CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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$Q [x] / (x^{2} - 1)$ is not a field: reducibility creates nonzero zero divisors

Statement refuted

The quotient by any nonconstant polynomial over a field is a field.

Counterexample

In the rational field $Q$ (The rationals form a field), take $R = Q [x] / (x^{2} - 1) .$ The two classes $x - 1 + (x^{2} - 1)$ and $x + 1 + (x^{2} - 1)$ are nonzero, but their product is zero.

Facts & Assumptions

Given: The quotient $R = Q [x] / (x^{2} - 1)$ .

[F1]

Division by a nonzero polynomial gives a unique remainder of smaller degree (Division algorithm for polynomials over a field).

[F2]

$F [x] / (p)$ is a field if and only if the nonconstant polynomial $p$ is irreducible (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible).

Verification

technique · counterexample

1.1

Neither $x - 1$ nor $x + 1$ lies in $(x^{2} - 1)$ : each already has degree below $2$ , so uniqueness of the remainder in [F1] would otherwise make it zero.

2.1

Yet $(x - 1) (x + 1) = x^{2} - 1$ , so the product of their nonzero classes is zero.

step 1.1algebra

3.1

Thus $R$ has nonzero zero divisors and is not a field, agreeing with [F2] because $x^{2} - 1$ is reducible.

F2step 2.1∎

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: 46 results over 11 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

T. Judson, Abstract Algebra: Theory and Applications, Extension Fields (standard reference, not scraped)

Q[x]/(x2−1)Q[x]/(x2−1) is not a field: reducibility creates nonzero zero divisors