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Yes. More generally, if $\mathcal{T}$ is a triangulated category and $\mathcal{S}$ is a thick subcategory, then any morphism $\varphi:M\to N$ of $\mathcal{T}$ that becomes zero in $\mathcal{T}/\mathca …

In the category of ungraded bimodules, the multiplication map $R\otimes_\mathbb{C}R\to R$ is not a projective cover. For example, the proper sub-bimodule of $R\otimes_\mathbb{C}R$ generated by $1\otim …

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 …

Not even for a Gorenstein ring. Let $R$ be $\mathbb{Z}_p$, the ring of $p$-adic integers. It is Gorenstein, and therefore its own dualizing module. Let $M$ be the direct sum of the two complexes $$\cd …

No, even if $S$ is commutative. There may be easier counterexamples, but ... There are commutative Noetherian local (and therefore semiperfect) rings $S$ with a finitely generated indecomposable modul …

Take $R=\mathbb{Z}_{(p)}$ for some prime $p$, with $x=p$, and $M=\mathbb{Q}\oplus R$. To show that this is a counterexample, the only nonobvious thing to show is that $\operatorname{Ext}^{1}_{R}(\math …

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.

The Baer-Specker group $B$, the direct product of countably many copies of $\mathbb{Z}$, is semi-free but not free. It is semi-free, because for any nonzero element $x\in B$ there is some projection $ …

There is at least one proof of Specker's theorem that can be adapted in an obvious way. I believe that the first half of this proof is due to Sąsiada, and the second half to Łoś. Let $A$ be a free $\m …

This is not a complete answer, but a construction that might give an answer. I'll start by constructing a ring with several objects (a.k.a. preadditive category) $\mathcal{C}$ by generators and relati …

I just stumbled across the answer to this in Fuchs' 2015 book on Abelian Groups. The papers Hill, Paul, Two problems of Fuchs concerning tor and hom, J. Algebra 19, 379-383 (1971). ZBL0228.20027. and …

Interpreting the various matrices as maps between free modules in the usual way, the question becomes: If $M$ is a submodule of $R^m$, and $\alpha,\beta: R^k\to M$ are epimorphisms, then must $\alpha$ …

For every $M$, $M\oplus R$ is virtually small, so your question is equivalent to the question: Does every finitely generated $R$-module embed in a finitely generated module of finite projective dimen …

Regarding the question about the Steenrod algebra, it is not true that every non-negatively graded module for the $\text{mod }2$ Steenrod algebra is a direct sum of indecomposable modules. I haven't c …

There's a nice, short proof in Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, …