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No. For example, let $R=\mathbb{Z}$, $M=\mathbb{Z}^2$, $N$ the subgroup generated by $(4,0)$ and $(2,1)$, and $L$ the subgroup generated by $(8,0)$ and $(0,2)$. Then $M/N$ and $N/L$ are both isomorph …

In which case this is proved in Corollary 5.9 of "Direct-sum representations of injective modules" by C. Faith and E.A. Walker, J. Algebra 5, 203-221 (1967). …

There's a clue to a counterexample at the end of Section 5 of "Direct-sum representations of injective modules" by C. Faith and E.A. Walker, J. Algebra 5, 203-221 (1967). … There are rings $R$ that have only three left ideals, $0$, $\text{rad}(R)$ and $R$, with $R/\text{rad}(R)\cong\text{rad}(R)$ as left modules, but which are not right Artinian. …

As @S.Carnahan shows. this is not true. Perhaps the statement you want is that the dual of $M[p^n]$ is $X/p^nX$? Take the exact sequence $$0\to M[p^n] \to M \to M \to M/p^nM \to 0,$$ where the mid …

Let $R$ be the four-dimensional algebra $k[x,y]/(x^2,y^2)$, where $k$ is an infinite field. For each $\lambda\in k$, $M_\lambda=R/(x-\lambda y)R$ is a two-dimensional module with one-dimensional radi …

Isn't the $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ $H$-supplemented but not $D2$?

Every simple module is trivially quasi-projective, and if every simple $R$-module is projective then $R$ is semisimple. So semisimple rings are the only rings for which quasi-projective implies projec …

They are equivalent. If any object of $\textrm{add}(X)$ satisfies (1) then so does $X$, and $A$ satisfies (1), so (2) implies (1). If $G$ satisfies (1) then let $I$ be the set of homomorphisms $\alp …

It's not true. Consider representations of the quiver $$\bullet\stackrel{\alpha}{\rightarrow}\bullet\stackrel{\beta}{\leftarrow}\bullet.$$ The representation $k \to k^2 \leftarrow k$, where the arr …

Arturo Magidin's answer is absolutely correct, but there's an earlier counterexample than Corner's. This question is Kaplansky's first "test problem" in his 1954 book on Infinite Abelian Groups, and …

For example, let $A$ be a finite-dimensional $R$-algebra, and $M\to N$ a non-split epimorphism of $A$-modules. … Then $\operatorname{Hom}_A(N,-)\to\operatorname{Hom}_A(M,-)$ is a non-split monomorphism of functors from finitely generated $A$-modules to finite-dimensional $R$-modules. …

(In fact $M\mapsto M_u$ is right adjoint to the functor $E\mapsto E(u)$ from representations to $R$-modules.) …

I think the paper "Whitehead Test Modules" by Jan Trlifaj (Trans. … In Lemma 2.4 he seems to prove that (if $R$ is not perfect) then for any cardinal $\kappa$, it is consistent for there to be a non-projective module $M$ such that $\operatorname{Ext}^1(M,N)=0$ for all modules …

No. Take $N=M=A$, where $A$ is any non-trivial algebra over a field $k$, and $B=k$.

By the second Brauer-Thrall conjecture, $\bar{k}\otimes_kA$ has infinitely many nonisomorphic indecomposable modules of some dimension, and so has some that are not defined over $k$ (i.e., not of the form …