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(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.
20
votes
Accepted
Is the homotopy category of an abelian model category abelian?
No. The projective model structure on chain complexes of modules over a ring is an abelian model category, and the homotopy category is the derived category, which is never abelian unless the ring is …
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 …
16
votes
Accepted
When the restriction of a derived functor to a subcategory is the derived functor of the res...
In the example where $\mathcal{D}$ is the category of abelian groups and $\mathcal{C}$ is the category of finite abelian groups, take $F(X)=X\otimes_\mathbb{Z}\mathbb{Q}/\mathbb{Z}$. Then the restrict …
14
votes
"Sums-compact" objects = f.g. objects in categories of modules?
If it's considered bad form to resurrect year-old threads, then please slap my wrist (gently, please; I'm new here!)
A fairly simple explicit example of a "sumpact" module that is not f.g. is as foll …
12
votes
Accepted
Non isomorphic two term complexes with isomorphic kernel, image and cokernel
Let $R=k[x,y,z]/\left((xy-1)z\right)$, for $k$ some field.
Let $A:R\to R$ be multiplication by $z$.
Let $B:R\to R$ be multiplication by $xz$.
Then $A$ and $B$ have the same image, since $z=xyz$, an …
11
votes
The cohomology group $H^{1}(GL_{2}(\mathbb{F}_{p}), M_{2}(\mathbb{F}_{p}))$
Here's a (sketch of a) proof that this cohomology group is always zero using the fact that $G=GL_2(\mathbb{F}_p)$ has a cyclic Sylow $p$-subgroup (and so it definitely doesn't generalize easily to $GL …
11
votes
Accepted
Must the inclusion of an indecomposable module in the direct sum of two copies always split?
Yes, it must be split.
Since $M$ is an indecomposable module for an Artin algebra, its endomorphism ring $E$ is a local ring with nilpotent Jacobson radical $J(E)$. Say $J(E)^n=0$.
Let the monomorphis …
11
votes
Quasi isomorphisms in a commutative diagram
No.
Let $X'=Y'=Z'=0$, let $\alpha:Y\to Z$ be any quasi-isomorphism between acyclic complexes whose kernel is not acyclic, and let $X=\ker(\alpha)$.
For example, if $R=\mathbb{Z}$, then $Y$ could be $\ …
11
votes
Accepted
Is a "smooth" finite-dimensional algebra separable modulo its radical?
Let $K$ be an algebraic closure of $k$.
The following lemma must surely be well-known, but I haven't found an explicit reference, so I'll include a proof at the end of this post.
Lemma. If $S$ is …
10
votes
When is a functor a right derived functor?
Here's a counterexample for unbounded derived categories (this doesn't answer the revised question with the "$t$-left exact" condition).
Suppose $\mathcal{A}$ is a Grothendieck category with enough p …
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 …
9
votes
Commutativity of $\operatorname{Hom}$ and $\varprojlim$
No.
$\varprojlim_{n\in\mathbb{N}}\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}_p$, the $p$-adic integers.
Take $I=\mathbb{Q}_p$, the $p$-adic rationals.
There is a nonzero map $\mathbb{Z}_p\to\mathbb{Q}_p$, the …
9
votes
Accepted
Endomorphism ring of trivial source modules for abelian p-groups
Representations of $B$ (or at least an equivalent category) are studied in the literature under the name of "cohomological Mackey functors".
Theorem 1.1 of
Bouc, Serge; Stancu, Radu; Webb, Peter, On t …
9
votes
Accepted
Local property of split exact sequence
Without extra finiteness assumptions, this is not true in general.
Even for $A=\mathbb{Z}$, there are infinitely generated $A$-modules $M$ that are locally free (in the sense that $M_\mathfrak{p}$ is …
9
votes
What should I call a "differential" which cubes, rather than squares, to zero?
Such objects, for more general values of $3$, or at least the graded version (i.e., $\mathbb{Z}$-graded objects where $D$ is a degree one map with $D^N=0$) have attracted some interest in the represen …