New answers tagged homological-algebra
6
votes
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Two spectral sequences arising from a simplicial spectrum
The precise relation between the skeleton filtration and the levelwise Postnikov filtration is that the décalage of the first is isomorphic to the second. This is explained in [Ariotta: Coherent ...
- 1,758
2
votes
Accepted
Using the mapping cone to show that a chain map defines a stable equivalence between two symmetric algebras
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'...
- 33.2k
3
votes
Accepted
What conditions on an Abelian category allow members of a direct sum to be determined entirely by their components?
In B. Mitchell. Theory of categories (1965) three properties of abelian categories were introduced.
An abelian category is termed $C_1$ if it is cocomplete with exact coproducts, $C_2$ if it is ...
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4
votes
Accepted
Adjunctions and inverse limits of derived categories
A reference for exactly this type of problem in general is a paper by Horev and Yanovski called "On conjugates and adjoint descent".
Given a diagram $C \to D_i$ of left adjoints $f_i$ with ...
- 12.6k
1
vote
Adjunctions and inverse limits of derived categories
Not an answer, but too long for a comment: there are many details to adjust. I am not familiar with derived categories of rings, but I guess you are looking for an explicit description of the limit of ...
- 1,856
3
votes
Accepted
How is the classification of groups extensions by $H^2$ related to Yoneda Ext?
As I mentioned in my comment above, Gruenberg gave a direct bijection between $\operatorname{Ext}^1_{\mathbb ZG}(I_G,A)$ and group extensions $1\to A\to H\to G\to 1$. The details can be found in ...
- 36.8k
2
votes
Vanishing of $\operatorname{Ext}_R^{1}(M,R)$ when $R$ is a Gorenstein local ring of dimension $1$ and $M$ is not finitely generated
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}(\...
- 33.2k
4
votes
Accepted
Is $\mathrm{Hom}_R(M,R)\neq \mathfrak m \mathrm{Hom}_R(M,R)$ if $M \neq \mathfrak m M$ and $\mathrm{Hom}_R(M,R)\neq 0$?
$\DeclareMathOperator\Hom{Hom}$Yes. Choose $f\neq 0$ in $\Hom_R(M,R)$. Choose $v\in M$ such that $f(v)\neq 0$. Let $v^\ast:\Hom_R(M,R)\to R$ be given by $v^\ast(g)= g(v)$. Since the image of $v^\ast $ ...
- 53.6k
8
votes
How is the classification of groups extensions by $H^2$ related to Yoneda Ext?
I can tell you what to do in one direction; then the problem is to prove that this is well defined and bijective. Suppose you're given an element of $\operatorname{\rm Ext}^2_{\mathbb{Z}G}(\mathbb{Z},...
- 10.3k
3
votes
Accepted
Why is this map a split monomorphism?
Let me elaborate on what Leo says.
The morphism $f\otimes k$ lives in the derived category of bounded complexes of finite-dimensional $k$-vector spaces, which is equivalent to the category of globally ...
- 14.9k
0
votes
Finitistic dimension equal to cofinistic dimension for QF-3?
I think I found a counterexample by accident:
Let A be the following algebra given by quiver and relations (in QPA):
...
3
votes
The Krull dimension of the tensor product of rings
You are looking at infinite tensor products so we really shouldn't expect the infinite product to have finite dimension in most cases. Your second example was simply an infinite tensor product of $\...
- 128
4
votes
Accepted
Pontryagin product on the homology of cyclic groups
I recommend always looking at the canonical reference: Ken Brown's "Cohomology of Groups". Here Chapter V.5 is literally titled "The Pontraygin product" and then the very next ...
- 16.9k
3
votes
Group homology for a metacyclic group
Everything follows from the p-primary decomposition theorem, which is nicely explained in Ken Brown's group cohomology bible (Kasper's answer is the restriction-corestriction argument written circa ...
- 16.9k
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