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? 

I assume that $R$ is commutative in your question.  As other answers have already pointed out, the answer to your question is false for trivial reasons.  But here is a closely related statement which is true.  (It relies on the fact that a direct sum of two Noetherian modules is again Noetherian; I leave this for you to investigate, but feel free to leave a comment if you would like more information.)
Theorem: If a ring $R$ has a faithful Noetherian module $M$, then $R$ is Noetherian.
Proof: Because $M$ is Noetherian, it is finitely generated. Say $M = \sum Rm_i$ for some finite generating set $m_1, \dots, m_n$. Consider the homomorphism $f \colon R \to M^n$ given by $f(r) = (rm_1, \dots, rm_n)$. This is injective because the annihilator of $M$ is trivial: if $f(r) = 0$ then $rm_i = 0$ for all $i$, whence $rM = r \sum Rm_i = \sum Rrm_i = 0$.  Thus $R$ is isomorphic as an $R$-module to a submodule of the Noetherian module $M^n$.  It follows that $R$ is Noetherian.
Corollary: If $M$ is a Noetherian $R$-module, then $R/\mathrm{ann}(M)$ is Noetherian.

What happens for noncommutative rings?  The results above no longer hold.  A ring may have a faithful simple (hence Noetherian) left module but fail to be either left or right Noetherian.  For instance, let $V$ be an infinite dimensional vector space over a field $k$, and let $R = \mathrm{End}_k(V)$ be the ring of $k$-linear endomorphisms of $V$. Then $V$ is a simple left $R$-module, but it can be shown that $R$ is neither left nor right Noetherian. 
