Alphabeta Math
CorollaryStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck 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 RR be a commutative ring, aRa\in R, and fR[x]f\in R[x]. Then f(a)=0f(a)=0 if and only if xax-a divides ff in R[x]R[x].

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

Facts & Assumptions

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

[L1]

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

[L2]

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

Proof

technique · direct
1.1

If RR is the zero ring then R[x]R[x] has one element, f=0=0(xa)+f(a)f=0=0\cdot(x-a)+f(a), and the conclusion holds with q=0q=0; this case is separated because xa=0x-a=0 there, which has no leading coefficient and so is not monic, leaving [L1] inapplicable. Otherwise 101\ne0, so xax-a is monic of degree one. Apply [L1]; the remainder rr is zero or constant, and applying [L3] to f=q(xa)+rf=q(x-a)+r gives f(a)=q(a)(aa)+r=rf(a)=q(a)(a-a)+r=r, so f=q(xa)+f(a)f=q(x-a)+f(a).

givenL1L2L3
2.1

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

step 1.1L3algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 26 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources