The Integers ($\mathbb{Z}$)

Introduction

This is the first article of the Integers system in Part V, and it occupies the algebra slot of that system: this article constructs $\mathbb{Z}$ from $\mathbb{N}$, establishes its universal property, records its arithmetic and its divisibility, and describes its place among the other systems. The construction is the group completion of the additive monoid of the naturals: $\mathbb{Z}$ is the Grothendieck group of $(\mathbb{N}, +)$, and it is the smallest ring of characteristic $0$. Equivalently, $\mathbb{Z}$ is the free group on one generator for the additive structure, and the initial object in the category of commutative rings with identity.

This article carries out the construction by pairs, verifies that the operations are well defined and satisfy the ring axioms, develops the order and the absolute value, describes the divisibility structure — units, primes, greatest common divisors, the Euclidean algorithm, the principal ideal domain and unique factorisation — and records the quotients $\mathbb{Z}/n\mathbb{Z}$. The system supports an algebra slot and an applications slot: there is no geometry of the integers, no interval structure beyond the discrete order, no analysis, and the arithmetic applications are the modular arithmetic, which is written after this article and is not used here. The divisibility theory of The Natural Numbers is the base of the divisibility theory below, and the construction of $\mathbb{Q}$ as the fraction field of $\mathbb{Z}$ belongs to The Rational Numbers.

Throughout, $\mathbb{Z}$ is the ring of integers, its elements are written as differences $a - b$ of naturals or as the cosets of the diagonal in $\mathbb{N} \times \mathbb{N}$, $n \mapsto \bar n$ is the embedding $\mathbb{N} \hookrightarrow \mathbb{Z}$, the units are $\pm 1$, and the order is the unique total order making $\mathbb{Z}$ an ordered ring. The corpus's default base is the commutative ring with identity, and $\mathbb{Z}$ is the initial such ring; every result below is stated for that base. The theory of rings, ideals, principal ideal domains and Euclidean domains is from Rings and Unique Factorisation Domains; the order-theoretic background is from Order Theory and Lattices.

The Construction of $\mathbb{Z}$

The Grothendieck Group

Definition. On $\mathbb{N} \times \mathbb{N}$ define

$$ (a, b) \sim (c, d) \iff a + d = b + c . $$

Write $[(a,b)]$ for the class of $(a,b)$ and define

$$ [(a,b)] + [(c,d)] = [(a+c, b+d)], \qquad [(a,b)] \cdot [(c,d)] = [(ac + bd, ad + bc)] . $$

Theorem. The relation $\sim$ is an equivalence relation and the two operations are well defined on the quotient $\mathbb{Z} = (\mathbb{N}\times\mathbb{N})/{\sim}$. With these operations, with $0 = [(0,0)]$ and $1 = [(1,0)]$, the quotient is a commutative ring with identity, and the map

$$ \iota : \mathbb{N} \to \mathbb{Z}, \qquad \iota(n) = [(n,0)], $$

is an injective homomorphism of semirings whose image is the set of nonnegative elements.

Proof. The relation is reflexive, symmetric and transitive: transitivity is the implication $a+d = b+c$ and $c+f = d+e$ imply $a+f = b+e$, which follows from the cancellation law of The Natural Numbers after adding the two equations. Well-definedness of addition and multiplication is checked by replacing $(a,b)$ by an equivalent pair and using cancellation again; the ring axioms are inherited from $\mathbb{N}$ componentwise, and the multiplicative identity is $[(1,0)]$ because $(ac+bd, ad+bc)$ with $(c,d) = (1,0)$ gives $(a,b)$. The embedding $\iota$ is injective because $(n,0) \sim (m,0)$ gives $n = m$, and it preserves $+, \cdot, 0, 1$.

Definition. The negation is $-[(a,b)] = [(b,a)]$, and one writes $[(a,b)] = \iota(a) - \iota(b)$; the absolute value is

$$ \lvert n \rvert = \begin{cases} n, & n \geq 0, \\ -n, & n < 0. \end{cases} $$

Theorem (universal property). Let $A$ be a commutative ring with identity and let $f : \mathbb{N} \to A$ be a homomorphism of semirings, that is, a map preserving $+, \cdot, 0, 1$. Then there is a unique ring homomorphism $F : \mathbb{Z} \to A$ with $F \circ \iota = f$. In particular $\mathbb{Z}$ is the initial object in the category of commutative rings with identity: for every such ring $A$ there is a unique unital homomorphism $\mathbb{Z} \to A$, sending $n$ to $n \cdot 1_A$, and this map is injective exactly when $A$ has characteristic $0$.

Proof. Uniqueness is forced, since $\mathbb{Z}$ is additively generated by $\iota(\mathbb{N})$ together with negatives. For existence, put $F([(a,b)]) = f(a) - f(b)$, using the additive inverses of $A$; the definition does not depend on the representative because $(a,b) \sim (c,d)$ gives $f(a) + f(d) = f(b) + f(c)$, and it preserves the operations by direct verification. The initial statement is the case $f$ the universal map of The Natural Numbers; the last clause is the criterion for characteristic $0$.

Corollary. $\mathbb{Z}$ is the free group on one generator for its additive structure: $(\mathbb{Z}, +)$ is generated by $1$ with no relations, so for every abelian group $A$ and every $a \in A$ there is a unique group homomorphism $\mathbb{Z} \to A$ with $1 \mapsto a$. Equivalently, the category of abelian groups is the category of $\mathbb{Z}$-modules.

Arithmetic in $\mathbb{Z}$

The Ring Structure

Theorem. $\mathbb{Z}$ is a commutative ring with identity of characteristic zero, and it is an integral domain: if $nm = 0$ then $n = 0$ or $m = 0$.

Proof. The ring axioms were verified in the construction. If $n = \iota(a) - \iota(b)$ and $m = \iota(c) - \iota(d)$ with $nm = 0$, then $\iota(ac + bd) = \iota(ad + bc)$, so $ac + bd = ad + bc$ in $\mathbb{N}$. If $a \geq b$, write $a = b + k$ with $k \in \mathbb{N}$; substituting gives $bc + kc + bd = bd + kd + bc$, hence $kc = kd$ and $k(c-d) = 0$, so $k = 0$ or $c = d$ by the integrality of $\mathbb{N}$, that is, $n = 0$ or $m = 0$. If $a < b$, the same computation with $b = a + k$ gives $k(d-c) = 0$, so again $n = 0$ or $m = 0$. The characteristic is zero because $\iota$ is injective and $k \cdot 1 = \iota(k) \neq 0$ for $k > 0$.

Theorem (cancellation). If $nm = nk$ and $n \neq 0$ then $m = k$. Consequently the multiplicative monoid $\mathbb{Z}\setminus\{0\}$ is cancellative and its group of units is $\{1, -1\}$.

Proof. $n(m-k) = 0$ and the integrality gives $m - k = 0$. The units are the divisors of $1$; an integer $u$ with $uv = 1$ has $\lvert u \rvert \lvert v \rvert = 1$ in $\mathbb{N}$, so $\lvert u\rvert = 1$.

Theorem. The division algorithm holds in $\mathbb{Z}$: for all $n, m \in \mathbb{Z}$ with $m \neq 0$ there are $q, r \in \mathbb{Z}$ with

$$ n = qm + r, \qquad 0 \leq r < \lvert m \rvert , $$

and $q, r$ are unique. Thus $\mathbb{Z}$ is a Euclidean domain with Euclidean function $n \mapsto \lvert n \rvert$.

Proof. Reduce to the nonnegative case: if $n \geq 0$ and $m > 0$, apply the division algorithm of The Natural Numbers; if $n < 0$ or $m < 0$, adjust signs. Uniqueness is the uniqueness of the natural division algorithm.

The Order

Theorem. There is a unique total order $\leq$ on $\mathbb{Z}$ extending the order of $\mathbb{N}$ under $\iota$ and making $\mathbb{Z}$ an ordered ring, that is, such that for all $n, m, k$,

$$ n \leq m \implies n + k \leq m + k, \qquad 0 \leq n,\ 0 \leq m \implies 0 \leq nm . $$

With this order, $\mathbb{Z}$ is a discrete ordered ring: every element has an immediate successor $n+1$ and an immediate predecessor $n-1$, and there is no element strictly between $n$ and $n+1$.

Proof. Define $n \geq 0$ if $n = \iota(k)$ for some $k$, and $n \leq m$ if $m - n \geq 0$. The order axioms follow from the corresponding facts in $\mathbb{N}$ and the translation invariance of the differences. Discreteness is immediate from the definition of the successor.

Theorem. $\mathbb{Z}$ is Archimedean as an ordered group: for $n > 0$ and any $m$ there is $k \in \mathbb{N}$ with $kn > m$. Every nonempty subset of $\mathbb{Z}$ bounded below has a least element, and every nonempty subset bounded above has a greatest element.

Proof. Choose $j$ with $m \leq \iota(j)$, which is possible because $m$ is an integer. Since $n > 0$ in $\mathbb{Z}$ we have $n \geq 1$, so $\iota(j+1) \leq n\,\iota(j+1)$ and $j+1$ satisfies $(j+1)n \geq \iota(j+1) > m$. For the second statement, let $A$ be nonempty and bounded below by $m$; then $\{n - m : n \in A\}$ is a nonempty subset of $\mathbb{N}$, so it has a least element $n_0 - m$ by the well ordering of $\mathbb{N}$, and $n_0$ is the least element of $A$.

Remark. The order type of $(\mathbb{Z}, <)$ is $\zeta$, the order type of the integers; it is a countable discrete order without endpoints, and it is the unique countable discrete order without endpoints up to isomorphism. This is in contrast with $\mathbb{N}$, whose order type $\omega$ has a least element.

Divisibility and Factorisation

Units, Primes and gcd

Definition. For $n, m \in \mathbb{Z}$, $n \mid m$ if $m = nk$ for some $k$. Two integers are associates if each divides the other, which in $\mathbb{Z}$ means $n = \pm m$. An element $p$ is irreducible if $p \neq 0$, $p$ is not a unit, and $p = nm$ forces $n$ or $m$ to be a unit; $p$ is prime if $p$ is not a unit and $p \mid nm$ implies $p \mid n$ or $p \mid m$. A greatest common divisor of $n$ and $m$ is a common divisor divisible by every common divisor.

Theorem. In $\mathbb{Z}$ an element is irreducible if and only if it is prime, and the irreducibles are the numbers $\pm p$ with $p$ a prime of $\mathbb{N}$.

Proof. An irreducible is prime in every unique factorisation domain; conversely a prime is irreducible, since $p = nm$ with $p$ prime gives $p \mid n$ or $p \mid m$, so one of $n, m$ is an associate of $p$ and the other is a unit. The irreducibles of $\mathbb{N}$ are its primes, and adjoining the sign gives the irreducibles of $\mathbb{Z}$.

Theorem (Bézout). Any two integers $n, m$, not both zero, have a greatest common divisor $d$, unique up to sign, and there are $u, v \in \mathbb{Z}$ with

$$ un + vm = d . $$

The Euclidean algorithm computes $d$, and $d$ is determined up to sign by this property: it is the least positive element of the ideal generated by $n$ and $m$.

Proof. The set $I = \{un + vm : u, v \in \mathbb{Z}\}$ is the ideal generated by $n$ and $m$; it contains a least positive element $d$ by the order properties, and division of $n$ by $d$ leaves remainder $r = n - qd \in I$ with $0 \leq r < d$, whence $r = 0$ and $d \mid n$; similarly $d \mid m$, and every common divisor of $n,m$ divides every element of $I$, hence $d$. The Euclidean algorithm computes $I$ by decreasing remainders, terminating by the well ordering from below.

Principal Ideals and Unique Factorisation

Theorem. Every ideal of $\mathbb{Z}$ is principal: $I = (n)$ for some $n \geq 0$, and $\mathbb{Z}$ is a principal ideal domain. The ideal $(n)$ is prime if and only if $n$ is $0$ or a prime, and maximal if and only if $n$ is a prime.

Proof. If $I \neq 0$, let $n$ be the least positive element of $I$; the division algorithm gives $I = (n)$. The ideal $(n)$ is prime exactly when $n$ has no proper divisors other than units, by the definition of primality; prime ideals of a principal ideal domain are maximal because $\mathbb{Z}/(p)$ is a field for $p$ prime.

Theorem (fundamental theorem of arithmetic). Every nonzero non-unit of $\mathbb{Z}$ is a product of irreducibles, uniquely up to order and up to multiplication by units. Equivalently, the multiplicative monoid $\mathbb{Z}\setminus\{0\}$ is the free commutative monoid on the primes $\pm p$ modulo the relation $-1$; every nonzero integer has a unique expression

$$ n = \pm \prod_{p} p^{v_p(n)}, \qquad v_p(n) \in \mathbb{N}, $$

with finitely many nonzero exponents. Thus $\mathbb{Z}$ is a unique factorisation domain.

Proof. Existence and uniqueness are the corresponding statements of The Natural Numbers applied to $\lvert n \rvert$, together with the sign of $n$.

Corollary. The greatest common divisor and least common multiple of $n$ and $m$ are given by

$$ \gcd(n,m) = \prod_p p^{\min(v_p(n), v_p(m))}, \qquad \operatorname{lcm}(n,m) = \prod_p p^{\max(v_p(n), v_p(m))}, $$

up to sign, so that $\gcd(n,m) \cdot \operatorname{lcm}(n,m) = \lvert nm \rvert$.

Quotients and Embeddings

The Ring of Residues

Definition. For $n \geq 1$ let $\mathbb{Z}/n\mathbb{Z}$ or $\mathbb{Z}_n$ be the quotient ring of $\mathbb{Z}$ by the ideal $(n)$; its elements are the residue classes modulo $n$, and the quotient is the ring of residues modulo $n$. Write $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ for $p$ prime.

Theorem. $\mathbb{Z}/n\mathbb{Z}$ has exactly $n$ elements, it is a commutative ring with identity, and it is a field if and only if $n$ is prime, in which case it is $\mathbb{F}_p$. For every $n$, the canonical map $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ is a surjective ring homomorphism with kernel $(n)$, and $\mathbb{Z}/n\mathbb{Z}$ is the initial object among rings $A$ with $n \cdot 1_A = 0$.

Proof. The classes are $0, 1, \dots, n-1$ and are distinct, so there are $n$ of them. If $n$ is prime then every nonzero class is a unit by Bézout: $up + vm = 1$ gives an inverse of $m$; if $n = m_1 m_2$ with $1 < m_1, m_2 < n$, then $m_1$ is a zero divisor and not a unit. The universal property is the standard one of the quotient by the kernel of the map $\mathbb{Z} \to A$.

Corollary (Chinese remainder). If $m$ and $n$ are coprime then $\mathbb{Z}/mn\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ as rings.

Proof. The map $k \mapsto (k \bmod m, k \bmod n)$ has kernel $mn\mathbb{Z}$ and is surjective by Bézout; the first isomorphism theorem gives the result.

The Place of $\mathbb{Z}$ among the Systems

Theorem. The natural numbers embed in $\mathbb{Z}$ as the nonnegative elements, and $\mathbb{Z}$ embeds in $\mathbb{Q}$, in $\mathbb{R}$ and in $\mathbb{C}$ as a subring. The ring $\mathbb{Z}$ is initial among commutative rings with identity, and $\mathbb{Q}$ is its fraction field.

Proof. The inclusion $\mathbb{N} \subset \mathbb{Z}$ is the construction; the fraction field of the integral domain $\mathbb{Z}$ is the field of fractions $\mathbb{Q}$ of The Rational Numbers; the inclusions into $\mathbb{R}$ and $\mathbb{C}$ are the images of the initial homomorphism.

System	Adjoined structure	What fails in $\mathbb{Z}$
$\mathbb{Q}$	multiplicative inverses	no solution to $2n = 1$
$\mathbb{R}$	order-completeness	no supremum for $\{n : n^2 < 2\}$
$\mathbb{C}$	algebraic closure	no root of $n^2 + 1$
$\mathbb{Z}/n\mathbb{Z}$	quotient by $(n)$	the residue classes collapse divisibility

Theorem. $\mathbb{Z}$ is countable of cardinality $\aleph_0$, is not well ordered, since $\mathbb{Z}$ itself has no least element, but every nonempty subset bounded below has a least element. It is a Euclidean domain, hence a principal ideal domain, hence a unique factorisation domain, and it is integrally closed in $\mathbb{Q}$.

Proof. The countability is the countability of $\mathbb{N} \times \mathbb{N}$; the failures of well ordering are witnessed by $\mathbb{Z}$ itself and by the subset of negative integers. The implication Euclidean $\Rightarrow$ principal $\Rightarrow$ unique factorisation is the standard chain of Unique Factorisation Domains, and integral closedness follows because a rational root of a monic integer polynomial is an integer.

Endomorphisms and the Initial Position of $\mathbb{Z}$

The Universal Properties

Theorem (initial object). $\mathbb{Z}$ is the initial object in the category of unital rings: for every unital ring $R$ there is exactly one unital ring homomorphism $\varphi : \mathbb{Z} \to R$, given by $\varphi(n) = n \cdot 1_R$, and it is injective if and only if $R$ has characteristic $0$.

Proof. A unital homomorphism must send $1$ to $1_R$ and therefore, by additivity, $n$ to $n \cdot 1_R$; this determines the map and it is a homomorphism by the distributive law. Its kernel is the principal ideal generated by the characteristic of $R$, so it is injective exactly in characteristic $0$.

Corollary. The correspondence $R \mapsto \operatorname{Hom}(\mathbb{Z}, R)$ is a bijection onto the underlying set of $R$; the image of $\varphi$ is the prime ring of $R$, and $\mathbb{Z}$ is the prime ring of every ring of characteristic $0$.

Theorem (endomorphisms and automorphisms). The unital endomorphisms of $\mathbb{Z}$ reduce to the identity, the ring endomorphisms to the identity and the zero map, and the additive group endomorphisms to the multiplications by integers: $\operatorname{End}_{\mathrm{ring}}(\mathbb{Z}) = \{\mathrm{id}, 0\}$, $\operatorname{Aut}_{\mathrm{ring}}(\mathbb{Z}) = \{\mathrm{id}\}$, and $\operatorname{End}(\mathbb{Z}, +) \cong \mathbb{Z}$ with $\operatorname{Aut}(\mathbb{Z}, +) = \{\pm\mathrm{id}\}$. Every derivation of $\mathbb{Z}$ is zero.

Proof. A unital endomorphism is determined by the image of $1$, which must be $1$, so it is the identity; a nonunital endomorphism sends $1$ to an idempotent, and the only idempotents of $\mathbb{Z}$ are $0$ and $1$, giving the zero map and the identity. An additive endomorphism is multiplication by its value at $1$, and it is bijective exactly when that value is $\pm 1$. A derivation $D$ satisfies $D(1) = D(1 \cdot 1) = 2D(1)$, hence $D(1) = 0$ and $D = 0$.

Theorem (the subgroups). Every subgroup of $(\mathbb{Z}, +)$ is cyclic, and the subgroups are exactly the $n\mathbb{Z}$; the subgroup lattice of $\mathbb{Z}$ is the divisibility order of the nonnegative integers, reversed, and the quotient $\mathbb{Z}/n\mathbb{Z}$ is the ring of residues.

Proof. A subgroup $H$ is the set of multiples of its least positive element, by the division algorithm; the containment $m\mathbb{Z} \subseteq n\mathbb{Z}$ is equivalent to $n$ dividing $m$.

The Place of the Universal Properties

Remark. The universal property of $\mathbb{Z}$ as the initial unital ring is the categorical expression of the fact that the integers are the arithmetic of counting, before any other structure is imposed; the same property read for groups says that $\mathbb{Z}$ is the free group on one generator, and read for ordered rings it says that $\mathbb{Z}$ is the initial object of the category of totally ordered rings. These three universal properties, together with the Euclidean algorithm and the unique factorisation that follow from the order, are the whole of the algebraic content of the system $\mathbb{Z}$; the analytic and geometric slots of the system do not exist, because the integers support neither a nontrivial distance that interacts with the arithmetic nor a geometry of their own, and the residue rings $\mathbb{Z}/n\mathbb{Z}$ are the first place where the arithmetic of $\mathbb{Z}$ produces a new structure rather than a property of $\mathbb{Z}$ itself.

Summary

The integers are the Grothendieck group of the additive monoid of the naturals: $\mathbb{Z} = (\mathbb{N}\times\mathbb{N})/{\sim}$ with $(a,b) \sim (c,d) \iff a+d = b+c$, with componentwise addition and the multiplication $(a,b)(c,d) = (ac+bd, ad+bc)$. The quotient is a commutative ring with identity, the map $n \mapsto [(n,0)]$ is an injective semiring homomorphism, and $\mathbb{Z}$ is the initial object among commutative rings with identity: every such ring receives a unique unital homomorphism from $\mathbb{Z}$, injective exactly in characteristic $0$. Additively, $\mathbb{Z}$ is the free group on one generator, and the abelian groups are its modules.

$\mathbb{Z}$ is an integral domain of characteristic zero with units $\pm 1$, in which cancellation holds and the division algorithm makes $\lvert\cdot\rvert$ a Euclidean function. Its order is the unique total order making it an ordered ring; the order is discrete, Archimedean and well ordered from below, with order type $\zeta$. Divisibility is governed by the associates $n = \pm m$, the irreducibles are the $\pm p$ with $p$ prime in $\mathbb{N}$, irreducible and prime coincide, the greatest common divisor exists and satisfies the Bézout identity $un + vm = d$, and every ideal is principal, so that $\mathbb{Z}$ is a Euclidean domain, a principal ideal domain and a unique factorisation domain with $n = \pm\prod p^{v_p(n)}$. The quotients $\mathbb{Z}/n\mathbb{Z}$ have $n$ elements and are fields exactly for $n$ prime, and the Chinese remainder theorem decomposes $\mathbb{Z}/mn\mathbb{Z}$ for coprime $m, n$. The naturals embed in $\mathbb{Z}$ as the nonnegative elements and $\mathbb{Z}$ embeds in $\mathbb{Q}$ as its fraction field, in $\mathbb{R}$ and in $\mathbb{C}$; $\mathbb{Z}$ is countable, of cardinality $\aleph_0$.

Summary of Notation

Symbol	Meaning
$\mathbb{Z}$	The ring of integers, $(\mathbb{N}\times\mathbb{N})/{\sim}$
$\iota$	Embedding $\mathbb{N} \hookrightarrow \mathbb{Z}$, $n \mapsto [(n,0)]$
$[(a,b)]$	Class of $(a,b)$, equal to $a - b$
$\lvert n\rvert$	Absolute value
$\leq$	The order of $\mathbb{Z}$, extending that of $\mathbb{N}$
$n \mid m$	Divisibility
$\gcd(n,m)$, $\operatorname{lcm}(n,m)$	Greatest common divisor and least common multiple
$p$	A prime
$v_p(n)$	Exponent of $p$ in $n$
$(n)$	Principal ideal generated by $n$
$\mathbb{Z}/n\mathbb{Z}$, $\mathbb{F}_p$	Ring of residues, field with $p$ elements
$\mathbb{Q}$	The field of fractions of $\mathbb{Z}$, the rational numbers
$\aleph_0$	Cardinality of $\mathbb{Z}$