn can only factor if gi0 ◦fn(0) is a dith power for some di ∈ D,
34
so only if there is a dji0 ∈ D such that es−i0 is divisible by dji0 . Then gi0 ◦ fn(x)
will be irreducible for all n, so the sequence {g1, g2, . . . } will terminate with gi0 .
As this sequence is finite, all such sequences must be finite, and f is eventually
stable over Q(ζm), and therefore over Q.
3.1.3 Maximality
Let Kn be the splitting field of g◦fn(x) over K := Q(ζm), and Gn := Gal(Kn/K).
Let Hn be the subgroup of Gn fixing Kn−1. We wish to define what it means for
Hn to be maximal in this context. Let β1, . . . , βr be the roots of g ◦ fn−1. These
roots are distinct, as g◦fn−1 is irreducible. We obtain Kn from Kn−1 by adjoining
to Kn−1 the roots of f(x)− βi for each i. Let Li be the splitting field of f(x)− βiover Kn−1, and Ji := Gal (Li/Kn−1). Note that Ji ∼= Jj for all i, j.
Definition 3.13. With the above notation, Hn is maximal if #Hn = #Jri .
Lemma 3.14. Let K = Q(ζm) and f(x) = xm + c. Suppose that g | fk for some
k ≥ 0 and g ◦ fn is irreducible for all n ≥ 0. Suppose further that the translated
orbit of zero, {g(0), g ◦ f(0), . . . }, is infinite. Then Hn is maximal for infinitely
many n.
Proof. Let D := {d ∈ Z | d divides m and d 6= 1} as in the proof of Proposition
3.7. Let {pn}n≥1 be an infinite sequence of primes of Z. Let g(0) = rs
with r, s ∈ Z
and gcd(r, s) = 1. Let cn = g ◦ fk(n−1)(0), so cpn = g ◦ fk(pn−1)(0). Let 1rcpn = bn.
Let S = {p ∈MK | vp(c) < 0} ∪M∞. Then bn is an S-integer for each n.
Let di ∈ D and φdi be the curve given by g◦f 2(x) = rydi . If bn is a dith power,
then (fk(pn−1)−2(0), di√bn) will be an S-integer point on φdi . Let πx : φdi → P1(C)
be projection on the x-coordinate. By the Riemann Hurwitz formula, the genus
g of φdi satisfies
g = −1 +1
2
∑P∈C
P ramifies
(eP − 1).
35
As g ◦ f 2(x) is irreducible over Q, it is separable. The degree of g ◦ f 2(x) is at
least m2 ≥ 4, so there are at least m2 points ramifying to degree 2. Hence, φdi
has positive genus.
Suppose that bn is an S-unit. By Dirichlet’s Unit Theorem, the unit group U
of RS is finitely generated. Thus there are finitely many cosets uiUm comprising
U . Consider the curve ψi given by uig ◦ f 2(x) = rym where ui is a coset rep-
resentative. Suppose that bn lies in the coset ujUm and that ui = u−1
j . Then
(fk(pn−1)−2(0), m√uibn) is an S-integer point on ψi. As with φdi , ψi has positive
genus.
By Siegel’s Theorem, (see, for example [1, Theorem 7.3.9, Remark 7.3.10 ]), a
curve of positive genus can only contain a finite number of S-integer points.
Two facts follow. First, bn can not be an S-unit for infinitely many n. Second,
there is only a finite number of n such that bn is a dith power.
Definition 3.15. Let {an}n≥1 be a sequence in a number field K. If there is a
prime q of OK such that, for some index k, vq(ak) = e > 0, but vq(aj) = 0 for all
j < k, then we call q a primitive prime divisor of ak.
As the cn form a rigid divisibility sequence and pn is prime, any prime q such
that q divides the numerator of bn is a primitive prime divisor of cpn . We know
there exists N such that bn is not an S-unit or a dith power for n > N . Hence, for
n > N , cpn has a primitive prime divisor q. Further vq(bn) 6= di for any di in D.
Thus there infinitely many n such that the sequence {g ◦ fn(0)}n≥0 = {an}n≥1 ⊃
cn ⊃ cpn has a primitive prime divisor whose exponent in an is not di for any di
in D.
Suppose that q ∈ Z is a primitive prime divisor of an = g◦fn(0) Let g◦fn(0) =
qksq where sq ∈ Q and both the numerator and denominator of bq are relatively
prime to q. Suppose further that k 6= di for any di ∈ D and that p - m, and,
for any root θ of g and any extension q of q in Q(ζm), g′(θ) /∈ q. We have only
36
ruled out finitely many primes of OK , so there are infinitely many n for which
such q exists. Let t equal the degree of Gn−1 acting as a permutation group. Let
β1, . . . , βt be the roots of g ◦ fn−1.
The field Kn is obtained from Kn−1 by adjoining m√f(0)− βi for each i. Thus(
t∏i=1
(f(0)− βi)
) 1m
= qkm s
1mq ∈ Kn.
As gcd(k,m) = 1, we have that m√q ∈ Kn. Thus q ramifies to degree m in Kn.
Let qi be extensions of q in Q(ζm). Then each qi ramifies to degree m in Kn,
because only primes dividing m ramify in Q(ζm).
We show that q is unramified in Kn−1. As in the function field case, if a prime
q of K ramifies in Kn−1, it ramifies in one of the K(α) ∼= K[x]/g ◦ fn−1(x) where
α is a root of g ◦ fn−1. Suppose P of OK(α) is ramified over K. Then it contains
(g ◦ fn−1)′(α) = g′(fn−1(α)
) n∏j=2
f ′(fn−j(α)),
because OK(α)/P must be a separable extension of K/(P ∩ K). For any j,
fn−j(α) ∈ {(g ◦ f j−1)−1(0)}. If P contains fn−j(α), then P contains g ◦ f j−1(0).
As j ranges between 2 and n, primes containing∏n
j=2 f′(fn−j(α)) are those whose
intersections with K contain one of g ◦ f(0), . . . , g ◦ fn−1(0). As q is a primitive
prime divisor of g ◦ fn(0), g ◦ fn−j(0) /∈ q for any j > 0. We have assumed that
g′(fn−1(α)) = g′(θ) where θ is a root of g is not in q for any extension of q in OK0 .
Hence q is unramified in Kn−1.
Let Pi be an extension in OKn−1 of q, and suppose without loss of generality
that f(0) − βi ∈ Pi. We know that Gn−1 acts transitively on the βj, so it must
be that, for each j ∈ {1, . . . , s}, f(0)− βj ∈ Pj where Pj also extends q.
Let Li = Kn−1(αi) where αi = m√f(0)− βi. If a prime M of Li ramifies over
Kn−1, then it contains f ′(αi) = mαm−1i , so its intersection with Kn−1 contains
f(0) − βi. Hence, Pi ramifies in Li, but not in Lj for j 6= i, as f(0) − βj /∈ Pi.
37
If f(0)− αj ∈ Pi, then k
√∏ti=1(f(0)− βi) ∈ P2
i , in which case q would ramify in
Kn−1.
For each i, Pi ramifies to degree m in Li, as q ramifies to degree m in Kn, and
Kn is a Galois extension. We see that [Li : Kn−1] = m, as Pi ramifies to degree
m, and [Li : OKn−1 ] is at most m. Let L =∏
j 6=i Lj, the compositum. As Pi is
unramified in L, [Kn : L] = m, as [Kn : L] is at most m. As this is true for all i,
[Kn : Kn−1] = mr.
3.1.4 Probability
Jones [11] uses probability theory to measure the proportion of elements in suc-
cessive Galois groups of iterated quadratics that fix a root. We use his technique
with some adjustments for the different degree of our polynomial, and show that
this proportion approaches zero as n increases.
As before, we will let f(x) = xm + c. Let g be a polynomial divisor of fk for
some k ≥ 0, and suppose further that g ◦ fn is irreducible over K := Q(ζm) for
all n ≥ 0. Let Kn be the splitting field of g ◦ fn over K and Gn the Galois group
of Kn over K. Recall Kn ⊂ Kn+1. This subset relation gives an inverse system of
groups with restriction maps. That is, let σj ∈ Gj, and suppose j > i. Then let
ψi,j be the restriction of gj ∈ Gj to Ki. Let
G(f, g) = lim←−n∈N
Gn
equipped with the profinite topology.
Then G(f, g) is a compact, Hausdorff topological group, and so it has a Haar
measure. Let P be the normalized Haar measure on G(f, g), so P(G(f, g)) = 1.
Let Ψn be the projection on nth coordinate of ~σ ∈ G(f, g)and B be the Borel
σ-algebra on G(f, g). Then the triple (G(f, p),B,P) is a probability space. We
38
assign to it a set of random variables, Xn, where, for σ ∈ G(f, g), Xn(σ) = t if
Ψn(σ) fixes exactly t roots of g ◦ fn. Then
P(Xn = t) =#{σ ∈ Gn | σ fixes t roots of g ◦ fn}
#Gn
.
The sequence of random variables gives a stochastic process Jones calls the Galois
process, and labels GP (f, g).
Definition 3.16. [3, Definitions 3.4,3.5] A stochastic process X1, X2, . . . is a
martingale with respect to a filtration F1,F2, . . . if
(i) Xn is integrable for each n = 1, 2 . . . ;
(ii) X1, X2, . . . is adapted to F1,F2, . . . ;
(iii) E(Xn+1 | Fn) = Xn for each n = 1, 2 . . . ; The process is a supermartingale
if the third condition reads E(Xn+1 | Fn) ≤ Xn for each n = 1, 2 . . . .
The sequence of σ-fields generated by Xn, denoted σ(Xn) gives a filtration,
and the sequence Xn is adapted to it [3, Example 3.3]. As each Gn is finite, Xn
is integrable. Here we show that the third criterion also holds.
Theorem 3.17. [12, Theorem 2.5] Let f, g ∈ K[x] be such that g ◦fn is separable
for all n ≥ 0. Suppose that for n ≥ 1 and every root β of g ◦ fn−1, the polynomial
f(x)− β is irreducible over Kn−1. Then GP (f, g) is a Martingale.
We refer the reader to [12, p.6] for the proof.
Claim 3.18. With g, f, and K as in Theorem 3.2, GP (f, g) is a martingale.
Proof. We have assumed that g ◦fn is irreducible over Q for all n. Hence g ◦fn is
separable. By Capelli’s Lemma, we know that for each βi, f(x)− βi is irreducible
over K(βi). For each βi := β and α where f(α) = β, we have the following
diagram of field extensions.
39
K
K(β)
Kn−1(α)
K(β)(α) Kn−1
By Capelli’s Lemma, as g ◦ fn−1 and g ◦ fn are irreducible, so is f(x) −
β over K(β). So [K(β)(α) : K(β)] = m. Let [Kn−1 : K(β)] = d1. Then
Kn−1 = K(β)(x1, . . . , xd1). Then Kn−1(α) ⊆ K(β)(α)(x1, . . . , xd1), so [Kn−1(α) :
K(β)(α)] = d2 ≤ d1. Hence [Kn−1(α) : Kn−1] ≥ m. As [Kn−1(α) : Kn−1] is at
most m, it must be that f(x)− β is irreducible over Kn−1.
Martingales are of interest to us, because they converge almost surely. Note
that a martingale is always a supermartingale.
Theorem 3.19. [3, Theorem 4.2, Doob’s Martingale Convergence Theorem] Sup-
pose that X1, X2, . . . is a supermartingale (with respect to a filtration F1,F2, . . . )
such that
supnE(| Xn |) <∞.
Then there is an integrable random variable X such that
limn→∞
Xn = X almost surely.
We have that | Xn |= Xn, and, as {Xn}n≥1 is a martingale, E(Xn) = E(X1) <
∞ [3, p.50]. Thus supnE(| Xn |) < ∞. Let X = limn→∞Xn. Let t ∈ N and
suppose that Pr{X = t} > 0. As the Xn are integer-valued, it must be that there
exists m ∈ N such that
Pr (∩i≥m{Xi = t}) = r > 0
40
for some fixed r ∈ Q>0. We will show that, if t > 0 and Hn is maximal for
infinitely-many n, then Pr (∩i≥m{Xi = t}) < r for any fixed r > 0.
We make a counting argument to adjust for the difference between the degree
of our polynomial and a quadratic one. Of course, our argument reduces to the
quadratic case if m = 2.
Lemma 3.20. Suppose that Hn is maximal. Let t ∈ N. Then P(Xn = t | Xn−1 =
t,Xn−2 = t, . . . , Xn−k = t) ≤ 12.
Proof. Let the degree of Gn−1 acting as a permutation group be r and β1, . . . , βr
the roots of g ◦ fn−1. Thus #Hn = mr, and #Gn = mr ·#Gn−1. Just as in the
proof of Lemma 2.11, we know that Gn ⊆ Gn−1[Sm], and as ζm ∈ K, and σ ∈ Gn
fixes ζm. The rest follows immediately as in Lemma 2.11, and we may assume
that Gn∼= Gn−1[Cm]r. Let
P(Xn−1 = t, . . . , Xn−k = t) =k
#Gn−1
.
As, for n ≥ 1, t is a multiple of m, we may replace t with mt; this substitution
serves to ease the notation in the calculations below. As always, we assume m ≥ 2.
Suppose that Xn−1(σ) = mt. The elements τ of Gn such that ψn−1,n(τ) = σ
have the form τ = (σ; π1, . . . , πr) with πi ∈ Cm. If σ(βi) 6= βi, then τ will not
fix any inverse image of βi, so πi may be any of the m elements of Cm. Thus we
have remaining r −mt coordinates. We choose t of these, and place the identity
in those positions. The other mt − t positions have m − 1 possibilities for πj, as
we may not place the identity at those positions. This gives us(mt
t
)(m− 1)mt−tmr−mt
automorphisms τ of Gn that restrict to any particular automorphism of Gn−1
fixing mt roots.
We consider the conditional probability:
41
P(Xn = mt | Xn−1 = mt, . . . , Xn−k = mt)
= Pr[(Xn=mt)∩(Xn−1=mt,...,Xn−k=mt)]
P(Xn−1=mt,...,Xn−k=mt)
=
(k(mtt )(m−1)mt−tmr−mt
#Gn
)(#Gn−1
k
)=(mtt
) (m−1m
)mt(m− 1)−t
=(
mtt(m−1)
)(mt−1
(t−1)(m−1)
). . .(
mt−(t−1)(t−(t−1))(m−1)
) (m−1m
)mtFor fixed k < t, let R(m, t) = mt−k
(t−k)(m−1). Both ∂(R)
∂mand ∂(R)
∂tare negative, and(
m−1m
)mtis also decreasing as m, t increase. Thus
(mtt
) (m−1m
)mt(m− 1)−t takes its
maximum at the minimum values for m, t, that is m = 2 and t = 1. Hence,(mt
t
)(m− 1
m
)mt(m− 1)−t ≤
(2
1
)(2− 1
2
)2
(2− 1)−1 =1
2.
Lemma 3.21. If Hn is maximal infinitely often, then
limn→∞
P(Xn > 0) = 0.
Proof. Let X = limn→∞Xn. Let t ∈ N and suppose that P{X = t} > 0. There
exists m ∈ N and r ∈ Q>0 such that
P (∩i≥m{Xi = t}) = r > 0,
because the Xn are integer-valued. We fix t ∈ N. Let Ci = {Xi = t}.
r ≤ P (∩i≥mCi) ≤ P(∩ki=mCi
)for any integer k > m.
P(∩ki=mCi) = P(Ck | ∩mi=k−1Ci) · P(Ck−1 | ∩mi=k−2Ci) . . .P(Cm)
Suppose that Hn is maximal for s of the iterates between the mth and kth. Then
r < P(∩ki=mXi = t) ≤ 12
s. We let k go to infinity, and, since Hn is maximal for
infinitely-many n, s goes to infinity as well. Then
r < lims→∞
(1
2
)s.
This conclusion is clearly false, therefore P{X = t} = 0 for all t > 0.
42
3.2 Proof of Theorem 3.2
Proof. Let Pf,m,a := {p ∈ Em | fn(a) ∈ (p)RS for some n ∈ N}. By Proposition
3.7, we know that there is an N ∈ N such that, for all n ≥ N we have
fn(x) =s∏i=1
gi ◦ fni(x)
and gi ◦ f is irreducible for all n ≥ 0.
We fix gi and let an = gi ◦ fn(a) for some rational starting point a. Let
Pi,a = {p ∈ Em | an ∈ (p)RS for some n ∈ N}, and Pi,a,N = {p ∈ Pi,a | an ∈
(p)RS for some n ≥ N}. Let GN = Gal (KN/Q) and G′N = Gal (KN/Q(ζm)).
Let CN = {σ ∈ G′N | σ fixes at least one root of g ◦ fN}. Let FrobN(p) be the
Frobenius class of p in GN . Let PN = {primes p | Frob(p) ⊂ CN}.
Let D(S) be the natural density of set S.
Let σ ∈ CN , and τ = γσγ−1 for some γ ∈ GN . Then τ(ζm) = ζm, and, if σ
fixes α, τ fixes γ(α). Hence, τ ∈ CN , so CN is a union of conjugacy classes in GN .
Then, with the triangle inequality, me may apply Chebotarev’s Density Theorem
to show that D(PN) = #CN/#GN .
We claim that Pi,a,N ⊂ PN . If q ∈ Em, but q /∈ PN , then elements in Frob(q)
have no fixed point is KN . The restriction ψN,N+` maps FrobN+`(q) to FrobN(q).
If FrobN(q) has no fixed points in KN , then FrobN+`(q) has none in KN+`. But
q ∈ Pi,a,N contains aN+` for some `, so if q ∈ Pi,a,N , then q ∈ PN .
Then D(Pi,a,N) < D(PN). Further, there is a finite number of primes in
Pi,a that ramify in KN or contain an for some n < N . Therefore, D(Pi,a,N) =
D(Pi,a) < D(PN).
By Lemma 3.21 there is N ∈ N so
D(Pi,a) < D(PN) = #CN/#GN < #CN/#G′N <
ε
s
for any ε > 0.
43
As
Pf,m,a = ∪si=1Pi,a
we know that D(Pf,m,a) = 0.
3.3 Quadratics
We have shown that, if f(x) = x2 + c where c ∈ Q, Of (0) is infinite, and where
some prime of Z divides the numerator of c, then f is eventually stable. We would
like to be able to say something about when this is not the case, that is, when
c = 1s
for some s ∈ Z. It turns out that there are large classes of s where we may
assert that f is in fact stable. We hope that some of the techniques illustrated
here may be of use in showing that f is stable under wider conditions.
Theorem 3.22. Let f(x) = x2 + 1s
where s ∈ Z. Suppose that s < −1 or that
s ≥ 4 and s is odd or a power of 2. Suppose further that f is irreducible. Then f
is stable.
Proof. Suppose that s < −1, and f is irreducible. Then fn(0) < 0 for n ≥ 2, so
fn(0) is not a positive square. By Lemma 3.8, we know that fn is then irreducible.
Thus we need only concern ourselves with positive s.
Suppose for the remainder of this Section that s > 0. Let {an}n∈N be the
sequence of numerators of fn(0). We note that
an = a2n−1 + s2n−1−1.
We would like to establish that, under certain conditions, an is non-square. The
denominator of fn(0) is always an even power of s for n ≥ 2. Thus the squareness
of fn(0), and so the reducibility of iterates of f , depends only on the an.
We begin with a few useful facts.
44
Claim 3.23. When s > 1, the second iterate of f(x) = x2 + 1s
factors if and only
if s has the form 4t2(t2 − 1) where t ∈ Z and t > 1.
Proof. From Lemma 3.8, we know that if f 2 is not irreducible over Q, then
f 2(x) = x4 +2
sx2 +
s+ 1
s2= (x2 + ax+ b)(x2 − ax+ b)
where a, b ∈ Q, b =√s+1s
, and 2b − a2 = 2s. Using the fact that s = (
√s+ 1 −
1)(√s+ 1 + 1), we find that a2 = 2√
s+1+1, so 2
a2∈ Z. Let 1
a2= t2. We know that
t ∈ Z, as the numerator of a2 divides 2. Then 2t2 =√s+ 1+1, so s = 4t2(t2−1).
Conversely, we see that if s = 4t2(t2 − 1), then
f 2(x) = (x2 +1
tx+
2t2 − 1
4t2(t2 − 1)) · (x2 − 1
tx+
2t2 − 1
4t2(t2 − 1)).
Note that Fact 3.23 implies that f 2 is irreducible if s is odd or a power of 2.
We include an example of what such factorization might look like.
Example 3.24. Let f(x) = x2 + 148
f 2(x) = x4 +1
24x2 +
49
2304
=
(x2 +
1
2x+
7
48
)(x2 − 1
2x+
7
48
)= g1(x)g2(x)
f 3(x) = x8 +1
12x6 +
17
384x4 +
49
27648x2 +
112993
5308416
=
(x2 − 1
2x+
19
48
)(x2 +
1
2x+
19
48
)(x4 − 11
24x2 +
313
2304
)= h1,1(x)h1,2(x)g2(f(x))
f 4(x) =
(x4 − 11
24x2 +
7 · 127
482
)(x4 +
13
24x2 +
937
482
)·(
x8 +1
12x6 − 175
384x4 − 527
27648x2 +
7 · 102871
484
)= h1,1(f(x))h1,2(f(x))g2(f 2(x))
45
This example serves to illustrate the difficulties that arise once factoring begins
if f(0) has no prime of positive valuation. We see that if gi(x) = (x2 ± 12x + 7
48),
then gi ◦ f 2n(0) is non-square for every n, because v7(gi ◦ f 2n(0)) = 1. However,
this gives us no control outside the rigid divisibility sequence {g ◦ f 2n(0)}n≥0, as
we can see by the factorization of f 4. At f 4 we have begun a rigid divisibility
sequence {h1,2 ◦ f 3n(0)}n≥0 in which 7 will never appear.
Claim 3.25. If f and f 2 are irreducible, then so is f 3.
Proof. Suppose there are integers s and y so that f 3(0) = y2. Then we get an
integer point on the elliptic curve given by x3 +x2 +2x+1 = y2. The only rational
points on this curve are (0, 1) and (0,−1) and the point at infinity.
Claim 3.26. If s ≥ 4, then fn(0) < s−2s(s−3)
.
Proof. Note that f(0) = 1s< 1
s(s−2)(s−3)
as s−2s−3
> 1. If fn−1(0) < s−2s(s−3)
, then
fn(0) <(s− 2)2
s2(s− 3)2+
1
s=
s− 2
s(s− 3)·s2 − 3s− 1 + 2
s−2
s(s− 3)<
s− 2
s(s− 3)
as 2s−2− 1 < 0.
From the above fact, we see that an ≤ (s−2)(s−3)
s2n−1−1. Suppose that an is a
square in Z. Then s2n−1−1 = (√an − an−1)(
√an + an−1). It turns out that, if
s2n−1−1 has such a factorization in integers, and s ≥ 4 is odd or a power of 2, then
we can show that an >(s−2)(s−3)
s2n−1−1.
Claim 3.27. If gcd((√an − an−1), (
√an + an−1)) = 1 if s is odd, and 2 if s is
even.
Proof. We show by induction that for all n ≥ 1, gcd(s, an) = 1. We see that a1 =
1. Suppose that gcd(an−1, s) = 1. If r = gcd(s, an), then r | (an−s2n−1−1) = a2n−1.
So r = 1. This means that an, an−1, and s are pair-wise relatively prime. Suppose
d = gcd((√an − an−1), (
√an + an−1)). Then d | 2√an and d | 2an−1. So d = 1 or
46
d = 2. If s is even, then both√an and an−1 are odd, so their sum and difference
is even. Then d = 2. If s is odd, then, as d | s, d=1.
Claim 3.28. Suppose that s is odd. Then, if an is a square,
an =
(r2n−1−1
2
)2
+
(k2n−1−1
2
)2
+s2n−1−1
2
where k, r are relatively prime integers with k > r.
Proof. By Fact 3.27, s2n−1−1 has a factorization into relatively prime integers:
s2n−1−1 = (√an − an−1)(
√an + an−1). Let s = rk where gcd(r, k) = 1, r2n−1−1 =
(√an − an−1), and k2n−1−1 = (
√an + an−1). We allow the possibility that r = 1.
Then√an = r2
n−1−1+k2n−1−1
2and
an =
(r2n−1−1
2
)2
+
(k2n−1−1
2
)2
+s2n−1−1
2.
With these facts in hand we will continue with the proof of 3.22.
Case 1:
Suppose that s is odd and s > 4. Then we know f is irreducible, and by Fact
3.23, we know that f 2 is as well. From Fact 3.28, we know that if an is a square,
there are relatively prime integers r, k so that
an =
(r2n−1−1
2
)2
+
(k2n−1−1
2
)2
+s2n−1−1
2.
From Fact 3.26, we know that(r2n−1−1
2
)2
+
(k2n−1−1
2
)2
+s2n−1−1
2≤ (s− 2)
(s− 3)· s2n−1−1.
Dividing by s2n−1−1 we find that
1
4
(( rk
)2n−1−1
+
(k
r
)2n−1−1)
+1
2<s− 2
s− 3
47
Thus ( rk
)2n−1−1
+
(k
r
)2n−1−1
<2(s− 1)
(s− 3)(3.28.1)
Let g(x, y, n) =(x+yx
)2n−1−1+(
xx+y
)2n−1−1
− 2(x2+xy−1)(x2+xy−3)
. We will show that g
is non-negative for x, y ≥ 1 and n ≥ 3. That would show that an is non-square,
for if an were a square, there would be integers x and y, with r = x and k = x+ y
and s = rk, such that g would be less than zero.
As n increases,(x+yx
)2n−1−1+(
xx+y
)2n−1−1
increases. Since n > 3,(x+ y
x
)3
+
(x
x+ y
)3
− 2(x2 + xy − 1)
x2 + xy − 3= g(x, y, 3) ≤ g(x, y, n)
for all n.
Suppose that y > x. Then(x+yx
)3> 23, whereas, for s ≥ 5, 2(s−1)
s−3≤ 4. Then,
if y > x, g(x, y, 3) > 0. Suppose x > y. We consider partials. We find that ∂g(x,y,3)∂x
is negative for s ≥ 5 in the cases where it is possible to factor s as x(x+ y) with
x > y. For any fixed y, we also see that the limit as x goes to infinity of g(x, y, 3)
is zero. Thus, for fixed y, g decreases toward zero as x increases, and so g(x, y, 3)
is always non-negative, which implies that g(x, y, n) is also.
Case 2:
Suppose that s = 2k for some integer k ≥ 2. Then, if an is a square, s2n−1−1 =
2k(2n−1−1) = (√an − an−1)(
√an + an−1) = 2 · 2k(2n−1−1)−1. Thus
√an = 1 +
2k(2n−1−1)−2, and
an = 1 + 22k(2n−1−1)−4 + 2k(2n−1−1)−1 (3.28.2)
Dividing by s2n−1−1 as before, and subtracting 12
from both sides as before, we
find that 2k(2n−1−1)−4 < 2(s−1)s−3
≤ 6 if an is a square. Then we would have that
k(2n−1 − 1)− 4 < 3. But k is at least 2, and n is at least 4.
We might name other classes of positive integers s such that we may show
f(x) = x2 + 1s
is stable by using the same techniques. For example, if s = 2p
48
where p is an odd prime, then, if an were a square, we would have that(p2
)2n−1−1<
2(s−1)s−3
< 4, which is clearly false, as(p2
)2n−1−1>(
32
)7. However, our goal is to show
that f is stable where s is any integer except −1, and excepting the cases where
f or f 2 factors, which we conjecture is true. Thus eliminating specific classes of s
in an ad hoc manner does not seem useful.
Work in rational orbits comes from research originally done in collaboration
with Adam Towsley.
49
4 Conclusion
We answered two questions about periodicity and orbits under the same map in
different arithmetically interesting sets. The link between these two questions has
been the method of proof. We use the fact that large Galois groups coming from
wreath products have few elements that fix points.
Odoni showed that the iteration of generic polynomials gives rise to Galois
groups that are wreath powers of the full symmetric groups on the degree of the
polynomial. This is the most general case, and shows what ”should” happen. We
have shown that field extensions that arise from the iteration of certain polyno-
mials can at least contain subgroups that grow in a way that is similar enough to
the growth of wreath powers. Below we consider areas where similar techniques
to Odoni’s might be applied, and also questions that naturally arise, not from the
proof technique, but from our research.
Opportunities for Further Research
Graph Components and Fixed Characteristic
In examples 2.2 and 2.1 one might ask about the components of the orbit graph.
The permutation has two, and the three-to-one map has only one. We could
50
provide example with several components in both settings; multiple components
simply come from multiple periodic cycles. Experimenting in a haphazard way,
one might be led to think that the occurrence of large numbers of components is
very rare. For example, x 7→ x5 + 2 over F101 has exactly one periodic cycle with
four elements in it.
Suppose that a reasonable heuristic for why orbit graphs for maps with a lot of
preperiodic points appear to have few components is that, as we move backward
through the graph away from periodic cycles, we must experience branching due
to the fact that the map is m-to-one. Would it then be reasonable to suppose
that over a field where the same map is a permutation we may expect to see more
components than over fields where it is not a permutation?
In the function field setting, we fixed the polynomial f and varied the char-
acteristic of the fields, p, while keeping our constant field as Fp. One might ask
what would happen if, rather than change the characteristic to larger and larger
primes, we instead took larger and larger extensions of Fp.
This is a different question from the one we proved, and one wonders to what
extent Galois theoretic methods will help. For example, suppose we considered a
the cubic map x 7→ x3 + 2. If p ≡ 1 mod 3, then we conjecture that the Galois
method would work, as there will be no constant field extension. However in
characteristic 5, we get a constant field extension. That is, over F5, our map is a
permutation, but in F25 it is not a permutation. Yet, when we move from F5 to
F25, we keep the permutation of F5, as F5 ⊂ F25.
Example 4.1. Let f(x) = x3 + 2 and consider first F5, then F25. Initially we get
two periodic cycles.
0 2 1 3
4
51
The same map over F25 is 3-to-1, except at 2, where the map ramifies. That
dictates where eight of the new points must go–they must fill in the additional
two pre-images for every point except 2.
0 2 1 3
4◦
◦
◦
◦
◦
◦
◦ ◦
However, can we provide a reasonable heuristic to justify the eventual picture?
0 2 1 3
4◦
◦
◦
◦
◦
◦
◦ ◦
◦◦◦◦◦◦
◦ ◦◦
◦
◦
◦
Eventual Stability and Finite Critical Orbit
In the rational case, our proof depended on two properties of our map. We required
that our map have infinite critical orbit and that it be eventually stable.
The Galois groups of iterations of the stable polynomial f(x) = x2−2 are never
maximal, as the map has finite critical orbit. An even more difficult example is
f(x) = x2 − 1. Here f enjoys neither infinite critical orbit nor eventual stability.
Work has been done studying this map in the context of iterated monodromy
groups (see, for example [17]). The Galois group of the infinite binary tree graph
given by the splitting fields of iterations of x 7→ x2−1 is called the basilica group.
This group is generated by a finite automaton. These two examples provide a
fruitful area for continued research (see for example [2]).
52
If a map has the property that all of its critical orbits are finite, such as
x 7→ x2 − 1, it is called post-critically finite. Post-critically finite rational maps
are a subject of study in themselves. Ingram [9] shows that the set of conjugacy
classes of post-critically finite polynomials of degree d with coefficients of algebraic
degree less than any fixed integer is a finite and effectively computable set.
More General Maps
In our work we considered a map, x 7→ xm+c, whose critical orbit furnished a suf-
ficient number of primitive primes. Suppose we consider more general polynomial
maps, in particular ones with multiple critical orbits. If the orbits had non-zero
intersection, then, even if we had a primitive prime in one critical orbit, it may
have appeared in another critical orbit at an earlier iterate. Hence, a primitive
in a critical orbit would not guarantee that we get enough new ramification from
critical points.
Suppose a map has at least one infinite critical orbit that does not intersect the
others and this critical orbit provides an infinite list of primitive primes. Suppose
further that the splitting fields of iterations of this map do not have extensions
of the constant field. Then is seems reasonable to conjecture that the Galois
groups of successive iterations of the map will end up being wreath products, and
therefore will have few elements with fixed points. It seems likely that Galois
theoretic methods would apply to this problem.
53
Glossary
This glossary is intended as an aid to readers from outside mathematics, so
we have organized it by page number rather than alphabetically. For the same
reason, we have omitted definitions of notation if that notation is defined in the
text.
• p.1
Fp: the finite field of p elements. This is the set {0, 1, . . . , p − 1} with
operations addition and multiplication modulo p.
• p.2
OK : the ring of integers of the field K. Unfortunately, this coincides with
the traditional orbit notation. The ring of integers of a number field is the
set of algebraic integers in that field. Such a ring is Dedekind–its ideals have
unique factorization, and primes are maximal. (See [10].)
vp(an0): vp(c), where c is a rational number, means the power to which p
appears in the factorization of the number:
v5(10) = 1, v5(50) = 2, v5(11) = 0, v5(25) = −1.
We can extend this definition to primes in other rings, by saying vp(c) = e
if c ∈ pe and c /∈ pe+1. This gives us a valuation, | · |p, where
| x |p= evp(x)
for some e ∈ (0, 1).
54
• p.3
Gal(L/K): The Galois group of a field L over a field K.
K: the algebraic closure of a field K.
• p.4
separable polynomial: A separable polynomial f(x) is irreducible, and
all its roots are distinct.
degree of a permutation group: When a group is considered as acting
on a set, the number of elements in the set is the degree of the group as
a permutation group. The Galois group of a separable polynomial acts by
permuting the roots of the polynomial, so its degree as a group is the degree
of the polynomial.
semi-direct product: A product of two groups, one of which acts on the
other. Suppose H acts on G, and denote the action of an element of H on
an element of G by gh. Then we could form the product of the two groups
with the following multiplication:
(g1, h1)(g2, h2) = (g1gh−11
2 , h1h2)
• p.7
unramified: Let L over K be a finite extension of fields. Then primes of
the ring of integers of K might factor in L as a product of primes. Suppose
pOL =d∏i=1
Pei1 .
If one of the ei > 1, we say p is ramified in L, and that Pi ramifies over K.
55
extension of a prime: In the above factorization, the primes P are said
to extend p. They all have the property that P ∩ OK = p.
decomposition group: If G = Gal(L/K), and p is a prime of OK , then
G permutes the primes extending p. The subgroup of elements σ ∈ G such
that σ(P) = P is the decomposition group of P. Note that σ may not fix
P element-wise, it just maps the set P onto the set P.
conjugate: Two elements of a group γ, τ are conjugate if γ = στσ−1 for
some σ in the group. Of extreme importance here is that conjugate elements
of a permutation group have the same cycle types. So if a permutation has
a fixed point, so do all it conjugates.
• p.8
conjugacy class : Conjugacy partitions a group. The sets in the partition
are conjugacy classes.
primitive mth root of unity: A number ζm such that ζmm = 1, and ζkm 6= 1
for any integer k < m.
where the map ramifies This means that a map that we would expect to
be k-to-1 is not k-to-1 over some point. It coincides with other notions of
ramification.
• p.12
Fp(t): The field of rational functions with coefficients on Fp. That is, ratios
of elements of Fp[t].
pα = (t − α): The ideal generated by the polynomial t − α. This ideal is
prime, because t− α is irreducible.
pα = {(t− α)g(t) | g(t) ∈ Fp[t]}
56
principal: A principal ideal is generated by one element.
genus: The genus of a curve is an important invariant. For a topologist, the
genus of a surface is the number of handles it has. A donut has genus one,
hence the famous joke about the donut, the coffee cup, and the topologist.
• p.15
compositum: The compositum of two field K,L, is the smallest field con-
taining both.
• p.16(OKn−1
)Pi
: The localization of a ring R at a maximal ideal m, denoted Rm
is the subset of the field of fraction of R whose denominators are NOT in m.
Such a ring is local–it has only one maximal ideal. However, whenever we
name a multiplicative subset of R, say S, and allow elements of S to be de-
nominators, we call that localization as well, although it doesn’t necessarily
create a local ring. It is hopefully clear from context which is meant. If a
ring is Dedekind, all its localizations at maximal ideals are discrete valuation
rings. These have the property that their maximal ideals are principal, and
anything in the ring can be written as a power of the generator times a unit,
like uαki , where αi generates the maximal ideal of the local ring.
• p.20
e(P ′ | P ): This is the ramification index of P ′ over P .
critical orbit: The orbit of a critical number.
tame ramification: Ramification is tame if it does not divide the charac-
teristic of the field.
• p.23
57
M∞: MK refers to the set of equivalent valuations, or primes or places of a
number field K. There are two kinds, the finite, which are extensions of the
p-adic valuations on Q, and infinite, which come from embeddings of K in
C, one for each real embedding and one for each conjugate pair of complex
embeddings. The latter are referred to as M∞.
• p.25
ultrametric property: We have that vp(·) satisfies a stronger version of the
triangle inequality. That is, vp(a+ b) ≥ min {vp(a), vp(b)}. If vp(a) > vp(b),
then vp(a+ b) = vp(b).
• p.27
Q: The algebraic closure of Q. That is, C without the transcendentals.
• p.30
Galois conjugates: Roots of the same irreducible polynomial.
• p.32
S-integer: defined p.22
• p.34
unit group A number field may have infinitely-many units, but Dirichlet
showed that the group of units is finitely generated. Then, if you take a
quotient of the group by the subgroup of units that are mth powers, the
number of cosets will be finite.
• p. 36 Probability Section
inverse system: A set of groups and homomorphisms as described. An
element in the inverse limit of this system will be a sequence {αi}i≥1 with
58
the property that for any αj in the sequence, ψi,j(αj) = αi. Then we can
think of Gn as a truncation of every sequence in G(f, g) at the nth place.
compact, Hausdorff topological group: A topological group is one that
is also a space. The group operation and the map sending an element to its
inverse are continuous. The important fact here is that a compact, Hausdorff
space has a Haar measure.
See [3], Chapter 2 for a discussion of martingales.
• p. 40
D(S): Natural density defined on p.2.
59
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