The Stacks project

Proof. By replacing $M$ by the image of $M \to N$, we may assume $M \subset N$. Choose a filtration $0 = N_0 \subset N_1 \subset \ldots \subset N_ t = N$ such that $N_ i/N_{i - 1} = B/\mathfrak q_ i$ for some prime ideal $\mathfrak q_ i \subset B$, see Algebra, Lemma 10.62.1. Set $M_ i = M \cap N_ i$. Then $Q = N/M$ has a filtration by the submodules $Q_ i = N_ i/M_ i$. It suffices to prove $Q_ i/Q_{i - 1}$ becomes flat after localizing at a nonzero element of $f$ (since extensions of flat modules are flat by Algebra, Lemma 10.39.13). Since $Q_ i/Q_{i - 1}$ is isomorphic to the cokernel of the map $M_ i/M_{i - 1} \to N_ i/N_{i - 1}$, we reduce to the case discussed in the next paragraph.

Assume $B$ is a domain and $M \subset N = B$. After replacing $A$ by the image of $A$ in $B$ we may assume $A \subset B$. By generic flatness, we may assume $A$ and $B$ are flat over $R$ (Algebra, Lemma 10.118.1). It now suffices to show $M \to B$ becomes $R$-universally injective after replacing $R$ by a principal localization (Algebra, Lemma 10.82.7). By generic freeness, we can find a nonzero $g \in A$ such that $B_ g$ is a free $A_ g$-module (Algebra, Lemma 10.118.1). Thus we may choose a direct summand $M' \subset B_ g$ as an $A_ g$-module, which is finite free as an $A_ g$-module, and such that $M \to B \to B_ g$ factors through $M'$. Clearly, it suffices to show that $M \to M'$ becomes $R$-universally injective after replacing $R$ by a principal localization.

Say $M' = A_ g^{\oplus n}$. Since $M \subset M'$ is a finite $A$-module, we see that $M$ is contained in $(1/g^ m)A^{\oplus n}$ for some $m \geq 0$. After changing our basis for $M'$ we may assume $M \subset A^{\oplus n}$. Then it suffices to show that $A^{\oplus n}/M$ and $A_ g/A$ become $R$-flat after replacing $R$ by a principal localization. Namely, then $M' \to A^{\oplus n}$ and $A^{\oplus n} \to A_ g^{\oplus n}$ are universally injective by Algebra, Lemma 10.39.12 and consequently so is the composition $M \to M' = A_ g^{\oplus n}$.

By generic flatness (see reference above), we may assume the module $A^{\oplus n}/M$ is $R$-flat. For the quotient $A_ g/A$ we use the fact that

and the module $A/g^ mA$ has a filtration of length $m$ whose successive quotients are isomorphic to $A/gA$. Again by generic flatness we may assume $A/gA$ is $R$-flat and hence each $A/g^ mA$ is $R$-flat, and hence so is $A_ g/A$. $\square$