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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 …

The answer is "no" unless $A=k$. Let $a\in A\setminus k$, and let $l_2$ and $l_1$ be the first and second rows of the commutative diagram $$\require{AMScd} \begin{CD} 0@>>>A@>\begin{pmatrix}1\\0\end{p …

Such rings were called "$F$-semiperfect", and more recently (thanks to rschweib for the information) "semiregular". One characterization is that these are the rings $R$ such that $\bar{R}=R/J(R)$, the …

No, not necessarily. Consider the case $B=kH$, $C=kG$ of finite group algebras over a field $k$, where $H\leq G$. I'll write $\downarrow$ and $\uparrow$ for restriction and induction. This case has a …

If $X$ is an $R$-module, there is a natural map $M\otimes_RX\to\text{Hom}_R\left(\text{Hom}_R(X,R),M\right)$ given by $m\otimes x\mapsto[\varphi\mapsto m\varphi(x)]$ that is easily checked to be an is …

Bass showed, in Corollary 4.5 of Bass, H., Big projective modules are free, Ill. J. Math. 7, 24-31 (1963). ZBL0115.26003, that every projective module for a connected Noetherian commutative ring is …

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 …

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 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 …

The motivation for the definition was an attempt to explain structurally the phenomenon of an "isotypy". This makes sense for arbitrary blocks, but let's stick to principal blocks for simplicity. Sup …

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 …

In the last paragraph of Kaplansky's proof, the construction could yield an infinite number of non-zero $x_{ij}$, even if each $M_i$ is finitely generated. The infinite matrix he produces will have fi …

I think the paper "Whitehead Test Modules" by Jan Trlifaj (Trans. AMS 348 (1996), 1521-1554) and the references in it, answer your question positively for perfect rings, but show that a negative answe …

No. A ring $R$ for which $R$ is projective as a left $R\otimes_{\mathbb Z}R^{op}$-module is sometimes called separable over ${\mathbb Z}$. This is equivalent to the splitting, as a surjection of $R\o …