TITLE: Noetherian module implies Noetherian ring?
QUESTION [7 upvotes]: 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?

REPLY [11 votes]: It is not clear to me exactly what you are asking in your main question.  If you are asking:
$\bullet$ If a ring $R$ admits a Noetherian module, must $R$ be Noetherian?
Then this is trivially false: Alex Youcis and Thomas Andrews have each shown that every commutative ring admits Noetherian modules.  (As a general rule, if you are looking for a counterexample to an assertion about modules and you haven't checked the zero module, you haven't looked hard enough.  Also looking at modules of the form $R/I$ is something to try early on.)  
If you are asking
$\bullet$ If for a ring $R$ every finitely generated $R$-module is Noetherian, must $R$ be Noetherian?
Then this is trivially true, as $R$ is a finitely generated $R$-module.
A less trivial statement is the following:
Lemma (Kaplansky): A ring is Noetherian iff it admits a faithful Noetherian module.
Another result vaguely along these lines is:
Theorem (Eakin-Nagata) Let $R \subset S$ be a ring extension such that $S$ is finitely generated as an $R$-module.  Then $R$ is Noetherian iff $S$ is Noetherian.  
Proofs of these and other results which are (even more) vaguely related to your question can be found in $\S 8.8$ of my commutative algebra notes.