Flatness is equivalent to preserving injections and to the ideal and finitely generated ideal tests

Claim 1 implies claim 2 by applying [L1] to $0\to K\to N$; claim 2 implies claim 3 by taking the inclusion $I\hookrightarrow R$ and using $R{\otimes }_{R}M\cong M$; claim 3 implies claim 4 by restriction to finitely generated ideals.

Assume claim 4. Claim 3 follows: an element $x\in I{\otimes }_{R}M$ is represented by finitely many coefficients from $I$, hence comes from $I_0{\otimes }_{R}M$ for the finitely generated ideal $I_0$ they generate; if its image in $M$ is zero, injectivity for $I_0$ makes its representative zero, so $x=0$.

Let $K\hookrightarrow N$ be injective and let $x\in K{\otimes }_{R}M$ map to zero. By [L6], the tensor $x$ uses finitely many elements of $K$, and a derivation of its zero image uses only finitely many generators and defining relations in $N{\otimes }_{R}M$. Hence there are finitely generated submodules $K_0\subseteq K$ and $N_0\subseteq N$ with $K_0\subseteq N_0$ for which $x$ comes from an element killed in $N_0{\otimes }_{R}M$.

Under claim 3, for every submodule $L\subseteq R^n$, the map $L{\otimes }_{R}M\to M^n$ is injective, by induction on $n$. For $n=0$ it is the unique map $0\to 0$; for $n=1$ it is claim 3 by [L5].

Present $N_0=R^n\mathrm{/}L$ using [L4], and let $L^{\mathrm{\prime }}\subseteq R^n$ be the inverse image of $K_0$. Right exactness identifies $N_0{\otimes }_{R}M$ with $M^n\mathrm{/}\mathrm{im}(L{\otimes }_{R}M)$ and $K_0{\otimes }_{R}M$ with the quotient of $L^{\mathrm{\prime }}{\otimes }_{R}M$ by $\mathrm{im}(L{\otimes }_{R}M)$.

For the induction step, let $Q\subseteq R^n$, put $Q^{\mathrm{\prime }}:=Q\cap (R\oplus 0^{n-1})$, and let $Q^{\mathrm{\prime }\mathrm{\prime }}$ be the image of $Q$ in $R^{n-1}$. Tensoring the exact rows $0\to Q^{\mathrm{\prime }}\to Q\to Q^{\mathrm{\prime }\mathrm{\prime }}\to 0$ and $0\to R\to R^n\to R^{n-1}\to 0$ gives right-exact rows by [L2]; the left and right vertical maps are injective by the $n=1$ case and the induction hypothesis from step 2.1.

If $z\in Q{\otimes }_{R}M$ maps to zero in $M^n$, its image in $Q^{\mathrm{\prime }\mathrm{\prime }}{\otimes }_{R}M$ maps to zero in $M^{n-1}$ and hence is zero by the right vertical injection in step 3.1. Right exactness of the top row lifts $z$ from some $y\in Q^{\mathrm{\prime }}{\otimes }_{R}M$. The image of $y$ in $R{\otimes }_{R}M$ maps to the zero image of $z$ in $M^n$; the left vertical injection in step 3.1 makes $y=0$, and therefore $z=0$. This completes the induction of step 2.1.

By step 4.1, both $L{\otimes }_{R}M$ and $L^{\mathrm{\prime }}{\otimes }_{R}M$ inject into $M^n$. Therefore the induced map of the quotients in step 2.2 is injective, so $x=0$. This proves claim 2 from claim 4.

Finally claim 2 and right exactness [L2] imply claim 1: for an exact sequence, replace the left map by the injection of its image into the middle term; tensoring preserves that injection by claim 2 and preserves the remaining image and cokernel statements by right exactness.

Steps 1.1 through 6.1 prove the cycle of equivalences. The zero ideal, $n=0$, and zero module cases occur explicitly in steps 1.2 and 2.1; no choice is made, and both directions of every equivalence have been supplied.