Noetherian module implies Noetherian ring? I know that a finitely generated $R$-module $M$ over a Noetherian ring $R$ is Noetherian. I wonder about the converse. I believe it has to be false and I am looking for counterexamples. 
Also I wonder if $M$ Noetherian imply that $R$ is Noetherian is true? And if $M$ Noetherian implies $M$ finitely generated is true? 
That is, do both implications fail or only one of them? 

Let $R$ be a commutative non-Noetherian and let $\mathcal m$ be a maximal ideal. Then $R/\mathcal m$ is finitely generated and Noetherian - it only has two sub-$R$-modules.
Note that, even if $R$ isn't Noetherian, it contains a maximal ideal, by Krull's Theorem. 
