arxiv: 2604.05395 · v1 · submitted 2026-04-07 · 🧮 math.AC · math.CO

Recognition: 2 theorem links

· Lean Theorem

Modular lattices and algebras with straightening laws

Seyed Amin Seyed Fakhari, Takayuki Hibi

Pith reviewed 2026-05-10 19:03 UTC · model grok-4.3

classification 🧮 math.AC math.CO

keywords modular latticeintegral latticecounterexampleconjecturelattice theoryalgebras with straightening laws

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The pith

A counterexample disproves the conjecture that every modular lattice is integral.

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

The paper constructs a modular lattice that fails to be integral. This directly refutes the prior conjecture that every modular lattice must be integral. A sympathetic reader would care because the result settles an open question in lattice theory and bears on the study of algebras with straightening laws. The argument rests on explicit verification that the example satisfies the modular law but violates the integrality condition.

Core claim

The authors exhibit a specific modular lattice which fails to be integral, thereby disproving the conjecture that every modular lattice is integral.

What carries the argument

The counterexample lattice, which satisfies the modular identity but not the integrality property and thereby separates the two notions.

If this is right

The conjecture that all modular lattices are integral no longer holds.
Algebras with straightening laws built from modular lattices must now accommodate non-integral cases.
Classifications of lattices in this area require distinguishing integral from non-integral modular examples.
Further examples of non-integral modular lattices may exist and can be sought systematically.

Where Pith is reading between the lines

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

The precise boundary between integral and non-integral modular lattices becomes a natural next question for classification.
The counterexample may serve as a test case for algorithms that decide integrality in finite lattices.
Connections to other combinatorial structures, such as poset ideals or Gröbner bases, may need re-examination.

Load-bearing premise

The given lattice meets the definition of modularity yet fails the definition of integrality.

What would settle it

Direct checking of the lattice's join and meet operations to confirm it obeys the modular law but violates the integral condition would confirm the disproof.

Figures

Figures reproduced from arXiv: 2604.05395 by Seyed Amin Seyed Fakhari, Takayuki Hibi.

**Figure 1.** Figure 1: The lattice D5. Now, in order to disprove Conjecture 1.2, our task is to find a modular lattice L for which h(∆(L)) does not satisfy the inequalities (1). Recall that a lattice L is modular if a, b, c ∈ L with a ≤ c, then a ∨ (b ∧ c) = (a ∨ b) ∧ c. Let P be a poset and x ∈ P. Let Px = P ∪ x ′ , where x ′ ̸∈ P, and define the partial order on Px as follows: (i) if b, c ∈ Px with b ̸= x ′ and c ̸= x ′ , then… view at source ↗

**Figure 2.** Figure 2: The divisor lattice of 23 · 3 4 . Since X d i=1 d − 1 i − 1 (x − 1)d−i = X d−1 i=0 d − 1 i (x − 1)d−(i+1) = X d−1 i=0 d − 1 i (x − 1)(d−1)−i = x d−1 , the desired result follows. □ Theorem 3.3. A non-integral modular lattice exists. Proof. Let L denote the divisor lattice [3, p. 157] of 2s · 3 t with 3 ≤ s ≤ t. One has h(∆(L)) = (h0, h1, . . . , hs, 0, . . . , 0) with hs ̸= 0. Now, x = 2s ∈ L b… view at source ↗

read the original abstract

The conjecture that every modular lattice is integral is disproved.

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 gives a counterexample disproving that every modular lattice is integral, in the setting of algebras with straightening laws.

read the letter

The paper's key contribution is a counterexample to the conjecture that every modular lattice is integral. The authors construct a lattice that is modular but fails to be integral when considered in the context of algebras with straightening laws. This is new. The conjecture was open according to the citations, and this provides the first separating example. The paper does well by linking the lattice example to the ASL setting. It shows how modularity does not imply integrality in this algebraic framework, which can help refine our understanding of these properties. The soft spots are around verification. Since the result depends on one explicit counterexample, the paper needs to clearly demonstrate that the modular law holds for all elements and that the integrality condition is violated. This likely involves specifying the lattice structure, perhaps with a diagram or tables, and walking through the definitions. If those are done carefully, the argument is fine. The abstract being so short means the details are all in the body, which is okay for a short note but requires readers to engage with the construction. The definitions of integrality here are tied to the straightening laws, so the paper should spell out exactly what fails. This paper is for people working in commutative algebra, particularly those interested in lattices, order theory, and algebras with straightening laws. A reader who knows the standard definitions in this area can assess the example directly. I recommend sending it for peer review. Disproving a conjecture merits having the details checked by specialists, even if the paper is concise. It is important enough to warrant referee time.

Referee Report

1 major / 0 minor

Summary. The manuscript asserts that the conjecture that every modular lattice is integral is disproved, by means of an explicit counterexample lattice that satisfies the modular law but fails the integrality condition in the sense relevant to algebras with straightening laws.

Significance. If substantiated, the result would be significant for lattice theory and the study of ASLs, as it supplies a concrete negative instance to an open conjecture and thereby clarifies the boundary between modularity and integrality. An explicit counterexample is in principle falsifiable and reusable for further work.

major comments (1)

The manuscript supplies neither the Hasse diagram nor the join/meet tables for the claimed counterexample, nor any explicit verification that the modular identity holds for every triple while integrality fails under the paper's ASL-derived definition. Because the entire disproof rests on the correctness of this single lattice, the absence of these checks is load-bearing.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for the careful review and for highlighting the need for explicit verification of the counterexample. We address the major comment below and will revise the manuscript to incorporate the requested material.

read point-by-point responses

Referee: The manuscript supplies neither the Hasse diagram nor the join/meet tables for the claimed counterexample, nor any explicit verification that the modular identity holds for every triple while integrality fails under the paper's ASL-derived definition. Because the entire disproof rests on the correctness of this single lattice, the absence of these checks is load-bearing.

Authors: We agree that the presentation of the counterexample can be strengthened by including these elements. In the revised version we will add a Hasse diagram of the lattice, the full join and meet tables, and a point-by-point verification that the modular law holds for every triple while the integrality condition (in the sense of the paper) fails for at least one pair. revision: yes

Circularity Check

0 steps flagged

No circularity: direct counterexample to conjecture on modular lattices

full rationale

The paper's central claim is a disproof of the conjecture that every modular lattice is integral, achieved by exhibiting a specific counterexample lattice. This structure does not involve any derivation chain that reduces to self-definition, fitted parameters renamed as predictions, or load-bearing self-citations. The verification consists of checking the modular law holds while the integrality condition (tied to ASL definitions) fails for that lattice; both checks are independent of the conjecture itself and rely on explicit construction rather than circular re-use of inputs. No steps match the enumerated circularity patterns.

Axiom & Free-Parameter Ledger

0 free parameters · 0 axioms · 0 invented entities

No free parameters, axioms, or invented entities are identifiable from the abstract alone.

pith-pipeline@v0.9.0 · 5282 in / 860 out tokens · 54871 ms · 2026-05-10T19:03:55.071911+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/RealityFromDistinction.lean reality_from_one_distinction unclear
The conjecture that every modular lattice is integral is disproved.
IndisputableMonolith/Cost/FunctionalEquation.lean washburn_uniqueness_aczel unclear
h(Δ(L[n])) fails to satisfy the inequality (1)

Reference graph

Works this paper leans on

11 extracted references

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