= Right exactness of the tensor product
{title2=$U\otimes_AE\longrightarrow V\otimes_AE\longrightarrow W\otimes_AE\longrightarrow0$}
For a <short exact sequence> $0\to U\xrightarrow{i}V\xrightarrow{q}W\to0$ and any <module> $E$, the sequence $U\otimes_AE\to V\otimes_AE\to W\otimes_AE\to0$ is exact. The last map is surjective since every pure tensor $w\otimes e$ can be lifted using a preimage of $w$. Put $T=(V\otimes_AE)/\operatorname{im}(i\otimes1_E)$. 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 $W\otimes_AE\to T$, inverse to the map induced by $q\otimes1_E$. Thus the kernel of $q\otimes1_E$ is exactly the image of $i\otimes1_E$. Consequently a module is flat exactly when tensoring with it also preserves injections.
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