5 problems

Converse conjecture. If every semibrick is left finite, then $A$ is brick finite.

Let $A$ be a finite-dimensional algebra over a field $K$. A brick is a module $M\in\operatorname{\mathsf{mod}} A$ whose endomorphism algebra $\operatorname{\mathsf{End}}_A(M)$ is a…

Let $A$ be an algebra, and call a set of pairwise Hom-orthogonal bricks in $\operatorname{mod} A$ a semibrick. Call $A$ brick-infinite if it has infinitely many bricks up to isomor…

Let $\Lambda$ be a $\tau$-tilting finite algebra, and let $\mathcal{D} \sqcup \mathcal{U}[1]$ be a semibrick pair, meaning a collection of bricks in the indicated degrees satisfyin…

Let $\Lambda$ be a finite-dimensional algebra, and let $\mathcal{S}_p,\mathcal{S}_n\in\mathsf{sbrick}\Lambda$. A semibrick pair is $\mathcal{S}_p\sqcup\mathcal{S}_n[1]$ satisfying…