Suppose $R$ is a domain and let $Q$ be its field of fractions. Then $ 0 \to R \to Q $ is exact. Now suppose that $M$ is a torsion free $R$ module.
Is it necessary that $ 0 \to M \to Q \otimes M $ is exact ?
This is true when $R$ is a PID. But I am not sure whether assuming that it is a domain suffices. If it is not please give a counter example. Thanks
On
Maybe this works: It's easy to see that both $Q(R) \otimes_{R}(-)$ and $(R-0)^{-1}(-)$ are two functors that works as left-adjoint to the forgetful functor from $Q(R)$ modules to $R$ modules. This immediately tells you that the two functors are isomorphic. So this tells you that $Q(R) \otimes_{R}(M)$ and $(R-0)^{-1}(M)$ are naturally isomorphic. Now you have by construction that $\frac{m}{1}=0$ in $(R-0)^{-1}M$ iff there is $a \neq 0$ in R such that $am=0$. That is, iff m is a torsion element. So for torsion free elements, your map is injective. Hope it helps and hope it works
Yes, $R$ being an integral domain suffices. See Corollary 4.27 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf.