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I would like to understand the Tame-Wild dichotomy, and in particular why an algebra cannot be tame and (semi-)wild at the same time. I've looked in the papers by Drozd and Crawley-Boevey [D80, CB88]. …

In Tame and wild matrix problems, Drozd defines an algebra $A$ to be wild if there is an $A-k\langle x,y \rangle$-bimodule $M$, such that ${M \otimes}-$ reflects isomorphisms. Usually I've seen the de …

A few such examples are constructed in Bekkert, Viktor; Drozd, Yuriy; Futorny, Vyacheslav, Derived tame local and two-point algebras, J. Algebra 322, No. 7, 2433-2448 (2009). ZBL1191.16017.

Following the terminology of Drozd, Yuriy A., Derived tame and derived wild algebras, Algebra Discrete Math. 2004, No. 1, 57-74 (2004). ZBL1067.16028. let $A$ and $R$ be algebras over a field $k$. A s …

Let $\Lambda$ be the path algebra of the quiver with relations $(a^2, ac, ba, cbc)$. Then I claim $\operatorname{findim}(\Lambda) \geq 1$, while $\operatorname{findim}(\Lambda^{op})=0$. The projectiv …

Let $\Lambda$ be finite dimensional algebra over a field $k$. The (left) finitistic dimension of a finite dimensional algebra is defined as $$\operatorname{findim}(\Lambda)=\sup\{\operatorname{pd}M | …