CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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Division with a degree-small remainder can fail over $\mathbb Z$ 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 $f=qg+r$ with $r=0$ or $\deg r<\deg g$ .

Facts & Assumptions

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

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

Counterexample

technique · contradiction

1.1

Suppose for contradiction that $x=q(2x+1)+r$ with $r=0$ or $\deg r<1$ . If $q$ had positive degree $m$ and leading coefficient $c\ne0$ , then $q(2x+1)$ would have nonzero coefficient $2c$ 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 $1=2q$ , which has no integer solution because $2$ is not a unit by [L3]. This does not contradict the monic-division theorem [L1], since $2x+1$ is not monic; hence the claimed division statement fails.

step 1.1L1L2L3discharge-contradiction∎

Depends on

Used by

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Sources

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

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