CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-11

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

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 $(Z / 6) [x]$ and $g = x^{2} - 1$ in $(Z / 8) [x]$ .

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

Counterexample

technique · direct

1.1givenL2L3algebra

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

2.1givenL1L2L3algebra∎

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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, Section 23, root-bound counterexample (standard reference, not scraped)

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