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Results tagged with homological-algebra
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user 1409
(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.
1
vote
Accepted
cap products and injective abelian groups
The isomorphism, at least, exists. Take $M\leftarrow Y$ and $\mathbb Z\leftarrow X$ projective resolutions of $M$ and of $\mathbb Z$ as $G$-modules. The double complex $\hom_G(X,\hom_{\mathbb Z}(Y,D …
- 47.3k
4
votes
Accepted
An exercise in group cohomology
Fix the signs as Wilberd suggests in the comments, check that you get a natural automorphism of the complex computing cohomology, see that it induces in fact an automorphism of the universal $\delta$- …
- 47.3k
23
votes
Heuristic behind $A_{\infty}$ - algebras
If you have a differential algebra $A$, a complex $B$ and a isomorphism of complexes $f:A\to B$, then you can transport the structure (i.e., the multiplication) on $A$ to turn $B$ into a differential …
- 47.3k
10
votes
Accepted
What is the projective dimension of the ring $\mathbb{Z}_l[T,T^{-1}]$ ? l is a prime number
$\mathbb Z_p$ is a principal ideal domain, so it is Noetherian and its global dimension is $1$. Now, there is a general theorem that tells you that for all right Noetherian rings $R$ one has $$\operat …
- 47.3k
13
votes
Does the derived category remember the homological dimension?
There are lots of examples to be found in the theory of tilted algebras.
A tilted algebra $B$ is an algebra of the form $\operatorname{End}_A(T)$ with $A$ an hereditary finite dimensional algebra (ov …
- 47.3k
2
votes
Accepted
Do Gorenstein rings necessarily have a finite projective dimension (as a module over itself)?
Take $A=k[x]/(x^2)$ for a field $k$. This is a self-injective $k$-algebra (that is, it is an injective module over itself), so it is Gorenstein. Yet the residue field is of infinite projective dimensi …
- 47.3k
4
votes
Commutative Ring of Finite Global Dimension
An artinian ring $R$ is a finite direct product of local artinian rings. If $R$ is of finite global dimension, so are the factors, and then they are regular local by Serre's theorem. As regular local …
- 47.3k
7
votes
Accepted
Ostensibly different products on Ext-groups
Show that each of those products distributes over the others, and use Hilton-Eckmann (over a field, or for $A$ projective; in general, I don't know...) This ends up being then an exercise in using nat …
- 47.3k
3
votes
Accepted
Is $\mathbb{Z}_p$ flat $\mathbb{Z}_pG$-module for a finite $p$-group $G$?
(This is answering a comment to the main question)
$\newcommand\ZZ{\mathbb Z}$
If $G$ is cyclic of order $p$, then there is a resolution of $\ZZ$ looking like $$\cdots\to\ZZ G\xrightarrow{d_{\mathrm{ …
- 47.3k
4
votes
Accepted
Why is Ext^n(k,M) a vector space over k?
The action of an element $r\in R$ on $\mbox{Ext}_R^n(k,M)$ is the map $\mbox{Ext}_R^n(k,M)\to \mbox{Ext}_R^n(k,M)$ which is induced by either the map $k\to k$ given by multiplication by $r$, or by the …
- 47.3k
4
votes
Accepted
Generators of a certain ideal
A polynomial $f\in K[\underline Y]$ is in the kernel of your map iff $f$ is zero in the quotient $$\frac{k[X,Y]}{\bigl((X_i-X_j)Y_{i,j}-1:1\leq i<j\leq n\bigr)}.$$In other words, your kernel is the in …
- 47.3k
7
votes
Accepted
Does a finite dimensional algebra having a Cartan matrix with determinant 1 imply finite glo...
In general, no. See [Burgess, W. D.; Fuller, K. R.; Voss, E. R.; Zimmermann-Huisgen, B. The Cartan matrix as an indicator of finite global dimension for Artinian rings. Proc. Amer. Math. Soc. 95 (19 …
- 47.3k
15
votes
Accepted
How does one get the short exact sequence in a two-column spectral sequence?
This follows precisely from the very definition of convergence of the spectral sequence, once one has identified the $\infty$-term. It is done with some details in McLeary's User Guide---which is, in …
- 47.3k
9
votes
2
answers
460
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$N$-step simplicial complexes
Recently, answering a question here, Dror Bar-Natan observed that «way too often two-step complexes have a natural extension to become many-step complexes». By such a thing I mean (and I think Dror me …
- 47.3k
8
votes
Accepted
Hochschild homology of quiver algebras
The Hochschild homology of all path algebras $A=k\Gamma$ is known. I will assume the quiver is finite.
They are hereditary algebras, so $HH_p(A)$ is zero as soon as $p>0$ provided $A$ is finite dimen …
- 47.3k