Let $R$ be a non-commutative ring which is not noetherian. Let $I$ be an ideal in $R$. What are some conditions on $I$ such that the quotient ring $R/I$ is noetherian? Does this happen for example if $I$ is principal, or even just finitely generated?
Don't even know where to start to be honest, so this may also be treated as a reference request.
This statement is stolen from one of the exercise of Atiyah & Macdonald "Introduction to commutative algebra" (Chapter 6, Exercise 4). I assume that $R$ is commutative however (but maybe this assumption can be dropped).
Let $M$ be a noetherian module. Then take $I=(0:M)=\{a\in R: aM=0\}$. Then $R/I$ is noetherian. Conversely suppose that $R/I$ is noetherian. Then it is noetherian as an $R$-module (the $R$-action on $R/I$ is the same as the $R/I$-action) and $(0:R/I)=I$.
In total we get the characterization: $R/I$ is noetherian iff $I$ is the anhilitator of a noetherian $R$-module.
A proof of $\Leftarrow$ goes as follows: as $M$ is noetherian it is finitely generated by $x_1,\ldots,x_n\in M$, say. Consider the map $$ \phi:R \to \bigoplus_{i=1}^n M, \quad a\mapsto (ax_1,\ldots,ax_n).$$ By definition $\ker \phi = (0:M)$. Thus $\phi$ induces an injection $R/(0:M)\hookrightarrow \bigoplus_{i=1}^n M$ and the module on the right is noetherian, so $R/(0:M)$ is noetherian as it identifies with a submodule of a noetherian module.