The Stacks project

Proof. Let $I \subset R$ be an ideal. By Lemma 10.39.5 it suffices to show that $I \otimes _ R M \to M$ is injective. By Remark 10.75.9 we see that this kernel is equal to $\text{Tor}_1^ R(M, R/I)$. By Lemma 10.99.6 we see that $J \otimes _ R M \to M$ is injective for all ideals of finite colength.

Choose $n >> 0$ and consider the following short exact sequence

This is a sub sequence of the short exact sequence $0 \to R \to R^{\oplus 2} \to R \to 0$. Thus we get the diagram

Note that $I + \mathfrak m^ n$ and $\mathfrak m^ n$ are ideals of finite colength. Thus a diagram chase shows that $\mathop{\mathrm{Ker}}((I \cap \mathfrak m^ n)\otimes _ R M \to M) \to \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is surjective. We conclude in particular that $K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is contained in the image of $(I \cap \mathfrak m^ n) \otimes _ R M$ in $I \otimes _ R M$. By Artin-Rees, Lemma 10.51.2 we see that $K$ is contained in $\mathfrak m^{n-c}(I \otimes _ R M)$ for some $c > 0$ and all $n >> 0$. Since $I \otimes _ R M$ is a finite $S$-module (!) and since $S$ is Noetherian, we see that this implies $K = 0$. Namely, the above implies $K$ maps to zero in the $\mathfrak mS$-adic completion of $I \otimes _ R M$. But the map from $S$ to its $\mathfrak mS$-adic completion is faithfully flat by Lemma 10.97.3. Hence $K = 0$, as desired. $\square$