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Results tagged with homological-algebra
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user 22989
(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.
8
votes
Accepted
Left exact functor $F$ preserves quasi-isomorphism between $F$-acyclics
It's not true, without boundedness conditions, that a left exact functor always preserves quasi-isomorphisms between complexes of $F$-acyclic objects.
As alluded to in the question, a chain map is a q …
- 35.2k
10
votes
Accepted
Reference request: locally erasable delta-functor is universal
This is Proposition 4.2 in Buchsbaum’s Satellites and universal functors, Annals of Mathematics 71(2), pp. 199–209 (1960).
Well, to be precise, that is the dual result (for contravariant functors). Bu …
- 35.2k
7
votes
Accepted
A particular morphism being zero in the singularity category
Yes.
More generally, if $\mathcal{T}$ is a triangulated category and $\mathcal{S}$ is a thick subcategory, then any morphism $\varphi:M\to N$ of $\mathcal{T}$ that becomes zero in $\mathcal{T}/\mathca …
- 35.2k
7
votes
Accepted
Minimality of the Koszul resolution
In the category of ungraded bimodules, the multiplication map $R\otimes_\mathbb{C}R\to R$ is not a projective cover. For example, the proper sub-bimodule of $R\otimes_\mathbb{C}R$ generated by $1\otim …
- 35.2k
6
votes
Accepted
Does hereditary and connected imply that the underlying ring $k$ of a $k$-algebra is a field?
$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need …
- 35.2k
3
votes
A formula for the projective dimension of finite dimensional algebras
There are very easy examples (e.g., the path algebra of an $A_2$ quiver) where there is a nonzero projective module $P$ with $\operatorname{Hom}_A(P,\underline{A})=0$, so I assume you mean $\sup(\empt …
- 35.2k
3
votes
Comparing stabilization of stable category modulo injectives and a Verdier localization
This follows by applying Theorem 3.8 of Beligiannis' 2000 paper to the opposite
categories.
$\mathcal{I}$ is a full additive subcategory of $\mathcal{A}$, closed under
direct summands. It is covariant …
- 35.2k
5
votes
Accepted
Are module finite algebras over semiperfect rings again semiperfect?
No, even if $S$ is commutative. There may be easier counterexamples, but ...
There are commutative Noetherian local (and therefore semiperfect) rings $S$ with a finitely generated indecomposable modul …
- 35.2k
2
votes
Accepted
Using the mapping cone to show that a chain map defines a stable equivalence between two sym...
I'll give three answers, which basically say: (A) it doesn't matter,
(B) it's not true, and (C) here's (a sketch of) a proof.
But before that, there are a couple of relevant conditions in Linckelmann' …
- 35.2k
2
votes
Accepted
Vanishing of $\operatorname{Ext}_R^{1}(M,R)$ when $R$ is a Gorenstein local ring of dimensio...
Take $R=\mathbb{Z}_{(p)}$ for some prime $p$, with $x=p$, and
$M=\mathbb{Q}\oplus R$.
To show that this is a counterexample, the only nonobvious thing to show is that
$\operatorname{Ext}^{1}_{R}(\math …
- 35.2k
7
votes
Accepted
Semi-projective complexes of modules over a finite group
I think I have a counterexample.
Let $\operatorname{char}(k)=3$ and let $G$ be the symmetric group $S_{3}$.
Then $kG$ has two simple modules: the trivial module $k$ and another
one-dimensional module …
- 35.2k
6
votes
Accepted
Decompose an unbounded (cochain) complex in the homotopy category
Yes. Let
$$\tau^{\leq0}A^\bullet:= \cdots\to A^{-2}\to A^{-1}\to\ker(d^0)\to0\to\cdots$$
be the usual truncation of $A^\bullet$. Then the mapping cone of the inclusion map $\tau^{\leq0}A^\bullet\to A^ …
- 35.2k
6
votes
Unbounded acyclic resolutions
I'm afraid this is not very close to the case that you say you're most interested in ... maybe you want $F$ to preserve products?
But let $A=k[x]/(x^2)$, and let $\mathscr{A}$ be the category $\operat …
- 35.2k
2
votes
Accepted
Finitely generated module, which is a virtually small complex, embeds into a module of finit...
For every $M$, $M\oplus R$ is virtually small, so your question is equivalent to the question: Does every finitely generated $R$-module embed in a finitely generated module of finite projective dimen …
- 35.2k
6
votes
Injective modules
Yes.
Let $M$ be any $A$-module. Then its socle is a direct sum of simple modules: $\operatorname{soc}A=\bigoplus_iS_i$.
$A$ is a finite dimensional algebra, so the dual $\mathrm{Hom}_k(A,k)$ of $A$ is …
- 35.2k