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If $A$ happens to be (left) Noetherian then to show $f$ makes $B$ a faithfully flat right $A$-module it is enough to check that $B\otimes_A S\neq 0$ for every simple (left) $A$-module $S$ since, in th …

Theorem 4.2 of On separable algebras over a commutative ring says that the answer is always yes.

You might as well assume that $\mathrm{Ann}_R(R/I)=0$ since if $S=R/\mathrm{Ann}_R(R/I)$ then $R/I\cong S$ as $R$-modules if and only if $S/(I/\mathrm{Ann}_R(R/I))\cong S$ as $S$-modules. So now the q …

Following my nose gave the following argument. Writing $I$ be the left ideal generated by $x\partial^2$ and $x^3$ and using $\cdot$ to stress multiplication we get $$ x^2 \cdot x\partial^2 - \partia …

Copied from comments as requested. There isn't enough context in the question but if you know that $B$ is a projective left $R$-module and nothing else there is a fair chance that you want $\mathrm{H …

It is not true that the right ring of fractions always exists in the setting you describe. An example where it does not can be found in section 2.1.7 of McConnell and Robson's book Noncommutative Noet …

One way to think about this is that $D$ is the universal enveloping algebra $U(R,L)$ of the $(k,R)$ Lie-Rinehart algebra $\mathrm{Der}_k(R,R)$. Whenever one has such an enveloping algebra one may perf …

Here are a few observations that are too long for a comment but don't provide a full answer by any means. As Peter Samuelson observed left-right questions aren't significant in this setting since $\m …

Suppose you have an Artinian but not Noetherian finitely generated $R$ module $M$. Let $0\leq M_1\leq M_2\leq \cdots \leq M_n=M$ be a finite chain of $R$-modules such that each composition factor $M_i …