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Alexander polynomials from Seifert matrices — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

The knot-normalized Alexander fraction need not be a Laurent polynomial

Statement refuted

Assume the full Axiom of Choice and the existing knot convention DK(t)=ΔK(t)/(1−t). The fraction DK(t) is a Laurent polynomial in Λ=Z[t±1] for every smooth oriented knot K⊂S3.

Facts & Assumptions

Given: Full AC and the standard unknot U in the oriented three-sphere.

[A1]

Full AC is the choice-function axiom used by the Alexander-module and Seifert-matrix conventions. (The Axiom of Choice)

[F1]

The standard unknot is the oriented unit circle in a fixed coordinate two-plane of S3⊂R4. (Oriented connected sum and prime knot)

[F2]

A Seifert surface is a connected compact smooth oriented embedded surface with the given knot as oriented boundary; a disk has Euler characteristic 1 and genus 0. (Oriented Seifert surfaces in the three-sphere)

[F3]

A nonempty contractible space has the singular homology of a point. The singular chain complex of a point has ∂1=0 and ∂2=1, so H1(∗;Z)=0. (Contractible nonempty spaces have the homology of a point)

[F4]

The positive-push Seifert matrix is defined on an ordered basis of H1(F;Z), and the determinant of the empty 0×0 matrix is 1. (Seifert pairing and matrix of an oriented knot)

[F5]

The positive-meridian module is presented by V−tVT, the square presentation generates E0(AK) by its determinant, and that determinant is associated to ΔK(t). (The one-variable Alexander module of an oriented link, A Seifert matrix presents the one-variable Alexander module, The square Seifert presentation generates the knot Alexander polynomial, The Seifert determinant is the knot Alexander polynomial)

[F6]

For a knot, the separate knot-normalized fraction is DK(t)=ΔK(t)/(1−t). (The Alexander polynomial from the zeroth elementary ideal)

[F7]

The assignment t↦1 extends to a ring homomorphism ev⁡1:Λ→Z, since 1 is invertible in the Laurent ring. (The Laurent polynomial ring as the principal localisation of Z[t] at t)

[F8]

Connected-sum multiplicativity gives ΔK#J≐ΔKΔJ. (The knot Alexander polynomial multiplies under connected sum)

[F9]

Full AC implies the Axiom of Countable Choice, the stated hypothesis of the Seifert-surface definition. (AC implies DC implies countable choice, The Axiom of Countable Choice (ACω))

[F10]

Every generator ΔK satisfies ΔK(1)=±1. (Knot Alexander normalization, symmetry, and genus bound)

[F11]

The Alexander polynomial is unchanged up to a unit under oriented ambient isotopy, and a transported Seifert matrix is related by unimodular congruence. (The Alexander polynomial is an oriented link invariant, Seifert matrices of ambient-isotopic pairs are S-equivalent)

Counterexample

1.1A1F1F2F3F4F5F9F10F11givenconstruct

By [A1] and [F9], the choice hypothesis of [F2] holds. The standard unknot U bounds a hemisphere disk D in a coordinate 2-sphere in S3, oriented to induce the chosen orientation of U. The disk contracts linearly to its center, so [F3] gives H1(D;Z)=0. Its basis is empty, so its Seifert matrix is 0×0 and its determinant is 1 by [F4]. The matrix theorem and local E0 lemma in [F5] give ΔU≐1; by [F10] choose the representative ΔU=1. By [F11] the computation applies to the oriented unknot type, independently of the chosen representative.

2.1F6F7step 1.1algebra

By [F6], DU(t)=1/(1−t). If DU belonged to Λ, there would be q(t)∈Λ with (1−t)q(t)=1. Applying [F7] gives 0=1 in Z, a contradiction. Thus DU is not a Laurent polynomial.

3.1F7F8F11step 1.1step 2.1algebra∎

By [F8], ΔU#U≐ΔU2≐1. Any representative of DU#U therefore has the form ±tk/(1−t), whose numerator evaluates to ±1 at t=1; applying [F7] as in step 2.1 shows it is not in Λ.

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