Right exactness of the tensor product ($U{\otimes }_{A}E⟶V{\otimes }_{A}E⟶W{\otimes }_{A}E⟶0$)

For a short exact sequence $0\to U\stackrel{i}{\to }V\stackrel{q}{\to }W\to 0$ and any module $E$, the sequence $U{\otimes }_{A}E\to V{\otimes }_{A}E\to W{\otimes }_{A}E\to 0$ is exact. The last map is surjective since every pure tensor $w\otimes e$ can be lifted using a preimage of $w$. Put $T=(V{\otimes }_{A}E)/\mathrm{im}(i\otimes 1_E)$. There is a well-defined balanced map $W\times E\to T$, sending $(q(v),e)$ to the class of $v\otimes e$: replacing $v$ by $v+i(u)$ changes it by an element of the discarded image. This induces $W{\otimes }_{A}E\to T$, inverse to the map induced by $q\otimes 1_E$. Thus the kernel of $q\otimes 1_E$ is exactly the image of $i\otimes 1_E$. Consequently a module is flat exactly when tensoring with it also preserves injections.