CounterexampleConstruction: Literature-sourcedVerification: 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.

Division with a degree-small remainder can fail over $Z$ when the leading coefficient of the divisor is not a unit

Statement refuted

For every commutative ring $R$ and every nonzero $g \in R [x]$ , each $f \in R [x]$ can be written $f = q g + r$ with $r = 0$ or $\deg r < \deg g$ .

Facts & Assumptions

Given: The polynomials $f = x$ and $g = 2 x + 1$ in $Z [x]$ .

[L1]

Division is guaranteed over a commutative ring when the divisor is monic (Division by a monic polynomial over a commutative ring).

[L2]

The integers form a commutative ring (The integers form a commutative ring).

[L3]

The only units of $Z$ are $1$ and $- 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

Counterexample

technique · contradiction

1.1

Suppose for contradiction that $x = q (2 x + 1) + r$ with $r = 0$ or $\deg r < 1$ . If $q$ had positive degree $m$ and leading coefficient $c \neq 0$ , then $q (2 x + 1)$ would have nonzero coefficient $2 c$ in degree $m + 1 > 1$ , which the constant remainder could not cancel. Thus $q$ is a constant integer.

assume-contragivenL2algebra

2.1

Comparing coefficients of $x$ then gives $1 = 2 q$ , which has no integer solution because $2$ is not a unit by [L3]. This does not contradict the monic-division theorem [L1], since $2 x + 1$ is not monic; hence the claimed division statement fails.

step 1.1L1L2L3discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

33 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

Neil Donaldson, Math 120B Notes, discussion after Theorem 23.14 (standard reference, not scraped)

Division with a degree-small remainder can fail over ZZ when the leading coefficient of the divisor is not a unit