CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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If a prime $p$ divides a finite product $\prod_{i < n} a_{i}$ of integers then $p ∣ a_{i}$ for some $i < n$ ; at $n = 0$ the product is $1$ and the hypothesis cannot hold

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), let $n \in N$ , and let $a : N \to Z$ , with the finite product $\prod_{i < n} a_{i}$ taken 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. If

$p ∣ \prod_{i < n} a_{i},$

then $p ∣ a_{i}$ for some $i < n$ .

At $n = 0$ the product is the empty product $1$ , and $p ∣ 1$ is impossible for a prime, so the hypothesis is never satisfied there and the assertion is true rather than excluded. Since $\prod_{i < n} a_{i}$ depends only on $a_{0}, \dots, a_{n - 1}$ (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), the statement applies verbatim to a finite list $a : n \to Z$ .

Facts & Assumptions

Given: A prime $p$ ; the property $P (n)$ : "for every family $a : N \to Z$ , if $p ∣ \prod_{i < n} a_{i}$ then $p ∣ a_{i}$ for some $i < n$ ".

[L1]

$(Z, \cdot, 1)$ is a commutative monoid, and $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ( $(Z, \cdot, 1)$ 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]

In a monoid, $\prod_{i < 0} g_{i} = e$ and $\prod_{i < σ (n)} g_{i} = (\prod_{i < n} g_{i}) \cdot g_{n}$ ; the value depends only on $g_{0}, \dots, g_{n - 1}$ (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).

[L3]

If $q$ is prime and $q ∣ u v$ then $q ∣ u$ or $q ∣ v$ (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ ).

[L4]

$p > 1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L5]

Induction: a property holding at $0$ and inherited by successors holds at every natural number (The principle of mathematical induction).

[L6]

On $N$ : $n < σ (n)$ , and $m < σ (n)$ exactly when $m \leq n$ (On $N$ the order is membership: $m < n ⟺ m \in n$ , The natural numbers $N$ (von Neumann)).

[L7]

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); $Z$ is a commutative ring (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L8]

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

[L9]

$d ∣ u$ means $u = d q$ for some $q \in Z$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ ).

Proof

technique · induction

1.1

$0 < 1$ , since $1 = ι (1)$ is nonnegative and $1 \neq 0$ by injectivity of $ι$ ; adding $- 1$ gives $- 1 < 0$ . Hence $- 1 < 0 < 1 < p$ , so $p \neq 1$ and $p \neq - 1$ .

L4L7L8

1.2

Let $n \in N$ and assume $P (n)$ .

1.3

Let $a : N \to Z$ and suppose $p ∣ \prod_{i < σ (n)} a_{i}$ . By the recursion clause this product equals $(\prod_{i < n} a_{i}) \cdot a_{n}$ , so [L3] gives $p ∣ \prod_{i < n} a_{i}$ or $p ∣ a_{n}$ .

L1L2L3L9

2.1

$P (0)$ holds: $\prod_{i < 0} a_{i} = 1$ for every family $a$ , and $p ∣ 1$ would force $p = 1$ or $p = - 1$ , which step 1.1 excludes. So the hypothesis of $P (0)$ is never satisfied and $P (0)$ is true vacuously.

basestep 1.1L1L2

2.2

In the first case step 1.2 supplies $i < n$ with $p ∣ a_{i}$ , and $i < n < σ (n)$ gives $i < σ (n)$ ; in the second case $i : = n$ works, since $n < σ (n)$ . Either way some $i < σ (n)$ has $p ∣ a_{i}$ , so $P (σ (n))$ holds.

step 1.2step 1.3L6

3.1

$P (0)$ holds and $P (n)$ implies $P (σ (n))$ for every $n$ , so $P (n)$ holds for every $n \in N$ .

step 2.1step 2.2L5discharge-induction∎

Remarks

The case $n = 0$ is real and is not an exception. The empty product is $1$ by 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, and a prime never divides $1$ , so the implication holds with a false hypothesis. Stating the corollary from $n = 1$ upwards would be a weaker theorem for no gain, and would have to be repaired at every use.
No commutativity is used. The proof consumes only the recursion clause $\prod_{i < σ (n)} a_{i} = (\prod_{i < n} a_{i}) a_{n}$ and Euclid's lemma; the monoid is named commutative only because $(Z, \cdot, 1)$ is, and because later items on this page do use commutativity.
A prime dividing a product of primes equals one of them. If every $a_{i}$ is prime and $p ∣ \prod_{i < n} a_{i}$ , then $p ∣ a_{k}$ for some $k$ , and then $p$ is a positive divisor of the prime $a_{k}$ other than $1$ , so $p = a_{k}$ by Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ . That one-line consequence is the engine of the uniqueness half of 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 $π \in Sym (r)$ .

Depends on

Used by

Dependency tree · two levels

56 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Euclid's lemma (Wikipedia) (standard reference, not scraped)
Number Theory: In Context — Fundamental Theorem of Arithmetic (standard reference, not scraped)

If a prime pp divides a finite product ∏i<nai∏i<n​ai​ of integers then p∣aip∣ai​ for some i<ni<n; at n=0n=0 the product is 11 and the hypothesis cannot hold