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.$$ If $R$ is right hereditary then $aR$, as a right ideal of $R$, is projective, and so this short exact sequence splits (as a sequence of right $R$-modules). …

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 $ …

If the answer to Question 1 is ``no'', then this answers the supplementary question in the OP about modules. As for abelian groups: Clearly $R$ is countable. Question 2. Is $R$ torsion free? …

$\operatorname{Hom}_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ means that $\operatorname{Hom}_{\mathcal{T}}\left(X, \Sigma^i(Y)\right) = 0$ for all objects $X,Y$ of $\mathcal{M}$ and all integer …

So you could take $\mathcal{C}$ to be the full subcategory of finite direct sums of pure-injective modules. … By induction, $\mathcal{C}$ contains $X/\operatorname{rad}X$, which is an infinite direct sum of simple modules. …

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 …

(The structure of projective modules.), Invent. Math. 13, 295-304 (1971). ZBL0232.16020. Azumaya, Goro, F-semi-perfect modules, J. Algebra 136, No. 1, 73-85 (1991). ZBL0717.16005. …

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 …

There is quite a lot of literature to be found on semi-artinian rings and modules. … Dung; Smith, Patrick F., On semi-artinian $V$-modules, J. Pure Appl. Algebra 82, No. 1, 27-37 (1992). …

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

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 …

So we just need to find simple modules $S$ for $kH$ and $T$ for $kG$ for which $S$ is a composition factor of $T\!\downarrow$ but $T$ is not a composition factor of $S\!\uparrow$. … Then $kG$ has a $4$-dimensional simple projective module $T$ and $kH$ has three one-dimensional simple modules: take $S$ to be one of the non-trivial ones. …

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 …

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 …

$ 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 …