Equivalence of definition for faithful functor?

Question

My professor gave me the following definition of a faithfully flat module. A module $M$ that is flat and such that the functor $N \mapsto M \otimes N$ defined on any $R-$module is faithful. However, he said that the functor being faithful means that if $M \otimes N = 0$ then $N = 0.$ I don't see how this is equivalent to the definition listed on Wikipedia for a faithful functor. That we should have $Hom(N, N') \rightarrow Hom(M \otimes N, M \otimes N')$ is injective. I guess if $M \otimes N = 0$ then $Hom(N, N') = 0$ for every module $N'.$ In particular for $N' = N$ Hence, $Hom(N, N)$ which has at least 2 different morphisms for $N \neq 0$ has 1 morphism. That is to say, $N = 0.$ So I guess how I would prove the other direction. Or is my professor saying that this is one result from the fact that the functor is faithful?

Answer 1

Well, here you proved that if the functor $N\mapsto M\otimes N$ is faithful, then for all $N$, $M\otimes N = 0 \implies N = 0$.

For the converse : Assume that $M\otimes N = 0 \implies N=0$.

Let $N,N'$ be two modules, and assume that $f,g : N\to N'$ are such that $id_M \otimes f = id_M \otimes g$.

Without loss of generality (consider $f-g$) we may assume $g=0$. Then let $K= \mathrm{Im}f$. Clearly $M\otimes K = 0$ (because $id_M \otimes f = id_M \otimes 0 = 0$). By the property, $K=0$, that is, $f$ is the zero map, and so $f=g$. Therefore, the functor is faithful.