Showing that if $R$ is local and $M$ an $R$-module, then $M \otimes_R (R/\mathfrak m) \cong M / \mathfrak m M$.

Let $R$ be a local ring, and let $\mathfrak m$ be the maximal ideal of $R$. Let $M$ be an $R$-module. I understand that $M \otimes_R (R / \mathfrak m)$ is isomorphic to $M / \mathfrak m M$, but I verified this directly by defining a map $M \to M \otimes_R (R / \mathfrak m)$ with kernel $\mathfrak m M$. However I have heard that there is a way to show these are isomorphic using exact sequences and using exactness properties of the tensor product, but I am not sure how to do this. Can anyone explain this approach?

Also can the statement $M \otimes_R (R / \mathfrak m) \cong M / \mathfrak m M$ be generalised at all to non-local rings?

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Just to emphasize: there is no need for $R$ to be local or for $\mathfrak m$ to be anything other than an ideal. On the other hand, what you end up with in this case is a vector space, so that's nice. Note that you cannot assume in general that $I \otimes M \approx IM$; if this held for all $I$ then the module would be flat. – Dylan Moreland May 26 '12 at 18:13
In connection with the end of Dylan's comment, see Critch's answer here. – KCd May 26 '12 at 21:27
The book Skew Fields by P.K.Draxl might just help you. – awllower May 27 '12 at 7:57

If $R$ is any commutative ring and $I \subset R$ is an ideal, then $M \otimes_R R/I \cong M/IM$. Consider the short exact sequence of $R$-modules $0 \to I \to R \to R/I \to 0$ and tensor with $M$ over $R$ to obtain the exact sequence $M \otimes_R I \to M \to M \otimes_R R/I \to 0$. The image of the first map is $IM$, so by the first isomorphism theorem we obtain $M/IM \cong M \otimes_R R/I$ as desired.
Morover, let $I$ be a right ideal of a ring $R$ (noncommutative ring) and $M$ a left $R$-module, then $M/IM\cong R/I\otimes_R M$.