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 750

(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.

6 votes
1 answer
334 views

A potential resolution of $R/r$

The DGA For $k$ some field, let $R$ be a $k$-algebra, and let $r\in R$. Define a differential graded algebra $\mathbf{R}_r$ as follows. As a graded algebra, it is isomorphic to $R\langle t\rangle$, t …
Greg Muller's user avatar
17 votes

Can a module be an extension in two really different ways?

I believe this is a counter example. Let $R=\mathbb{C}[x]$, and consider finite-dimensional modules (ie, f.d. vector spaces equipped with a distinguished endomorphism). For convenience, I will iden …
Greg Muller's user avatar
9 votes

Free resolution dimension?

When requiring finitely-generatedness of the resolution, then the free dimension of a projective module can be infinite. As a simple example, take the ring $R=k\oplus k$, and let $e_1=(1,0)$ and $e_2= …
Greg Muller's user avatar
14 votes

Elementary $\mathrm{Ext}^1$ intuition

It seems like you already can see this, but $Ext^1(M,N)$ is measuring all the ways to form distinct short exact sequences $0\to N\to ?\to M\to 0$ If you are looking for intuition, what you should do i …
Greg Muller's user avatar
3 votes

Graded or stacky Serre duality

Theres graded local duality which works just like local duality; however, it requires that $A_0$ is a field. I've had some luck making things work when A_0 is not a field, but then the local duality …
Greg Muller's user avatar
17 votes
3 answers
1k views

Freyd-Mitchell for triangulated categories?

Is there a nice analog of the Freyd-Mitchell theorem for triangulated categories (potentially with some requirements)? Freyd-Mitchell is the theorem which says that any small abelian category is a fu …
Greg Muller's user avatar