TITLE: Embedding a module into its quotient module
QUESTION [6 upvotes]: I've got a very basic question on tensor products.

Let $R$ be a commutative integral domain, $K$ its quotient field and let $M$ be a $R$-module. Is the map $M \rightarrow K\otimes_R M$ given by $m\mapsto 1\otimes m$ an embedding of $M$ into the tensor product? What is the most general condition that $R$ and $M$ must satisfy for there to exist such an embedding? 

Many thanks!

REPLY [9 votes]: Note that $K\otimes_{R} M$ is isomorphic as an $R$-module to the localization of $M$ at the multiplicative subset of $R$ consisting of all non-zero elements. (I think this point of view would be helpful if you're familiar with the notion of localization of modules.)
The kernel of the map $M\to K\otimes_{R} M$ is equal to the $R$-submodule of $M$ consisting of elements annihilated by some non-zero element of $R$. (Exercise!) Therefore, $M\to K\otimes_{R} M$ is an embedding if and only if $M$ is torsion-free as an $R$-module.
I hope this helps!

REPLY [7 votes]: I guess you need $M$ to be torsion-free. If $0 \ne m \in M$ and $0 \ne r \in R$ are such that $r m = 0$, then
$$
1 \otimes m = (r^{-1} r) \otimes m  = r^{-1} \otimes r m = 0.  
$$