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.

Factor theorem over a commutative ring

Statement

Let $R$ be a commutative ring, $a \in R$ , and $f \in R [x]$ . Then $f (a) = 0$ if and only if $x - a$ divides $f$ in $R [x]$ .

More precisely, there is a unique $q \in R [x]$ such that $f = q (x - a) + f (a)$ .

Facts & Assumptions

Given: A commutative ring $R$ , an element $a \in R$ , and a polynomial $f \in R [x]$ .

[L1]

Division by the monic polynomial $x - a$ gives unique $q, r$ with $f = q (x - a) + r$ and $r = 0$ or $\deg r < 1$ (Division by a monic polynomial over a commutative ring).

[L2]

Evaluation at $a$ is the finite coefficient sum defining $f (a)$ (Evaluation and roots of a polynomial in a commutative target ring).

[L3]

Evaluation at $a$ is a unital ring homomorphism and sends $x$ to $a$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

Proof

technique · direct

1.1givenL1L2L3

If $R$ is the zero ring then $R [x]$ has one element, $f = 0 = 0 \cdot (x - a) + f (a)$ , and the conclusion holds with $q = 0$ ; this case is separated because $x - a = 0$ there, which has no leading coefficient and so is not monic, leaving [L1] inapplicable. Otherwise $1 \neq 0$ , so $x - a$ is monic of degree one. Apply [L1]; the remainder $r$ is zero or constant, and applying [L3] to $f = q (x - a) + r$ gives $f (a) = q (a) (a - a) + r = r$ , so $f = q (x - a) + f (a)$ .

2.1step 1.1L3algebra∎

If $f (a) = 0$ , step 1.1 gives $f = q (x - a)$ ; conversely, if $f = q (x - a)$ , applying [L3] gives $f (a) = q (a) (a - a) = 0$ , proving the biconditional.

Factor theorem over a commutative ring

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources