$M$ is a flat $R$-module if and only if its character module, i.e. $\hom_{\mathbb{Z}}(M,\mathbb{Q/Z})$, is injective. $M$ is a flat $R$-module if and only if $\hom_{\mathbb{Z}}(M,\mathbb{Q/Z})$ is injective.
One direction is easy. Suppose $M$ is flat. We know that
$$ \hom_\mathbb{Z}(-\otimes M, \mathbb{Q/Z}) \cong_{\mathbb{Z}} \hom_{\mathbb{Z}}(-,\hom_{\mathbb{Z}}(M,\mathbb{Q/Z}))$$
Since $- \otimes M$ is exact and $\mathbb{Q/Z}$ is injective, the left functor is exact which shows that the right functor is exact, i.e. $\hom_{\mathbb{Z}}(M,\mathbb{Q/Z})$ is injective.
The other direction looks difficult and indeed I haven't been able to prove it attacking the problem by considering short exact sequences. The thing is I don't see why $\mathbb{Q/Z}$ is special in this problem and I think I need to figure that out to be able to come up with a proof. Any help is appreciated.
 A: In order to prove your claim, you need to use that  $\mathbb{Q}/\mathbb{Z}$ is a cogenerator, i.e. $\text{Hom}(-,\mathbb{Q}/\mathbb{Z})$ is a faithful functor. Since $- \otimes_RM$ is a right exact functor, all we need to show is that, given an injective morphism of modules $f: A \to B$ we obtain an injective morphism $f \otimes_R M : A \otimes_R M \to B \otimes_R M$.
By the left exactness of $\text{Hom}(-, M^*)$, where $M^*$ stands for $\text{Hom}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$, we get a surjective morphism $f^*:\text{Hom}(B, M^*) \to \text{Hom}(A, M^*)$. By the tensor-hom adjunction we have a surjective morphism $\psi = \text{Hom}(f \otimes_R M,\mathbb{Q}/\mathbb{Z}):\text{Hom}(B \otimes_R M, \mathbb{Q}/\mathbb{Z}) \to \text{Hom}(A \otimes_R M,\mathbb{Q}/\mathbb{Z})$.
Now, if we consider the exact sequence $$0 \to \ker(f \otimes_RM) \to A \otimes_RM \xrightarrow{f \otimes_R M} B \otimes_RM $$ and we  apply the functor $\text{Hom}(-,\mathbb{Q}/\mathbb{Z})$, we get the exact sequence $$\text{Hom}(B \otimes_RM,\mathbb{Q}/\mathbb{Z}) \xrightarrow{\psi} \text{Hom}(A \otimes_RM,\mathbb{Q}/\mathbb{Z}) \to \text{Hom}(\ker(f \otimes_RM),\mathbb{Q}/\mathbb{Z}) \to 0 $$ hence, by the surjectivity of $\psi$, we have $\text{Hom}(\ker(f \otimes_RM),\mathbb{Q}/\mathbb{Z})=0$ and finally we get $\ker(f \otimes_RM)=0$, because $\mathbb{Q}/\mathbb{Z}$ is a cogenerator. This completes the proof.
