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CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-11
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Factor theorem over a commutative ring

Statement

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

More precisely, there is a unique q∈R[x] such that f=q(x−a)+f(a).

Facts & Assumptions

Given: A commutative ring R, an element a∈R, and a polynomial f∈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).

Proof

technique · direct
1.1

If R is the zero ring then R[x] has one element, f=0=0⋅(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≠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).

givenL1L2L3
2.1

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.

step 1.1L3algebra∎

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