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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 …

Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions. …

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.

Countable torsion abelian groups are better behaved than uncountable ones. For example, Kaplansky’s “test problems” If $G$ and $H$ are isomorphic to direct summands of each other, is $G\cong H$? If …

Let $A=k[x]$ and $B=k[x]/(x^2)$, let $X$ be the complex $\hskip{.1in}\dots\to 0 \to B\stackrel{x}{\to} B\to 0\to \dots$, and let $Y$ be $\hskip{.1in}\dots\to 0\to k\stackrel{0}{\to}k\to 0\to\dots$. Th …

This is way too addictive, so I'm going to try to quit, and I'll just leave my thoughts here in case they're useful for other addicts. This is based on the ideas in the previous non-answer that I post …

There seems to be quite some literature about rings with this property, sometimes under the name "FGC domains". From Googling, not personal knowledge: In Theorem 14 of Kaplansky, Irving, Modules ove …

Let $R=k[x,y,z]/\left((xy-1)z\right)$, for $k$ some field. Let $A:R\to R$ be multiplication by $z$. Let $B:R\to R$ be multiplication by $xz$. Then $A$ and $B$ have the same image, since $z=xyz$, an …

A few months after the last activity on this question, Neena Gupta gave a proof that over a field $k$ of positive characteristic, a retract of a polynomial algebra need not be a polynomial algebra: ht …

There seems to be a classification of representations for the example you mention (and more generally for representations of $C_p$ over $\mathbb{Z}/p^s\mathbb{Z}$) in V. S. Drobotenko, E. S. Drobo …

Yes, it must be split. Since $M$ is an indecomposable module for an Artin algebra, its endomorphism ring $E$ is a local ring with nilpotent Jacobson radical $J(E)$. Say $J(E)^n=0$. Let the monomorphis …

In "Rings of continuous functions in which every finitely generated ideal is principal" by L. Gillman and M. Henriksen (Trans. Amer. Math. Soc. 82 (1956), 366-391 link), Example 3.4 is of a topologica …

My previous attempt was completely wrong, as Jason Starr politely pointed out. But I think the idea I was grasping for does work, in this example: Let $R=k[x,y]$ for a field $k$, and let $$M=\frac{ …

Take $R=\mathbb{Z}_p$, the $p$-adic integers, and $A=\mathbb{Q}_p/\mathbb{Z}_p$. Then $A$ is an Artinian $R$-module, but doesn't have a projective cover. [Since $R$ is local, projectives are free. I …

Without extra finiteness assumptions, this is not true in general. Even for $A=\mathbb{Z}$, there are infinitely generated $A$-modules $M$ that are locally free (in the sense that $M_\mathfrak{p}$ is …