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If $m\neq0$ then pick $0\neq x\in m$. The map $r\mapsto xr$ is a nonzero right module homomorphism from $R$ to $m$. So $R/m$ is a subquotient (as right module) of $m$ and so $|m|\geq |R/m|$. But $|R| …

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 …

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 …

Take $R=\mathbb{Z}_{(p)}$ for some prime $p$, with $x=p$, and $M=\mathbb{Q}\oplus R$. To show that this is a counterexample, the only nonobvious thing to show is that $\operatorname{Ext}^{1}_{R}(\math …

No, even if $S$ is commutative. There may be easier counterexamples, but ... There are commutative Noetherian local (and therefore semiperfect) rings $S$ with a finitely generated indecomposable modul …