LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A reduced Laurent plane curve has a principal equation

Statement

Assume the Axiom of Choice. Let $C$ be a finite union of distinct irreducible one-dimensional closed algebraic subsets of $(C^{*})^{2}$ . Suppose its radical vanishing ideal in $C [M^{\pm 1}, L^{\pm 1}]$ is the scalar extension of a radical ideal in $Q [M^{\pm 1}, L^{\pm 1}]$ ; this is what it means here for the reduced union to be defined over $Q$ . Then its complex vanishing ideal is generated by a square-free Laurent polynomial $F$ with primitive integer coefficients. Any two such primitive integer-coefficient generators differ by multiplication by $\pm M^{a} L^{b}$ , where $a, b \in Z$ .

Statement references

Milne, Algebraic Geometry, v6.10, §§1–3

The Stacks Project, Section 10.9

Facts & Assumptions

Given: Full AC, the two-dimensional algebraic torus $(C^{*})^{2}$ , and the reduced finite union $C$ of distinct irreducible one-dimensional closed subvarieties, with a rational structure on the union.

[F1]

Every prime ideal of height one in a finite-variable polynomial ring over a field is principal and generated by an irreducible element (Every finite-variable polynomial ring over a field is a UFD, with prime irreducibles and principal height-one primes).

[F2]

For an affine domain of finite type over a field, its Krull dimension is the transcendence degree of its fraction field (Affine-domain dimension equals transcendence degree).

[F3]

Geometric dimension of an affine algebraic set over an algebraically closed field equals Krull dimension of its coordinate ring, under full AC (Affine geometric dimension equals ring dimension; The Axiom of Choice). The torus is affine: $(M, L) \mapsto (M, L, M^{- 1}, L^{- 1})$ identifies it with ${(M, L, U, V) : M U = L V = 1} \subset C^{4}$ .

[F4]

For a finite-type domain $A$ and prime $P$ , full AC gives $ht (P) + \dim (A / P) = \dim A$ (Height plus quotient dimension equals ambient dimension in an affine domain; The Axiom of Choice).

[F5]

Prime ideals disjoint from a localization set correspond by extension and contraction, preserving inclusions (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

[F6]

The Laurent polynomial ring is the localization of $C [M, L]$ obtained by inverting $M$ and $L$ ; its units are nonzero scalars times monomials, as follows from polynomial factorization in [F1].

[F7]

The fixed field of all automorphisms of $C$ fixing $Q$ is $Q$ (The fixed field of Aut(C/Q) is Q).

Proof

1.1F2F3F4given

Put $R = C [M, L]$ and $S = {M^{a} L^{b} : a, b \geq 0}$ . The coordinate ring of $(C^{*})^{2}$ is $A = S^{- 1} R$ , a finite-type domain with fraction field $C (M, L)$ of transcendence degree two. Each irreducible curve $Y$ in $C$ has prime ideal $P_{Y} \subset A$ , and its quotient coordinate ring has dimension one by [F3]. Thus [F2] gives $\dim A = 2$ and $\dim (A / P_{Y}) = 1$ ; [F4] yields $ht (P_{Y}) = 1$ .

2.1F1F5step 1.1algebra

Contract $P_{Y}$ to $p_{Y} = P_{Y} \cap R$ . It avoids $S$ because every element of $S$ is a unit in $A$ . By [F5], contraction and extension identify all primes below $P_{Y}$ with primes below $p_{Y}$ , so $p_{Y}$ has height one. By [F1], $p_{Y} = (f_{Y})$ for an irreducible polynomial $f_{Y} \in R$ . Since $p_{Y}$ avoids $M$ and $L$ , neither variable divides $f_{Y}$ . In the localization of the UFD $R$ , an irreducible remains prime unless it divides an inverted element; therefore $f_{Y} / 1$ is a nonunit prime and $P_{Y} = (f_{Y} / 1)$ .

3.1F1F6step 2.1algebra

Distinct components have distinct prime ideals, hence their irreducible generators $f_{Y}$ are pairwise nonassociate in the Laurent ring. A Laurent polynomial belongs to every $(f_{Y})$ exactly when each pairwise nonassociate prime factor divides it; unique factorization then makes it divisible by their product. If $C$ is empty, the vanishing ideal is the unit ideal and the empty product is $F_{0} = 1$ ; otherwise the same product over the components generates the vanishing ideal. In either case the generator is square-free, and any other generator differs by a Laurent unit.

4.1F6F7givenstep 3.1algebra

By the stated rational structure, every automorphism of $C$ fixing $Q$ preserves the vanishing ideal. Shift the finite support of $F_{0}$ so its lexicographically least exponent pair is $(0, 0)$ , then scale its coefficient there to $1$ . This gives a unique normalized generator: by [F6] any two generators differ by a Laurent unit, and the support shift and specified coefficient remove that unit. Every such field automorphism preserves the normalization, hence fixes each coefficient. By [F7] the coefficients are rational. Clearing denominators and dividing by their integer gcd gives a primitive integer Laurent polynomial. Any two primitive integer-coefficient generators differ by a Laurent unit with rational scalar part; primitivity forces that scalar to be $\pm 1$ .

5.1step 1.1step 2.1step 3.1step 4.1∎

Steps 1.1, 2.1, 3.1, 4.1 prove existence of the square-free primitive integer equation and show that its only ambiguity is $\pm M^{a} L^{b}$ .

Depends on

Used by

The ordinary knot A-polynomial Definition
Boundary eigenvalue curve elimination for the A-polynomial Lemma

Dependency tree · two levels

37 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

J. S. Milne, Algebraic Geometry v6.10, §§1–3 (standard reference, not scraped)
The Stacks Project, Section 10.9 (Tag 00CM), Localization (standard reference, not scraped)