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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.
5
votes
Accepted
Do levelwise quasi-isomorphisms of bicomplexes induce a quasi-isomorphism between the total ...
For the $\prod$ version the answer is no.
Take $C^{p,q}$ to be
$$\begin{array}{ccccccccccc}
\mathbb{Z}&\to&\mathbb{Z}&\to&0&\to&0&\to&0&\to&\dots\\
\uparrow&&\uparrow&&\uparrow&&\uparrow&&\uparrow&& …
- 35.2k
1
vote
Accepted
Explicit indecomposable monomorphism of finitely generated non-indecomposable Abelian groups.
How about the map $f:\mathbb{Z}\to\mathbb{Z}\oplus(\mathbb{Z}/2\mathbb{Z})$ given by $f(n)=(2n,n)$?
Or for a finite example, the same idea works with a map $\mathbb{Z}/4\mathbb{Z}\to(\mathbb{Z}/8\mat …
- 35.2k
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 …
- 35.2k
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 $\ …
- 35.2k
7
votes
Accepted
Is such a map null-homotopic?
Probably there's a much less artificial example, and even more probably there's a much simpler one, but ...
Let $A=k[\varepsilon]/(\varepsilon^2)$, where $k$ is a field of characteristic two (purely …
- 35.2k
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 …
5
votes
Accepted
Can you test flatness on $FP_3$-modules?
Answering my own question in comments:
Let $k$ be a field, and $A=k\oplus V$, where $V$ is an infinite-dimensional square-zero ideal. Then I think $A$ has no non-projective $FP_3$-modules (using your …
- 35.2k
8
votes
Accepted
Tensor product of monomorphisms is a monomorphism?
Let $F$ be a field, and $k=F[x,y]/(x^2,xy,y^2)$. Since $k$ is a finite-dimensional $F$-algebra, flat=projective, and for $k$-modules $M,N$, there is a natural isomorphism $\operatorname{Hom}_k(M,N^\as …
- 35.2k
4
votes
Accepted
Finite universal delta-functors
I don't think $F^{d-\bullet}$ will often be universal, even when $\mathcal{A}$ does have enough projectives.
Let $A$ be any ring with finite global dimension that is not hereditary, so that there is …
- 35.2k
6
votes
Accepted
Is a inverse limit of indecomposable again indecomposable?
Suppose $M$ and $M'$ are two $R$-modules with a common submodule $N$ that has a descending chain of submodules
$$N=N_0\supseteq N_1\supseteq N_2\supseteq\dots$$
with zero intersection.
Then if $X^\b …
- 35.2k
1
vote
Accepted
Finite add(N)-resolution
I think that this is possible if and only if $\operatorname{Ext}^1_A(M,M)=0$, and so the question reduces to another question Ext^1 for a local finite dimensional selfinjective algebra that you have a …
- 35.2k
2
votes
Accepted
On some modules with bounded syzygies
For (1), take a quiver with two vertices, an arrow $\alpha$ from vertex $1$ to vertex $2$, a loop $\beta$ at vertex $2$, and relations $\alpha\beta=0$ and $\beta^2=0$.
For (2), take a quiver with fou …
- 35.2k
3
votes
Accepted
Direct sum of K-injectives over a noetherian ring
No.
The $K$-injective complexes form a triangulated subcategory of the homotopy category of complexes, and so if they were also closed under coproducts, the homotopy colimit of a sequence of $K$-inje …
- 35.2k
5
votes
When for every module $M$, $|E(M)| = |M|$
If $R$ is a finite-dimensional algebra (of dimension $d$, say) over a field $k$, and $M$ is a left $R$-module, then
$$\dim(M)\leq\dim\left(E(M)\right)\leq d.\dim(M),$$
and so if $k$ is infinite then $ …
- 35.2k
5
votes
Accepted
Is 'the' homotopy colimit of a sequence of exact triangles an exact triangle?
I'll assume that you at least want your triangulated category to have the property that a (countable) coproduct of exact triangles is an exact triangle?
Even then, I think this is probably not true i …
- 35.2k