Why is the field of fractions of a domain $R$ a flat $R$ module?

Question

I want to show this lemma:

Let $R$ be a domain. If $A$ is a torsion $R$-module, then $\operatorname{Tor}_1^R (K,A)\cong A$ where $\operatorname{Frac}(R)=Q$ and $K=Q/R$.

When I was reading this proof need to show $Q$ is flat and it back to this problem

if $R$ is a domain with $Q=\operatorname{Frac}(R)$, then $Q$ is flat $R$-module?

I don't know why $Q$ is flat. Can you help please?
thank you

Answer 1

If you have a short exact sequence of modules and localize it, it remains exact.

What is the relation between localizing a module and tensoring it with the field of fractions?

Answer 2

Mariano Suárez-Alvarez's answer is correct. I only add that in general $S^{-1}$ is exact:
Atiyah-Macdonald has:

and

The particular case is field of fraction of a domain