# “Two ways” to reduce a module

Let $M$ be a module over a principal ideal domain $R$ and $\mathfrak{m}$ a maximal ideal of $R$ with residue field $R/\mathfrak{m}=k$ of characteristic $p$.

Under what circumstances are the modules $$M\otimes_R k\quad\mbox{and}\quad M/\mathfrak{m}M$$ isomorphic?

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They are always (naturally) isomorphic, and you do not need the hypothesis that $R$ is a PID or that $\mathfrak{m}$ is maximal. Are you familiar with the universal properties of both of these constructions? –  Qiaochu Yuan Sep 7 '12 at 4:48

Let $M$ be a module over a commutative ring $R$. Let $I$ be an ideal of $R$.

The following sequence of $R$-modules is exact.

$$0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$$

Since the functor $M\otimes_R -$ is right exact, the following sequence of $R$-modules is exact.

$$M\otimes_R I \rightarrow M \rightarrow M\otimes_R (R/I) \rightarrow 0$$

Hence $M\otimes_R (R/I)$ is isomorphic to $M/IM$.

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