Lemma 10.99.1. Suppose that $R \to S$ is a local homomorphism of Noetherian local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Let $M$ be a flat $R$-module and $N$ a finite $S$-module. Let $u : N \to M$ be a map of $R$-modules. If $\overline{u} : N/\mathfrak m N \to M/\mathfrak m M$ is injective then $u$ is injective. In this case $M/u(N)$ is flat over $R$.

Proof. First we claim that $u_ n : N/{\mathfrak m}^ nN \to M/{\mathfrak m}^ nM$ is injective for all $n \geq 1$. We proceed by induction, the base case is that $\overline{u} = u_1$ is injective. By our assumption that $M$ is flat over $R$ we have a short exact sequence $0 \to M \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} \to M/{\mathfrak m}^{n + 1}M \to M/{\mathfrak m}^ n M \to 0$. Also, $M \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} = M/{\mathfrak m}M \otimes _{R/{\mathfrak m}} {\mathfrak m}^ n/{\mathfrak m}^{n + 1}$. We have a similar exact sequence $N \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} \to N/{\mathfrak m}^{n + 1}N \to N/{\mathfrak m}^ n N \to 0$ for $N$ except we do not have the zero on the left. We also have $N \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} = N/{\mathfrak m}N \otimes _{R/{\mathfrak m}} {\mathfrak m}^ n/{\mathfrak m}^{n + 1}$. Thus the map $u_{n + 1}$ is injective as both $u_ n$ and the map $\overline{u} \otimes \text{id}_{{\mathfrak m}^ n/{\mathfrak m}^{n + 1}}$ are.

By Krull's intersection theorem (Lemma 10.51.4) applied to $N$ over the ring $S$ and the ideal $\mathfrak mS$ we have $\bigcap \mathfrak m^ nN = 0$. Thus the injectivity of $u_ n$ for all $n$ implies $u$ is injective.

To show that $M/u(N)$ is flat over $R$, it suffices to show that $\text{Tor}_1^ R(M/u(N), R/I) = 0$ for every ideal $I \subset R$, see Lemma 10.75.8. From the short exact sequence

$0 \to N \xrightarrow {u} M \to M/u(N) \to 0$

and the flatness of $M$ we obtain an exact sequence of Tors

$0 \to \text{Tor}_1^ R(M/u(N), R/I) \to N/IN \to M/IM$

See Lemma 10.75.2. Thus it suffices to show that $N/IN$ injects into $M/IM$. Note that $R/I \to S/IS$ is a local homomorphism of Noetherian local rings, $N/IN \to M/IM$ is a map of $R/I$-modules, $N/IN$ is finite over $S/IS$, and $M/IM$ is flat over $R/I$ and $u \bmod I : N/IN \to M/IM$ is injective modulo $\mathfrak m$. Thus we may apply the first part of the proof to $u \bmod I$ and we conclude. $\square$

Comment #3817 by Alapan Mukhopadhyay on

In the diagram $\frac{M}{IN+u(N)}$ is to be replaced by $\frac{M}{IM+u(N)}$.

Comment #6515 by Jefferson Baudin on

After showing that $u$ is injective, I believe there is a less "technical-looking" proof that "it is enough to show that $N/IN \to M/IM$ is injective" :

To show flatness of $M/u(N)$, it is enough to show that $Tor_1^R(M/u(N), R/I) = 0$ for all ideals $I$ of $R$. We have the short exact sequence which, after tensoring by $R/I$, gives where we used flatness of $M$. Hence we get what we wanted.

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