$M$ be $\mathfrak m^{-1}R$ module where $\mathfrak m$ be maximal ideal in a ring $R$
I want to show that: $\mathfrak m^{-1}R \otimes_R M$ is isomorphic to $M$
First: I define a map
$f$: $\mathfrak m^{-1}R \otimes_R M$ $\to$ $M$ by ($\frac{r}{s}\otimes $m$ $) $\to$ $\frac{r}{s}$m
and the inverse map by
$g$: $M$ $\to$ $\mathfrak m^{-1}R \otimes_R M$ by m $\to$ $\frac{1}{1}$ $\otimes$m
However $f \circ g$ is identity, but $g \circ f$ is not, since $\frac{r}{s}$ $\otimes$m $\to$ $\frac{r}{s}$m $\to$ $\frac{1}{1}$ $\otimes$ $\frac{r}{s}$m
Note that if we define the $\mathfrak m^{-1}R \otimes_R M$ be $\mathfrak m^{-1}R$ module by $\frac{r_1}{s_1}$ ($\frac{r}{s}\otimes $m$ $)=$\frac{r_1r}{s_1s}$ $\otimes$ m, we have no condition to show that $\frac{1}{1}$ $\otimes$ $\frac{r}{s}$m = ($\frac{r}{s}\otimes $m$ $) since we are tesoring with R module, can any one help me ? thanks
