A short exact sequence with flat quotient remains short exact after tensoring

By [L4], choose a surjection $\epsilon :F\to N$ from a free module and let $K=\ker \epsilon$, so $0\to K\to F\to N\to 0$ is short exact.

Tensor the sequence in step 1.1 with each of $A,B,C$. By [L2], the three resulting columns $X{\otimes }_{R}K\to X{\otimes }_{R}F\to X{\otimes }_{R}N\to 0$ are right exact. The map $C{\otimes }_{R}K\to C{\otimes }_{R}F$ is injective by flatness of $C$ and [L1].

Tensor the given short exact sequence with $F$. Since $F$ is flat by [L3], the middle row $0\to A{\otimes }_{R}F\to B{\otimes }_{R}F\to C{\otimes }_{R}F\to 0$ is short exact. Tensoring it with $K$ and $N$ gives right-exact bottom and top rows by [L2].

Let $x\in A{\otimes }_{R}N$ map to zero in $B{\otimes }_{R}N$. By right exactness of the $A$-column, lift $x$ to $y\in A{\otimes }_{R}F$. Its image $y_B\in B{\otimes }_{R}F$ maps to zero in $B{\otimes }_{R}N$, so right exactness of the $B$-column gives $z\in B{\otimes }_{R}K$ mapping to $y_B$.

The image of $z$ in $C{\otimes }_{R}K$ maps in $C{\otimes }_{R}F$ to the image of $y_B$, which is zero because $y$ came from $A{\otimes }_{R}F$. The injectivity in step 2.1 therefore makes the image of $z$ in $C{\otimes }_{R}K$ zero.

By right exactness of the bottom row in step 2.2, choose $w\in A{\otimes }_{R}K$ mapping to $z$. In $B{\otimes }_{R}F$, the images of $y$ and of $w$ are both $y_B$; injectivity of $A{\otimes }_{R}F\to B{\otimes }_{R}F$ from step 2.2 makes $y$ the image of $w$.

The composite $A{\otimes }_{R}K\to A{\otimes }_{R}F\to A{\otimes }_{R}N$ is zero because $K\to F\to N$ is zero. Hence step 5.1 gives $x=0$, proving $i\otimes 1$ injective.

Right exactness [L2] already gives exactness at $B{\otimes }_{R}N$, surjectivity onto $C{\otimes }_{R}N$, and the terminal zero. Together with step 6.1, the tensored sequence is short exact.