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Let $k$ be a field, let $R$ be the path algebra over $k$ of the quiver $$1\xrightarrow{\gamma}2\xrightarrow{\delta}3$$ modulo the relation $\gamma\delta=0$, and let $e=e_2$ be the idempotent associate …

There's a four-dimensional counterexample over any field. $A$ has basis $\{e,a,b,c\}$, with all products of basis elements zero except for $$e^2=e,\quad ab=c,\quad ea=a,\quad ec=c,\quad be=b,\quad ce= …

No. Let $R$ be the ring of eventually constant sequences of integers, and let $e_i\in R$ be the sequence whose $i^{th}$ term is 1, with all other terms zero. Then if $I$ is the annihilator of an eleme …

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 …

I'm no expert, but I think it follows from Theorem 11 and the rest of the discussion in Section 6 of Bergman, George M.; Lewin, Jacques, The semigroup of ideals of a fir is (usually) free, J. Lond. Ma …

There’s a natural injective module homomorphism $$R\to\bigoplus_iR/I_i$$ that takes $x$ into the semisimple submodule $\bigoplus_i\text{soc}(R/I_i)$, so the right ideal generated by $x$ is semisimple …

If $\Gamma$ acts $2$-transitively on an infinite set $X$, then the permutation module $\mathbb{Q}[X]$ will be a counterexample. For example, take an action of the free group of rank $2$ on a countabl …

There are abelian groups $A$ such that $A\cong A\oplus A \oplus A$ but $A\not\cong A\oplus A$. Let $E=\operatorname{End}(A)$. The functor $F=\operatorname{Hom}(A,-)$ is an equivalence from the catego …

In this answer, I was assuming the naive definition of a simple module $M$ for a nonunital ring $R$ as one with no proper nonzero submodules. It seems to be common to also insist that $RM\neq0$, in wh …

If $\Lambda$ is a ring, then the isomorphism classes of finitely generated projective $\Lambda$-modules form a commutative monoid $(A,+)$, with $[P]+[Q]=[P\oplus Q]$. This monoid contains a distinguis …

The answer to (2) is "no" even for hereditary rings. For example, if $S=T$ is the algebra of upper triangular $2\times 2$ matrices (or, more generally, pretty much any finite dimensional hereditary al …

Since $S$ has infinite projective dimension, there is some indecomposable summand $M$ of $\Omega^n(S)$ that has infinite projective dimension. For simple modules $T$ of finite injective dimension, $\ …

The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book Prest, M …

In the representation theory of finite dimensional algebras, at least, it's called a "trivial extension algebra" (although that sometimes refers to the special case where $N$ is the vector space dual …

This isn't an area that I'm expert on, and it's quite possible there's a much more elementary and/or more general answer. But if the Bass Conjecture on Hattori-Stallings ranks for group rings is true …