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For questions about the derived categories of various abelian categories and questions regarding the derived category construction itself.
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 …
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 …
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 …
2
votes
Accepted
A perfect complex over a local Cohen--Macaulay ring whose canonical dual is concentrated in ...
Not even for a Gorenstein ring.
Let $R$ be $\mathbb{Z}_p$, the ring of $p$-adic integers. It is Gorenstein, and therefore its own dualizing module.
Let $M$ be the direct sum of the two complexes
$$\cd …
20
votes
Accepted
Is the functor from the unbounded derived category of coherent sheaves into the derived cate...
No, not always.
In
Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, Article ID …
6
votes
Accepted
Decompose an unbounded (cochain) complex in the homotopy category
Yes. Let
$$\tau^{\leq0}A^\bullet:= \cdots\to A^{-2}\to A^{-1}\to\ker(d^0)\to0\to\cdots$$
be the usual truncation of $A^\bullet$. Then the mapping cone of the inclusion map $\tau^{\leq0}A^\bullet\to A^ …
6
votes
Unbounded acyclic resolutions
I'm afraid this is not very close to the case that you say you're most interested in ... maybe you want $F$ to preserve products?
But let $A=k[x]/(x^2)$, and let $\mathscr{A}$ be the category $\operat …
2
votes
Accepted
Finitely generated module, which is a virtually small complex, embeds into a module of finit...
For every $M$, $M\oplus R$ is virtually small, so your question is equivalent to the question: Does every finitely generated $R$-module embed in a finitely generated module of finite projective dimen …
4
votes
Yoneda extensions in derived categories
There is such a sequence, but it's not very interesting.
Given an element of $\text{Hom}_{D^b(\mathcal{A})}(E,F[i])$, then in the same way you describe, this gives a distinguished triangle
$$F\to Z_{i …
20
votes
Accepted
Recovering an abelian category from the Ext of its simple objects
Here's a counterexample that appears in nature.
Fix a prime $p$ and a field $k$ of characteristic $p$, and let
$G=C_{p^{n}}$ be a cyclic group of order $p^{n}$ (where $n\geq1$ if
$p$ is odd, and $n\ge …
5
votes
Accepted
What is the smallest group for which Broué's abelian defect group conjecture has not yet bee...
I don't know the group of smallest order for which the conjecture has not been verified.
But certainly it is known to be true for all groups of order less than 200. There are general results that deal …
6
votes
Accepted
Rickard's strengthening of Broué's abelian defect group conjecture and the lifting of some e...
In almost all cases I know of where people have proved derived equivalences between blocks of finite groups, the proof hasn't really gone that way (i.e., finding a virtual bimodule and refining it to …
3
votes
Accepted
Smallness condition for augmented algebras
No.
Let $k$ be a field, and let $A$ be the algebra of upper triangular $2\times 2$ matrices over $k$, with augmentation map $\pmatrix{a&b\\0&c}\mapsto a$.
$A$ and $A^e$ have finite global dimension, s …
4
votes
Faithfully flat modules over a group algebra
Let $G$ be infinite cyclic, generated by $x$.
Let $M_\bullet$ be a free resolution of the $\mathbb{Z}[G]$-module $U=\mathbb{Z}/3\mathbb{Z}$ with $x$ acting by multiplication by $-1$. For example, tak …
4
votes
Accepted
Indecomposable objects in bounded derived category of $\mathbb C[x]/x^2$-mod
Up to shifts, every indecomposable object is of one of the forms described in the question.
I don't know an explicit reference, but here's a sketch of a proof.
By induction on the length, it's not h …