Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted 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.

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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 $\ …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …
Jeremy Rickard's user avatar
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 …

1
2 3 4 5
9
15 30 50 per page