Isomorphism between tensor product of modules and quotient module

Question

I'm trying to show that $M\otimes_{R}R/\mathbf{m}$ and $M/\mathbf{m}M$ are isomorphic as $R$-modules, where $M$ is an arbitrary $R$-Module, $R=k[x_{1},\ldots,x_{n}]$ with a field $k$, $\mathbf{m}=(x_{1},\ldots,x_{n})$ and $R/\mathbf{m}$ is regarded as an $R$-module by setting $x_{i}=0$. I think an isomorphism is given by $m\otimes r\mapsto r\overline{m}$. While the surjectivity is clear, I have difficulty doing to show injectivity. If $r\overline{m}=0$, then it means $rm\in\mathbf{m}M$. But why should $m\otimes r$ be $0$ then? And what does $0\in M\otimes_{R} R/\mathbf{m}$ mean?

Answer 1

I think this is a general fact about modules and ideals, and you might be getting boggled with particular details about $k[x_1,\ldots,x_n]$ which might be irrelevant to the problem in hand. I'd like to offer two solutions:

• In case you're familiar with category theory: define a functor $F_{M,{\bf m}}\colon R\mbox{-}{\sf mod} \to {\sf Set}$ by $$F_{M,{\rm m}}(N) = \{f\colon M \to N \mid f|_{{\bf m}M} = 0\},$$and show that $F_{M,{\bf m}}$ is naturally isomorphic to both ${\rm Hom}(M\otimes_R R/{\bf m},\_)$ and ${\rm Hom}(M/{\bf m}M,\_)$. Then conclude that $M\otimes_R R/{\bf m} \cong M/{\bf m}M$ by Yoneda's Lemma.

• A more pedestrian approach is, instead of trying to check injectivity for maps defined in tensor products (which can be difficult), exhibit the inverse directly. Check that $$M\times (R/{\bf m}) \ni (m, r+{\bf m}R) \mapsto rm + {\bf m}M \in M/{\bf m}M$$is well-defined. This is $R$-balanced, so induces a map $M\otimes_R (R/{\bf m}) \to M/{\bf m}M$. The inverse is defined by passing $$M \ni m \mapsto m\otimes (1+{\bf m}R) \in M\otimes_R(R/{\bf m}M)$$to the quotient as a map $M/{\bf m}M \to M \otimes_R(R/{\bf m}M)$.