Search Results
| 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 homological-algebra
Search options not deleted
user 2083
(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.
54
votes
Accepted
Do all exact $1 \to A \to A \times B \to B \to 1$ split for finite groups?
This is true (1). It was extended to finitely generated profinite groups here (2). Surprisingly, it is also true in the category of finitely generated modules over a Noetherian commutative ring (3).
…
- 30.6k
42
votes
4
answers
8k
views
Serre intersection formula and derived algebraic geometry?
Let $X$ be a regular scheme (all local rings are regular). Let $Y,Z$ be two closed subschemes defined by ideals sheaves $\mathcal I,\mathcal J$. Serre gave a beautiful formula to count the intersectio …
- 30.6k
26
votes
Short exact sequences every mathematician should know
Given a finitely generated module $M$ over a commutative Noetherian ring $R$, there is a short exact sequence $$0\to M_1 \to R^n \to M\to 0$$
where you map $1$ in each $R$ to a generator of $M$ and $M …
21
votes
Serre's theorem about regularity and homological dimension
ADDED: There is an account written by Buchsbaum (see page 1 and 2 of number 23 here) which described in more details what they wrote in [1]. So the localization problem for regular rings was definitel …
- 30.6k
21
votes
6
answers
3k
views
A ring such that all projectives are stably free but not all projectives are free?
This question is motivated by this recent question. Suppose $R$ is commutative, Noetherian ring and $M$ a finitely generated $R$-module. Let $FD(M)$ and $PD(M)$ be the shortest length of free and proj …
- 30.6k
20
votes
Accepted
Can a module be an extension in two really different ways?
It is worth noting some very interesting cases when the answer is yes. An amazing result by Miyata states that if $R$ is Noetherian and commutative, $M,N$ are finitely generated and $E \cong M\oplus …
- 30.6k
14
votes
Accepted
Why do modules with small support have high Exts?
To understand what "nice" is in your sense has been a very interesting question in commutative algebra.
In the following discussion I will assume, unless otherwise notice, that $(R,m,k)$ is Noetheria …
- 30.6k
13
votes
0
answers
494
views
Are the supports of $Ext^i(M,N)$ eventually periodic?
Let $R$ be a Noetherian, commutative ring and $M,N$ be finitely generated $R$-modules.
Question: Do the sets of minimal primes of $\text{Ext}^i_R(M,N)$ (for a fixed pair of $M,N$) become periodic fo …
- 30.6k
12
votes
Accepted
Differential graded structures on free resolution?
It is true if the projective dimension of $M$ over $A$ is at most $3$, and counter examples exist when the projective dimension is $4$.
The first counter example was given in Lucho Avramov's paper
" …
- 30.6k
12
votes
Accepted
example of Local cohomology
Take $M$ to be the second syzygy of $k$ over $S=k[x_1,x_2,x_3]$. Then a graded version of local duality tells us that $H^2_m(M)$ is dual to $Ext^1(M,R)= Ext^3(k,R)$, the last one is $k$ either by dir …
- 30.6k
12
votes
Accepted
Projective & injective dimensions
To complement Mariano's answer: If finite projective dimension implies finite injective dimension for any module $M$, then $R$ better have finite injective dimension (the converse is also quite easy). …
- 30.6k
11
votes
Accepted
Commutative algebras with modules of small complexity
There are no other examples. This property is equivalent to $A$ being a hypersurface (see Avramov's note "Infinite Free Resolutions"). By Cohen Structure Theorem, an Artinian local hypersurface (which …
- 30.6k
11
votes
Tor and projective dimension
Since your question is really about projective dimension of flat modules, it is worth noting the following result (see Raynaud-Gruson MR0308104, Cor 3.3.2 or Jensen MR0407091, Thm 5.8) which compleme …
- 30.6k
10
votes
Accepted
Can we say anything about the Krull dimension of a localization?
The dimension of $R[1/v]$ is the biggest height of some prime ideal $P$ such that $v\notin P$. So, let $I_{d-1}$ be the intersection of all primes of height at least $d-1$ ($d= \dim R$), then
$\di …
- 30.6k
8
votes
Accepted
Equivalence of definitions of Cohen-Macaulay type
For the equivalence you need two more assumptions: (a) $M$ to have finite projective dimension and (b) $R$ to be Gorenstein. (a) is implicitly required in the second condition, otherwise you don't hav …
- 30.6k