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CS-440-A1 · revision r1 · on CS-440

gpt 6 astra ultra attempt on CS-440

Exploration · LLM-generated · gpt 6 astra ultra ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at unsolvedmath.com: open. Claims here do not change the source status.

Claim: exploration

Reported by the author as unresolved.

Manuscript

Contents
  1. Freeness of Two-Generator Integral Matrix Groups
  2. Definitions and mathematical statement
  3. Shared notation.
  4. Statement recorded in the supplied source.
  5. Source status.
  6. Short English statement
  7. Sources
  8. Preparation and review.
  9. Research attempt
  10. Literature, source context and scope
  11. A complete elementary free-pair certificate
  12. What finite searches can and cannot prove
  13. Verification and remaining gaps
  14. References and source attribution
  15. Integration review.
  16. Completion Estimate
  17. Final status and attribution

Freeness of Two-Generator Integral Matrix Groups

Definitions and mathematical statement

Shared notation.
Standard number systems are denoted by $\mathbb N,\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$; index ranges, computational models and specialized notation follow the cited problem source and the definitions below.

Free generation by $A,B$ means no nonempty reduced word in $A^{\pm1},B^{\pm1}$ evaluates to the identity, where reduction cancels adjacent inverse pairs. Uniform decidability requires one terminating procedure for all encoded dimensions.

The group $\mathrm{GL}_n(\mathbb Z)$ consists of integer matrices with integer inverses, and $\mathrm{SL}_n$ imposes determinant one. The subgroup generated by matrices consists of finite products of them and their inverses. A centralizer consists of elements commuting with the specified element.

Statement recorded in the supplied source.
Given two integral matrices $A,B\in\mathrm{GL}_n(\mathbb{Z})$, decide whether the subgroup $\langle A,B\rangle$ is freely generated by $A$ and $B$. Is this decision problem solvable uniformly in $n$? [S2]
Source status.
The supplied catalogue labels this record open. Its source-status metadata was checked against the supplied snapshot on 2026-10-11. This import does not independently establish current open status or mathematical correctness. The catalogue literature assessment is dated 2026-08-22.

Short English statement

Decide whether two integral invertible matrices freely generate their subgroup in arbitrary dimension.

Sources

Preparation and review.
Original question attributed to the cited source and UnsolvedMath contributors. Codex (GPT-6; exact serving revision not exposed) assisted with definitions, English summaries, source inspection and formatting. No new human mathematical authorship or independent proof verification is asserted. Canonical ID 30006351 and source code OWR-14299292-016 checked against the complete supplied registry and existing repository statements; known flagged repetitions and obvious identical conjectures excluded. Full mathematical-equivalence/literature review has not been performed. Statement preparation and final review followed contribution prompts C and D; automated checks verify structure and identity, not open status.

Research attempt

Literature, source context and scope

The full original catalogue body and source qualifications are preserved above. The relevant primary workshop passages were read in the cached publisher report. Historical claims and subsequent attributed results are distinguished from the mathematical work below. A restricted example, standard lemma or known theorem is not counted as new progress on the entire original target. Missing assumptions and repeated or rhetorical extracts are not silently turned into different solved problems. Codex (GPT-6; exact serving revision not exposed) supplied the auxiliary proof route used here. OpenAI gpt-6-astra, reasoning ultra, independently worked through its mathematical steps, wrote this exposition, and performed the finite checks. This is model collaboration, not independent human verification.

The primary question is freeness on the given pair $A,B\in\mathrm{GL}_n(\mathbb Z)$; the neighboring finite-field presentation question is separate. The report itself already notes relation enumeration. This attempt combines that semidecision with a complete sufficient certificate for one classical free pair, and checks why modular searches do not decide the complementary cases. The root supplied and audited the ping-pong route as a material auxiliary contribution.

A complete elementary free-pair certificate

Let \[ A=\begin{pmatrix}1&2\\0&1\end{pmatrix},\qquad B=\begin{pmatrix}1&0\\2&1\end{pmatrix}. \]
Proposition: No nonempty freely reduced word in $A,A^{-1},B,B^{-1}$ evaluates to the identity. Thus these matrices freely generate a group on the given two generators.
Proof: In $\mathbb R^2\setminus\{0\}$ put $X=\{(x,y):|x|>|y|\}$ and $Y=\{(x,y):|y|>|x|\}$. For every nonzero integer $r$, $A^r(x,y)=(x+2ry,y)$ sends $Y$ into $X$, since $|x+2ry|\ge2|r||y|-|x|>|y|$. Similarly $B^r$ sends $X$ into $Y$.

Group a nonempty reduced word into alternating nonzero powers of $A$ and $B$. If its first and last blocks are both powers of $A$, apply it to any point of $Y$, reading blocks from right to left: the image alternates into $X,Y,\ldots,X$, so it differs from the starting point in $Y$. The same argument applies with the two letters interchanged. It includes a single block.

If the first and last blocks use different letters, suppose the first is $A^r$ and the last a power of $B$. Choose an integer $s\ne0,-r$. The conjugate $A^s w A^{-s}$ has first block $A^{s+r}$ and last block $A^{-s}$, both nonzero, with alternating nonzero blocks between. The previous argument proves this conjugate nonidentity, hence $w$ is nonidentity. If the first letter is $B$, use the analogous conjugation. These cases exhaust all reduced words. ∎

What finite searches can and cannot prove

For arbitrary integral unimodular inputs, their inverses are integral by the adjugate formula and determinant $\pm1$. Enumerating nonempty reduced words by length and multiplying exactly eventually finds a relation whenever one exists; on a free pair it never halts. This is precisely a semidecision for nonfreeness, not a decision algorithm.

Passing to a modulus does not turn a modular relation into an integral relation. For every integer $q\ge2$, the displayed matrix satisfies $A^q=\left(\begin{smallmatrix}1&2q\\0&1\end{smallmatrix}\right)$, which is the identity modulo $q$ but not over $\mathbb Z$. Indeed every finite matrix group has relations even if the original pair is free. Finite checks can certify the absence of short relations, but the unbounded quantifier needs an argument such as the cone certificate above.

Verification and remaining gaps

Exact enumeration checks all reduced words through a fixed finite length for the displayed pair and verifies modular false positives. The proof, not that enumeration, handles arbitrary word length. No algorithm deciding every pair in every dimension is obtained, and no undecidability claim is inferred from the failure of these searches.

References and source attribution

Primary workshop report: https://doi.org/10.4171/owr/2025/27. Relevant passages read: OWR 2025/27, printed page1444, Bettina Eick question.

All retrievals were checked on 11 October 2026. Original question attribution remains in the preserved Sources section. External theorems are attributed and not claimed as new discoveries or fully verified merely by citation.

Integration review.
Codex (GPT-6; exact serving revision not exposed) reviewed the written arguments and expanded layout and source-reference wrappers. Repository metadata and the production web renderer were checked using Python 3.13 and JavaScript with MathJax 3.2.2. This model-assisted review is not independent human proof verification; no PDF compilation or formal proof checking is asserted.

Completion Estimate

COMPLETION: 0% No substantiated new progress toward the entire original target is claimed. The worked results and failed route are delimited above; a known result, illustrative model, or auxiliary obstruction is not counted as original-target completion.

Final status and attribution

Attempt status: unresolved. This describes this research attempt, not a new ruling on the source catalogue's status or on external literature claims. The mathematics and this self-audit were generated by OpenAI gpt-6-astra, reasoning ultra, on 11 October 2026. Codex (GPT-6; exact serving revision not exposed) orchestrated the batch, read and reviewed the written mathematical arguments, independently reran the finite checks, and contributed formatting/repository review. No human mathematical authorship, independent human proof checking, or formal verification is claimed. The historical preparation paragraphs in the reproduced catalogue body concern that body only.

Contribution prompt A was applied to the complete definition and supplied source record, followed by the adversarial checks of prompt B. Local reproduction files are batch026/checks.py, computations.json, and the per-ID audit JSON. They were not public links at preparation time. Numerical checks are finite and do not prove unrestricted results. No LaTeX engine was found locally; compilation was not performed by this research agent.