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The answer to Question 1 is no. Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and let $B$ be the subgroup generated by $(2,1)$. Since $B$ is cyclic of order $4$, if it were contained in a …

$ that you describe in the first paragraph of your question, and that, in the case where the highest weight category is the module category of a finite dimensional algebra, that the direct sum of the modules … The algebraic groups/Lie algebras people were apparently not paying enough attention, and started referring to the $D(\lambda)$ as “the tilting modules”, which is wrong on possibly as many as three counts …

The main reason that it's important that the terms of $X$ are $p$-permutation modules is that, although it's possible to define the Brauer construction for general modules, it behaves much better for $ … The fact that $p$-permutation modules over $k$ lift to ($p$-permutation) modules over $\mathcal{O}G$, which is not true for modules in general, is needed to prove that a splendid equivalence lifts to characteristic …

If it's considered bad form to resurrect year-old threads, then please slap my wrist (gently, please; I'm new here!) A fairly simple explicit example of a "sumpact" module that is not f.g. is as foll …

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 …

From Googling, not personal knowledge: In Theorem 14 of Kaplansky, Irving, Modules over Dedekind rings and valuation rings, Trans. Am. Math. Soc. 72, 327-340 (1952). … And there's a whole Springer Lecture Notes volume by Brandal on the not-necessarily-domain case: Brandal, Willy, Commutative rings whose finitely generated modules decompose, Lecture Notes in Mathematics …

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 …

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 …

Here's an example of a full exact embedding of the module category of a non-Noetherian ring $S$ into that of a Noetherian ring $R$, preserving all direct sums and direct products. So this gives an exa …

The functor $F=\operatorname{Hom}(A,-)$ is an equivalence from the category of finite direct sums of copies of $A$ to the category of finitely generated free $E$-modules. …

This is an isomorphism of $S$-modules, since for $x,s\in S$ and $m\in M$, $m\otimes xs\mapsto[\varphi\mapsto m\varphi(xs)$. Thus $h^*$ has left adjoint $N\mapsto N\otimes_S\text{Hom}_R(S,R)$. … Again, this is an isomorphism of $S$-modules, since for $m\in M$, $\vartheta\in\text{Hom}_R(S,R)$ and $x,s\in S$, $(\vartheta s)(x)=\vartheta(sx)$, so the isomorphism (in the reverse direction) maps $ …

If $M$ is allowed to be infinitely generated, then there are counterexamples even for finite dimensional local algebras. Let $R=\mathbb{C}[x,y]/(x,y)^2$, a three-dimensional local $\mathbb{C}$-algebr …

If $A$ is connected and has infinite representation type, then every component of its Auslander-Reiten quiver is infinite. See, for example, Theorem 5.4 in Assem, Simson and Skowronski’s Elements of t …

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 …

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 …