# Showing that if $R$ is a commutative ring 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?

• 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. Commented May 26, 2012 at 18:13
• In connection with the end of Dylan's comment, see Critch's answer here.
– KCd
Commented May 26, 2012 at 21:27
• The book Skew Fields by P.K.Draxl might just help you. Commented May 27, 2012 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.
• @stevengerrard: The first map sends $m\otimes i$ to $im$, so its image is the submodule of $M$ generated by elements of the form $im$ for $i\in I$ and $m\in M$, That's $IM$. Commented Aug 2, 2020 at 2:34
Morover, let $I$ be a right ideal of a ring $R$ (noncommutative ring) and $M$ a left $R$-module, then $M/IM\cong R/I\otimes_R M$.
• For this to be an isomorphism of $R/I$-modules I think $I\vartriangleleft R$ needs to be two sided. Otherwise I am not sure $M/IM$ has an obvious $R/I$-module structure. Commented May 5, 2018 at 14:37