CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-11

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.

A polynomial ring over an integral domain is an integral domain

If $R$ is an integral domain, then $R [x]$ is an integral domain.

Given: An integral domain $R$ .

[L1]

Polynomial convolution makes $R [x]$ a commutative ring and embeds $R$ injectively as the constant polynomials (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring).

[L2]

[L3]

An integral domain is a commutative ring with distinct zero and one and no zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors).

technique · direct

1.1givenL1

By [L1], $R [x]$ is a commutative ring and its zero and one are distinct because the constant embedding is injective.

2.1step 1.1L2L3∎

By [L2], two nonzero polynomials have nonzero product, so [L3] applied with step 1.1 makes $R [x]$ an integral domain.

Depends on