Search Results

This is Proposition 4.2 in Buchsbaum’s Satellites and universal functors, Annals of Mathematics 71(2), pp. 199–209 (1960). Well, to be precise, that is the dual result (for contravariant functors). Bu …

I posted the following example of a Grothendieck category with no simple objects in answer to a question on math.stackexchange about seven and a half years ago. I seem to have said at the time that it …

Let $k$ be a field, and let $A$ be the $3$-dimensional commutative $k$-algebra $k[x,y]/(x^2,xy,y^2)$. Then in the category of $A$-modules there is a unique indecomposable injective, namely the dual $D …

I'll give three answers, which basically say: (A) it doesn't matter, (B) it's not true, and (C) here's (a sketch of) a proof. But before that, there are a couple of relevant conditions in Linckelmann' …

The category of finite abelian $p$-groups (where $p$ is your favourite prime) is an abelian category with no proper nonzero Serre subcategories, but not every short exact sequence splits.

No. The third condition implies that $U$ is countable, and so has countably many finite subsets, and so has countably many finitely generated subgroups. But there are uncountably many finitely generat …

I'm afraid this is not very close to the case that you say you're most interested in ... maybe you want $F$ to preserve products? But let $A=k[x]/(x^2)$, and let $\mathscr{A}$ be the category $\operat …

Here's another example for Question 2 that I encountered in nature. In the derived category of modules for a ring, pick one module $M_i$ for each $i\in\mathbb{Z}$, and let $A_i=M_i[i]$. Then the natur …

Assuming that the question is about finitely generated modules, I think that the following gives a finite dimensional Frobenius algebra $A$ that is a counterexample. In fact, for any non-projective fi …

There is such a sequence, but it's not very interesting. Given an element of $\text{Hom}_{D^b(\mathcal{A})}(E,F[i])$, then in the same way you describe, this gives a distinguished triangle $$F\to Z_{i …

If $R$ is any (necessarily noncommutative if it is nonzero) ring that does not have the Invariant Basis Number property, then free $R$-modules on different numbers of generators can be isomorphic.

If $\mathcal{T}_{1}=\langle M_{1}\rangle_{d_{1}+1}$ and $\mathcal{T}_{2}=\langle M_{2}\rangle_{d_{2}+1}$, then $\mathcal{T}_{1}\ast\mathcal{T}_{2}\subseteq\langle M_{1}\oplus M_{2}\rangle_{d_{1}+d_{2} …

To answer the first question, there is a larger Krull-Schmidt category than $\text{mod}\,A$ unless $A$ has finite representation type. Every indecomposable pure-injective module has local endomorphism …

Here's a counterexample that appears in nature. Fix a prime $p$ and a field $k$ of characteristic $p$, and let $G=C_{p^{n}}$ be a cyclic group of order $p^{n}$ (where $n\geq1$ if $p$ is odd, and $n\ge …

I don't know the group of smallest order for which the conjecture has not been verified. But certainly it is known to be true for all groups of order less than 200. There are general results that deal …