Noetherian modules over noetherian rings are faithful I must show that a Noetherian module $M$ over a Noetherian ring $A$ is faithful.
I've been thinking for a while, but I didn't came up with nothing. Basically I tried to prove that if $A$ is noetherian the kernel of $A\to M^r:a\mapsto (am_1,\dots ,am_k)$ is zero, where $m_1,\dots ,m_k$ are the generators of $M$. The problem is that I don't know how to use that $A$ is finitely generated: I could say that I have an $A$-linear map $A^k\to M^r$, but it seems not useful at all because the kernel of this map is far from zero. Can you only give a hint?
 A: Your statement is quite clearly not true. The zero module is obviously noetherian over any ring, but never faithful (except over the zero ring).
And that is not a cheesy counter-example, you can also take the module $\mathbb{Z}/n\mathbb{Z}$ (which is finite, so very noetherian) over the ring $\mathbb{Z}$ for instance. In short, it is very much not true, and the counter-examples are quite numerous.
A: 
This sounds like a garbling of (a true statement for commutative rings) "If  has a faithful Noetherian module, then  is Noetherian," or else a failed attempt at concocting a converse. –
rschwieb


@rschwieb that's a failed attempt at concocting a converse, I wrote the precise text of the exercise below. –
Dorian


actually the text of the exercise required to prove that  noetherian and faithful imply  noetherian (and this is clear), and to show that the faithfulness in necessary. – Dorian

Every ring has a Noetherian module because every ring has a simple module. This works even for non-Noetherian rings. So obviously faithfulness takes a very big role in making the theorem "if $A$ has a faithful Noetherian module, then $A$ is Noetherian" work.
None of these potential converses go anywhere: Noetherian modules over Noetherian rings need not be faithful (e.g. $R/I$ for a nontrivial ideal $I$). Faithful modules over Noetherian rings need not be Noetherian (e.g. $\oplus_{i\in\mathbb N}R$)
But a Noetherian ring obviously has at least one faithful Noetherian module: i.e. $R_R$.
