CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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Every nonzero integer $n$ is $u \prod_{i < r} p_{i}$ with $u \in {1, - 1}$ and every $p_{i}$ prime; $u$ and $r$ are determined by $n$ , and the list is determined up to a permutation

Statement

Let $n \in Z$ with $n \neq 0$ , and take finite products in the commutative monoid $(Z, \cdot, 1)$ of $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , as in The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity.

Existence. There are $u \in {1, - 1}$ , $r \in N$ and a list $p : r \to Z$ of primes (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) with

$n = u \prod_{i < r} p_{i} .$
Uniqueness. If also $n = u^{'} \prod_{j < s} q_{j}$ with $u^{'} \in {1, - 1}$ and $q : s \to Z$ a list of primes, then $u = u^{'}$ , $r = s$ , and $q_{i} = p_{π (i)}$ for every $i < r$ , for some $π \in Sym (r)$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

Facts & Assumptions

Given: A nonzero integer $n$ .

[L1]

Every integer $m \geq 1$ is $\prod_{i < r} p_{i}$ for some $r \in N$ and some list $p : r \to Z$ of primes; and conversely every such product is $\geq 1$ (Every integer $n \geq 1$ is a finite product of primes: there are $r \in N$ and a list $p : r \to Z$ of primes with $n = \prod_{i < r} p_{i}$ , the case $n = 1$ being the empty product, The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , Semigroup and monoid).

[L2]

If $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ for lists of primes, then $r = s$ and $q_{i} = p_{π (i)}$ for all $i < r$ , for some $π \in Sym (r)$ (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ , The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[L3]

$∣ x ∣ = x$ when $x \geq 0$ and $∣ x ∣ = - x$ when $x < 0$ (The absolute value $∣ a ∣$ of an integer); $∣ x ∣ \geq 0$ , $∣ x ∣ = 0$ exactly when $x = 0$ , and $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ (Absolute value in $Z$ : $∣ a ∣ \geq 0$ ; $∣ a ∣ = 0$ exactly when $a = 0$ ; $∣ - a ∣ = ∣ a ∣$ ; $∣ a b ∣ = ∣ a ∣ ∣ b ∣$ ; $- ∣ a ∣ \leq a \leq ∣ a ∣$ ; and $∣ a ∣ \leq c$ exactly when $- c \leq a \leq c$ ).

[L4]

If $x z = y z$ and $z \neq 0$ then $x = y$ (The integers have no zero divisors; multiplicative cancellation).

[L5]

$Z$ is a commutative ring: multiplication is associative and commutative, $x \cdot 1 = x$ , $x \cdot (- 1) = - x$ , and every $x$ has an additive inverse, with $- (- x) = x$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L6]

The order on $Z$ is total, antisymmetric and transitive and is compatible with addition (The integers form a totally ordered ring, Order on the integers).

[L7]

$ι : N \to Z$ is injective, preserves the order, and has as image exactly the nonnegative integers, with $ι (0) = 0$ and $ι (1) = 1$ (The naturals embed in the integers).

[L8]

On $N$ : $0 \leq k$ for every $k$ (Order on the natural numbers); $m < k$ exactly when $σ (m) \leq k$ (Discreteness: $σ (n)$ is the immediate successor); $1 = σ (0)$ (The natural numbers $N$ (von Neumann)).

Proof

technique · direct

1.1

$0 < 1$ , since $1 = ι (1)$ is nonnegative and differs from $0 = ι (0)$ by injectivity; and if $0 < x$ then $1 \leq x$ , because $x = ι (k)$ with $k \neq 0$ , so $1 = σ (0) \leq k$ and $ι$ preserves the order.

L7L8

2.1

$∣ n ∣ \geq 0$ and $∣ n ∣ \neq 0$ , so $∣ n ∣ > 0$ and hence $∣ n ∣ \geq 1$ .

step 1.1L3L6

2.2

For uniqueness, suppose $n = u P = u^{'} P^{'}$ where $P : = \prod_{i < r} p_{i}$ and $P^{'} : = \prod_{j < s} q_{j}$ and $u, u^{'} \in {1, - 1}$ . By [L1] both $P \geq 1$ and $P^{'} \geq 1$ , so both are positive and $∣ P ∣ = P$ , $∣ P^{'} ∣ = P^{'}$ .

step 1.1L1L3L6

3.1

By [L1] there are $r \in N$ and a list $p$ of primes of length $r$ with $∣ n ∣ = \prod_{i < r} p_{i}$ .

step 2.1L1choose

3.2

Taking absolute values, $∣ n ∣ = ∣ u ∣ ∣ P ∣ = P$ and likewise $∣ n ∣ = ∣ u^{'} ∣ ∣ P^{'} ∣ = P^{'}$ , since $∣ 1 ∣ = 1$ and $∣ - 1 ∣ = 1$ . Hence $P = P^{'}$ .

step 2.2L3L5

4.1

The order is total and $n \neq 0$ , so $n > 0$ or $n < 0$ . If $n > 0$ then $∣ n ∣ = n$ and $n = 1 \cdot \prod_{i < r} p_{i}$ ; if $n < 0$ then $∣ n ∣ = - n$ , so $n = - (- n) = - ∣ n ∣ = (- 1) \prod_{i < r} p_{i}$ . In both cases clause 1 holds, with $u = 1$ and $u = - 1$ respectively.

step 3.1L3L5L6

4.2

By [L2] applied to $P = P^{'}$ we get $r = s$ and a permutation $π \in Sym (r)$ with $q_{i} = p_{π (i)}$ for every $i < r$ .

step 3.2L2

4.3

And $u P = u^{'} P$ with $P \neq 0$ , since $P \geq 1 > 0$ ; cancellation gives $u = u^{'}$ .

step 2.2step 3.2L4L6

5.1

Clause 1 is step 4.1 and clause 2 is steps 4.2 and 4.3.

step 4.1step 4.2step 4.3∎

Remarks

This is why Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ can insist on $p > 1$ without loss. The sign of $n$ is carried by the unit $u$ , not by the primes, so admitting negative primes would buy nothing and would break uniqueness, since $(- 2) (- 3)$ and $2 \cdot 3$ would be different lists with the same product.
The uniqueness of $u$ needs $P \neq 0$ , and that is why the empty product is harmless. At $r = 0$ the statement reads $n = u$ , so the nonzero integers with an empty prime list are exactly $1$ and $- 1$ — which is $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ again, from a different direction.
Nothing is claimed at $n = 0$ . Zero is divisible by every prime and is not a product of primes times a unit at all, since every such product is $1$ or $- 1$ times a positive integer. It is excluded by hypothesis, not overlooked.

Depends on

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Sources

Fundamental theorem of arithmetic (Wikipedia) (standard reference, not scraped)
Inquiry into Advanced Algebra: Division, primes, and factorisation (standard reference, not scraped)

Every nonzero integer nn is u∏i<rpiu∏i<r​pi​ with u∈{1,−1}u∈{1,−1} and every pipi​ prime; uu and rr are determined by nn, and the list is determined up to a permutation