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? 

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.
