The Factoradic Integers

THE FACTORADIC INTEGERS

Joshua S. Brinseld

Thesis submitted to the Faculty of the Virginia Polytechnic Institute and State University in partial fulllment of the requirements for the degree of

Master of Science in Mathematics

William J. Floyd, Chair Ezra A. Brown Leonardo C. Mihalcea

May 5, 2016 Blacksburg, Virginia

Joshua S. Brinseld

(ABSTRACT)

The arithmetic progressions, considered as functions x 7→ λx+φ equipped with addition and composition, are examined from an algebraic standpoint as left outer-distributive rings, i.e. objects satisfying the ring axioms with a weakened left-distributive law: namely, x(y + z) = xy + xz + D(x), where D is a function dependent only on x. Under appropriate constraints on λ and φ, the images of these functions give bases for topologies on Z (or any unique factorization domain); the ring of factoradic integers Z¯ is dened as the completion of the topology generated on Z by the constraints λ ∈ N, φ ∈ Z (i.e. the evenly spaced integer topology) under the metric 1 , maximal such that . It is shown d!(n, m) = N! N N! | (n − m) that ¯ is ring-isomorphic to the direct product of the -adic integers over all primes , i.e. Z p Zp p to the pronite completion of the integers. Analogies between Z¯ and Z are exploited to allow for general factoradic integers to be used as exponents; this results in a unique factorization theorem which completely characterizes multiplication in Z¯, giving the multiplicative group of units as ∼ Q , where the direct product is over all odd primes U = Z/2Z×Z2 × Z/(p−1)Z×Zp p p. Similarly, bounds of summation are extended to take general values in Z¯, and continuous functions are examined in terms of these factoradic series and the nite dierence operator. It is found that a continuous function is equal to its own Newton series if and only if it can be decomposed as a direct product of functions over all primes . The relationship Zp → Zp p between Z¯ and certain completions of polynomial rings is then examined, and it is shown that ¯ ∼= R/xR, where R = lim [x]/x(x + 1)...(x + n) [x]. Z ←− Z Z The Factoradic Integers

Joshua S. Brinseld

(GENERAL ABSTRACT)

The arithmetic progressions under addition and composition satisfy the usual rules of arithmetic with a modied distributive law. The basic algebra of such mathematical structures is examined; this leads to the consideration of the integers as a metric space under the factoradic metric, i.e., the integers equipped with a distance function dened by d(n, m) = 1/N!, where N is the largest positive integer such that N! divides n − m. Via the process of metric completion, the integers are then extended to a larger set of numbers, the factoradic integers. The properties of the factoradic integers are developed in detail, with particular attention to prime factorization, exponentiation, innite series, and continuous functions, as well as to polynomials and their extensions. The structure of the factoradic integers is highly dependent upon the distribution of the prime numbers and relates to various topics in algebra, number theory, and non-standard analysis. Contents

Introduction 1 Motivation 1 Notation and Background 1 Topology 1 Algebra 3 Modular Arithmetic 4 P-Adic Numbers 5 Factoradic Integers 6

Z¯ and Related Constructions 9

Part 1. Left Outer-Distributive Rings 11 1. Denition & General Properties 11 2. Wave Spaces 13

Part 2. The Factoradic Integers 17 3. Construction 17 4. The Factoradic Rationals 22 5. The Digit-Flip Function 23 6. Sequences & Series 25 6.1. Innite Series 25 6.2. Factoradic Limits 26

6.3. Factoradic Sequences in Z¯ 27 7. Exponentiation 29 8. The Units 30 8.1. Units and Zero-Divisors 30

8.2. Topological Z¯-Modules and the Multiplicative Group of Units 32 8.3. The Total Logarithm 35 9. Continuous Functions 35 9.1. Some Useful Functions 35 9.2. Finite Dierences 37 iv 9.3. Direct & Indirect Continuity 39 9.4. Factoradic Series 41 9.5. Geometric Series of Units 45 10. Projective Limits 45 10.1. Projective Limits 45

10.2. Direct Limits & Prüfer p-Groups 47

Part 3. Projective Limits and Polynomial Completions 48

11. Power Series in Z¯ 48 12. F[x] 49 13. lim [x]/x!n [x] 50 ←−Z Z

Bibliography 53 References 53

v Introduction

Motivation

The prime numbers are easily dened in terms of multiplication, but their additive distribution is extremely complex; a similar situation holds for many sets of integers of number- theoretic interest. Indeed, multiplying a number by a sum is trivial in light of the distributive law a(b+c) = ab+ac, but there is no such general formula for adding a number to a product; intuitively, addition is too fundamental to be so easily described. The structure of the integers, however, is entirely determined by the interplay between addition and multiplication. The simplest functions involving both addition and multiplication are the waves, i.e. the functions f(x) = λx + φ for some xed λ, φ ∈ Z, whose images are fundamental periodic sets which correspond to the bidirectionally innite arithmetic progressions. It is the object of this paper to examine these waves from two perspectives. First, it is observed that as functions under addition and composition they satisfy a modication of the ring axioms with an altered distributive law, and the basic algebra of such structures is investigated. The case of waves over unique factorization domains is given special attention, resulting in a collection of topologies over any unique factorization domain R whose basis elements are wave images, i.e. sets of the form . Second, the topology generated {λx + φ}x∈R by all non-singleton wave images over the integers is given a particular metric and completed, extending Z to the larger topological ring of factoradic integers Z¯, and the properties of Z¯ are investigated. In particular, a unique factorization theorem is proven for Z¯ which characterizes multiplication in Z¯ completely, including the structure of the multiplicative group of units, and innite summations over Z¯ are developed in some detail. As a ring, Z¯ is found to be isomorphic to the pronite completion of the integers, i.e. to the direct product of the p-adic integers over all primes p; several alternative constructions are given, including the pronite construction via projective limits and a construction as a quotient of a certain completion of the polynomial ring Z[x].

Notation and Background

Notationally, the natural numbers include all positive integers but not ; denotes N 0 N0 N ∪ {0}. As usual, Z denotes the set of integers, Q the set of rational numbers, and R the set of real numbers. When the meaning is clear, subscripts may be used to modify sets in obvious ways, e.g. denotes the nonzero integers, denotes the negative real numbers, Z6=0 R<0 etc. The symbol P always denotes the set of prime numbers, i.e. the positive integers each of which is divisible only by itself and by 1. The number 1 itself does not count as prime. Basic facts and denitions from algebra, modular arithmetic, and topology are assumed as common knowledge; this section elucidates some standard denitions and basic results from these elds.

Topology. Given a set X, a topology is a collection T of subsets of X such that X, ∅ ∈ T , T is closed under nite intersections of elements, and T is closed under arbitrary unions of 1 elements; the sets in T are called open sets, and their complements are called closed sets. If a set is both open and closed, it is called clopen. A basis for a topology T is a set B ⊂ T such that every element of T is a union of elements of B; a set B is a basis for a topology on X whenever ∀x ∈ X ∃b ∈ B such that x ∈ b, and ∀a, b ∈ B if x ∈ a ∩ b then ∃c ∈ B such that x ∈ c ⊂ a ∩ b. A set X equipped with a topology T is called a topological space. Given two topological spaces X and Y , a function f : X → Y is continuous if preimages of open sets are open; that is, if whenever S ⊂ Y is an open set in Y , f −1(S) = {x ∈ X : f(x) ∈ S} is an open set in X. If there exists a continuous bijection f : X → Y which has a continuous inverse, we say f is a homeomorphism between X and Y , which are consequently homeomorphic topological spaces. A metric space is a set together with a function , called a metric, X X d: X × X → R≥0 which satises the following:

(1) (Identity) ∀x, y ∈ X d(x, y) = 0 if and only if x = y. (2) (Symmetry) ∀x, y ∈ X d(x, y) = d(y, x). (3) (Triangle Inequality) ∀x, y, z ∈ X d(x, z) ≤ d(x, y) + d(y, z).

An ultrametric is a metric satisfying d(x, z) ≤ max {d(x, y), d(y, z)}, a stronger form of the triangle inequality. If X is a metric space, we can dene the open metric balls as follows: if and , then the open metric ball of radius centered at is r ∈ R>0 x ∈ X r x Br(x) := {y ∈ X : d(x, y) < r}. It is also convenient to consider the closed metric balls, which allow d(x, y) = r; the closed metric ball of radius r centered at x is Br[x] = {y ∈ X : d(x, y) ≤ r}. Now the set of all open metric balls is always a basis for a topology {Br(x): r ∈ R>0, x ∈ X} on X, and we say the metric d induces this topology; thus a metric space is always a topological space. A Cauchy sequence is a sequence {x } in a metric space X such that ∀ > 0 ∃N ∈ n n∈N N such that n, m > N −→ d(n, m) < . That is, going far enough out in the sequence we nd that all terms are arbitrarily close to one another. If there is an x ∈ X such that ∀ > 0 d(x, xn) < for suciently large n, we say x is the limit of {xn} and {xn} converges to x as n → ∞, and we write lim xn. n→∞ A metric space in which every Cauchy sequence converges is called a complete metric space. If a metric space X is not complete, it can be extended to a complete metric space via metric completion: rst, put the metric d ({xn} , {yn}) = lim d(xn, yn) on the set of Cauchy n→∞ sequences in X. This is not a true metric, because two Cauchy sequences may be distinct and yet have distance 0; however, the Cauchy sequences at distance 0 from one another form an equivalence class, so letting {xn} ∼ {yn} ⇐⇒ d ({xn} , {yn}) = 0 we have that ∼ is an equivalence relation; if the set of Cauchy sequences is designated C, then C/ ∼, i.e. the set of Cauchy sequences with {xn} considered equal to {yn} whenever {xn} ∼ {yn}, gives the metric completion X¯ of X. If X is a ring with addition and multiplication continuous, X¯ will be as well. 2 There are several other useful denitions from topology which may be utilized in this paper; given a topological space X we have the following:

(1) Neighborhood: A neighborhood of a point x ∈ X is any open set containing x. (2) Limit point: The set of limit points of a subset S ⊂ X is denoted by S0 and dened as the set of all x ∈ X such that every neighborhood of x contains an element of S\{x}. A closed set contains all of its own limit points. (3) Closure: The closure of a subset S ⊂ X is dened by S¯ = S ∪ S0. (4) Continuous function: On a metric space, there is an alternative characterization of continuous functions. If X and Y are metric spaces and f : X → Y , then f is continuous at x0 ∈ X if ∀ > 0 ∃δ > 0 such that ∀x ∈ X with 0 < d(x, x0) < δ we have d(f(x), f(x0)) < . If for each there exists a valid choice of δ which holds for all x, then f is uniformly continuous. (5) Hausdor: X is Hausdor if ∀x, y ∈ X with x 6= y there exist neighborhoods N,M of x and y, respectively, such that N ∩ M = ∅. (6) Open covering: An open covering of a subset S ⊂ X is a collection C of open sets in X such that S ⊂ S c. Given an open covering C of S, a subcovering is a subset of c∈C C such that we still have S ⊂ S c. c∈C (7) Compact: A subset S ⊂ X is compact if every open covering of S has a nite subcovering. X itself is compact if it satises this criterion; every closed subset of a compact set is compact.

Algebra. A group is a set G equipped with a binary operation ∗ such that:

(1) ∗ is associative. That is, ∀a, b, c ∈ G we have a ∗ (b ∗ c) = (a ∗ b) ∗ c. (2) There is an identity on G. That is, ∃i ∈ G such that ∀g ∈ G g ∗ i = i ∗ g = g. (3) Every element has an inverse. That is, ∀g ∈ G ∃h ∈ G such that g ∗ h = h ∗ g = i.

If G is commutative under ∗ (that is, g ∗ h = h ∗ g ∀g, h ∈ G), we say that G is abelian, and we will normally write + for the binary operation and 0 for the identity, denoting the additive inverse of a group element g by −g. Similarly, a ring is an abelian group G written additively and equipped with a second binary operation corresponding to multiplication, such that:

(1) Multiplication is associative. That is, ∀a, b, c ∈ R we have a(bc) = (ab)c. (2) There is a multiplicative identity. That is, ∃1 ∈ R such that ∀r ∈ R r1 = 1r. (3) The right distributive law holds. That is, ∀a, b, c ∈ R (a + b)c = ac + bc. (4) The left distributive law holds. That is, ∀a, b, c ∈ R a(b + c) = ab + ac. A ring in which multiplication is commutative is called a commutative ring. An integral domain is a commutative ring R in which the only zero-divisor is 0, where x ∈ R is called a zero divisor if there exists y 6= 0 in R such that xy = 0; a eld is a commutative ring R in which every nonzero element is a unit, where x ∈ R is a unit if it has a multiplicative 3 inverse, i.e. if ∃y ∈ R such that xy = 1. Any eld is thus also an integral domain, because units and zero divisors are mutually exclusive. An element x of an integral domain is called irreducible if x = yz implies that either y or z must be a unit. An element x of an integral domain is called prime if x = yz implies x | y or x | z. An integral domain X is a unique factorization domain (UFD) if every element x ∈ X can be written as a product of irreducible elements and a unit; in a UFD, all irreducibles are prime, and vice versa. An ideal of a commutative ring R is a subset I of R such that I under addition is a subgroup of R, and such that for all r ∈ R and i ∈ I we have ir ∈ I. An ideal I is called principal if it is generated by a single element, i.e. if ∃i ∈ I such that I = iR; if every ideal in an integer domain is principal, it is called a principal ideal domain (PID); every PID is a UFD. Given an ideal I of a ring R, the quotient ring R/I is composed of elements of the form {r + I : r ∈ R}, where r + I = {r + i : i ∈ I}, which form a ring. Finally, given rings R and S, a ring homomorphism is any function f : R → S such that f(r1) + f(r2) = f(r1 + r2) and f(r1)f(r2) = f(r1r2) ∀r1, r2 ∈ R. If a ring homomorphism is also a bijection, it is called an isomorphism, and R and S are said to be isomorphic, and we write R ∼= S. The kernel of a homomorphism R → S is ker(f) = {x ∈ R : f(x) = 0 in S}. Note that there are also group homomorphisms, which are functions between groups G → H which have f(x) ∗ f(y) = f(x ∗ y); in this case the kernel of f is the set of x ∈ G such that f(x) is the identity in H. The First Homomorphism Theorem asserts that for a ring homomorphism f : R → S, the quotient ring R/ker(f) is isomorphic to the image f(R) of f, which is a subring of S.

Modular Arithmetic. Z is a principal ideal domain; the quotient rings Z/nZ for n a xed natural number are known as the integers modulo n. If two integers x, y map natually via x 7→ x + nZ, y 7→ y + nZ to the same element of Z/nZ, we say that x and y are congruent modulo n, and write x ≡ y (mod n). Two integers are congruent modulo n if and only if they give the same remainder when divided by n; every integer can thus be associated with its remainder modulo n, and these give equivalence classes corresponding to the elements of Z/nZ. Thus there are n elements of Z/nZ, asociated with the integers {0, 1, .., n − 1}, and x ≡ 0 (mod n) if and only if n | x. x is congruent to a unit modulo Z/nZ if and only if gcd(x, n) = 1. As an additive group, Z/nZ is cyclic, i.e. it can be generated by repeated addition of a single element. The multiplicative structure of Z/nZ is elucidated by the Chinese Remainder Theorem and the existence of primitive roots, as follows:

Theorem. (Chinese Remainder Theorem) If {n1, ....nN } are integers and {m1, ..., mN } are integers which are pairwise coprime, then the system of congruences {x ≡ nk (mod mk)} has N Q an integer solution; in particular, it has a unique solution X modulo M = mk, and an k=1 integer is a solution if and only if it is ≡ X (mod M). 4 N This can be recast in algebraic language: let Q xn be the prime factorization of a natural pn n=1 number ; then ∼ x1 x2 x3 , with an explicit isomorphism ν Z/νZ = (Z/p1 Z) × (Z/p2 Z) × ... × (Z/p3 Z) given by x1 x2 xN . This form of the statement holds f(x + νZ) = (x + p1 Z, x + p2 Z, ..., x + pN Z) in any principal ideal domain P , if the pn are replaced by prime elements of P . From the Chinese Remainder Theorem we conclude that the residues of integers modulo powers of distinct primes are independent, i.e. an integer's residue modulo a power of 5 puts no constraints whatsoever on its residues modulo powers of 7. In considering the multiplicative structure of Z/nZ it is thus sucient to assume n is a prime power. The existence of primitive roots modulo all powers of odd primes was proven by Gauss in 1801 (in Article 57 of his Disquisitiones Arithmeticae) [4]. A primitive root modulo pn is a number γ that generates all the units modulo pn by repeated multiplication. The multiplicative group of units modulo pn is always cyclic, and a primitive root is the element such that every unit moduo pn corresponds to some element of {γ, γ2, ....}. The order of the multiplicative group of units modulo pn is always given by Euler's totient function φ, dened so that φ(n) gives the number of naturals coprime to n and not greater than n . Thus a primitive root γ has γx ≡ 1 (mod pn) if and only if φ(n) = (p − 1)pn−1 | x. Note that an integer which is a primitive root modulo p2 is a primitive root modulo pn ∀n. Note also that there is no primitive root modulo 2n when n > 2; however, for each n > 2 a unit can be represented as the product of a unique power of −1 and a unique power of an element of order 2n−2. We can x an integer choice for this second element valid modulo 16 and it will be valid for all higher powers of 2.

P-Adic Numbers

The elds of p-adic numbers were introduced by Kurt Hensel in his 1897 paper Über eine neue Begründung der Theorie der algebraischen Zahlen [5]. They have become common tools in number theory, largely because they allow power series methods to be brought to bear on number-theoretical problems. The p in the term p-adic is a placeholder for a choice of prime; for any choice of there is a corresponding eld of -adic numbers. The p ∈ P Qp p simplest approach to their construction is to rst consider the rings of -adic integers, Zp p which are more immediately relevant and which relate to in the same way that relates Qp Z to (specically, is the eld of fractions of ). Q Qp Zp For a xed prime , is the completion of under the -adic metric, dened by p Zp Z p dp(x, y) = ( 0 : x = y , where is the largest integer such that n . Thus two -adic numbers 1 n p | (x−y) p pn : x 6= y are considered close to one another if their dierence is divisible by a large power of p. We can dene an absolute value on by 1 , where is the maximal integer such that Zp |x|p := pn n n and where , and we will have ; this is called the p | x |0|p = 0 d(x, y) = |x − y|p ∀x, y ∈ Zp p-adic norm. 5 Now since 1 is equivalent for to n , we see that if and dp(x, y) ≤ pn x, y ∈ Z x ≡ y (mod p ) x are nonnegative integers then 1 where is the largest number such that y dp(x, y) = pN N x and y share their rst N digits when written in base-p notation. More formally, writing ∞ ∞ x = P d pn and y = P δ pn for {d } and {δ } sequences in {0, 1, ..., p − 1} which n n n n∈N n n∈N n=0 n=0 ∞ P n are both 0 for suciently large n, we have x − y = (dn − δn)p ; if dn − δn = 0 for all n n=0 N N+1 in {0, ..., N − 1} but not for n = N, then x − y = p (dN − δN ) + p (dN+1 − δN+1) + ..., which is divisible by N but but not N+1, so 1 . p p dp(x, y) = pN This property holds true in general: the p-adic integers correspond bijectively with series of ∞ P n the form dnp with each dn ∈ {0, ..., p − 1}, i.e. to base-p integers allowed to extend n=0 innitely far to the left of the radix point. Addition and multiplication can be performed up to any desired precision according to the usual rules (addition can be performed by adding digits term-by-term and carrying ones, and multiplication can be performed by hand via base-p long multiplication or formally via the Cauchy product). The p-adic norm is simply 1 , where is the number of leading s in the base- representation of . |x|p = pn n 0 p x ∞ It is worth noting that P (p−1)pn = −1, as can easily be seen by adding 1 and observing that n=0 every digit carries. This is in fact the crucial dierence between base-p notation extended to the left ( ) and base- notation extended to the right ): the more familiar real series Zp p (R ∞ P (p − 1)p−n is the base-p analogue of 0.999... in base 10, and is a second representation of n=1 1p0 + 0p−1 + 0p−2 + ... = 1; unlike in R, where a number with a terminating base-p digit sequence has a second base- representation, the digit sequence for an element of is truly p Zp unique. A similar phenomenon will occur in Z¯.

A p-adic integer x is a unit if and only if |x|p = 1; thus every p-adic integer (and thus every integer) is invertible in unless it is divisible by . We also have that for any Zp p x ∈ Zp\{0} there is some unique and some unique unit such that n. Now is n ∈ N0 u ∈ Zp x = up Qp the eld of fractions of , so any nonzero -adic number can therefore be written in Zp p q ∈ Qp the form n, where is a unit in and is a (possibly negative) integer. In base- q = up u Zp n p ∞ representation, this means that consists of elements of the form P n, where Qp x = dnp N n=N can be any xed integer; that is, in we are permitted a nite number of digits to the Qp right of the radix as well as innitely many to the left. One may also construct directly Qp by completing Q under the p-adic metric.

Factoradic Integers

The construction of the factoradic integers proceeds similarly to that of the p-adic integers, but makes use of the properties of factorials rather than prime powers. The earliest directly related result is Hillel Furstenberg's topological proof of the innitude of the primes, which 6 was published during his undergraduate studies in 1955 [3]. This short and elegant proof of Euclid's classical result relies on the introduction of a topology on the integers with basis the bidirectionally innite arithmetic progressions (i.e. wave images), sometimes called the evenly-spaced integer topology. Theorem. (Furstenberg, 1955) There are innitely many primes.

Proof. Endow Z with the topology whose basis is the set of bidirectionally innite arithmetic progressions. Each basis element is clopen, since Z\{λx + φ : x ∈ Z} is equal to S {λx + ψ : x ∈ Z}; thus any nite union of basis elements is closed. Now consider ψ6≡φ (mod λ) S S := {px + 0 : x ∈ Z}. The complement Z\S = {−1, 1} is clearly not a union of arith- p∈P metic progressions and therefore is not open, so S is not closed. Therefore S is not the union of nitely many arithmetic progressions; so there are innitely many primes. As Furstenberg mentions in passing, this topological space is metrizable; the most nat-

ural metric is in many ways the factoradic metric. We dene this metric by d!(x, y) = ( 0 : x = y , where is the largest integer such that , and we denote by ¯ the 1 n n! | (x − y) Z n! : x 6= y corresponding metric completion of Z. The metric balls are thus the bidirectionally innite arithmetic progressions of factorial common dierence, i.e. {n!x + φ : x ∈ Z} for xed φ and n (and every bidirectionally innite arithmetic progression is clearly a union of these). Now just as every element has a unique radix expansion in any base, so also is there a x ∈ N0 ∞ P unique representation of the form x = n!dn with each dn ∈ {0, 1, ..., n}. These are called n=1 factoradic representations, whence the term factoradic integers for ¯, and the are the Z dn called the factoradic digits of . On these representations terminate with for x N0 dn = 0 all suciently large n; similarly to the p-adic case, we have that the metric ball of radius 1 consists of all such series which share the same values for , since this gives N! {d1, ..., dN−1} ∞ ∞ P - P . Since is clearly dense, in n!dn n!δn = N!(dN − δN ) + (N + 1)!(dN+1 − δN+1) + .... N n=1 n=1 ∞ the metric completion ¯ we thus obtain a canonical series P for each element of ¯, Z n!dn Z n=1 and the distance between two elements of ¯ is given by 1 , where is the least Z d!(x, y) = N! N natural such that the Nth respective digits dN and δN of x and y dier. (Note that we only have agreement of the rst N − 1 digits, not the rst N, because unlike in the p-adic case there is no d0.) Moreover, addition and multiplication are continuous functions, and so we can multiply and add such series in the usual way, as in the p-adic case performing carries whenever digits become too large.

∞ P In Z¯ we have the identity −1 = n!n, which is readily veried since 1 + (1!1 + 2!2 + 3!3...) n=1 = 1!2 + 2!2 + 3!3... = 2!3 + 3!3 + ... = 3!4 + ..., a sequence which has limit 1!0 + 2!0 + 3!0 + ... = 0. Similarly to the p-adic case, we can represent real numbers in [0, 1] by series 7 ∞ of the form P dn with each , and these representations are unique unless (n+1)! dn ∈ {0, ..., n} n=1 they terminate in a string of 0 digits, in which case if dn is the last nonzero digit we have d1 d2 dn−1 dn d1 d2 dn−1 dn−1 n+1 n+2 , i.e. there 2! + 3! + ... + n! + (n+1)! = 2! + 3! + ... + n! + (n+1)! + (n+2)! + (n+3)! + .... are two representations. In this case, however, with the exceptions of 0 and 1 the numbers in [0, 1] with terminating factoradic representations are exactly the rational numbers (just write p p(q−1)! and then perform digit carries as many times as necessary). Some attention q = q! is given to this phenomenon in Section 5, where it is used to show that Z¯\Z is homeomorphic to [0, 1]\Q. The factoradic integers preserve all of modular arithmetic, whereas the p-adic integers preserve only arithmetic modulo powers of p; that is, no natural number except for 1 is a ∼ unit, and Z¯/nZ¯ = Z/nZ ∀n ∈ N. This is evident from the form of canonical series; ∞ N−1 P is always congruent to P , and since elements of ¯ are limits x = n!dn n!dn (mod N!) Z n=1 n=1 of sequences in Z which are eventually constant modulo any factorial (and hence modulo any natural number), properties of addition and multiplication modulo n which hold true in Z also hold true in Z¯. Indeed, if the sequence of partial sums of each canonical series is identied with a sequence of residues modulo increasing factorials, then the set of all such sequences is maximal such that the Chinese Remainder Theorem and the consis- tency law, x ≡ c (mod pn) −→ ∃d such that x ≡ c + pnd (mod pn+1), hold. This is equivalent to the characterization of Z¯ algebraically as the pronite completion of the integers in Section 10. Consequently, because for a natural with prime factorization Q xn we have pn n Q xn ∼ x1 x2 by the Chinese Remainder Theorem, we obtain Z/ pn Z = (Z/p1 Z) × (Z/p2 Z) × .... n ¯ ∼ Q . This establishes the non-unit half of a unique factorization theorem characteriz- Z = Zp p∈P ing multiplication in Z¯; the characterization of the multiplicative group of units is performed along similar lines by considering primitive roots modulo powers of primes, but requires rst the establishment of a notion of exponentiation to factoradic powers. Not only exponentiation but summation can also be extended; for a large class of functions it b P is possible to continuously extend f(n) from a, b ∈ Z with a ≤ b to a, b general factoradic n=a integers. These factoradic series behave analogously to nite series in most ways, and in some ways behave like line integrals. For instance, if ∆ is the forward dierence operator, i.e. (∆f)(x) = f(x + 1) − f(x), then for any continuous function we have f(x) = f(0) + x b P (∆f)(n); likewise, when f is summable we have path-independence, i.e. P f(n) + n=1 n=a+1 c c a P P P f(n) = f(n) ∀a, b, c ∈ Z¯, and therefore closed paths f(n) sum to zero. The n=b+1 n=a+1 n=a+1 theory of nite dierences thus has a deep relationship to Z¯; indeed, it is shown in Theorem 13 that a continuous function f : Z¯ → Z¯ decomposes into a direct product of functions

8 over all if and only if it is equal to its own Newton series, which is true if Zp → Zp p ∈ P and only if (∆nf)(0) → 0 as n → ∞. These considerations lead via the rising and falling factorials to the consideration of polynomial ring completions. A ring of polynomials F[x] over a eld F is a principal ideal domain, and consequently its competion under the topology with basis all wave images decomposes into a direct product over the polynomial analogues of the rings of p-adic integers, just as ¯ ∼ Q . It is also shown that ¯ is ring-isomorphic to the quotient ring , where is Z = Zp Z R/xR R p∈P the projective limit lim /x(x + 1)...(x + n) , n ∈ ordered by magnitude. Elements of this ←− Z Z N ∞ ring are given by series of the form P with all coecients in . a0 + an+1(x + 0)...(x + n) Z n=0

Z¯ and Related Constructions

There are two notions of innite integers to which Z¯ is deeply related and which provide excellent intuitions for working with Z¯. The rst is the set of supernatural numbers, or Steinitz numbers. These were introduced to eld theory by Ernst Steinitz in 1910 [11]; they generalize the prime factorizations of the natural numbers, allowing an innite number of prime factors as well as factors of p∞. Supernatural numbers cannot be added, but can be multiplied, and the gcd and lcm operations extend naturally to them; they are useful extensions of N for describing the orders and indices of elements of pronite groups [10]. These correspond exactly to the elements of Z¯/U = zU : z ∈ Z¯ , where U is the multiplicative group of units in Z¯, which are used in Section 8.2 to characterize the multiplicative group of units in Z¯ as a product of topological Z¯-modules. The second notion is that of the hyperintegers from non-standard analysis. Non-standard analysis was developed by Abraham Robinson in the 1960s; it is based on a foundation of mathematical logic which gives extensions to the standard model of arithmetic in order to obtain a rigorous theory of innite and innitesimal numbers. The most relevant result here is the extension of the integers Z to the uncountably innite set ∗Z of hyperintegers, which is done in such a way that all rst-order propositions that hold in Z continue to hold in ∗Z, and vice versa. The hypernatural numbers contain innitely large but well-dened ∗N = ∗Z>0 numbers, and prime factorizations of hypernaturals correspondingly can involve innitely many primes, possibly raised to general hypernatural powers; indeed, there are innitely large primes in ∗N distinct from the primes in N, and there are hypernaturals all prime factors of which are innite. Sets in ∗N can be divided into two types: there are internal sets, and statements of the form for all sets of numbers... or there exists a set of numbers such that... true in N are also true in ∗N if set is replaced by internal set; and there are also external sets, to which such statements need not extend. N itself is an external set within ∗N, as is its complement ∗N\N. Note that the well-ordering property of N extends a priori only to internal sets: in particular, there is no least element of ∗N\N. [9] Carter Waid in 1974 published a short article entitled On Dirichlet's Theorem and Innite Primes which is of great interest [12]. Therein he follows Robinson in constructing a ring 9 T isomorphic to Z¯ from ∗N. Let µ = n(∗Z), so that µ is the set of all hypernatural integers n∈N divisible by every natural number; then µ is an ideal of ∗Z and is an external set. In fact, ∗Z/µ is isomorphic to Z¯. In this sense the elements of Z¯ are equivalence classes of all hyperintegers with the same nite part; it is important to note, however, that µ is not a principal idealthere is no least nonzero element m of µ, just as there is no least element of ∗N\N. The main results of the paper, however, deal with the relationship between ∗N, Z¯, and Dirichlet's Theorem. Dirichlet's Theorem on Arithmetic Progressions is a celebrated result in number theory; it had long been conjectured, but was rst proven by Johann Dirichlet in 1837 [6]. The statement of the theorem is this: every arithmetic progression {λn + φ : n ∈ N0} with gcd(λ, φ) = 1 contains an innite number of primes. In Z¯ this statement corresponds to the assertion P¯ = P∪U, where¯denotes topological closure and U denotes the multiplicative group of units in Z¯. Waid proves two theorems, giving three important results. Firstly, every innite prime p ∈ ∗N\N maps via the natural homomorphism to a unit in Z¯. Secondly, given Dirichlet's Theorem, every unit in Z¯ is the image of an innite prime. Lastly, any proof that every unit in Z¯ is the image of an innite prime would also constitute a proof of Dirichlet's Theorem.

10 Part 1. Left Outer-Distributive Rings

The ring axioms are nearly symmetrical in the constraints they place upon addition and multiplication. There are only two asymmetries, namely: (1) the axioms require additive inverses to exist for all elements, but do not require multiplicative inverses; and (2) the distributive law xes an asymmetrical relationship between addition and multiplication. The rst asymmetry is merely a lack of constraint, and ring theory includes the study of rings with all possible multiplicative groups of units; but the second asymmetry is also mutable, i.e. there are consistent sets of axioms which are identical to the ring axioms except for the form of the distributive law. As shall be shown, the arithmetic progressions, considered as increasing functions Z → Z and equipped with addition via (f + g)(x) = f(x) + g(x) and with multiplication dened as composition, i.e. (fg)(x) = f(g(x)), give a simple example of such an algebraic object. It will therefore be useful to examine the basic principles of algebra on rings with a modication of the distributive law akin to that found in the case of the arithmetic progressions; these are the left outer-distributive rings. Their most striking feature is the behavior of the additive identity: no longer does 0 absorb all elements under both left- and right-multiplication, but instead its action from the right encodes information about the distributive law, and it absorbs only when acting from the left.

1. Definition & General Properties

Denition 1. We say a set L equipped with binary operations +, · is a left outer-distributive ring (LODR) if:

(1) L is an abelian group under +; (2) · is associative; (3) ∀x, y, z ∈ L (x + y)z = xz + yz; (4) ∀x, y, z ∈ L x(y + z) = xy + xz + D(x) for some function D, called the distributor, dependent only on x.

We will write R, or simply , for the additive identity in L. Proposition 1. Let L be a LODR; then D(x) = −x.

Proof. x = x( + ) = x + x + D(x) =⇒ D(x) = −x. Proposition 2. Let L be a LODR; then x = ∀x.

Proof. x = ( + )x = x + x =⇒ x = . Corollary. Let L be a LODR; then 2 = . Proposition 3. Let L be a LODR; then the following negativity relations hold:

(1) (−x)y = −xy (2) x(−y) = −xy + x + x 11 Proof. For each part:

(1) Suppose (−x)y = −xy + f(x, y). Then f(x, y) = (−x)y + xy = (−x + x)y = y = . (2) Suppose x(−y) = −xy + f(x, y). Then f(x, y) = x(−y) + xy = x(−y + y) + x = x + x.

Corollary. Let L be a LODR and let x ∈ L; then D(D(x)) = −D(x).

2 Proof. D(D(x)) = −(−x) = −(−x ) = −D(x). Proposition 4. Let L be a LODR; then X = {x ∈ L : −x = } is a ring under the inherited LODR operations. If there is a left-identity in L, it is in X.

Proof. X consists precisely of those x whose distributor is . Therefore multiplication is properly distributive within X. X is additively closed since ∀x, y ∈ X ∀a, b ∈ L (x + y)(a + b) = x(a + b) + y(a + b) = xa + ya + xb + yb = (x + y)a + (x + y)b, hence D(x + y) = so (x + y) ∈ X, and X is multiplicatively closed since ∀x, y ∈ X ∀a, b ∈ L xy(a + b) = x(ya + yb) = xya + xyb. Additive inverses of elements of X are in X since ∀x ∈ X ∀a, b ∈ L (−x)(a + b) = −(x(a + b)) = −xa − xb by Proposition 3, and ∈ X since D() = −2 = . If there is a left-identity in , then , so . e− L e−(a + b) = a + b = e−a + e−b ∀a, b ∈ L e− ∈ X Denition 2. Suppose L is a LODR; then X as in Proposition 4 is called the distributive subring of L and is denoted DSR(L). If DSR(L) is commutative we say L is DSR-commutative. Proposition 5. Let L be a LODR; then ∀x ∈ L, x − x ∈ DSR(L).

Proof. D(x − x) = −(x − x) = −x − (−x) = −x + x = . Denition 3. Suppose L is a LODR. Then a left L-module is an abelian group A and an operation · : L × A → A such that ∀r, ρ ∈ L and ∀a, b ∈ A the following hold:

(1) r · (a + b) = r · a + r · b − r · 0A (2) (r + ρ) · a = r · a + ρ · a (3) (rρ) · a = r · (ρ · a) (4) e− · a = a, if e− is a left-identity in L. Proposition 6. Let L be a LODR and let M be a left L-module. Then ∀m ∈ M, ∀r ∈ L we have:

(1) R · m = 0M (2) (−r) · m = −(r · m)

Proof. Proceeding for each part:

(1) · m = ( + ) · m = · m + · m; (2) 0 = · m = (r − r) · m = r · m + (−r) · m

12 2. Wave Spaces

Denition 4. Suppose L is a LODR. Then we dene the LODR of waves over L to be the set DSR^(L) := DSR(L) × L equipped with the operations (λ, φ) + (ρ, ψ) = (λ + ρ, φ + ψ) L and (λ, φ)(ρ, ψ) = (λρ, λψ + φ); the elements of a LODR of waves are called waves. We will utilize the notation , and so we have and . λe := (λ, φ) λe+ρe = λ]+ ρ λeρe = λρf ∀λe, ρe ∈ DSR^(L) φ φ ψ φ+ψ φψ λψ+φ φ ψ L In general, if Λ and Φ are subsets of a LODR L, then we write Λe := λe : λ ∈ Λ, φ ∈ Φ ⊂ Φ φ DSR^(L). L

Proposition 7. Let L be a LODR. Then DSR^(L) is a LODR. L

Proof. We verify the axioms:

(1) Additivity: Under addition DSR^(L) ∼= DSR(L)⊕L, so it is clearly an abelian group. L (2) Associativity: λe ρµ = λe ρµ = λρµg and λeρ µ = λρf µ = λρµg , and φ ee φ f φ e e e ψz ρz+ψ λ(ρz+ψ)+φ ψ z λψ+φz λρz+λψ+φ these are equal precisely when λ ∈ DSR(L). (3) Right-Distributivity: λe + ρ µ = λ]+ ρµ = (λ^+ ρ)µ = λµ^+ ρµ since (λ + ρ) ∈ φ e e e ψ z φ+ψ z (λ+ρ)z+φ+ψ λz+ρz+φ+ψ DSR(L), which is = λµf + ρµ = λeµ + ρµ. f φ e ee λz+φ ρz+ψ z ψz (4) Left Outer-Distributivity: λe ρ + µ = λeρ]+ µ = λ(^ρ + µ) = λρ^+ λµ = λρf + λµf φ e e φ ψ z ψ+z λ(ψ+z)+φ λψ+λz+φ λψ+φ λz = λρf + λµf − . eφ λψ+φ λz+φ

Proposition 8. Let L be a (properly distributive) ring with identity. Then:

(1) 1fL is the identity in DSR^(L) = Le. 0L L L (2) λe is left-invertible in Le if and only if λ is left-invertible in L; if it is, then λg−1 λe = e1. φ L −λ−1φφ 0 (3) λe is right-invertible in Le if and only if λ is right-invertible in L; if it is, then λe λg−1 = φ L φ−λ−1φ e1. 0 13 Proof. λee1 = λf1 = λe = 1fλ = e1λe. If λ is left-invertible, λg−1 λe = λ]−1λ = e1, and if it φ 0 λ0+φ φ 1φ+0 0 φ −λ−1φφ λ−1φ−λ−1φ 0 is right-invertible, 1 = λλ]−1 = λe λg−1 . If λ is not right- invertible, λeρ = λρf which is e −1 −1 e 0 λ(−λ φ)+φ φ−λ φ φψ λψ+φ never since , and similarly is never if is not left-invertible. So is invertible e1 λρ 6= 1 ρeλe e1 λ λe 0 ψ φ 0 φ from a given side if and only if λ is. Denition 5. Let be a LODR. Then for each we dene the th order wave space L n ∈ N0 n n (or simply n) over inductively by setting 0 and setting n n−1 WL W R W = R W = DSR^(W ) Wn−1 for each n > 0, and we consider each Wn a Wn+1-module under the action λe · x = λx + φ. φ This does indeed make Wn into a Wn+1-module, as the following proposition shows. Proposition 9. Let be a LODR. Then , n−1 is a left n-module under the L ∀n ∈ N WL WL action λe · x = λx + φ. φ

Proof. n−1 is certainly an abelian group under addition, since it is a LODR; we verify the WL other axioms:

(1) Left Outer-Distributivity: λe(x+y) = λx+λy+φ = λx+φ+λy+φ−φ = λex+λey−λe0. φ φ φ φ (2) Right Proper-Distributivity: λe + ρe x = λ]+ ρx = (λ + ρ)x + φ + ψ = λx + φ + φ ψ φ+ψ . ρx + ψ = λex + ρex φ ψ (3) Associativity: . λeρe x = λρf x = λρx + λψ + φ = λ(ρx + ψ) + φ = λe ρex φψ λψ+φ φ ψ (4) Identity: If is a left-identity in n−1 then , so . a WL ae0 = e0 = e0 ax = ax + b = x eb eb x ax+b x eb

Remark. For any LODR , 1 is isomorphic to the set L WL {f : L → L such that ∃λ ∈ DSR(L) and φ ∈ L with f(x) = λx + φ} under the operations (f + g)(x) = f(x) + g(x) and (fg)(x) = f(g(x)).

Denition 6. Let L be a LODR, and let S ⊂ DSR^(L). Then we dene the set of wave L images over L by S/L = λeL : λe ∈ S . φ φ

Denition 7. Suppose L is a unique factorization domain (UFD). Then ΩΦ(L), or simply ΩΦ, denotes the set of coprime waves over L, dened by ΩΦ = λe : gcd(λ, φ) = 1 . φ 14 Remark. Note that if γ ∈ L6=0, γΩΦ = γλe : gcd(λ, φ) = 1 = λe : gcd(λ, φ) = γ , since φ φ gcd(λ, φ) = 1 =⇒ gcd(γλ, γφ) = γ and if gcd(λ, φ) = γ we can write λe = γ eλ . φ γ φ/γ

Theorem 1. (Wave Topology Theorem) Let L be a LODR such that DSR(L) is commutative, and let Λ = {λ ∈ DSR(L): λ is not a zero divisor}. Then Λe/L is a basis for a L topology on L. Moreover, if L is a UFD, then Λ = L6=0, and:

(1) If , , , and , then lcm . γ, κ ∈ L6=0 λe ∈ γΩΦ ρe ∈ κΩΦ µeL = λeL ∩ ρeL µe ∈ (γ, κ)ΩΦ φ ψ z φ ψ z (2) If Γ is a subset of L6=0 such that γ1, γ2 ∈ Γ −→ lcm(γ1, γ2) ∈ Γ, then: (a) S is a basis for a topology on S . γΩΦ/L γeL γ∈Γ γ∈Γ 0 (b) S is a basis for a topology on . γe/L L γ∈ΓR

Proof. First, we shall show that Λe/L is a basis for a topology on L. Λ ⊂ DSR(L) so for all L λ ∈ Λ and φ ∈ Φ we have φ = λe0, so every r ∈ L is in some element of Λe/L. Moreover, φ L suppose . Then such that , so , and since x ∈ λeL ∩ ρeL ∃n λn + φ = x λm + x = λn + λm + φ φ ψ this means , so ; similarly, . Therefore λ ∈ DSR(L) λm + x = λ(n + m) + φ λeL = λeL ρeL = ρeL φ x ψ x ; but subtracting , this is equivalent to . If is a UFD this x ∈ λeL∩ρeL = λeL∩ρeL x 0 ∈ λeL∩ρeL L x x φ ψ 0 0 intersection of principal ideals is itself a principal ideal , so ; if not, we µeL x ∈ µeL = λeL ∩ ρeL 0 x φ ψ can take µ = λρf; then since µL ⊂ λeL ∩ ρL, taking = x we have µL ⊂ λeL ∩ ρL = λeL ∩ ρL. e e 0 e z e x e e 0 0 0 0 z x φ ψ Now are not zero-divisors so ; therefore λ, ρ µ 6= 0 ∀λeL, ρeL ∈ Λe/L ∀x ∈ λeL ∩ ρeL ∃µeL ∈ Λe φ ψ L φ ψ z L such that , and so is a basis for a topology on . Now, suppose is a x ∈ µeL ⊂ λeL ∩ ρeL Λe/L L L z φ ψ L UFD. Then we will clearly have µ = lcm(λ, ρ). If, moreover, γ = gcd(x, λ) = gcd(x, ρ), then γ = gcd(x, lcm(λ, ρ)) = gcd(x, µ); so for each γ ∈ L6=0, γΩΦ/L is closed under nonempty intersection, and therefore gives a basis for a topology on γL by the argument above. Indeed, e0 lcm lcm , so if and , we gcd(x, (λ, ρ)) = (gcd(x, λ), gcd(x, ρ)) λeL ∈ γλΩΦ/L ρeR = L ∈ γρΩΦ/L φ ψ have lcm as claimed. So if is a subset of closed under λeL ∩ ρeL = µeL ∈ (γλ, γρ)Ωφ/L Γ L6=0 φ ψ z S taking least common multiples, then γΩΦ/L is closed under nonempty intersection, and γ∈Γ S therefore the above argument applies, and γΩΦ/L is a basis for a topology on the subset γ∈Γ 15 of L included in the basis elements, i.e. in the union S S γλeL. Since S λeL = L (it γ∈Γ φ φ λe∈ΩΦ λe∈ΩΦ φ φ would be , but ), we have S S S S . As for S , we see L6=0 e1 ∈ ΩΦ γλeL = γL = γeL γe/L 0 γ∈Γ φ γ∈Γ γ∈Γ 0 γ∈ΓR λe∈ΩΦ φ that it is closed under nonempty intersections as well, but that the union of its elements contains all of L. This establishes claims (2a) and (2b). Corollary. If is a DSR-commutative LODR, then n is a basis for a topol- L DSR^(L)/WL n WL ogy on n. In particular, if is a commutative (properly distributive) ring which is WL R an integral domain, then by denition and zero divisors , so W = Re DSR(R)\{ } = R-0  R     g is a basis for a topology on . We have Rf-0/W = Rf-0/Re = λe Re : λ ∈ R\{0}, ρe ∈ Re W R 0 R ψ R W Re   R  ρe   ψ  g λe µ = λµ^+ ρ = λefµ, so each basis element is of the form λefR for some λ ∈ R 0 and 0 e ρ α - z λz+ψ ρ e λez λeR ψ ψ β , i.e., it is of the form . Thus the basis elements are α, β ∈ R ρe : ρ ≡ α, ψ ≡ β (mod λ) ψ precisely the sets of waves with a common image in for some under the natural R/λR^ λ ∈ R-0 R/λR homomorphism . ρe 7→ ρ^+ λR ψ ψ+λR

Denition 8. Suppose L is a DSR-commutative LODR and T is a topology on S ⊂ Wn. We say T is a wave topology if ∃W ⊂ DSR^(L) such that W/L is a basis for T . We say T L is the standard wave topology (or simply the wave topology where unambiguous) on L if we may take W = DSR(L)\{^zero divisors}. L Theorem 2. Let T be a wave topology on a subset S of a UFD R, and let W be a subset of Re such that W/R is a basis for T . Then: R (1) T is the discrete topology if and only if for each s ∈ S there exists a nite set D ⊂ R such that Q D = 0 and deR ∈ W/R ∀d ∈ D. s (2) T is the trivial topology if and only if, given ∆S = {s2 − s1 : s1, s2 ∈ S} and n o Λ = λ : ∃s ∈ S with λeR ∈ W/R , we have lcm(Λ)ˆ | gcd(∆ˆS) for all nite subsets s Λˆ, ∆ˆS of Λ and ∆S, respectively.

Proof. Since R is a UFD, nonempty nite intersections of wave images are wave images; so T is the discrete topology if and only if for each s ∈ S there exists a nite set of wave images 16 T λfnR with intersection {s} = e0R. Since λfnR = lcm(λ^1, ..., λN )R, s s n∈{1,..,N} s n∈{1,..,N} s and since lcm(λ1, ..., λN ) = 0 ⇐⇒ λ1 ··· λN = 0, this occurs if and only if for each s ∈ S there exists a nite set D ⊂ R such that Q D = 0 and deR ∈ W/R ∀d ∈ D. Regarding the s trivial topology, and using the denitions of Λ and ∆S above, we have that T is the trivial T topology⇐⇒ ∀σ ∈ S ∀λeR ∈ Λ σ ∈ λeR⇐⇒ ∀σ ∈ SS ⊂ Λe/R ⇐⇒ λn | (s2 − s1) for φ φ σ ˆ ˆ each s1, s2 ∈ S and each λn ∈ Λ ⇐⇒ gcd(∆S) is divisible by lcm(Λ) for all nite subsets ˆ ˆ Λ ⊂ Λ and ∆S ⊂ ∆S. Corollary. Take S = R in the preceding theorem, and suppose W contains no wave of zero-divisor wavelength. Then:

(1) T is the discrete topology if and only if R is the zero ring. (2) T is the trivial topology if and only if every element of R is a zero-divisor or a unit. Denition 9. Given a UFD , let denote the set of primes in and let ≥n denote R PR R PR all elements of R which are the product of at least n primes, counted with multiplicity. If W ⊂ Re generates a topology on S W/R, we write that topology T (W ). We name the R topologies generated by the following bases accordingly (these are all bases for topologies on the union of the basis elements by the Wave Topology Theorem (Theorem 1); the even- numbered topologies below are all derived from part (2a), the odd-numbered from part (2b) of the theorem):

(1) (Standard Wave Topology) ≥0 ≥0 Rg6=0/R = R6=0ΩΦ/R = PR ΩΦ/R = PgR /R R R (2) (GCD=γ Wave Topology, Coprime Wave Topology) γΩΦ/R for γ ∈ R6=0 (3) (Composite Multiple Wave Topology of Order ) ≥n for n PgR /R n ∈ N0 R (4) (Coprimal Composite Multiple Wave Topology of Order ) ≥n for n PR ΩΦ/R n ∈ N0 (5) (r-adic Topology) rfN0 /R for r ∈ R6=0 R N0 (6) (Coprimal r-adic Topology) r ΩΦ/R for r ∈ R6=0

Part 2. The Factoradic Integers

3. Construction

In light of the vocabulary and notation introduced in the preceding section, it is here worth- while to repeat Furstenberg's 1955 proof of the innitude of the primes, which was presented in the introduction.

Theorem 3. (Furstenberg's Proof) There are innitely many primes. [3] 17 Proof. Consider Z with the wave topology. Each wave image λeZ is clopen since Z\λeZ = φ φ S S λeZ, so any nite union of wave images is closed. Let C = pZ; then Z\C = ψ e ψ6≡φ (mod λ) p∈P0 {−1, 1} is nite hence clearly not open, so C cannot be closed; therefore P is innite. Denition 10. We dene the factoradic metric by setting d! : Z → R≥0 d!(n, m) = ( 1 : n 6= m N! , where N = max {N ∈ N : N! | (n − m)}; to emphasize the metric, we may 0 : n = m write for the integers considered as a metric space under . Additionally, let denote Z! d! Br(x) the open ball {y ∈ X : d(x, y) < r} of radius r centered at x in a metric space X, and let Br[x] denote the closed ball {y ∈ X : d(x, y) ≤ r}.

Proposition 10. The factoradic metric is an ultrametric on Z; the induced topology is the wave topology (generated by basis Ne/Z), and the closed & open metric balls are given by Z B 1 [x] = B 1 (x) = Nf! . N! (N−1)! x Z

Proof. That d!(n, m) = d!(m, n) is obvious and that d!(n, n) = 0 follows by denition. Now for integers a, b, c, if Nab! is the greatest factorial dividing b−a and Nbc! is the greatest factorial dividing c−b, then write m = min{Nab,Nbc}, M = max{Nab,Nbc} and we have for some as follows: . There- k, j ∈ Z c−a = c−b+b−a = m!k+M!j = m!(k+(m+1)(Nab+2) ··· (M)j) fore m! | (c − a), so the greatest factorial Nac! dividing c − a has Nac ≥ m = min{Nab,Nbc}. 1 1 n 1 1 o Consequently d!(a, c) = ≤ = max , = max {d!(a, b), d!(b, c)}, and Nac! min{Nab,Nbc}! Nab! Nbc! 1 so d! is an ultrametric. The open ball B 1 (n) of radius centered at n ∈ is therefore N! N! Z 1 such that m ∈ Z : d!(n, m) < N! = {m ∈ Z : ∃k ∈ N (N + k)! | (n − m)}

= {m ∈ Z :(N + 1)! | (n − m)} = (N^+ 1)!Z. Had we used a closed ball we would have had n 1 {m ∈ : d!(n, m) ≤ } = Nf! . Since the open balls are a basis, and each open ball is in Z N! n Z Ne/Z, it remains to show only that the remaining elements of Ne/Z are unions of open balls; Z Z (λ−1)!−1 S but this is immediately clear, for λeZ = λe! Z for all waves λe ∈ Ne. φ k=0 φ+kλ φ Z Lemma 1. for some set of , which is unique ∀m ∈ N ∀φ ∈ N0 mf!Z = e2 e3 e4 e5 ··· mZ dk ∈ Z φ d1d2d3d4 dem under the constraint dk ∈ {0, 1, ..., k}.

Proof. e2 e3 e4 e5 ··· m ≡ 3!e e4 e5 ··· m ≡ 4!e e5 ··· m ≡ · · · ≡ mf! . The dk d d d d d e 2!d +d d d de 3!d +2!d +1!d d d e m−1 1 2 3 4 m−1 2 1 3 4 m 3 2 1 4 m−1 P k!dk k=1 exist and are unique because φ has a unique residue modulo 2! (hence a unique d1 ∈ {0, 1}), a unique residue modulo (hence a unique since is already xed), etc. 3! d2 ∈ {0, 1, 2} d1 18 Theorem 4. Put the metric d ({xn}, {yn}) = lim d!(xn, yn) on the set C of Cauchy se- n→∞ quences in , and dene an equivalence relation . Then Z {xn} ∼ {yn} ⇐⇒ d ({xn}, {yn}) = 0 C/ ∼ denes the metric completion Z¯ of Z under the factoradic metric, and the z ∈ Z¯ ∞ P correspond bijectively to innite series of the form n!dn : each dn ∈ {0, ..., n} via the n=1 ∞ N P P correspondence z = n!dn := lim n!dn. Two such series can be added and multiplied n=1 N→∞n=1 up to the Nth term and then truncated to yield the rst N terms of the series corresponding to their sum or product, respectively.

Proof. A sequence is Cauchy precisely if by omitting a nite number of terms from the beginning we can obtain a sequence contained in metric balls of arbitrarily small size; i.e. a sequence in is Cauchy if and only if such that we have {xn} Z! ∀N ∈ N ∃ν ∈ N ∀n > ν . For a particular , let denote the corresponding . Then we see from xn ∈ Nf!Z N φN xν xν Lemma 1 that for each there is a unique sequence with each N > 1 {dn}n∈N dn ∈ {0, ..., n} N−1 P such that φN ≡ n!dn (mod N!). Furthermore, we must have φN+1 ≡ φN (mod N!) ∀N n=1 since , and so if and and Nf!Z ∩ (N^+ 1)! 6= ∅ N,M ∈ N\{1} {dn}n∈{1,..,N−1} {δn}n∈{1,...,M−1} φN φN+1 are the unique sequences corresponding to N and M, respectively, we must have dn = δn N−1 . So x the sequence such that P . Since ∀n < min{N,M} {dn}n∈N ∀N φN ≡ n!dn (mod N!) n=1 the are metric balls with diameters approaching as , if the Nf!Z 0 N → ∞ d ({xn}, {yn}) 6= 0 φN same procedure applied to {yn} does not give the same φN ∀N, and d ({xn}, {yn}) = 0 if it does give the same . Consequently the sequences correspond bijectively to the φN ∀N {dn}n∈N equivalence classes of Cauchy series induced by , and we can take as a canonical ∼ {φN }N∈N representative of the equivalence class containing {xn}. So considering the equivalence classes N as elements of the metric completion ¯, we have P , which Z lim xn = lim φN = lim n!dn n→∞ N→∞ N→∞n=1 yields a canonical representative each z ∈ Z¯. That these series can be added and multiplied via truncation as claimed follows since we can add or multiply their respective φN (say φN ˆ and φN ) and extract the rst N − 1 terms; since for K ≥ N the rst N − 1 terms of the ˆ ˆ series for φK +φK and φK φK are completely determined by arithmetic modulo N!, and since we must have and ˆ ˆ modulo , this operation is well-dened. φK ≡ φN φK ≡ φN N! Denition 11. We dene the ring of factoradic integers to be the metric completion of Z under the factoradic metric and denote it by Z¯, as in Theorem 4. For a given z ∈ Z¯, the ∞ P canonical series for z is the unique series of the form n!dn with each dn ∈ {0, ..., n} which n=1 sums to , and for each we say that is the th factoradic digit (or simply th digit z n ∈ N dn n n when the meaning is clear) of Z¯. Proposition 11. Z¯ has the following properties: 19 (1) Z¯ is a complete, commutative topological ring; (2) N is dense in Z¯; (3) Z¯ has the cardinality of the continuum; (4) Z¯ is totally bounded, totally disconnected, and sequentially compact; (5) The maximum distance between any two points in Z¯ is 1; (6) The inherited topology on Z¯ is T Ne/Z¯ = T Ne/Z¯ , and the open metric balls are Z Z¯ ¯ ¯ B 1 (x) = (N^+ 1)!Z, i.e. the set of z ∈ Z such that the rst N factoradic digits of z N! x are the same as those of x.

Proof. Proceeding for each part:

(1) Z¯ is the metric completion of a commutative topological ring. (2) Every wave image contains positive integers. (3) As a set, Z¯ = {0, 1} × {0, 1, 2} × {0, 1, 2, 3} × .... (4) ¯ is totally bounded since the open metric balls of radius 1 are given by Nf!¯, Z (N−1)! x Z x ∈ {0, ..., N! − 1}, and cover Z¯. Z¯ is totally disconnected for the same reason: given any subset S ⊂ Z¯, choose s ∈ S and N ∈ N large enough that the metric ball Nf!¯ does not contain all of S. Then Nf!¯ is closed as well as open since ¯\Nf!¯ = s Z s Z Z s Z S Nf!Z¯ is an open set, so Nf!Z¯ and Z¯\Nf!Z¯ separate S. Z¯ is closed and totally φ6≡s (mod N!) φ s s bounded, so it is compact; it is a compact metric space, so it is sequentially compact. (5) By the denition of the factoradic metric, the distance is always or 1 for some 0 n! n ∈ N. (6) These claims follow directly from the construction of Z¯ as the metric completion of Z and the continuity of addition and multiplication by constants.

Proposition 12. S ⊂ Z¯ is clopen if and only if S is a nite union of wave images. ∞ Proposition 13. −1 = P n!n. n=1

Proof. Since n!(n + 1) = (n + 1)!1, adding 1 to the proposed canonical series causes every digit to carry, giving 1+1!1+2!2+3!3+... = 1!2+2!2+3!3+... = 2!3+3!3+... = 3!4+... → 0 ∞ P in the limit, so subtract 1 from both sides of 1 + n!n = 0. n=1 Denition 12. Suppose p ∈ . Then we dene p∞0 to be the unique element of ¯ such that ( P Z 0 (mod qn): q = p p∞0 ≡ . If z ∈ Z¯ and P is the set of primes dividing Z¯ (i.e., the set 1 (mod qn): q 6= p of such that ¯) then we dene ∞0 Q ∞0 (where we take a limit if the product is p z ∈ peZ z = p 0 p∈P innite). This is a sound denition, as the following proposition shows. 20 Proposition 14. Let p ∈ P. Then the given denition for p∞0 denes a unique element of Z¯, and products of innitely many such elements are always well-dened factoradic integers invariant under rearrangement of factors.

n−1 Proof. p∞0 := T pn ¯ ∩ T pn ¯ , where {p } is an enumeration of with p = p. e0 Z ek Z k k∈N0 P 0 n∈N 0 k=1 1 Since we always have n+1 ¯ n ¯, this is an intersection of nested closed sets and is hence pgi Z ⊂ pei Z φ φ N n−1 ^N−1 nonempty; moreover, we have for each nite intersection T n ¯ T n ¯ N Q N ¯, pe0 Z ∩ pek Z = p0 pk Z n=1 0 k=1 1 k=1 φN ( N where satises 0 (mod p ) (such exist since by the Chinese φN φN ≡ N φN 1 (mod pk ), k ∈ {1, ..., N} N−1 Remainder Theorem we can take each ). For each , dene N Q N and φN ∈ Z N λN = p0 pk k=1 n−1 we have for the th partial intersection ¯. Now we see that T n ¯ T n ¯ N IN := λfN Z pe0 Z ∩ pek Z φN n∈N 0 k=1 1 ∞ T uniquely denes a factoradic integer as follows: there exists an = In ∀κ ∈ N N ∈ N n=1 such that ∀n > N κ! | λn, since letting π(x) = (the number of primes ≤ x) and pk = π(κ) the th prime) we can write the prime factorization Q mk , and then choose ( k κ! = pk N > k=1 max{π(κ), m1, m2, ..., mπ(κ)}. Thus we need only the rst N intersections to fully determine ∞ T a single value of x (mod κ!) which must hold for all x ∈ In, i.e. the value of the rst n=1 κ−1 P κ − 1 terms i!di of the canonical series for x; since there exists an N for every κ, every i=1 digit of the canonical series is determined by the intersection. So the intersection is not only nonempty, but uniquely determines a single point in Z¯. Though this shows that a unique canonical series will correspond to the intersection, it does not give the canonical series directly. However, we do know which waves of prime-power order do and do not contain p∞0 ; ( 0 (mod qn): q = p in particular, for any q ∈ P and n ∈ N we have p∞0 ≡ . Now for any set 1 (mod qn): q 6= p ( N 0 (mod qn): q ∈ P enumerated by we have Q ∞0 Q ∞0 , P ⊂ P {pn} p = lim pn ≡ n ∀n ∈ N p∈P N→∞n=1 1 (mod q ): q∈ / P which clearly does not depend on the enumeration but rather is entirely determined modulo qn by whether P ∩ {q} = ∅; so the product over innite sets of this form always exists and is invariant under rearrangement of factors. 21 ∼ Q ∞0 Theorem 5. Z¯ = Z¯/p Z¯, where the isomorphism is of rings and the product is a direct p∈P product. Moreover, for each p ∈ P, Z¯/p∞0 Z¯ is isomorphic to the ring of p-adic integers, that is, ¯ ∞0 ¯ ∼ . Z/p Z = Zp ( ∞0 Denition 13. For each , let ¯ be dened by p (mod q ): q = p, where p ∈ P gp ∈ Z gp ≡ q 1 (mod q∞0 ): q 6= p  pn ≡ 0 (mod qN ): q = p, N ≤ n  ranges over . We clearly have n ∞0 since n n N lim gp = p gp ≡ p (mod q ): q = p, N > n P n→∞ 1 (mod qN ): q 6= p gives us n ∞0 n ; therefore we dene ∞ ∞0 ∞0 . gp ≡ p (mod q ) ∀q ∈ P gp = gp = p Proposition 15. Let z ∈ Z¯. Then there exists a unit ω and a unique prime-indexed sequence Q xp {x } of elements of ∪ {∞} such that z = ω gp . p p∈P N0 p∈P Remark. Proposition 15 is a unique factorization theorem, albeit in terms of an innite product. However, the familiar primes from have been replaced by ; for each , {p} N {gp} p is an associate of that is not in . Somewhat surprisingly, given that ¯ arises very gp p Z Z naturally from , we cannot use in place of , because modulo ∞0 for any N P {gp} q q ∈ P\{p} we will not in general have p ≡ 1. Indeed, the values of the primes modulo the powers of other primes vary widely, and consequently the innite product Q pxp will not converge for p∈P n all sequences of natural exponents; and unlike in the case of gp, we cannot rely on lim p to n→∞ converge either (though we do have convergence of n ∞0 ∞ . The are in this limp = p = gp ) gp n→0 sense a naturally privileged choice of prime associates, and the choice in our denition that they be ≡ 1 (mod q∞0 ) for q 6= p is not merely a notational convenience. Proposition 16. The gcd and lcm extend naturally to ¯ and are well-dened up to multi- ( ) Z Q xn,p plication by a unit. If Z = ωn gp is a set of factoradic integers expressed as in p∈ P n∈ Proposition 15, then: N

(1) Q xp where for each we have (and is gcd(Z) = U gp p ∈ P xp = min {xn,p : n ∈ N} ∞ p∈P the largest possible value of xn,p). (2) lcm Q xp where for each we have (and is (Z) = U gp p ∈ P xp = max {xn,p : n ∈ N} ∞ p∈P the largest possible value of xn,p).

4. The Factoradic Rationals

Denition 14. We shall denote by Qˆ the ring of factoradic rational numbers, i.e. the ring −1 −1 of fractions Z¯ Z¯\{zero divisors} . We denote by Q¯ the ring of fractions ZN¯ . ∼ Proposition 17. Q¯ = Z¯ ⊗ Q, where ⊗ denotes the tensor product. Z 22 Proposition 18. Each element of ˆ can be written Q np , where each and Q u p np ∈ Z ∪ {∞0 } p∈P u is a unit in Z¯.

5. The Digit-Flip Function

Proposition 19. Consider For each there is a sequence with each [0, 1] ⊂ R. r ∈ [0, 1] {dn} ∞ such that P dn . Moreover, the sequence corresponding to is dn ∈ {0, ..., n} r = (n+1)! {dn} r n=1 unique unless , in which case thare are exactly two corresponding sequences r ∈ Q\{0, 1} {dn} and {δn}, which are such that the following hold:

(1) d is zero except at a nite set of values of n; n  d : n < N  n (2) Let . Then . N = max{N ∈ N : dN 6= 0} δn = dn − 1 : n = N n : n > N

Denition 15. The digit-ip function is the function DF : Z¯ → [0, 1] dened on canonical ∞ ∞ series by P P dn . DF n!dn = (n+1)! n=1 n=1 Proposition 20. The digit-ip function is bijective restricted to

Z¯\ (Z\{−1, 0}) → [0, 1]\ (Q\{0, 1}), and is 2-to-1 restricted to Z\{−1, 0} → Q ∩ (0, 1). We have DF (−1) = 1 and DF (0) = 0.

Proof. This follows since the only elements of [0, 1] with two representations are the elements  d : n < N  n of , whose digit sequences are given by Proposition 19 as , Q\{0, 1} δn = dn − 1 : n = N n : n > N where dn is the nth digit of the terminating digit sequence and δn is the nth digit of the non- terminating equivalent, and . The two preimages of a rational N = max{N ∈ N : dN 6= 0} number in (0, 1) under the digit-ip function are thus a factoradic integer with terminating digit sequence (i.e. an element of ) and a factoradic integer with digit sequence terminating N0 in maximal digits (i.e. an element of . Clearly we have dn = n Z\N0) DF (Z) ⊂ (Q ∩ [0, 1]) as well as (Q ∩ [0, 1]) ⊂ DF (Z), so DF (Z) = (Q ∩ [0, 1]). 0 ∈ (Q ∩ [0, 1]) has the single ∞ ∞ representation P 0 and hence the single preimage P ¯, and has the single (n+1)! n!0 = 0 ∈ Z 1 n=1 n=1 ∞ ∞ representation P n and hence the single preimage P , but every other rational (n+1)! n!n = −1 n=1 n=1 in [0, 1] has two representations and hence two preimages. ∞ Lemma 2. −(n!) = P k!k. k=n 23 ∞ ∞ ∞ ∞ Proof. n! + P k!k = n!(1 + n) + P k!k = (n + 1)! + P k!k = ... = (n + m)! + P k!k k=n k=n+1 k=n+1 k=n+m for any m ∈ N. This goes to 0 as m → ∞ and so must be equal to 0; therefore −(n!) = ∞ P k!k. k=n

∞ Proposition 21. If P dn , each , is the terminating representation of an (n+1)! dn ∈ {0, ..., n} n=1 element of , let . Then −1 where x Q ∩ (0, 1) N := max {N ∈ N : dN 6= 0} DF (x) = {ν, τ} N P τ = n!dn and ν = τ − N!(N + 2). n=1

N N−1 ∞ Proof. The two representations of are P dn and P dn dN −1 P n , and x (n+1)! (n+1)! + (N+1)! + (n+1)! n=1 n=1 n=N+1 N P consequently the preimages of x under the digit-ip function are τ = n!dn and ν = n=1 N−1 ∞ ∞ P P P n!dn + N!(dN − 1) + n!n. This gives ν − τ = n!n − N!, which by Lemma n=1 n=N+1 n=N+1 2 is equal to −(N + 1)! − N! = −N!(N + 1 + 1) = −N!(N + 2). Proposition 22. The digit-ip function is continuous.

Proof. Metric balls of factoradic integers correspond to sets of factoradic integers sharing the same rst n digits, where n → ∞ as the radius of the metric ball goes to 0. For any z, ζ ∈ Z¯ sharing their rst n digits we have that DF (z) and DF (ζ) also share their rst n digits, and therefore n+1 n+2 1 , so is continuous. |DF (z) − DF (ζ)| ≤ (n+2)! + (n+3)! + ... = (n+1)! DF Proposition 23. Let ∼ be the equivalence relation which diers from identity only to equate the two digit-ip preimages of each rational in (0, 1). The inverse of the digit-ip function is continuous considered as a function [0, 1] → (Z¯/ ∼) with the quotient topology inherited from Z¯.

Proof. Like the previous proposition, this follows from the fact that in both [0, 1]\ (Q\{0, 1}) and ¯ a sequence approaches a point if and only if such that if Z\ (Z\{−1, 0}) {xn} x ∀N ∃K then the rst digits of agree with the rst digits of . For points in n > K N xn N x Q ∩ (0, 1) the two digit sequences representing a point correspond to the two points making up a given equivalence class in Z¯/ ∼, so any sequence of elements of [0, 1] approaching a chosen rational ¯ x will give a sequence via digit-ip approaching an element of Z/ ∼. Theorem 6. Under their respective subspace topologies, Z¯\Z is homeomorphic to [0, 1]\Q. More strongly, Z¯\ (Z\{−1, 0}) is homeomorphic to [0, 1]\ (Q\{0, 1}).

Proof. DF is a homeomorphism by the preceding propositions. 24 6. Sequences & Series

6.1. Innite Series.

∞ Proposition 24. Let {x } be a sequence in ¯. Then P x converges if and only if n n∈N Z n n=1 lim xn = 0. n→∞

∞ P Proof. For each n write the canonical series xn = k!dn,k, and suppose lim xn = 0. Then k=1 n→∞ we must have for each k only nitely many n with dn,k 6= 0; if we dene for each k the K value and write the partial sums P , we Nk = max{N ∈ N : dN,k 6= 0} SK = k!dn,k k=1 therefore have for all . Therefore SK ≡ Smax{Nk:k≤K} (mod (K + 1)!) K ≥ max{Nk : k ≤ K} we have , i.e. 1 . ∀n, m ≥ max{Nk : k ≤ K} Sn − Sm ≡ 0 (mod (K + 1)!) d(Sn,Sm) ≤ (K+1)! So {S } is Cauchy, and the series converges. For the other half of the claim, suppose n n∈N lim xn 6= 0. Then since xn = Sn − Sn−1 (setting S0 = 0), lim Sn − lim Sn−1 6= 0; but if the n→∞ n→∞ n→∞ converged to both of these limits would be , and we have a contradiction. Sn L L ¯ Proposition 25. Let {xn} be a sequence in with lim xn = 0, and let σ be any per- n∈N Z n→∞ ∞ ∞ mutation of . Then P x = P x . Furthermore, let {x } be a sequence in ¯ with N σn n n n∈N Z n=1 n=1 P lim xn = 0, and let {Kn}n∈ be a sequence of sets that partition , and let xk denote n→∞ N N k∈Kn M ∞ ∞ P P P P lim yn,i for {yn,i} some xed enumeration of Kn. Then xk = xn. M→∞ i∈N i=1 n=1k∈Kn n=1

∞ P Proof. Since xn → 0, the Nth factoradic digit of xn depends on only the nitely many n=1 terms of {xn} that have xn 6≡ 0 (mod (N + 1)!). Under any permutation of the sequence the Nth digit of the sum will still depend only on this same nite set of numbers, and ∞ ∞ P P therefore xσn converges to a sum with the same canonical series as xn, as claimed. n=1 n=1 Similarly, if is any sequence of sets that partion , then the elements of give {Kn}n∈ N {Kn} a disjoint cover of theN nite set , and since this applies S = {n ∈ N : xn 6≡ 0 (mod (N + 1)!)} ∞ and therefore we obtain the same factoradic digit sequence for both sums, P P ∀N ∈ N xk n=1k∈Kn ∞ converges to the same value as P regardless of the order of summation over the . xn Kn n=1 P Corollary. If S is a countable subset of Z¯ and the only limit point of S is 0, then s con- s∈S verges to a well-dened factoradic integer z invariant under all rearrangements and groupings of terms. 25 Corollary. In particular, if and is a bijection 2 , lim xn = 0 [a, b] {(a, b) ∈ 0 : a ≥ b} → n→∞ N N ∞ a ∞ ∞ P P P P then x[a,b] = x[a,b]. a=0b=0 b=0a=b Proposition 26. Let {f } be a sequence of continuous functions ¯ → ¯ such that f n n∈N0 Z Z n ∞ converges uniformly to the zero-function as , and let ¯. Then P is a n → ∞ z ∈ Z fn(x) n=0 continuous function Z¯ → Z¯.

Proof. The series converges for any x since the fn converge to 0, so the claim is equiva- ∞ ∞ ∞ ∞ lent to the assertion that ¯ P P . Now P P ∀z ∈ Z lim fn(x) = fn(z) lim fn(x) − fn(z) x→zn=0 n=0 x→zn=0 n=0 ∞ P . Note that 1 . Given , let = lim (fn(x) − fn(z)) d(x, y) ≤ m! ⇐⇒ m! | (y − x) m! N ∈ N x→zn=0 such that ∀n > N m! | fn(x) ∀x, which is possible since fn → 0 uniformly. Then modulo ∞ N P P m!, (fn(x) − fn(z)) = (fn(x) − fn(z)). So choose n=0 n=0 , and ¯ we have δ = max {δ ∈ N : δ! | (z − x) −→ m! | (fn(z) − fn(x))} ∀x ∈ δe!Z fn(x) − n∈N z N P fn(z) ≡ 0 (mod m) ∀n ∈ {0, ..., N} and hence 0 ≡ (fn(x) − fn(z)) n=0 ∞ ∞ P . In other words, 1 implies P ≡ (fn(x) − fn(z)) (mod m) d(x, z) ≤ δ! d(0, (fn(x) − fn(z))) ≤ n=0 n=0 ∞ ∞ ∞ . Therefore we have P , so P P . lim (fn(x) − fn(z)) = 0 lim fn(x) = fn(z) x→zn=0 x→zn=0 n=0

6.2. Factoradic Limits. Denition 16. Let be a metric space and let be a sequence in . We say X {xn}n∈ X {xn} ¯N converges factoradically to L ∈ X as n approaches z ∈ , and write limxn = L, if limxn = L Z n→z n→z when is considered a function ¯ . That is, if such that xn xn : N! ⊂ Z → X ∀ > 0 ∃δ > 0 , or equivalently, such that ∀n ∈ N 0 < dZ¯ (n, z) < δ −→ dX (xn,L) < ∀ > 0 ∃N ∈ N ¯ ∀n ∈ Nf! dX (xn,L) < . z Z Proposition 27. Let X be a metric space and let {x } be a sequence in X, let L ∈ X, n n∈N and let ¯. Then if and only if for every subsequence z ∈ Z limxn = L lim xnk = L {xnk } n→z k→∞ k∈N with lim nk = z. k→∞

Proof. This follows immediately from the denition. Theorem 7. Suppose X is a metric space, L ∈ X, and {x } is a sequence in X. Then n n∈N if and only if ¯. lim xn = L limxn = L ∀z ∈ Z n→∞ n→z 26 Proof. lim xn = L implies that every subsequence of {xn} converges to L, and so in particular n→∞ ¯ ¯ any subsequence {xn } where nk → z ∈ converges to L; thus limxn = L ∀z ∈ . In k Z n→z Z ¯ ¯ the other direction, suppose limxn = L ∀z ∈ . Since is sequentially compact, every n→z Z Z sequence {n } in ¯ has a convergent subsequence; thus every subsequence {x } of k k∈N Z nk k∈N {xn} has a subsequence converging to L. Now suppose lim xn 6= L; then there exists n→∞ some subsequence {x } of {x } and some > 0 such that d (x ,L) > ∀k. But then nk k∈N n X nk some subsequence of converges factoradically, and therefore the corresponding nkj j∈ {nk} n o N subsequence x of {x } converges to L. But this is a contradiction, for d (x ,L) > nkj nk X nk j∈N ∀k; therefore lim xn = L. n→∞ Denition 17. Suppose {x } is a sequence in a metric space X. If Z ⊂ ¯ and ∀z ∈ Z n n∈N Z limxn exists, we say {xn} is convergent on Z. Moreover: n→z

• If {xn} converges on with lim xn = xN , we say {xn} is consistent. Equivalently, N n→N is consistent if the function ¯ dened by is continuous. {xn} f : N! → Z f(n) = xn If converges on ¯, we say is universally convergent. • {xn} Z {xn} Let ¯ be the maximal set whereupon is convergent; then the function • Z ⊂ Z {xn} f : Z → X induced by {xn} is dened by f(z) = limxn. n→z Proposition 28. If {x } is a sequence in a metric space X convergent on Z ⊂ ¯, then n n∈N Z the induced function f(z) = limxn is continuous Z → X. n→z

Proof. Let z ∈ Z and let {z } be a sequence in Z converging strictly monotonically n n∈N to z as n → ∞, i.e. m > n implies d!(zn, z) > d!(zm, z). (If this is not possible, z is an isolated point so f(z) satises the denition of continuity vacuously.) For each n let {m } be a sequence in converging to n as k → ∞ and let N ∈ such that ∀k ≥ N n,k k∈N N n N 1 . Then and is convergent on so by denition d(f(zn), xm ) < n lim mn,N = z ∈ Z {xn} Z n,k 2 n→∞ n converges to . Therefore , since 1 lim xm f(z) lim f(zn) = f(z) d(xm,N , f(zn)) < n → 0 n→∞ n,Nn n→∞ n 2 as n → ∞.

6.3. Factoradic Sequences in Z¯. Remark. For convenience, let us rephrase the convergence condition in terms of , δ ∈ N ¯ ¯ rather than , δ ∈ >0: given a sequence {xn} in and L ∈ , limxn = L if and only if R n∈N Z Z n→z ¯ ¯ ∀ ∈ ∃δ ∈ such that n ∈ δe! −→ xn ∈ e! . N N z Z L Z Denition 18. Suppose is a sequence in ¯. We say that is modularly even- {xn}n∈ Z {xn} tually periodic (MEP) if N such that . If ∀κ ∈ N ∃λ, N ∈ N ∀n ≥ N xn ≡ xn+λ (mod κ) {xn} satises the stronger condition that such that , ∀κ ∈ N ∃λ ∈ N ∀n ∈ N xn ≡ xn+λ (mod κ) we say that {xn} is modularly periodic (MP). 27 Proposition 29. A sequence {x } in ¯ is modularly periodic if and only if it is periodic n n∈N Z modulo pm ∀p ∈ P, m ∈ N.

Proof. The Chinese Remainder Theorem says that for any κ ∈ N, Z/κZ is a nite direct product of rings Z/pmZ over some set of prime powers pm, so the least common multiple of the periods modulo pm suces as a period modulo κ. Thus periodicity modulo each pm implies periodicity modulo every natural number; likewise, periodicity modulo every natural m number implies periodicity modulo each p , so the two conditions are equivalent.

Proposition 30. Let {x } be a sequence in ¯ and let f : → ¯ be dened by f(n) = x . n n∈N Z N! Z n Then f is uniformly continuous if and only if {xn} is modularly periodic.

Proof. If f is uniformly continuous then ∀ ∈ N ∃δ ∈ N such that n ≡ m (mod δ!) =⇒ ; thus , so is modularly periodic. xn ≡ xm (mod !) ∀κ ∈ N ∀n ∈ N xn ≡ xn+δ! (mod κ) {xn} If, on the other hand, is modularly periodic, then such that {xn} ∀κ ∈ N ∃λ ∈ N ∀n ∈ N , so such that xn ≡ xn+λ (mod κ) ∀ ∈ N ∃λ ∈ N n ≡ m (mod λ) =⇒ xn ≡ xm (mod !) ⇐⇒ f(n) ≡ f(m)(mod !); take δ! = λ!, and f is uniformly continuous.

Proposition 31. Let be a universally convergent sequence in ¯. Then induces {xn}n∈ Z {xn} a uniformly continuous functionN if and only if it is MEP.

Proof. Let {y } be dened by y = f(n), where f : ¯ → ¯ is the function induced by n n∈N n Z Z {xn}. If is uniformly continuous then is uniformly continuous as a function ¯, and f {yn} N! → Z therefore is a modularly periodic sequence inducing f by Proposition 30. Thus the sequence {z } , where z = x − y ∀n, induces the zero function; so modulo κ ∈ the sequence n n∈N n n n N {xn} is identical to the sequence {yn} except at a nite number of indices (otherwise there would be an innite set of naturals n with zn = xn − yn taking on a particular value modulo , and this set of naturals would have a limit point in ¯, so the function induced by κ Z {zn} could not be universally 0). Since {yn} is MP, it follows that {xn} is MEP.

In the other direction, suppose {xn} is MEP. Then given a prime p and a natural N, modulo pN the sequence {x } is MP except at a nite number of points; so let {y } be a n N,n n∈N sequence which is MP and has for each that N for all but a nite p ∈ P yN,n ≡ xn (mod p ) N number of n. Clearly yn := lim yN,n exists for each n since yN+1,n ≡ yN,n (mod p ) ∀p ∈ , N→∞ P N ∈ ; thus we have a sequence {y } which is MP, and since {z := x − y } induces N n n∈N n n n n∈N the zero function (because , we have N at all but nitely many ∀p ∈ P N ∈ N xn ≡ yn (mod p ) n) it follows that {xn} induces the same function f as does the MP sequence {yn}, and so f must be uniformly continuous. 28 7. Exponentiation

∞ Considering −1 = P n!n in light of exponentiation, the immediate question is whether for a n=1 ∞ P n!n unit u we have the multiplicative inverse u−1 of u equal to lim un=1 . Happily, the answer N→∞ K is a resounding yes: xing a modulus κ ∈ , and writing the prime factorization κ = Q pxk N nk k=1 (where is a nite subsequence of ), we have ¯ ¯ {pnk } {pn} = {2, 3, 5, 7, 11, 13, ...} Z/NZ = Z/NZ K ∀N ∈ and we have by the Chinese Remainder Theorem ¯/κ¯ ∼ Q ¯/pxk ¯. Therefore N Z Z = Z nk Z k=1 K × × × ¯/κ¯ ∼ Q ¯/pxk ¯ , and a multiplicative group of units ¯/pxk ¯ with p prime is Z Z = Z nk Z Z nk Z nk k=1 ∞ cyclic of order φ(pxk ) = (p − 1)pxk−1, where φ is Euler's totient function. Since P n!n ≡ nk nk nk n=1 m−1 P n!n ≡ −1 (mod m) for every m ∈ N (because adding 1 will give 0 through digit carry) n=1 ∞ P n!n this holds in particular for xk−1 for each . Thus n=1 −1 xk for (pnk − 1)pn k lim u ≡ u (mod pn ) k N→∞ k ∞ P n!n each k and therefore lim un=1 ≡ u−1 (mod κ). Since this holds ∀κ it holds in particular N→∞ for κ any factorial, dening a unique canonical series. We expand this argument to dene general exponentition with factoradic exponents. To preserve the identity z0 = 1 ∀z ∈ Z¯ while also easily representing the quantity limzn, it will be necessary to make use of the n→0 symbol ∞0 . Denition 19. Suppose b ∈ ¯, x ∈ ¯. Then we dene b∞0 +x = limbn. We also dene bx to Z Z n→x be the product of b with itself x times if x ∈ N, or 1 if x = 0. When b is 0 or a unit modulo p∞0 for each prime p, we may simply write bx rather than b∞0 +x since, as shall be shown, the ( 0 : z = 0 denitions coincide. We dene ∞0 ∞0 = ∞0 , ∞0 + ∞0 = ∞0 , and ∀z ∈ Z¯ ∞0 z = . ∞0 : z 6= 0

Proposition 32. ∀b, x ∈ Z¯ the limit b∞0 +x is well-dened. Moreover, we have ∀b ∈ Z¯ that if ¯ then: x, y ∈ N0 ∪ z + ∞0 : z ∈ Z (1) 1x = 1 (2) bxby = bx+y (3) (bx)y = bxy

Proof. Modulo pN where p is prime, either b is a unit hence bn is modularly periodic hence uniformly continuous, or b is pku for some unit u, in which case lim bn ≡ 0 (mod p∞0 ), so n→∞ ∞0 +x is well-dened . Letting ¯ , we have: b ∀x x, y ∈ N0 ∪ z + ∞0 : z ∈ Z 29 (1) 1n = 1 ∀n ∈ N (2) If we have the result. Without loss of generality, if , x, y ∈ N0 x ∈ N0 y = ∞0 + υ then bxby = bx limbn = limbx+n = b∞0 +x+υ = bx+y. If x = ξ + ∞0 and y = υ + ∞0 , then n→υ n→υ bxby = limbn lim bm = lim lim bn+m = lim bn+m = b∞0 +ξ+υ = bx+y. n→ξ m→υ n→ξm→υ k→ξ+υ (3) If either or is then from (1) we get x y xy . If we have the x y 0 (b ) = b = 1 x, y ∈ N0 result. If x ∈ , y = ∞0 +υ then (bx)y = limbnx = bxυ+∞0 , and xy = x(υ+∞0 ) = xυ+∞0 . N n→υ If x = ∞0 + ξ, y ∈ N then (bx)y = limbny = bξy+∞0 = bxy. If x = ∞0 + ξ and y = ∞0 + υ n→ξ then (bx)y = lim lim bnm = bxy. n→ξm→υ

Proposition 33. ¯ the function ¯ ¯ dened by ∞0 +x is uniformly ∀x ∈ Z fx : Z → Z fx(z) = z continuous. ¯ the function ¯ ¯ dened by 0∞ +z is uniformly continuous. ∀b ∈ Z gb : Z → Z gb(z) = b

n n Proof. fx(z) = limfn(z) along n ∈ , and ∀n ∈ fn(z + N!k) − fn(z) = (z + N!k) − z is n→x N N ¯ ¯ ¯ in Nf! since every term will be divisible by N!k. Nf! is closed, so fx(z) ∈ Nf! , i.e. we can 0 Z 0 Z 0 Z take δ = and we have uniform continuity. As for gb(z), it is uniformly continuous because {g (n)} is modularly periodic; to see this, observe that for any chosen p ∈ and m ∈ b n∈N P N ( m ∞0 +n 0 (mod p ): p | b we will have ∀n ∈ N b ≡ , and both possibilities are periodic. bn (mod pm): p - b

8. The Units

8.1. Units and Zero-Divisors.

Theorem 8. Let z ∈ Z¯. Then:

(1) z is a zero-divisor if and only if p∞0 | z for some p ∈ P. (2) z is a unit if and only if p - z ∀p ∈ P.

Proof. If no p∞0 divides z, then xz = 0 implies x is divisible by p∞0 for every p ∈ P, i.e. x = 0; if p∞0 divides z, then (1 − p∞0 )z = 0. If p | z then z is not a unit since z ≡ 0 (mod p) and 0x is never 1 modulo p; if no prime divides z, then z is unit modulo every p∞0 and is therefore a unit. Denition 20. Let U denote the set of units in Z¯, and let D denote the set of zero-divisors. For any ¯ let be the unique unit and Q xp the unique product of values such z ∈ Z υz σz = gp gp p∈P ∞0 that z = υzσz (taking υz ≡ 1 (mod p ) where xp = ∞ to preserve uniqueness, since υz is actually unconstrained when xp = ∞). Then υz is called the unitary part of z and is written upart(z), and σz is called the zero signature of z and is written zsig(z). Observing that 30 ( ( 0 (mod p∞0 ): p | z 1 (mod p∞0 ): p | z z∞0 ≡ and 1 − z∞0 ≡ , we call z∞0 z the coprime 1 (mod p∞0 ): p - z 0 (mod p∞0 ): p - z part and (1 − z∞0 )z the noncoprime part of z. Note that z∞0 z + (1 − z∞0 )z = z. Proposition 34. z ∈ Z¯ is idempotent (i.e. z satises z2 = z and hence zx = z∞0 +x = z ∀x ∈ Z¯\{0}) if and only if z = ζ∞0 for some ζ ∈ Z¯.

Proof. z2 −z = z(z −1) ≡ 0 (mod p∞0 ). So either z ≡ 0 or z ≡ 1, since if neither z nor z −1 is 0, neither is divisible by p∞0 , so their product is divisible by at most a nite power of p, and ∞0 Q ∞0 ∞0 therefore nonzero. Letting P = {p ∈ P : z ≡ 0 (mod p )}, we thus have z = p = ζ for p∈P any ζ which is divisible by the primes in P and by no other primes. Proposition 35. z ∈ Z¯ is a zero-divisor if and only if ∃p ∈ P such that p∞0 | z. and yz = 0 if and only if ∀p ∈ P p∞0 | x or p∞0 | y.

Proof. If z is not divisible by any p∞0 , and y 6= 0, then y is divisible by at most a nite power of p for some p ∈ P, so zy 6≡ 0 (mod p∞0 ) hence zy 6= 0, so z is not a zero divisor. On the Q ∞0 ∞0 other hand, since p = 0, if P is the set of primes p with p dividing x, and Q = P\P , p∈P then for y divisible by q∞0 for every q ∈ Q we have xy = 0, and for y not divisible by some ∞0 ∞0 q with q ∈ Q we have xy 6≡ 0 (mod q ) hence xy 6= 0, which establishes the claim. Proposition 36. λeZ¯ ∈ Ne/Z¯ contains a unit if and only if λe is a coprime wave, i.e. φ Z¯ φ gcd(λ, φ) = 1. A coprime wave image contains innitely many units.

Proof. If λe is not coprime then any prime dividing gcd(λ, φ) divides each element of λeZ¯, so φ φ ( φ + λz (mod λ∞0 ) in particular λeZ¯ must be unit-free. If λe is coprime, x ≡ is a unit φ φ ψ (mod (1 − λ∞0 ) in λeZ¯ for any z ∈ Z¯ and ψ ∈ U. φ Proposition 37. U is perfect, closed, and compact.

Proof. ¯ S ¯, so is closed. ¯ is compact, so closed sets are compact. Moreover, U = Z\ peZ U Z p∈P0 every neighborhood of a unit contains a coprime wave, and every coprime wave contains innitely many units, so every point of U is a limit point of U. Therefore U is perfect. Theorem. (Dirichlet's Theorem) P¯ = P ∪ U. [6] Proof. The statement is equivalent to Dirichlet's Theorem, since each coprime wave contains innitely many primes if and only if each coprime wave contains a prime if and only if U ⊂ P¯, and has no limit points outside of since any sequence of primes has P U {pn} pN ≡ 0 (mod q) ⇐⇒q = pN =⇒ pn 6≡ 0 (mod q) ∀n > N, and therefore no sequence of primes converges to a limit divisible by any prime. 31 8.2. Topological Z¯-Modules and the Multiplicative Group of Units. Denition 21. LetR be a topological ring, and let M be an R-module. We say M is a topological R-module if M is endowed with a topology under which addition M × M → M and scalar multiplication R × M → M are continuous (where the domains are given their respective product topologies). M is a torsion R-module (or simply torsion) is ∀m ∈ M ∃r ∈ R\{0} such that r · m = 0.

Lemma 3. Let G be a torsion topological Z-module, i.e. ∀g ∈ G ∃n ∈ Z\{0} such that |g| = n, i.e. n is minimal such that n · g = 0. Then G is a Z¯-module under the action z · g = (z mod |g|) · g.

Proof. We verify the axioms: (1) z·(a+b) = (z mod |a+b|)·(a+b) = (z mod |a+b|)·a+(z mod |a+b|)·b. The order of a+b must be divisible by the orders of a and b so this is = (z mod |x|) · x + (z mod |y|) · y = z · x + z · y. (2) (z + ζ) · a = r · a + ρ · a since z + ζ + |a|Z¯ = z + |a|Z¯ + ζ + |a|Z¯, i.e. since the natural map Z¯ → Z¯/|a|Z¯ is an additive homomorphism. (3) (zζ) · a = z · (ζ · a) since the natural map Z¯ → Z¯/|a|Z¯ is a multiplicative homomorphism. (4) 1 · a = a holds since (1 mod |a|) · a = a.

Denition 22. Let M be a topological Z¯-module. Then we dene |M| = lcm {|m| : m ∈ M}, ( gcd{z ∈ ¯ : z · m = 0} : ∃z ∈ ¯\{0} with z · m = 0 where |m| = Z Z and the lcm of any set ∞ : otherwise including ∞ is ∞. Both |M| and |m| are dened only up to their zero-signatures, so |M| and |m| are not elements of {∞} ∪ Z¯ but of {∞} ∪ Z¯/U = {∞} ∪ {zU : z ∈ Z¯}. We say |m| is the order of m and |M| is the elemental order of M. Note that we have |M| = ∞ if M contains any element which is not Z¯-torsion, and |M| = 0 if every element of M is Z¯-torsion but ∀z ∈ Z¯\{0} ∃m ∈ M such that mz 6= 1. If |M| = ∞ we say M is of non-factoradic elemental order, and otherwise |M| ∈ Z¯ and we say M is of factoradic elemental order; so a Z¯-module of factoradic elemental order is precisely a torsion Z¯-module. If M is a topological Z¯-module of factoradic elemental order we say M is a factoradically nite group. If there is some g ∈ M such that ∀m ∈ M ∃z ∈ Z¯ such that m = z · g, then we say M is cyclic and write M =< g >= z · g : z ∈ Z¯ . Proposition 38. Let M be a factoradically nite group. Then ∀m ∈ M, |m| · m = 0.

Proof. Write |m| = µU for µ a zero signature, and let X = x ∈ Z¯ : x · m = 0 . Then by denition . For each let denote the element in divisible by the µU = gcd(X) p ∈ P xp X same maximal power of p as is µ (where ∞0 is considered the largest; such an element must P ∞0 ∞0 exist since µU = gcd(X)). Then (1−p )xp is ≡ xp (mod p ) for each p and so is divisible p∈P P ∞0 by the same maximal power of each prime as is µ, i.e. (1 − p )xp = µω for some unit ω. p∈ 32 P ! −1 −1 P ∞0 −1 P Then for any unit u, µu · m = uω · (µω · m) = uω · (1 − p )xp · m = uω · 0 p∈P p∈P = 0. Thus |m| · m = 0. Lemma 4. Let M =< g > be a cyclic factoradically nite group, and let d ∈ Z¯/U. If d - |M| then ∀m ∈ M |m|= 6 d, and if d | |M| then:

• If for some p ∈ P p∞0 | |M| and p∞0 - d, then no m ∈ M has |m| = d; • Otherwise,∃m ∈ M such that |m| = d.

Proof. If d - |M| then no element has order dU since |M| is the least common multiple of the orders. On the other hand, if d | |M| = |g|, choose c ∈ Z¯/U such that cd = |g|. Take m = c · g; then for n ∈ Z¯/U, n · m = 0 ⇐⇒ nc · g = 0 ⇐⇒ |g| | nc. Now choosing nU = dU we have |g| | nd. For the rst case, suppose there is some prime p such that p∞0 | |M| but p∞0 - d; then since cd = |M| and d is divisible by a nite maximal power of p, we have p∞0 | c.. Thus if δ is divisible by the same maximal powers of the same primes as d except that p - δ, δc = dc so δ · m = δc · g = 0, and d is not the order of m hence not the order of any element of M. For the second case, suppose there is no such prime, i.e. ∀p ∈ P either p∞0 - |M| or p∞0 | d. Then we are free to choose c such that c is divisible by only a nite maximal power of each prime, since if |M| = cd is divisible by p∞0 then so is d. Then d · m = 0; moreover, for any δ dividing d with δ 6= d, either for some prime p we have p∞0 | d but p∞0 - δ, in which case p∞0 - c so p∞0 | |M| but p∞0 - δc and hence δ · m = δc · g 6= 0, or some prime p divides d with a greater nite maximal power than δ, and divides c only to a nite power, so δc is a proper divisor of |M| and δ · m 6= 0. Therefore d = gcd(δ : δ · m = 0) = |m|, and both claims are established. Proposition 39. ∀κ ∈ Z¯ the additive group Z¯/κZ¯ is a cyclic factoradically nite group of order κ.

Proof. Z¯ acts naturally on Z¯/κZ¯ via z · (ζ + κZ¯) = zζ + κZ¯. Thus Z¯ · (1 + κZ¯) = Z¯/κZ¯ and ¯ ¯ z · (1 + κZ) = 0 + κZ if and only if κ | z, as claimed. Theorem 9. (Multiplicative Modular Groups) For all odd primes p and n ∈ N ∪ {∞0 }, n × n−1 z Z¯/p Z¯ is a cyclic factoradically nite group of order (p−1)p under the action z·u = u . (In the case of n = ∞0 , we have equivalently a group of order (p − 1)p∞0 −1 or (p − 1)p∞0 , since n × ∼ + n−2 + these are associates). For all n ∈ N ∪ {∞0 }, Z¯/2 Z¯ = Z¯/2Z¯ × Z¯/2 Z¯ unless × n = 1 in which case Z¯/2Z¯ is the trivial group {1}.

n × n Proof. Let p be an odd prime; then for all n ∈ N, Z¯/p Z¯ is cyclic of order φ(p ) = n−1 (so the result holds for ), and if is a primitive root modulo 2, it is (p − 1)p n ∈ N γp p also a primitive root modulo n . So x ¯ ∞0 ¯ such that is a primitive root p ∀n γp ∈ Z/p Z γp modulo 2. Now xn xm mod n if and only if mod n−1 , so x y p γp ≡ γp ( p ) xn ≡ xm ( (p − 1)p ) γp = γp if and only if x ≡ y (mod (p − 1)pn) ∀n ∈ N, i.e. if and only if x ≡ y (mod (p − 1)p∞0 ). ¯ Furthermore, Z contains all units: if ¯ ∞0 ¯ write xn mod n We must have γp u ∈ Z/p Z u ≡ γp ( p ). 33 ( x :(mod p − 1) x ≡ x (mod (p − 1)pn−1) ∀n, so in particular if ξ ≡ 1 and we write n+1 n p 1 : (mod p∞0 ) ( 1 : (mod p − 1) for some , then mod n−1 so there is xn = ξpΞp,n Ξp,n ≡ n−1 Ξp,n+1 ≡ Ξn ( p ) ∀n xn :(mod p ) n ξpΞp a p-adic Ξp with Ξp ≡ Ξp,n (mod p ) ∀n, and we have u = γp . This establishes the claim for odd primes. For , choose a primitive root such that and generate n ¯× p = 2 γ2 γ2 −1 Z/2 Z n × for all n. Then by the same argument as for odd primes, every element of Z/2 Z¯ is of the form 0 x or 1 x for some ¯ ∞0 ¯, and no two such expressions give the (−1) γ2 (−1) γ2 x ∈ Z/2 Z same element since must have order ∞0 . γ2 2 ( some primitive root (mod q∞0 ): q = p Denition 23. For each odd prime p, let γ ≡ , p 1 (mod q∞0 ): q 6= p ( −1 (mod q∞0 ): q = 2 where q ranges over the primes, let τ ≡ , and let γ be ≡ 1 (mod q∞0 ) 1 (mod q∞0 ): q 6= 2 2 for q 6= 2 and have multiplicative order 2∞0 U modulo 2∞0 , as in Theorem 9. Then we say is a choice of primitive roots, and we will freely use the symbol where its meaning {γp}p∈ τ is clear.P

Lemma 5. (Unit Factorization Lemma) Given a choice of primitive roots {γ } , for p p∈P each there is a unique choice of in ¯ ¯ ¯ ∞0 ¯ ω ∈ U (ξ2, Ξ2, ξ3, Ξ3, ξ5, Ξ5, ...) Z/2Z × Z/2 Z × ξ Ξ Q ¯ ¯ ¯ ∞0 ¯ such that ξ2 Ξ2 Q p p , where the and are in- Z/(p − 1)Z × Z/p Z ω = τ γ2 γp ξp Ξp p∈P\{2} p∈P\{2} ( ξ (mod p − 1) terpreted as factoradic integers by taking any representatives such that ξ ≡ p p 1 (mod p∞0 ) ( mod and 1 ( p − 1) for odd , and taking any representatives at all for , . In Ξp ≡ ∞0 p ξ2 Ξ2 Ξp (mod p ) Q particular, U under multiplication is therefore isomorphic to Z¯ × Z¯/2Z¯ × Z¯/(p − 1)Z¯ p∈P under addition.

Proof. This follows directly from Theorem 9.

Theorem 10. (Unique Factorization Theorem) For each p ∈ P let ¯ ¯ ¯ ∞0 ¯ and dene addition on componentwise, Xp = {∞} ∪ N0 × Z/(p − 1)Z × Z/p Z Xp with ∞ + x = ∞ ∀x. Then Z¯ under multiplication is isomorphic to the cartesian product Q under componentwise addition. That is, for each ¯, given a choice of primitive Xp z ∈ Z p∈P x ξ Ξ roots there is a unique factorization x2 ξ2 Ξ2 Q p p p with the th (γ2, γ3, γ5, ...) z = g2 τ γ2 gp γp p p∈ Q P component of the corresponding element of Xp given by (xp, ξp, Ξp) ∈ Xp, with the caveat 34 x ξ Ξ that if we must interpret p p p ∞0 for odd primes or x2 ξ2 Ξ2 ∞0 . (xp, ξp, Ξp) = ∞ gp γp := p g2 τ γ2 := 2 ( mod The product is to be interpreted as ξp ( p − 1). ξpΞp ≡ ∞0 Ξp (mod p )

Proof. This follows from Theorem 9 and Proposition 15. Theorem 11. (p − adic Roots of Unity) If p is an odd prime and n ∈ Z¯, then xn − 1 ≡ 0 (mod p∞0 ) has a nontrivial solution x = ν in Z¯/(p − 1)p∞0 Z¯ if and only if n | (p − 1) or n = mp∞0 for some m | (p − 1). Moreover:

(1) For , the roots of unity of order are generated by p∞0 . n = p − 1 p − 1 γp (2) For ∞0 , the roots of unity of order ∞0 are generated by p−1. n = p p γp 8.3. The Total Logarithm. Denition 24. We dene ¯ ¯ ¯ ∞0 ¯ and for each odd prime we L2 := Z × (Z/2Z) × (Z/2 Z) p dene ¯ ¯ ¯ ∞0 ¯ . We then dene Q , and all of these direct Lp := Z × (Z/(p − 1)Z) × (Z/p Z) L := Lp p∈P products we consider rings under componentwise addition and multiplication. For each prime , ˆ denotes extended by an element such that ˆ we have and p Lp Lp ∞ ∀x ∈ Lp x + ∞ = ∞ ( 0 : x = 0, and we denote ˆ Q ˆ . Given a xed choice of primitive roots x∞ = L := Lp ∞ : x 6= 0 p∈P {γ } , the total logarithm of base Q γ is the unique map log: ¯ → ˆ dened by the p p∈P p Z L p∈P Unique Factorization Theorem.

Proposition 40. Qˆ under multiplication is isomorphic to Lˆ under addition. Proof. This is an immediate consequence of the Unique Factorization Theorem and Propo- sition 18.

9. Continuous Functions

9.1. Some Useful Functions.

Denition 25. For n ∈ Z, d ∈ Z\{0}, we dene n%d to be the least representative of n modulo in , and we dene n to be the unique integer such that n . We d N0 b d c b d cd + n%d = n extend this denition to n ∈ Z¯ according to the following proposition: Proposition 41. For xed , n is a uniformly continuous function of , d ∈ Z\{0} b d c Z → Z n and therefore uniquely extends to a uniformly continuous function z taking ¯ to ¯. b d c z ∈ Z Z

Proof. Given , take and n+δ!k n δ! (+d)!k and since some number ! δ = + d b d c − b d c = d k = d in { + 1, ..., + d} is divisible by d, we certainly have ! | (+d)!k . Therefore the function is 35 d uniformly continuous on Z, and thus extends uniquely to a uniformly continuous function ¯ on the metric completion Z of Z. n−1 Denition 26. Dene the Pochhammer symbol Q , with (x)n := (x − k) ∀n ∈ N0 (x)0 = 1 k=0 the empty product. Note that (x)n is always a polynomial of degree n, hence uniformly continuous as a function of x. Proposition 42. {(x) } is a basis over for [x]. n n∈N0 Z Z Proof. A polynomial of zero degree f(x) = c can be written in exactly one way, namely c(x)0. Suppose all polynomials of degree less than n can be written in exactly one way, and N n 0 P let f(x) = anx + ... + a0x be a polynomial of degree n. Then if f(x) = bk(x)k with k=0 bN 6= 0, we must have N = n since (x)k is a polynomial of degree k ∀k and f is of degree n. Moreover, (x)k is always monic, so we must have bn = an. So f(x) − an(x)n is a polynomial of degree n−1, which by induction can be represented in exactly one way, so there is exactly one way to write f as a linear combination of Pochhammer symbols over Z. Lemma 6. ¯ . ∀z ∈ Z n! | (z)n ¯ Proof. It suces to consider z ∈ since is dense, (z)n is continuous, and ne! is closed. N N 0 Z We proceed by induction to show that for , mod ; equivalently, z ∈ N ∀n ∈ N (z)n ≡ 0 ( n!) the product of any n consecutive naturals is divisible by n!. Every natural is divisible by 1!, so the proposition holds in the case n = 1. Now suppose the proposition holds for all naturals less than n, and let Iφ = {φ, φ+1, ..., φ+(n−1)} be a set of n consecutive integers. We will proceed here as well by induction, this time on φ, in order to show that ∀φ ∈ N Q Q n! | Iφ. For φ = 1 the claim holds since Iφ = n!. Suppose φ > 1 and the result holds Q for all naturals less than φ. Then distributing the last term, Iφ = φ(φ + 1)...(φ − 1 + n) = (φ − 1)φ(φ + 1)...(φ + n − 2) + nφ(φ + 1)...(φ + n − 2). The rst expression is a product of n consecutive numbers starting from φ − 1, so by induction on φ it is ≡ 0 (mod n!); the second expression is n times a product of n − 1 consecutive numbers starting from φ, so by induction on it is mod . Therefore Q mod . Thus Q , as n ≡ 0 ( n!) Iφ ≡ 0 ( n!) n! | Iφ ∀φ ∈ N claimed, and thus the product of any n consecutive naturals is divisible by n!, as claimed. Thus by the continuity of with respect to , ¯ . (z)n z ∀z ∈ Z n! | (z)n ¯ Proposition 43. ∀z ∈ , lim (z)n = 0. Z n→∞ Proof. mod by Lemma 6. (z)n ≡ 0 ( n!) z Denition 27. For k ∈ , z ∈ ¯, we dene the binomial coecient := (z)k . Con- N0 Z k k! x sidered for xed k as a function of x, is called the binomial polynomial of order k. Fur- k thermore, let denote the ring of integer-valued polynomials, i.e. . Q[x]Z {f ∈ Q[x]: f(Z) ⊆ Z} Note that Q[x] is an abelian group and so may be considered a Z-module. Z 36 z Proposition 44. For xed k ∈ , f(z) := is a continuous function ¯ → ¯. N0 k Z Z

z Proof. ∀z ∈ ¯ (z) ≡ 0 (mod k!) so = (z)k is a factoradic integer ∀z. Furthermore, Z k k k! z + N!m z − = 1 ((z + N!m) − (z) ) and (z) is continuous so this → 0 as k k k! k k k N → ∞. n Proposition 45. For n ∈ , = 0 if k > n. N k

n n−1 Proof. (n)k and Q which includes a factor of if . = k! (n)k = (n−k) (n−n) = 0 k > n k k=0

9.2. Finite Dierences. Denition 28. Writing for the set of functions ¯ ¯, for ¯ we dene F Z → Z z ∈ Z ∆z : F → F by (∆zf)(x) = f(x + z) − f(x). If the subscript is omitted, we dene ∆ = ∆1. Similarly, (∇zf)(x) := f(x) − f(x − z) and ∇ = ∇1.

Proposition 46. Fixing k ∈ N, we have the following identities: ( 1 : k | (x + 1) (1) x ∆b k c = 0 : k - (x + 1) (2) ∆(x)n = n(x)n−1 x x (3) ∆ = k k − 1

Proof. Proceding for each part: ( b x c + 1 : k | (x + 1) (1) x+1 k . b k c = x b k c : k - (x + 1) (2) (x + 1)n − (x)n = (x + 1)(x)(x − 1)...(x + 2 − n) − (x)(x − 1)...(x + 1 − n) = (x)(x − 1)...(x + 2 − n)(x + 1 − (x + 1 − n)) = n(x)n−1. In the case n = 0 note that we still have ∆(x)0 = 1 − 1 = 0 = 0(x)−1 regardless of the denition of (x)−1. x + 1 x (3) − = 1 ((x + 1) − (x) ) = 1 ((x) (x + 1) − (x) (x + 1 − k)) k k k! k k k! k−1 k−1 x = k (x) = k! k−1 k − 1

Proposition 47. Given any function f : Z¯ → Z¯ and a natural number n, we have 37 n n (∆nf)(x) = P (−1)n−kf(x + k). k=0 k

1 Proof. For n = 1 the result holds since (∆f)(x) = f(x + 1) − f(x) = (−1)0f(x + 1) + 0 1 (−1)1f(x). Suppose the result holds for all naturals less than n; then (∆nf)(x) = 1 (∆n−1f)(x + 1) − (∆n−1f)(x) n−1 n − 1 n−1 n − 1 = P (−1)n−1−kf(x + 1 + k) − P (−1)n−1−kf(x + k) k=0 k k=0 k n n − 1 n−1 n − 1 = P (−1)n−kf(x + k) + P (−1)n−kf(x + k) k=1 k − 1 k=0 k n − 1 n−1 n − 1 n − 1 = (−1)nf(x) + P + (−1)n−kf(x + k) 0 k=1 k − 1 k n − 1 n − 1 n n − 1 n + (−1)0f(x + n), and since = 1 = , = 1 = , n − 1 0 0 n − 1 n n − 1 n − 1 n n n and + = , this is equal to P (−1)n−kf(x + k). So the k − 1 k k k=0 k result holds for n, and therefore ∀n ∈ N by induction. Proposition 48. (Newton Series on ) For any function ¯ there is a unique N f : N0 → Z ∞ x sequence of coecients {a } such that f(x) = P a , given by a = (∆nf)(0) ∀n. n n∈N n n n=0 n

Proof. We adapt a proof given by Alain M. Robert [8]. The an are unique as follows: suppose ∞ ∞ P x x P x f(x) = an. We have = 0 when n > x (Proposition 45), so an = n=0 n n n=0 n x 0 P x P 0 an. Thus f(0) = an = a0 uniquely determines a0. Now suppose that n=0 n n=0 n N P N f(0), ..., f(N − 1) uniquely determine a0, ...., aN−1. Then f(N) = an ⇐⇒aN = n=0 n N−1 P N N f(N) − an, since = 1, so f(0), ..., f(N) uniquely determine a0, ..., aN . n=0 n N Thus by induction the {aN } are uniquely determined by the values of f. Now we shall ∞ x show existence: set n . We have P x P x n . an = (∆ f)(0) ∀n ∈ N0 an = (∆ f)(0) n=0 n n=0 n x x At x = 0 this is (∆0f)(0) = f(0). Writing g(x) = f(x) − P (∆nf)(0), we thus n=0 n N N have g(0) = 0. Applying Proposition 47, ∆N g(0) = P (−1)N−kg(k), so we clearly k 38 k=0 have the implication (∀k ∈ {0, ..., N} g(k) = 0) −→ ∀k ∈ {0, ..., N} (∆kg)(0) = 0; moreover, (∀k ∈ {0, ..., N} g(k) = 0)∧∀k ∈ {0, ..., N} (∆kg)(0) = 0 −→ (g(N + 1) = 0) since if N+1 N + 1 we assume the antecedent then g(N +1) = f(N +1)− P (∆nf)(0) = f(N +1)− n=0 n N+1 N + 1 N + 1 (∆N+1f)(0) = f(N+1)− P (−1)N+1−kf(k) = f(N+1)− (−1)0f(N+ k=0 k N + 1 . Therefore since , by induction , and so 1) = 0 g(0) = 0 g(n) = 0 ∀n ∈ N0 f(x) = x P x n , as claimed. (∆ f)(0) ∀x ∈ N0 n=0 n Proposition 49. is a free -module with basis the binomial polynomials. Q[x]Z Z

9.3. Direct & Indirect Continuity.

Denition 29. Suppose f : Z¯ → Z¯ is continuous. Then we say f is directly continuous if there exists a function ¯ ∞0 ¯ ¯ ∞0 ¯ such that ∞0 ¯ mod ∞0 ∀p ∈ P fp : Z/p Z → Z/p Z f(x) ≡ f(x + p Z)( p ) ∀x ∈ Z¯, i.e. if f is a direct product over p ∈ P of continuous p-adic functions. Otherwise, we say f is indirectly continuous. Example.

(1) Every polynomial is directly continuous, since addition and multiplication can be performed independently modulo Z¯/p∞0 Z¯ ∀p. (2) Let u ∈ U. If there are exponents such that for some choice of primitive roots we (p−1)Ξ have ξ2 Ξ2 Q 2 , then x is directly continuous, since modulo u = τ γ2 γp f(x) = u p∈P\{2} ∞0 we have x ξ2x Ξ2x which depends only on mod ∞0 , and modulo any ∞0 we 2 u ≡ τ γ2 x 2 p x (p−1)Ξ2x ∞0 have u ≡ γp which depends only on x mod p . If there are no such exponents, then x is indirectly continuous, since if ξ mod ∞0 with , the f(x) = u u ≡ γp ( p ) (p − 1) - ξ value of f(x) mod p∞0 depends on x mod (p − 1) as well as on x mod p∞0 . Note that in both cases, f(x) is uniformly continuous by modular periodicity, so it is entirely possible for indirectly continuous functions to be uniformly continuous. (3) For ¯ and the th prime write the -adic expansion of mod ∞0 as x ∈ Z pn n pn x pn x ≡ ∞ P k . Considering the to be elements of , dene the function ac- dpn,kpn dpn,k N f(x) k=0  mod ∞0 0 ( p ): p = 2 cording to f(x) ≡ ∞ . Then f is continuous (its Pd pk (mod p∞0 ): p ∈ \{2}  pn−1,k P k=0 value modulo Q xn is determined by the value of modulo Q xn as ranges over pn x pn−1 n any set of naturals), but it clearly does not reduce to a direct product of p-adic functions; thus f is indirectly continuous. Proposition 50. An indirectly continuous function is not the pointwise limit of any sequence of directly continuous functions. 39 Proof. Let ¯ ¯, for each dene ¯ ¯ ∞0 ¯ by ∞0 ¯, and let f : Z → Z p ∈ P fp : Z → Z/p Z fp(x) = f(x) + p Z ¯ such that mod ∞0 and . Now suppose Xp = q ∈ P : ∃x, y ∈ Z x ≡ y ( (1 − q ) fp(x) 6= fp(y) is indirectly continuous, and x such that . Choose and f p ∈ P Xp\{p}= 6 ∅ q ∈ Xp\{p} choose ¯ such that mod ∞0 but . Then for all directly x, y ∈ Z x ≡ y ( (1 − q )) fp(x) 6= fp(y) continuous g : ¯ → ¯ we have g(x)+p∞0 ¯ = g(y)+p∞0 ¯. So if {F } is a sequence of directly Z Z Z Z n n∈N continuous functions and F (x) = lim Fn(x) ∀x, we have F (x) = F (y), but f(x) 6= f(y). n→∞ Therefore an indirectly continuous function is not the pointwise limit of any sequence of directly continuous functions.

There are two results particularly relevant to direct continuity in Z¯. The rst is known as Mahler's Theorem, and was proven by Kurt Mahler in 1958 [7]; Mahler used properties of quadratic extension of in his original proof, but there is a very nice proof by R. Bojanic in Zp 1974 based on the properties of factorials [2]. The second result is a renement of Mahler's theorem in the case , and was proven by R. Ahlswede and R. Bojanic in 1975 f : N0 → Zp [1]. These results follow: Theorem 12. (Mahler's Theorem, 1958) If is continuous, then we f : Zp → Qp ∀x ∈ Zp ∞ x have f(x) = P (∆kf)(0). [7][2] k=0 k Theorem. (Ahlswede & Bojanic, 1975) Let be any function and let denote f : N0 → Zp ||p n the p-adic norm. Then lim |(∆ f)(0)|p = 0 if and only if n→∞ t as . [1] max {|f(n + p ) − f(n)|p : n ∈ N0} → 0 t → ∞

So a continuous function f : Z¯ → Z¯ must have f(x + p∞0 +z) ≡ f(x)(mod p∞0 ) ∀z ∈ Z¯ and ∀p ∈ P in order to have (∆nf)(0) = 0 (since a sequence converges as n → ∞ if and only if it converges as ¯). Then writing the induced function ˆ ¯ as n → z ∀z ∈ Z f : Z → Zp fˆ(x) = f(x) + p∞0 Z¯, we see that fˆ(x + np∞0 ) = fˆ(x) whenever n ∈ N, since fˆ(x + p∞0 ) = fˆ(x + p∞0 + p∞0 ) = ...., and therefore by the continuity of f we will have fˆ(x + zp∞0 ) = f(x + zp∞0 ) + p∞0 Z¯ = f(x) + p∞0 Z¯ = fˆ(x) ∀z ∈ Z¯, and f is directly continuous. Mahler's Theorem provides the converse, resulting in the following theorem:

∞ P z k Theorem 13. Let f : Z¯ → Z¯ be continuous. Then f(z) = (∆ f)(0) ∀z ∈ Z¯ if and k=0 k only if f is directly continuous. Example. The most familiar and well-behaved indirectly continuous functions are the func- x (p−1) ∞0 tions f(x) = u for u some unit which fails to be a power of γp modulo p for at least one p ∈ P. For a unit u we have (∆f)(x) = ux+1 − ux = ux(u − 1), and by induction it is easily shown that (∆nf)(x) = ux(u − 1)n. So (∆nf)(0) = (u − 1)n goes to 0 as n → ∞ if and only if (u − 1)∞0 = 0, i.e. if and only if u − 1 is divisible by every prime, which holds if and only if mod , which holds if and only if for some set of we have u ≡ 1 ( p) ∀p ∈ P Ξp (p−1)Ξp 0∞ u ≡ γp (mod p ) ∀p ∈ P, which is exactly the condition required in order for f to be directly continuous. 40 9.4. Factoradic Series. Denition 30. Let z ∈ ¯ and let {x } be a sequence in ¯. Then we say the fac- Z n n∈N Z ∞0 +z N P P toradic series xn converges to ( or sums to) lim xn; if the limit does not exist we n=1 N→zn=1 say the series diverges at ∞0 + z. Note that this is a factoradic limit over N, i.e. N z ranges over ¯ . Omitting the , we have the slightly dierent denition P N ∩ Z\{z} ∞0 xn = n=1 ∞0 +z  P x : z∈ /  n N0 n=1 . This distinction is unfortunately necessary if we are to make x1 + x2 + ... + xz : z ∈ N  0 : z = 0

any sense of sequences of partial sums {Sn} that are not consistent, i.e. for which we do not ¯ ¯ ¯ have lim Sn = SN for N ∈ 0. In general, if Z ⊂ , z ∈ , f : Z → is continuous, and n→N N Z Z Z z a+N , then P P . A continuous function ¯ ¯ is called a + N0 ⊂ Z f(n) := lim f(n) f : Z → Z n=a N→(z−a) n=a ∞0 +z P summable if f(n) converges ∀z ∈ Z¯. As shall be shown, we may omit the ∞0 when f is n=1 known to be a continuous function of n. Remark. Observe that a factoradic series sums over exactly the same set of elements as the more familiar form of innite series; the dierence is in how the terms are grouped. For ex- ∞ ample, let be a sequence in ¯, let P be the canonical series for a factoradic integer {xn} Z n!dn n=1 N ∞0 +z ∞ Sk , and let , P be its partial sums; then either P P P or z S0 = 0 SN∈N = n!dn xn = xn n=1 n=1 k=1n=Sk−1+1 ∞0 +z P xn diverges. The sequence of partial sums of the canonical series in this example can be re- n=1 placed with any sequence converging to z, because either we have convergence for every such N P choice, and they all have the same value, or lim xn fails to exist, and the series diverges N→zn=1 at z. Thus we can dene an equivalence relation ∼ on the partitions of N into contiguous regions by P = {{1, ..., M1}, {M1 +1, ..., M2}, ...} ∼ Q = {{1, ..., T1}, {T1 +1, ..., T2}, ...} ⇐⇒ lim (Mk − Tk) = 0; now if a contiguous partition P has its maximal elements dened by k→∞ N P MN = n!dn, then any other partition with maximal element sequence converging to the n=1 same factoradic will be equivalent to P , and the sums of a summable sequence will agree over ∞ P any two equivalent partitions. On the other hand, for xn we have P = {{1}, {2}, {3}, ...}; n=1 when this converges the sum is invariant under reorderings and regroupings, so convergent innite series have sums invariant under changes to the underlying partition of N; i.e., a series in Z¯ converges over {{1}, {2}, {3}, ...} if and only if it converges to the same value over all partitions. 41 z P Proposition 51. ∀z ∈ Z¯ we have 1 = z. n=1

z P Proof. f(z) = z − 1 is identically 0 on N and therefore extends by continuity to f(z) = 0 n=1 everywhere. b−1 b−1 P P Proposition 52. Let f : Z¯ → Z¯ be summable, and let a, b ∈ Z¯. Then f(n) = f(n) − n=a n=0 a−1 b−1+∞0 P f(n) = P f(n). n=0 n=a

N P Proof. Dene the partial sum sequence SN = f(n) . Since {SN } is universally n=1 N∈ N ¯ ¯ convergent the induced function F (x) = lim SN is continuous → . Since for any xed N→x Z Z z+N P N ∈ N (F (z + N) − F (z)) − f(n) is continuous and equal to 0 on N, it is equal to n=z+1 a+N−1 P 0 on Z¯. Thus f(n) = F (a + N − 1) − F (a − 1). Since F is continuous, taking n=a b−1 b−1 a−1 the limit as N → b − a we obtain P f(n) = F (b − 1) − F (a − 1) = P f(n) − P f(n) n=a n=1 n=1 b−1 a−1 = P f(n) − P f(n), as claimed in the rst equality. For the second equality, note that n=0 n=0 F (z) − F (z − 1) = lim (SN − SN−1) = lim f(N) = f(z). Consequently F (b) − F (a − 1) → N→z N→z b−1 b−1+∞0 when , so we always have P P . f(a) + f(z + 1) + ... + f(b) b − a ∈ N0 f(n) = f(n) n=a n=a z P Theorem 14. Let f : Z¯ → Z¯ be summable. Then f(n) is continuous, and ∀a, b, c ∈ Z¯ n=1 the following hold:

b b−c P P (1) Reindexing Invariance: ∀a, b ∈ Z¯ f(n) = f(n + c) n=a n=a−c c b c P P P (2) Path-Invariance: ∀a, b, c ∈ Z¯ f(n) = f(n) + f(n) n=a+1 n=a+1 n=b+1 b−a b−a P P (3) Orientation Invariance: ∀a, b ∈ Z¯ f(a + n) = f(b − n) n=0 n=0

z N Proof. P f(n) is the function induced by P f(n) which is universally convergent by n=1 n=1 N∈N z assumption, so P f(n) is continuous. Reindexing invariance follows since the partial sums n=1 42 b b−c of P f(n) and P f(n + c) give exactly the same sequence, and therefore have the same n=a n=a−c c b c b c−b P P P P P limits. Now f(n) = f(n) + f(n) = f(n) + f(n + b) for all b ∈ N n=a+1 n=a+1 n=b+1 n=a+1 n=1 and so taking the limit as b approaches a factoradic it still holds, giving path-invariance. b−a P For orientation-invariance, if a, b ∈ N with a < b the left-hand sum f(a + n) evaluates n=0 b−a f at {a, ..., b} and the right-hand sum P f(b − n) evaluates f at {a, ..., b}, so taking b n=0 to a factoradic they remain equal by continuity; then we have for any summable g that b b a−1 P g(n) = P g(n) − P g(n), so we can also take a to a factoradic as well, and the two will n=a n=1 n=1 remain equal by continuity. Corollary. Let f : Z¯ → Z¯ be summable. Then f sums to 0 along closed paths, i.e. a P f(n) = 0 ∀a ∈ Z¯. We therefore have the useful (and equivalent) identities: n=a+1

b a P P (1) f(b) − f(a) = f(n) + f(n) ∀a, b ∈ Z¯. n=a n=b b a P P (2) f(n) = − f(n) ∀a, b ∈ Z¯. n=a+1 n=b+1

z−1 P Proposition 53. Suppose f : Z¯ → Z¯ is continuous. Then f(z) = f(0) + (∆f)(n). n=0

N−1 P Proof. ∀N ∈ N (∆f)(n) = (f(1) − f(0)) + (f(2) − f(1)) + ... + (f(N) − f(N − 1)) n=0 N−1 P = f(N) − f(0), so lim (∆f)(n) = f(z) − f(0). N→z n=0

Corollary. If f : Z¯ → Z¯ satises f(z) = (∆g)(z) for some continuous function g, then f is summable.

z−1 kn−2−1kn−1−1 z Lemma 7. ∀z ∈ ¯ ∀n ∈ P ... P P 1 = . Z N n k1=0 kn−1=0 kn=0

z Proof. We proceed by induction. We have 1 = . Now suppose the proposition 0 x−1 kn−2−1 x z−1 kn−1−1 holds for all naturals less than n; then P ... P 1 = ∀x, so P ... P 1 n − 1 k1=0 kn−1=0 k1=0 kn=0 z−1 k z−1 k z z = P 1 . Thus the claim follows if P = . We have ∆ = k =0 n − 1 k=0 n − 1 n n 1 43 z z z−1 k by Proposition 46, and is a continuous function of z, so P = n − 1 n k=0 n − 1 z−1 0 P z z + ∆ (k) = 0 + . n k=0 n n Proposition 54. Suppose f : Z¯ → Z¯ is continuous. Then ∀n ∈ N we have ∀z ∈ Z¯ f(z) = n−1 z z−1 kn−1−1 P (∆kf)(0) + P ... P (∆nf)(k ). k n k=0 k1=0 kn=0 Proof. We proceed by induction, invoking Proposition 53 repeatedly to prove the claim: n−1 z z−1 kn−1−1 ∀n ∈ f(z) = P (∆kf)(0) + P ... P (∆nf)(k ). This holds for n = 1 since N k n k=0 k1=0 kn=0 z−1 P f(z) = f(0)+ (∆f)(k1) by Proposition 53. Suppose the claim holds for n−1; then f(z) = k1=0 n−2 z z−1 kn−2−1 P (∆kf)(0) + P ... P (∆n−1f)(k ), so expanding (∆n−1f)(k ) according to k n−1 n−1 k=0 k1=0 kn−1=0 proposition Proposition 53 we have: ! n−2 z z−1 kn−2−1 kn−1−1 f(z) = P (∆kf)(0) + P ... P (∆n−1f)(0) + P (∆nf)(k ) k n k=0 k1=0 kn−1=0 kn=0

n−2 z z−1 kn−2−1 z−1 kn−1−1 = P (∆kf)(0) + P ... P (∆n−1f)(0) + P ... P (∆nf)(k ). Now (∆n−1f)(0) k n k=0 k1=0 kn−1=0 k1=0 kn=0 does not depend on any of the ki, so we have:

z−1 kn−2−1 z−1 kn−2−1 P ... P (∆n−1f)(0) = (∆n−1f)(0) P ... P (1) k1=0 kn−1=0 k1=0 kn−1=0

z n−1 z z−1 kn−2−1 = (∆n−1f)(0) by Lemma 7, so f(z) = P (∆kf)(0)+ P ... P (∆nf)(k ) n − 1 k n k=0 k1=0 kn=0 as claimed, completing the induction. Theorem 15. Suppose f : Z¯ → Z¯ is uniformly continuous. Then f is summable and z−1 s(z) := P f(n) is uniformly continuous. n=0

Proof. Since f is uniformly continuous, it is modularly periodic. Fix M ∈ N and let P ∈ N denote the period of modulo . Then for any , write N f(x) M N ∈ N N = b P cP + N%P N P N%P and we have P N P P mod . As approaches some ¯, f(n) ≡ b P c f(n) + f(n)( M) N z ∈ Z n=1 n=1 n=1 where z , suciently large will always have mod , giving z = b P cP + z%P N N ≡ z%P ( P ) P z%P N P P for the th partial sum, which clearly converges as since the b P c f(n) − f(n) N N → z n=1 n=1 quotient is uniformly continuous; so s(z) is summable. Given !, take δ = + P and we 44 P have z+(+P )!k z P (+P )! , which is clearly an integer s(z + δ!k) − s(z) = b P c − b P c f(n) = P k n=1 divisible by ! since one of ( + 1), ..., ( + P ) must be divisible by P . Thus s(z) is modularly periodic hence uniformly continuous.

9.5. Geometric Series of Units. x P n Denition 31. For ω ∈ U, let Σω denote in this section the function Σω(x) := ω . n=1

Proposition 55. ∀ω ∈ U, Σω is uniformly continuous.

Proof. is a factoradic sum over x, a uniformly continuous function. f(x) = Σω(x) g(x) = ω Proposition 56. , ¯ x . ∀ω ∈ U x ∈ Z Σω(x) + ω Σω(y) = Σω(x + y)

x y x y x x+y x+y P n x P n P n P x+n P n P n P n Proof. ω + ω ω = ω + ω = ω + ω = ω . n=1 n=1 n=1 n=1 n=1 n=x+1 n=1 ( −1 : x odd Proposition 57. For ω = 1, Σ (x) = x. For ω = −1, Σ (x) = . ω ω 0 : x even Proposition 58. If is not a zero divisor, the roots of are precisely ¯. ω − 1 Σω(x) x ∈ |ω|Z If ω − 1 is a zero divisor and P is the set of primes with p∞0 | (ω − 1), then for x ∈ |ω|Z¯ we ( 0 (mod p∞0 ): p∈ / P have , and the roots are ∞0 ¯. Σω(x) ≡ |ω|(ω − 1) Z x (mod p∞0 ): p ∈ P

z z z z+1 P n P n P n P n 1 z+1 Proof. If z ∈ Z¯ such that ω = 0 then 0 = (1 − ω) ω = ω − ω = ω − ω = n=1 n=1 n=1 n=2 ω(1 − ωz), so ωz = 1, hence z ∈ |ω|Z¯. In the other direction, if ωz = 1 then working z z P n P n ∞0 backwards we have (1 − ω) ω = 0. If ω 6≡ 0 (mod p ) for some p ∈ P then ω ≡ 1 so n=1 n=1 z P n ∞0 ∞0 ω − 1 is a zero-divisor and ω ≡ z (mod p ), so a root must be ≡ 0 (mod p ). n=1

10. Projective Limits

10.1. Projective Limits. Denition. A poset is a set S together with a partial order ≤, i.e. ≤: S × S → {T,F } which satises ∀a, b, c ∈ S:

(1) a ≤ a (2) a ≤ b ∧ b ≤ a → a = b (3) a ≤ b ∧ b ≤ c → a ≤ c 45 A poset S is directed if ∀a, b ∈ S ∃c ∈ S such that a ≤ c and b ≤ c. Denition 32. Let be a directed poset, and let be a collection of sets indexed by I {Si}i∈I I.A projective system is an ordered pair ({Si} , {fij}), where {fij} is an indexed family of functions such that is the identity on and {fij : Sj → Si} i, j ∈ I ∀i ∈ I fii Si i ≤ j ∀i, j, k ∈ I with i ≤ j ≤ k we have fik = fij ◦ fjk. The functions fij are called bonding maps.A projective system of groups (or rings, or modules) is a projective system

({Si} , {fij}) such that the bonding maps are group (or ring, or module) homomorphisms; likewise for a projective system of topological groups (or rings, or modules) the bonding maps are continuous homomorphisms. If i ≤ j implies Si is a quotient ring (or quotient group) of Sj and we do not specify the bonding maps, they are dened to be the natural homomorphisms, i.e. if Si = Sj/Q for Q an ideal (or a normal subgroup) in Sj, then . The projective limit of the projective system fij(sj ∈ Sj) = sj + Q {Si}i∈I , {fij}i≤j∈I Q is dened by limSi := (si)i∈I ∈ Si : ∀i, j ∈ I with i ≤ j, si = fij(sj) . If G is a group, ←− i∈I i∈I the pronite completion of G is limG/N, where N runs over all normal subgroups N of G ←− N ordered by inclusion, and the bonding maps are the natural homomorphisms. Proposition 59. A projective limit of (topological) groups/rings/modules is a (topological) group/ring/module.

Proposition 60. ¯ is ring-isomorphic to lim /n! , where is ordered by magnitude. Z ←− Z Z N n∈N

∞ P Proof. The set of canonical series n!dn truncated to n ∈ {1, ..., N − 1} correspond bijec- n=1 tively via their sums in to the elements of . Each bonding map Z/N!Z Z/N!Z fij : Z/j!Z → is dened by , so in terms of truncated canonical series this is simply Z/i!Z fij(x) = x + i!Z j−1 i−1 P P . Thus is the subset of the direct product Q such fij n!dn = n!dn limZ/n!Z Z/n!Z n=1 n=1 ←− n∈ n∈N N that for every element there is a canonical series (i.e. a factoradic integer) whose residue modulo n! is the nth entry of the element ∀n. Since multiplication and addition modulo the factorials completely determines multiplication and addition in Z¯, this correspondence is an isomorphism. Proposition 61. For a xed prime p, the p-adic integers Z¯/p∞0 Z¯ are ring-isomorphic to lim /pn , where is ordered by magnitude. ←− Z Z N n∈N

Proof. This follows by the same argument as Proposition 60; simply replace Z/n!Z with ∞ ∞ n and the set of canonical series P with the set of -adic series P n. Z/p Z n!dn p dnp n=1 46 n=1 Proposition 62. ¯ is ring-isomorphic to lim /n , where is ordered by divisibility, i.e. Z ←− Z Z N n∈N n ≤ m ⇐⇒ n | m. This is the pronite completion of the integers.

Q Proof. Z¯ is isomorphic to the subset of the direct product Z/nZ obeying (x + nmZ) ∈ n∈N (x+nZ), since an element of Z¯ is uniquely determined by its residues modulo natural numbers and, N being dense, any nite set of residues must also correspond to some element of N. This is exactly lim /n , since the natural bonding maps are exactly (x + nm ) 7→ (x + n ). ←− Z Z Z Z n∈N

10.2. Direct Limits & Prüfer p-Groups.

Denition 33. Let be a directed poset, and let be a collection of sets indexed I {Si}i∈I by I.A direct system is an ordered pair ({Si} , {fij}), where {fij} is an indexed family of functions such that is the identity on and {fij : Si → Sj} i, j ∈ I ∀i ∈ I fii Si ∀i, j, k ∈ I i ≤ j with i ≤ j ≤ k we have fik = fjk ◦ fij. The functions fij are called structure maps; note that the dierence between projective and direct systems is the direction of the maps, i.e.

the structure maps are fij : Si → Sj for direct systems whereas the bonding maps are fij : Sj → Si for projective systems. As in the case of projective systems, if the Si are (topological) groups/rings/modules and the fij are (continuous) homomorphisms, then we say ({Si} , {fij}) is a direct system of (topological) groups/rings/modules. The direct limit F F of the direct system ({Si} , {fij}) is dened by limSi = Si/ ∼, where denotes disjoint −→ i∈I i∈I union and ∼ is the equivalence relation dened by a ∈ Si ∼ b ∈ Sj ⇐⇒ ∃k ≥ i, j such that fik(a) = fjk(b). Informally, limSi is thus the set of all elements of all of the Si, except that −→ i∈I two elements are considered the same whenever the structure maps eventually take them to the same element.

Proposition 63. Let be the direct system of groups indexed by ordered by ({Si} , {fij}) N n + m−n magnitude such that each Sn = ( /p ) and fnm(x) = p x. Then limSn is isomorphic Z Z −→ n∈N to the Prüfer p-group, i.e. the subgroup of the circle group composed of all p-power roots of unity.

Proposition 64. For each prime , is isomorphic to the Prüfer -group. p Qp/Zp p Proposition 65. Let be the direct system of groups indexed by ordered by ({Si} , {fij}) N + magnitude such that each Sn = ( /n! ) and fnm(x) = x(n + 1)(n + 2)...(m). Then limSn Z Z −→ n∈N is isomorphic to the torsion subgroup of the circle group, i.e. to the additive group Q/Z. 47 Part 3. Projective Limits and Polynomial Completions

∞ ∞ The fact that power series P n often fail to converge in ¯ while factorial series P anx Z an(x)n n=0 n=0 always converge is no accident, but an artifact of the relationship between Z¯ and Z[x]. The Wave Topology Theorem at the end of Part I is phrased in terms of unique factorization domains, and generalizations of the method of construction of ¯ (or any of the ) are Z Zp applicable to polynomial rings over UFDs; of particular interest are the various completions of Z[x] and Q[x].

11. Power Series in Z¯

Z¯ is in many ways poorly suited for analytic methods based on power series. Intuitively, this is because Z¯ shares the characteristics of the factorials from which it was constructed: meaningful innite iterations in Z¯ must generally involve addition in an intimate way, since they must encode meaningful patterns in each , and these spaces are very dierent. The need Zp for the symbol ∞0 was the rst hint of this exponential-phobic tendency: except for units, which are the paragon of periodicity (and thus regularity) under repeated multiplication, and for units which have had some information destroyed already (that is, units multiplied by factoradics of the form z∞0 ), factoradic integers lose information when raised to unnatural powers, i.e. f(x) = bx cannot be extended continuously from to ¯. Hence limbn cannot N Z n→z ∞ z P n be written simply b , but must involve ∞0 ; thus the terms of power series anx are not n=0 in general continuous functions of n, even when each an is 1, and power series inherit the irregularity one might expect as a result. The hyperintegers ∗Z from non-standard analysis provide an interesting perspective on this phenomenon. Since Z¯ is the quotient ring of ∗Z which preserves only the residues of hyperintegers modulo naturals, and exponentiation works in ∗Z by analogy with exponentiation in Z, we see that the need for ∞0 is a result of loss of information in traversing the natural map ∗Z → Z¯. More specically, the hyperintegers allow natural primes to be raised to hypernatural powers, and the hypernaturals are totally ordered and contain innite elements, whereas in ¯ the zero-signature of a Z gp ¯ ∼ prime p can only distinctly be raised to 0 ∪ {∞} because = lim /n ; the destruction of N Z ←− Z Z information by the ∞0 symbol thus corresponds to the map taking hypernatural to factoradic exponents, which is the identity on but takes all elements of to when the base N0 ∗N\N ∞ is factoradically a zero signature. Denition 34. We designate by the symbol the quantity mod ∞0 , i.e. GP ≡ p ( p ) ∀p ∈ P Q . GP = gp p∈P ∞ Proposition 66. Power series P n with coecients in ¯ satisfy the following: anx Z n=0

∞ P n ¯ (1) anx converges ∀x ∈ if and only if lim an = 0. Z n→∞ n=0 48 (2) Every power series converges on ¯. GpZ

Proof. The rst claim follows since an innite series converges if and only if its terms go to

0, and at x ∈ U this occurs only if an → 0. The second claim follows by the same reasoning, since n as only when . z → 0 z → ∞ Gp | z ∞0 ∼ Proposition 67. ∀p ∈ P Z¯/p Z¯ = Z[[x]]/(p − x)Z[[x]]. Proposition 68. (Base Representations of Factoradic Integers) Let be the set of GP D sequences {d } in ¯ such that for each n ∈ and p ∈ we have d ≡ 0, 1, ..., p − n n∈N0 Z N0 P n ∞ 2, or p − 1 (mod p∞0 ). Then P d Gn takes D to ¯ bijectively. n P Z n=0

∞0 Proof. Modulo each p , the constraints on the dn are precisely those that apply to p-adic digits, and ; moreover, the values of modulo each ∞0 can be chosen independently. GP ≡ p dn p So every choice of p-adic integers over all p corresponds to a unique element of D, and vice versa.

12. F[x]

Wave actions on a principal ideal domain behave much like those on Z, i.e. if R is a principal ideal domain then Re/R = {{r ∈ R :(r − φ) ∈ I} : φ ∈ R and I an ideal in R}. As a con- R ! sequence, completions of metrizations of the standard wave topology T F[^x]\{0}/F[x] F[x] on a polynomial ring F[x] over a eld F bear a striking algebraic resemblance to Z¯. Denition 35. Let F be a eld, let U denote the set of units in F[x], and let M be the set of monic irreducible polynomials in F[x]. Then: (1) We dene the topological ring [x] = lim [x]/f(x) [x], where F is the set of all F ←− F F f∈F products of elements of M with multiplicity (i.e. all monic polynomials), ordered by ! divisibility, and the topology is T F[^x]\{0}/F[x] . F[x] (2) We dene for each m ∈ the topological ring [[m]] = lim [x]/mn [x], where M F ←− F F n∈ N N is ordered by magnitude and the topology is T mgN0 /F[[m]] . This is the m- F[[m]] adic completion of F[x]. Note that for u ∈ F, the denition readily extends to give ∼ F[[um]] = F[[m]]. Proposition 69. Let F be a eld, and let M be the set of monic, irreducible polynomials in ∼ Q F[x]. Then F[x] = F[[m]] as rings. m∈ M 49 Q Q Q n Proof. F[[m]] is by denition the subset of D = F[x]/m F[x] such that if x ∈ D m∈M m∈Mn∈N with the component of in n denoted we have n n+1 x F[x]/m F[x] xmn xmn+1 +m (F[x]/m F[x]) = for each choice of . is a principal ideal domain, so by the Chinese Remainder xmn m, n F[x] ∼ Theorem if f, g ∈ F[x] have no common factor in M then F[x]/fgF[x] = F[x]/fF[x] ⊕ F[x]/gF[x] via the isomorphism h + fgF[x] 7→ . Therefore, given dene for (h + f(x)(F[x]/fgF[x])) ⊕ (h + g(x)(F[x]/fgF[x])) x ∈ D xmn powers of monic irreducibles mn as before, and for each monic f ∈ F[x] write the unique Q nm factorization f = m (with all but nitely many nm equal to 0) and dene xf to be m∈M the unique element of such that nm . Then F[x]/fF[x] xf + m (F[x]/fF[x]) = xmnm ∀m ∈ M for all monic the mapping is given by the natural homomorphism, i.e. f, g ∈ F[x] xfg 7→ xf the bonding map F[x]/fgF[x] → F[x]/fF[x] used in the denition of F[x] as a projective limit; therefore the map Q Q dened by is a F[[m]] → F[x]/fF[x] x 7→ (xf )f monic m∈M f monic homomorphism Q . It is injective since for some , F[[m]] → F[x] x 6= 0 → xf n 6= 0 f ∈ M n ∈ m∈M , and it is surjective since for any , writing for the component of in , N ξ ∈ F[x] ξf ξ F[x]/fF[x] the indexed set will have to satisfy n n+1 and will (ξmn )m∈M,n∈N ξmn+1 + m (F[x]/m F[x]) = ξmn ∼ Q consequently be a preimage of ξ. Therefore it is an isomorphism, and F[x] = F[[m]]. m∈M Corollary. In particular, ∼ Q for . F[x] = F[[m]] F = Q, Qp, and Fp m∈M

13. lim [x]/x!n [x] ←− Z Z

n Denition 36. Dene the rising factorial !n Q , and let denote the x := (x + k) Z[x]x!n k=1 !n projective limit lim [x]/x [x], where 0 is ordered by magnitude. ←− Z Z N n∈N0 Proposition 70. To each element of there corresponds a unique sequence of inte- f Z[x]x!n ∞ gers {a } such that f(x) = P a x!n, and vice versa. n n∈N n n=0

Proof. Since the bonding maps Z[x]/x!N Z[x] → Z[x]/x!(N−1)Z[x] are the natural homomorphisms and the x!n are a basis over Z for Z[x], the bonding maps simply take polynomials in x!0, ..., x!N to those in x!0, ..., x!(N−1) by omitting the term corresponding to x!N , from which the claim follows immediately. ∞ Proposition 71. Let P !n and let ¯ ¯ be dened by f = anx ∈ Z[x]x!n φ: Z[x]x!n → Map(Z, Z) n=0 ∞ ˆ P !n as a function ¯ ¯. Then is an injective homomorphism of f := (φ(f)) (x) := anx Z → Z φ n=0 rings. 50 Proof. Let ˆ . Suppose ˆ ¯ and the are not all . Then for any f = φ(f) f(x) = 0 ∀x ∈ Z an 0 N ≥ 0 ∞ N !(N+1) P !n P !n we have (−N) = 0, so an(−N) = an(−N) , so in particular if N is the least n=0 n=0 ˆ !N natural such that an 6= 0 we have f(N) = aN (−N) 6= 0, a contradiction. So the kernel of φ is trivial; φ is clearly an additive homomorphism since we may add coecients termwise, and it is also a multiplicative homomorphism, because it clearly preserves the multiplicative ∞ ∞ ∞ !0 P !n P !n P !n !n identity 1x , and if anx bnx = cnx then since for n ≥ N we have z ≡ n=0 n=0 n=0 ∞ ∞ mod ¯ and !n mod , we must have ¯ that P !n P !n 0 ( N!) ∀z ∈ Z x ≡ 0 ( (x)N ) ∀z ∈ Z anx bnx ≡ n=0 n=0 ∞ P !n modulo every factorial, and so the two are equal. cnx n=0 Lemma 8. We have the following identities:

(1) ∆x!n = n(x + 1)!(n−1) (2) ∇x!n= nx!(n−1) ∞ ∞ P !n P !n (3) (∇ − 1) bnx = ((n + 1)bn+1 − bn) x n=0 n=0

Proof. Proceeding for each part: (1) ∆x!n = (x + 1)!n − x!n = n(x + 1)!(n−1) (2) ∇x!n = x!n − (x − 1)!n = x!(n−1)(x + n − 1 − (x − 1)) = nx!(n−1) ∞ ∞ ∞ ∞ P !n P !n−1 P !n P !n (3) (∇ − 1) bnx = bnnx − bnx = ((n + 1)bn+1 − bn) x n=0 n=0 n=0 n=0 Lemma 9. is a ring automorphism of . f 7→ (1 − ∇)f = f(x − 1) Z[x]x!n

Proof. From the preceding lemma we see that left-multiplication by −(∇ − 1) = 1 − ∇ is a function . It is a homomorphism since Z[x]x!n → Z[x]x!n (1 − ∇)(f)(x) + (1 − ∇)(g)(x)= f(x−1)+g(x−1) = (f +g)(x−1) = (1−∇)(f +g)(x), and ((1 − ∇)(f)(x)) ((1 − ∇)(g)(x)) = f(x − 1)g(x − 1) = (fg)(x − 1) = (1 − ∇)(fg)(x). It is surjective since any f(x) is equal to (1 − ∇)(f(x + 1)), and it is injective since f(x − 1) = 0 ∀x implies f(x) = 0 ∀x. Theorem 16. For each let ¯ be dened by , where z ∈ Z φz : Z[x]x!n → Z φz(f) = (φf)(z) ∞ ∞ ¯ ¯ is the natural map P !n P !n. Then: φ: Z[x]x!n → Map(Z, Z) anx 7→ anx n=0 n=0 (1) and is surjective; thus ¯ ∼ . ker(φ0) = xZ[x]x!n φ0 Z = Z[x]x!n /xZ[x]x!n (2) and is surjective; thus ¯ ∼ . ker(φ1) = (x − 1)Z[x]x!n φ1 Z = Z[x]x!n /(x − 1)Z[x]x!n ∞ (3) For general ¯, is the set of all elements P !n for which there exists z ∈ Z ker(φz) anx n=0 a sequence {aˆ } such that aˆ = 0, aˆ = a , and ∀n ∈ we have a z!n = n n∈N0 0 1 0 N0 n n! (n + 1)ˆan+1 − aˆn. 51 Proof. Fix z; φz is the composition of φ and evaluation at z, which are both ring homomor- ∞ P !n phisms; therefore φz is a ring homomorphism. Now suppose f = anx is in the kernel of n=0 ∞ ( P !n !n 1 : n = 0 φz, i.e. anz = 0. It is immediately clear from 0 = that f is in ker(φ0) n=0 0 : n > 0 exactly when , i.e. , and since every canonical series is in the a0 = 0 ker(φ0) = xZ[x]x!n image is surjective, establishing ∼ ¯. For general , is a root if and φ0 Z[x]x!n /xZ[x]x!n = Z z z N−1 only if P !n mod , so let us give an equivalent characterization of this anz ≡ 0 ( N!) ∀N ∈ N n=0 !n condition by examing the case corresponding to each N. Recall that n! | (z + n − 1)n = z for all choices of and . The choices of constants that follow can all be made in : n z aˆn Z

(1) Set aˆ0 = 0. 0 !0 P !n !0 (2) At N = 1 we have a0z = a0 ≡ 0 (mod 1), i.e. anz = a0z =a ˆ1 for aˆ1 =a ˆ0 + a0. n=0 1 !0 !2 !1 P !n !1 (3) At N = 2 we have a0z + a1z =a ˆ1 + a1z ≡ 0 (mod 2), i.e. anz =a ˆ1 + a1z = n=0 2â2 for some aˆ2. !2 (4) At we have !0 !2 !2 !2 z mod , N = 3 a0z + a1z + a2z = 2â1 + a2z = 2 aˆ2 + a2 2 ≡ 0 ( 3!) 2 i.e. such that z!2 , and we have P !n . ∃aˆ3 3â3 =a ˆ2 + a2 2 anz = 3!â3 n=0

Proceeding by induction, suppose such that z!K , and ∀K < N ∃aˆK+1 (K + 1)âK+1 =a ˆK +aK K! K N !N we have P !n . Then P !n !n z , so anx = (K + 1)!âK+1 anz = N!âN + aN z = N! aˆN + aN N! n=1 n=0 z!N mod so such that z!N , and we have aˆN + aN N! ≡ 0 ( (N + 1)!) ∃aˆN+1 (N + 1)âN+1 =a ˆN + aN N! N P !n anz = (N + 1)!âN+1, as claimed; so the proposition holds for all naturals. Therefore f n=0 is in the kernel if and only if there exists a sequence {aˆ } such that aˆ = a and ∀n ∈ n n∈N 1 0 N0 z!n . Now z!n is equivalently !n (n+1)ân+1 =a ˆn +an n! (n+1)ân+1 =a ˆn +an n! anz = (n+1)!ân+1 − !n n!ân. Suppose z = 1; then z = n!, so this is n!an = (n + 1)!ân+1 − n!ân, or equivalently !n !n !n an = (n+1)ân+1 −aˆn, so anx = (n+1)ân+1x −aˆnx . This is exactly the condition giving ∞ ∞ ∞ P !n P !n P !n (∇−1) aˆnx = ((n + 1)ân+1 − aˆn) x , by Lemma 8; since aˆ0 = 0 ⇐⇒ x | aˆnx , we n=0 n=0 n=0 thus have ker(φ1) = {(∇ − 1)(xf): f ∈ Z[x]x!n } = {(x − 1)((∇ − 1)f)(x): f ∈ Z[x]x!n } , so . Moreover, is clearly surjective = {(x − 1)f : f ∈ Z[x]x!n } ker(φ1) = (x − 1)Z[x]x!n φ1 ∞ !n !0 P !n since 1 = (n − 1)! except at n = 0 where 1 = 1, and hence series of the form an1 n=0 include all canonical series.

52 Bibliography

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