faithful module and tensor product Let $R$ be a ring with identity and suppose $M$ a left $R-$module and  $N$ a rignt $R-$module.
Is there result: if $M$ is a faithful, then $N\otimes M =0$ implies $N=0$ ?
If the result is right how to prove, or need any other conditions?
Any help appreciate.
 A: If I understand you aright, you are asking that if $M$ is a faithful $R$-module, does $N\otimes M=0$ imply $N=0$ for a general $N$-module.
There are simple counterexamples to this. Take $R=\Bbb Z$, and $M=\Bbb Q$.
If $N$ is a torsion $\Bbb Z$-module, then $N\otimes M=0$.
A: Call a functor $F : C \to D$ between abelian categories 


*

*weakly faithful if $F(c) = 0$ implies $c = 0$ where $c$ is an object.

*faithful if $F(f) = 0$ implies $f = 0$ where $f$ is a morphism.



Lemma: If $F$ is faithful, then it is weakly faithful. The converse holds if $F$ is exact.

Proof. In the forward direction, an object is the zero object iff its identity morphism is a zero morphism. 
Conversely, suppose $F$ is weakly faithful and exact, and let $f : c \to d$ be a morphism. $f$ has an image factorization $c \to \text{im}(f) \to d$ which is preserved by $F$ by exactness. If $F(f) = 0$ then $F(\text{im}(f)) = 0$ (since this is the image factorization of the zero morphism), so by weak faithfulness $\text{im}(f) = 0$, meaning $f = 0$. $\Box$
Hence if $M$ is a flat module, the functor $(-) \otimes M$ is weakly faithful iff it is faithful. If either of these conditions holds $M$ is said to be faithfully flat. Note that, somewhat confusingly, $M$ being faithful in the usual sense and also being flat does not imply that it is faithfully flat, as Lord Shark's example of $\mathbb{Q}$ shows. However, it is true that every faithfully flat module is faithful (and flat). 
