Possible Duplicate:
Showing that if $R$ is local and $M$ an $R$-module, then $M \otimes_R (R/\mathfrak m) \cong M / \mathfrak m M$.
In one of the answers to one of my previous questions the following claim was mentioned:
$R/I \otimes_R M \cong M / IM$
So I tried to prove it. Can you help me finish my proof? Thanks!
We recall that $M \otimes_R -$ is a covariant right-exact functor that it is exact if $M$ is flat and we observe that the following is an exact sequence: $ 0 \to IM \xrightarrow{i} M \xrightarrow{\pi} M / IM \to 0$
$R/I$ is an $R$ module since $R/I$ is a subring of $R$ closed with respect to multiplication from $R$ but it is not necessarily flat hence we only get exactness on one side:
$ (R/I) \otimes_R IM \xrightarrow{id \otimes i} (R/I) \otimes_R M \xrightarrow{id \otimes \pi} (R/I) \otimes_R M / IM \to 0$
Then $ \mathrm{Im(id \otimes \pi)} = (R/I) \otimes_R M / IM \cong (R/I) \otimes_R M / \mathrm{Ker} (id \otimes i) $
so we want to show two things:
(i) that $\mathrm{Ker} (id \otimes i) = \{0\}$. To this end let $r + I \otimes im \in (R/I) \otimes_R IM $ and assume $ id \otimes i(r + I \otimes im ) = r + I \otimes im = 0 + I \otimes 0$. I'm not sure how to proceed from here.
(ii) and that $M /IM \cong (R/I) \otimes_R M / IM $
Can you help me? Thanks.