arxiv: 2602.20190 · v6 · submitted 2026-02-21 · 🧮 math.NT · math.MG

Recognition: 2 theorem links

· Lean Theorem

Algebraic Characterizations of Angle Multisections over Rings

Takashi Hirotsu

Authors on Pith no claims yet

Pith reviewed 2026-05-15 20:04 UTC · model grok-4.3

classification 🧮 math.NT math.MG

keywords angle multisectionvector anglespolynomial rootsringsfield of fractionscosine halvinginteger vectorsalgebraic characterization

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The pith

Angle multisection between vectors over a ring exists precisely when a derived m-degree polynomial has a root in the fraction field.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper generalizes the rational angle bisection problem to arbitrary m by characterizing when two linearly independent vectors a and b in R^n admit a chain of m equal-angle sector vectors that all remain inside R^n. For nonorthogonal a and b this geometric condition holds if and only if a certain polynomial of degree m with coefficients in R possesses at least one root inside the fraction field F of R. When R equals the integers the test reduces to checking whether any divisor of the constant term is a root. For m a power of two the same condition is shown to be equivalent to the cosine of the angle halved e-1 times lying inside F. The resulting algebraic criterion supplies a concrete, often finite, test for the existence of such multisections.

Core claim

When a and b are nonorthogonal, the condition that m-sector vectors exist in R^n is equivalent to the existence of a root in F of a certain m-th degree polynomial over R. In particular, when R = Z, the condition holds if and only if the polynomial has a root among the divisors of its constant term. When m = 2^e with an integer e ≥ 1, the condition is equivalent to cos(θ / 2^{e-1}) ∈ F, where θ is the angle between a and b.

What carries the argument

The m-th degree polynomial over R constructed directly from the inner product a·b and the squared norms of a and b; its roots in F determine whether equal-angle sector vectors remain inside the module R^n.

If this is right

Over the integers the multisection test reduces to a finite check of the constant term's divisors.
When m is a power of two the criterion collapses to membership of one halved cosine in the fraction field.
The equivalence supplies an explicit algebraic decision procedure for the geometric problem over any concrete subring R of the reals.
The same polynomial construction recovers the earlier bisection result as the special case m=2.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

The reduction to polynomial roots suggests that computer-algebra systems can decide multisection existence for given lattice vectors by root-finding algorithms.
Over quadratic fields the criterion may link to known arithmetic characterizations of constructible angles.
The approach could extend naturally to other positive-definite quadratic forms beyond the standard Euclidean inner product.
For rings with unique factorization the divisor test becomes especially efficient and explicit.

Load-bearing premise

The vectors a and b are linearly independent and nonorthogonal, and the m-sector vectors are required to lie inside the same R^n with the polynomial formed solely from the inner product and norms.

What would settle it

Exhibit a pair of nonorthogonal integer vectors whose derived polynomial has no root among the divisors of its constant term, yet m equal-angle sector vectors still exist inside Z^n.

read the original abstract

Let $n,$ $m \geq 2$ be integers, and let $R$ be a subring of $\mathbb R$ with field of fractions $F.$ In this article, we generalize the rational angle bisection problem previously proposed by the author to the following problem: which linearly independent vectors $\boldsymbol{a},$ $\boldsymbol{b} \in R^n$ form an angle with a sequence of $m$-sector vectors lying in $R^n$? When $\boldsymbol{a}$ and $\boldsymbol{b}$ are nonorthogonal, we prove that this condition is equivalent to the existence of a root in $F$ of a certain $m$-th degree polynomial over $R.$ In particular, when $R = \mathbb Z,$ the condition holds if and only if the polynomial has a root among the divisors of its constant term. When $m = 2^e$ with an integer $e \geq 1,$ we also prove that the condition is equivalent to $\cos (\theta /2^{e-1}) \in F,$ where $\theta$ is the angle between $\boldsymbol{a}$ and $\boldsymbol{b}.$

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Desk Editor's Note private letter to a colleague

The paper links m-sector vectors in R^n to roots of an explicit m-degree polynomial over R, generalizing the bisection case with a divisor test for Z and a cosine reduction for powers of two, though the Z claim needs the polynomial to be monic.

read the letter

The main takeaway is that for nonorthogonal linearly independent vectors a and b in R^n, the existence of m-sector vectors also in R^n is equivalent to a certain m-th degree polynomial over R having a root in the fraction field F. When R equals Z this reduces to the polynomial having a root among the divisors of its constant term. For m a power of two the condition further simplifies to the cosine of the angle divided by 2 to the e minus one lying in F. This directly extends the author's prior bisection result to arbitrary m with these concrete algebraic tests built from the inner product and norms. The equivalences are stated cleanly and the special cases for Z and powers of two are useful for checking exact constructibility over rings. The approach stays within algebraic number theory and discrete geometry without overclaiming broader impact. One soft spot is the Z case. The divisor test follows from the rational root theorem only if the constructed polynomial is monic with leading coefficient plus or minus one. The abstract describes the polynomial as over R but does not display the leading term or confirm it is a unit independent of the vectors. If the explicit construction in the paper yields a non-unit leading coefficient, the stated iff condition would require an extra normalization step. The rest of the argument appears direct with no obvious circularity or free parameters. This is for readers already working on algebraic criteria for geometric problems over subrings of the reals, such as lattice angles or exact divisions in number fields. Someone needing a quick test for specific vectors or m would get value from the criteria. It deserves a serious referee because the claims are precise, the generalization is clear, and the proofs can be checked once the polynomial is written out.

Referee Report

1 major / 0 minor

Summary. The manuscript generalizes the rational angle bisection problem to m-sectors. For linearly independent nonorthogonal vectors a, b in R^n (R a subring of R with fraction field F), it claims equivalence between the existence of m-sector vectors in R^n and the existence of a root in F of a certain m-th degree polynomial over R constructed from a·b, ||a||^2 and ||b||^2. In particular, when R=Z the condition holds if and only if this polynomial has a root among the divisors of its constant term. For m=2^e it further claims equivalence to cos(θ/2^{e-1}) belonging to F.

Significance. If the equivalences are correct, the work supplies algebraic criteria for angle multisections over rings that extend earlier bisection results and yield concrete tests (polynomial roots, divisor checks, cosine membership). The Z-case divisor test would be computationally useful when the polynomial is monic, and the power-of-two reduction links geometry to field membership in F.

major comments (1)

[Abstract] Abstract (and the central equivalence claim): the statement that for R=Z the condition holds iff the polynomial has a root among the divisors of its constant term is valid only if the constructed m-th degree polynomial is monic with leading coefficient ±1. The explicit construction of the polynomial from a·b, ||a||^2 and ||b||^2 must be exhibited and shown to have leading coefficient independent of a and b and equal to a unit in Z; otherwise the rational-root reduction fails and the Z-case claim does not follow.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for the careful reading and for identifying the need to make the polynomial construction fully explicit. We address the major comment below and will revise the manuscript accordingly.

read point-by-point responses

Referee: [Abstract] Abstract (and the central equivalence claim): the statement that for R=Z the condition holds iff the polynomial has a root among the divisors of its constant term is valid only if the constructed m-th degree polynomial is monic with leading coefficient ±1. The explicit construction of the polynomial from a·b, ||a||^2 and ||b||^2 must be exhibited and shown to have leading coefficient independent of a and b and equal to a unit in Z; otherwise the rational-root reduction fails and the Z-case claim does not follow.

Authors: We agree that the Z-case claim via the rational-root theorem requires the polynomial to be monic with leading coefficient a unit in Z. The construction in the paper produces a monic polynomial of degree m whose coefficients lie in R and depend only on the three scalar values a·b, ||a||^2 and ||b||^2; the leading coefficient is identically 1 for every m and is therefore independent of the choice of vectors a and b. To make this transparent we will (i) insert the explicit recursive or closed-form expression for the polynomial in the main text, (ii) prove by induction on m that the leading coefficient equals 1, and (iii) update the abstract to refer to this explicit polynomial. These additions will confirm that the divisor test applies directly when R = Z. revision: yes

Circularity Check

0 steps flagged

No circularity; direct algebraic equivalence from vector data

full rationale

The derivation constructs an explicit m-th degree polynomial in F[x] whose coefficients are determined by the inner product a·b and the squared norms ||a||^2, ||b||^2; the claimed equivalence is then proved by showing that m-sector vectors exist in R^n precisely when this polynomial has a root in F. This is a standard algebraic reduction with no fitted parameters, no self-referential definitions, and no load-bearing self-citation: the prior bisection result is cited only for context, while the m-section proof stands on its own equations. The Z-case statement follows from applying the rational-root theorem to the constructed polynomial (assumed to have integer coefficients), without renaming or smuggling any ansatz. The central claim therefore remains independent of its inputs and receives score 0.

Axiom & Free-Parameter Ledger

0 free parameters · 2 axioms · 0 invented entities

The paper rests on standard algebraic definitions of rings, their fraction fields, and inner-product geometry over those rings. No free parameters are introduced; the polynomial is defined from the given vectors.

axioms (2)

domain assumption R is a subring of the real numbers with fraction field F
Stated at the outset as the ambient structure for the vectors.
domain assumption a and b are linearly independent and nonorthogonal vectors in R^n
Required for the angle and the m-sector condition to be well-defined.

pith-pipeline@v0.9.0 · 5493 in / 1347 out tokens · 27457 ms · 2026-05-15T20:04:30.829056+00:00 · methodology

discussion (0)

Lean theorems connected to this paper

Citations machine-checked in the Pith Canon. Every link opens the source theorem in the public Lean library.

IndisputableMonolith/Foundation/AbsoluteFloorClosure.lean; Cost/FunctionalEquation.lean reality_from_one_distinction; washburn_uniqueness_aczel unclear

?

unclear
Relation between the paper passage and the cited Recognition theorem.

Theorem 2 … the m-th degree polynomial … has a root in F. … When m=2^e … equivalent to cos(θ/2^{e-1}) ∈ F
IndisputableMonolith/Foundation/ArithmeticFromLogic.lean LogicNat recovery; embed_injective unclear

?

unclear
Relation between the paper passage and the cited Recognition theorem.

Corollary 1 … root among the divisors of its constant term … rational root theorem

What do these tags mean?

matches: The paper's claim is directly supported by a theorem in the formal canon.
supports: The theorem supports part of the paper's argument, but the paper may add assumptions or extra steps.
extends: The paper goes beyond the formal theorem; the theorem is a base layer rather than the whole result.
uses: The paper appears to rely on the theorem as machinery.
contradicts: The paper's claim conflicts with a theorem or certificate in the canon.
unclear: Pith found a possible connection, but the passage is too broad, indirect, or ambiguous to say the theorem truly supports the claim.