LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every prime element of an integral domain is irreducible

Statement

Every prime element of an integral domain is irreducible.

Facts & Assumptions

Given: An integral domain $R$ and a prime element $p \in R$ .

[L1]

A prime element is nonzero and not a unit, and divides one factor of every product it divides; an irreducible element is nonzero and not a unit and has a unit factor in each of its factorizations (Irreducible and prime elements of an integral domain).

[L2]

Multiplicative cancellation by a nonzero element holds in an integral domain (Cancellation characterises domains: in a commutative ring with $1 \neq 0$ , the implication $a b = a c$ and $a \neq 0$ imply $b = c$ holds if and only if the ring has no zero divisors).

[L3]

An element is a unit when it has a two-sided multiplicative inverse (Left inverse, right inverse, and invertible element of a monoid).

Proof

technique · direct

1.1

Let $p = a b$ . Since $p ∣ p = a b$ , primality gives $p ∣ a$ or $p ∣ b$ .

L1given

2.1

If $p ∣ a$ , write $a = p c$ . Then $p = a b = p c b$ , and cancellation by nonzero $p$ gives $1 = c b = b c$ ; thus $b$ is a unit.

step 1.1L2L3given

2.2

If $p ∣ b$ , the symmetric argument gives that $a$ is a unit.

step 1.1L2L3given

3.1

Thus every factorization $p = a b$ has a unit factor, so $p$ is irreducible.

step 2.1step 2.2L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Sharifi, Abstract Algebra, Advanced Ring Theory (standard reference, not scraped)