Let $R$ be a subring of a number field $K$. Assume $R$ is finitely generated as a $\mathbb Z$ module. One can argue that the rank of $R$ as a $\mathbb Z$-module is equal too $[Q(R):\mathbb Q]$, where $Q(R)$ is the quotient field of $R$, from
$$
R\otimes_{\mathbb Z}\mathbb Q=Q(R).
$$
I would like to know why this equality holds. The inclusion $\subset$ is clear, but the inclusion $\supset$ not yet. If I can argue that $R\otimes_\mathbb Z\mathbb Q$ is a field, then I'd be done