CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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Quadratics can have four roots over $\mathbb Z/6$ and $\mathbb Z/8$

Statement refuted

Every nonzero quadratic over a commutative ring has at most two distinct roots.

Facts & Assumptions

Given: The polynomials $f=x^2-x$ in $(\mathbb Z/6)[x]$ and $g=x^2-1$ in $(\mathbb Z/8)[x]$ .

[L1]

The degree root bound holds for nonzero polynomials over integral domains (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L2]

Over any commutative ring, $a$ is a root exactly when $x-a$ divides the polynomial (Factor theorem over a commutative ring).

[L3]

The rings $\mathbb Z/6$ and $\mathbb Z/8$ are the corresponding quotient rings, so computations are modulo $6$ and $8$ (For every $n\in\mathbb N$ , the congruence-class ring $\mathbb Z/n$ is the quotient ring $\mathbb Z/n\mathbb Z$ ).

Counterexample

technique · direct

1.1

Modulo $6$ , the four distinct residues $0,1,3,4$ satisfy $a(a-1)=0$ , so they are roots of the nonzero quadratic $f$ .

givenL2L3algebra

2.1

Modulo $8$ , the four distinct odd residues $1,3,5,7$ have square congruent to $1$ , so they are roots of the nonzero quadratic $g$ ; these rings have zero divisors, so the domain hypothesis in [L1] fails and the statement is refuted.

givenL1L2L3algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 15 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

Neil Donaldson, Math 120B Notes, Section 23, root-bound counterexample (standard reference, not scraped)

Quadratics can have four roots over Z/6\mathbb Z/6Z/6 and Z/8\mathbb Z/8Z/8