Abstract:Let $A$ be a finite dimensional algebra over an algebraically closed field. Motivated by some foundational interactions between bricks and $\tau$-rigid modules, we prove, in full generality, that if all but finitely many bricks of the algebra $A$ are $\tau$-rigid, then $A$ is brick-finite. Equivalently, any brick-infinite algebra admits infinitely many bricks which are not $\tau$-rigid. Because $\tau$-rigidity implies rigidity, our result verifies a weaker version of an open conjecture which states that if (almost) all bricks over $A$ are rigid, then $A$ should be brick-finite. In retrospect, this work strengthens some previous results and contributes to the recent studies of a series of challenging problems, all tied to the $2$nd brick-Brauer-Thrall conjecture. More specifically, without any tameness assumption, we settle a question that was previously known only for $E$-tame algebras.
From: Charles Paquette [view email]
[v1]
Thu, 25 Jun 2026 14:08:04 UTC (12 KB)
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