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Results tagged with homological-algebra
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user 157483
(Co)chain complexes, abelian Categories, (pre)sheaves, (co)homology in various (possibly highly generalized) settings, spectra, derived functors, resolutions, spectral sequences, homotopy categories. Chain complexes in an abelian category form the heart of homological algebra.
6
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3
answers
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Is there a finite dimensional algebra with left finitistic dimension different from its righ...
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 | …
- 113
3
votes
1
answer
185
views
Explicit proof that algebra is derived wild
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 …
- 497
1
vote
Accepted
Explicit proof that algebra is derived wild
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.
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2
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Is there a finite dimensional algebra with left finitistic dimension different from its righ...
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 …
- 497
15
votes
1
answer
2k
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Has anyone seen this generalization of the snake lemma? Is it useful?
I originally posted this question on MSE (link), but was suggested to post here instead.
While learning about spectral sequences a friend of mine found a proof of the snake lemma using spectral seque …
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