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Jeremy Rickard's user avatar
Jeremy Rickard's user avatar
Jeremy Rickard's user avatar
Jeremy Rickard
  • Member for 12 years, 8 months
  • Last seen this week
  • Bristol, United Kingdom
7 votes
Accepted

Given a representation-infinite algebra, when is every AR component infinite?

7 votes
Accepted

Derived invariance of the Cartan determinant

7 votes

Whether a partial tilting complex has a complement

7 votes
Accepted

Why does every chain complex have a map into its cone?

7 votes
Accepted

Given a filtration of a finitely generated module over a noetherian ring that "looks" split, is it split?

7 votes
Accepted

Number of generators for the Schur multiplier of a finite group

6 votes

Two abelian groups, each being direct factor of the other

6 votes
Accepted

A simple colimit in the derived category?

6 votes
Accepted

Are all algebras Igusa-Todorov?

6 votes
Accepted

How to recognize different types of irreps

6 votes
Accepted

Number of homomorphism, or number of solution to equations, in finite groups

6 votes
Accepted

Does every finite abelian $p$-group $G$ admit a local ring structure with residue field of the same rank as $G$?

6 votes

New class of finite groups?

6 votes
Accepted

Simple object of $k[X,Y]/(Y^2)$

6 votes

When is $N^{*} \otimes_K M$ projective for a local Hopf algebra?

6 votes

A Hom-Tensor identity - $\text{Hom}_{R}(P,B)\otimes _SC \cong \text{Hom}_{R}(P,B \otimes_S C) $

6 votes
Accepted

Is flatness preserved under exterior power

6 votes

the relation between projective and quasi-projective modules

6 votes
Accepted

Cardinality of factors of infinite non-abelian groups

6 votes
Accepted

Is the following module over a group ring necessarily infinitely generated?

6 votes
Accepted

Is a inverse limit of indecomposable again indecomposable?

6 votes

If an abelian category $\mathcal{A}$ has enough injectives then so does $\mathrm{Ch}^{\geq 0}(\mathcal{A})$

6 votes
Accepted

Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding

6 votes

Is the image of a idempotent morphism in $\mathcal{K}(\mathcal{A})$ defined in the naive way?

6 votes
Accepted

Intersection of free objects

6 votes

free action on contractible spaces

6 votes
Accepted

Why can't one modify Kaplansky's proof to conclude that every projective module is a direct sum of its finitely generated projetive submodules?

6 votes

What is "van Dyck's theorem"

6 votes

global dimension

6 votes

Baer's criterion for projective modules

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