A left module is flat exactly when Tor one against every right module vanishes

If $M$ is flat, tensor a full projective resolution $Q_{\bullet }\to N$ with $M$. Exactness of $-{\otimes }_{R}M$ preserves its augmented exactness, so $H_i(Q_{\bullet }{\otimes }_{R}M)=0$ for every $i>0$. The right-resolution definition and The balanced Tor bifunctor give ${\mathrm{Tor}}_1^R(N,M)=0$.

For right modules $A\to B\to C\to 0$, the universal property Universal property of the tensor product for balanced maps into abelian groups gives $C{\otimes }_{R}M\cong (B{\otimes }_{R}M)\mathrm{/}\mathrm{im}(A{\otimes }_{R}M)$: a balanced map on $B\times M$ kills that image exactly when it factors through $C\times M$. Thus tensoring is right exact over the arbitrary ring $R$. In particular $H_0(Q_{\bullet }{\otimes }_{R}M)\cong N{\otimes }_{R}M$, naturally under augmentation-preserving maps.

Conversely, let $K\hookrightarrow P$ be an inclusion of right modules. Apply the projective horseshoe construction to $0\to K\to P\to P\mathrm{/}K\to 0$, tensor its degreewise split resolution sequence with $M$, and take the long exact homology sequence. Its degree-zero boundary identifies the kernel of $K\otimes M\to P\otimes M$ with the image of ${\mathrm{Tor}}_1(P\mathrm{/}K,M)$.

The assumed vanishing makes every such tensor map injective; together with right exactness this gives exactness on all short exact sequences, hence flatness.