If $M$ is torsionfree then $M$ embeds in $M \otimes Q$ This question arose as a small step in a larger proof. It certainly looks easy, but I can't prove it:

Suppose that $R$ is a semiprime right Goldie ring with Goldie quotient ring $Q$, and that $M$ is a torsionfree right $R$-module. Show that the natural map 
  $$M \to M \otimes_R Q, \quad m \mapsto m \otimes 1$$
  is an embedding.

Here, the Goldie quotient ring is obtained by inverting all regular (i.e. has trivial left and right annihilator) elements of $R$, and a module $M$ is torsionfree if the only element of of $M$ annihilated by some regular element is $0$.
I've tried to use the universal property of tensor products, but to no avail.
 A: You want to prove that $m\otimes 1=0$ only if $m=0$. So using the universal property of the tensor product, you want to find an $R$-module $N$ with an $R$-balanced map from $M\times Q$ to $N$ such that $(m,1)$ does not map to zero.
$Q$ is the localisation of $R$ by the set $S$ of all regular elements, i.e. $Q=S^{-1}R$. So take $N$ to be the set of all pairs $(m,s)$, $m\in M$, $s\in S$, quotiented by the equivalence relation $(m,s)\equiv (n,t)$ if $am=rn$ where $as=rt$ for some $a,r\in S$, we know such elements exist using the first Ore condition of $S$ and properties of regular elements.
The operations on $N$ are given by:
$(m,s)+(n,t)=(vm+un,vs)$ where $vs=ut$ and $v, u\in S$, and
$a(m,s)=(bm,t)$ where $ta=bs$. 
It's not difficult to show that these operations are well defined, and define an $R$-module with identity $(0,1)$.
Therefore, we have an $R$-balanced map from $M\times Q$ to $N$, where $(m,s^{-1}a)$ maps to the equivalence class containing $(am,s)$, and so $(m,1)$ maps to $(m,1)$.
But if $(m,1)$ is zero in $N$, i.e. $(m,1)\equiv (0,1)$, then it follows easily that $rm=0$ for some $r\in S$, contradicting the torsion free property if $m\neq 0$.
