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17 votes
3 answers
2k views

Characterising categories of vector spaces

Consider the category $FdVect_k$ of finite dimensional $k$-vector spaces, for some given field. It is abelian, semisimple, in that each object is a finite sum of simple objects (of which there is only ...
David Roberts's user avatar
  • 35.4k
13 votes
1 answer
5k views

What are tame and wild hereditary algebras?

What are tame and wild hereditary algebras? Are they related to hereditary rings? (Those are rings for which every left (resp. right) ideal is projective, equivalently, for which every left (resp. ...
Jose Brox's user avatar
  • 2,992
12 votes
1 answer
922 views

Does this algebra have finite global dimension ? (Human vs computer)

Usually computers can calculate the global dimension of a finite dimensional quiver algebra much faster than humans. But in this case a high end computer (calculating for 3 weeks) was not able to ...
Mare's user avatar
  • 26.5k
6 votes
1 answer
1k views

Solid rings and Tor

A solid ring is a ring $R$ such that the multiplication $R\otimes_{\mathbb{Z}} R \to R$ is an isomorphism. These were classified by Bousfield and Kan; they are subrings of $\mathbb{Q}$, $\mathbb{Z}/...
Jeff Strom's user avatar
  • 12.5k
34 votes
1 answer
5k views

Freyd-Mitchell's embedding theorem

Freyd–Mitchell's embedding theorem states that: if $A$ is a small abelian category, then there exists a ring R and a full, faithful and exact functor $F\colon A \to R\text{-}\mathrm{Mod}$. I have been ...
Bruno Stonek's user avatar
  • 3,004
20 votes
3 answers
2k views

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

(Edit: I've realized that there was an error in my reasoning when I was convincing myself that these two formulations are equivalent. Hailong has given a beautiful affirmative answer to my first ...
Anton Geraschenko's user avatar
7 votes
2 answers
484 views

Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring

Just recently I've stumbled across Warren Dicks' book Groups, trees and projective modules (1980) and I was pretty stunned. I know nothing of group cohomology, but I gather the "tree" ...
rschwieb's user avatar
  • 1,507
4 votes
1 answer
683 views

Quadratic algebras and Koszul algebras

Let $A$ be a quadratic algebra and $B$ the Ext-algebra of $A$. In case $A$ is a Koszul algebra, we should have that the global dimension of $A$ plus one is equal to the Loewy length of $B$ (is there a ...
Mare's user avatar
  • 26.5k
4 votes
1 answer
614 views

Characterisation of reflexive modules

Let $A$ be a semiperfect noetherian ring. A module $M$ is called reflexive in case the canonical map $f_M: M^{**} \cong M$ is an isomorphism, when $(-)^{*}:=Hom_A(-,A)$. This is equivalent to say that ...
Mare's user avatar
  • 26.5k
4 votes
0 answers
228 views

Question on $n$-torsionless modules

Let $A$ be a finite dimensional algebra. Recall that a module $M$ is called $n$-torsionfree in case $Ext_A^i(D(A),\tau(M))=0$ for all $i=1,...,n$ when $\tau$ denotes the Auslander-Reiten translate. ...
Mare's user avatar
  • 26.5k
32 votes
7 answers
4k views

"Sums-compact" objects = f.g. objects in categories of modules?

Hello, Let us call an object of an additive category sumpact (contraction of "sums" and "compact") if taking $Hom$ from it (considered as functor from the category to $Ab$) commutes with coproducts. ...
Sasha's user avatar
  • 5,562
19 votes
3 answers
13k views

Künneth formula for cohomology

$\DeclareMathOperator\Hom{Hom}$Is there an algebraic Künneth formula for cohomology? More precisely assume $A_{*}, B_{*}$ are chain complexes of free $R$-modules ($R$ is a $PID$) and $M, N$ are $R$-...
Axel's user avatar
  • 1,357
15 votes
2 answers
2k views

When is bar-cobar duality an equivalence?

Let $A$ be an augmented differential graded algebra over a field $k$. I will write $BA$ for its bar construction (whose homology is $Tor^A(k, k)$). This is a co-augmented differential graded ...
Craig Westerland's user avatar
10 votes
1 answer
1k views

Is there a way to define a prime ideal object via diagrams in the category of rings?

I like to think in terms of commutative diagrams rather than referring to elements. So to me a group is really a group object, i.e. an object with some maps satisfying certain commutative diagrams. ...
David White's user avatar
  • 30.3k
9 votes
2 answers
1k views

Extensions of an infinite product of copies of Z by Z

The question is simple: Let $P$ be an infinite direct product of copies of $\mathbb Z$. Do there exist any nontrivial extensions $$0 \to \mathbb Z \to E \to P \to 0$$ in the category of commutative ...
Xandi Tuni's user avatar
  • 4,015
8 votes
0 answers
4k views

Kunneth spectral sequence

In Rotman's Homological Algebra, 1st edition, there is written: Is every detail of 11.31-11.35 correct? Isn't the spectral sequence in 11.35 1st quadrant and not 3rd quadrant? Do 11.34-35 also hold ...
Leo's user avatar
  • 1,589
6 votes
0 answers
255 views

Homotopy quotient of groups

Suppose $0\to A \stackrel{\iota}{\to} B \stackrel{\pi}{\to} C \to 0$ is a short exact sequence of groups. We have an induced map $k[\iota] : k[A] \to k[B]$ of group algebras over a field $k$. What ...
Lukas Woike's user avatar
  • 1,382
6 votes
0 answers
182 views

On properties of an algebra as a bimodule

Let $A$ be a two-sided artinian ring. Recall that a module $M$ is said to have dominant dimension at least $n$ in case the terms $I_i$ in the minimal injective coresolution of $M$ are projective for $...
Mare's user avatar
  • 26.5k
5 votes
1 answer
353 views

Existence of non-trivial reflexive modules

Recall that a module $M$ over a ring $R$ is reflexive in case the natural evaluation map $f_M:M \rightarrow M^{**}$ (where $M^{*}=Hom_R(M,R)$) is an isomorphism, where $f_M(m)=g$ with $g(h)=h(m)$, ...
Mare's user avatar
  • 26.5k
5 votes
1 answer
367 views

Reference request: locally erasable delta-functor is universal

It is well-documented that an erasable delta-functor $(F^n,\delta)$ is universal. However, in 'small' abelian categories (in the technical sense or otherwise), there aren't always enough objects to ...
R. van Dobben de Bruyn's user avatar
5 votes
1 answer
750 views

What's the relationship between a $E_2$-Hochschild Cohomology module and a D-module?

Let's say for simplicity $A$ is a smooth algebra over a field $k$ ($A$ and $k$ are discrete commutative rings but from now on we are fully derived), and we will consider the $E_2$ algebra $HH^{\bullet}...
Andy Jiang's user avatar
  • 2,346
5 votes
1 answer
829 views

Rigid monoidal and closed monoidal categories

I am trying to understand the relationship between rigid monoidal categories and closed monoidal categories. First every rigid monoidal category is closed, with an adjoint to the functor $X \otimes -$ ...
Jake Wetlock's user avatar
  • 1,144
4 votes
1 answer
655 views

Are all algebras Igusa-Todorov?

A finite dimensional algebra A is called (n-)Igusa-Todorov in case there exists a module V such that for any module M there is an exact sequence: $0 \rightarrow V_2 \rightarrow V_1 \rightarrow \Omega^{...
Mare's user avatar
  • 26.5k
3 votes
1 answer
336 views

Why are relations of degree 3 or less enough in a presentation of the polynomial current Lie algebra g[t]?

Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$. The polynomial current Lie algebra $\mathfrak{g}[t] = \mathfrak{g} \otimes \mathbb{C} [t]$ has the bracket $$[xt^r, yt^...
B.R.J.'s user avatar
  • 138
1 vote
1 answer
195 views

Strongly graded algebras with no zero divisors

Let $A = \bigoplus_{i \in \mathbb{Z}} A_i$ be a strongly graded unital algebra over $\mathbb{C}$, with no zero divisors. Is it always true that $$ m: A_i \otimes_{A_0} A_j \to A_{i+j} $$ is an ...
Fofi Konstantopoulou's user avatar