The Stacks project

Proof. By Algebra, Lemma 10.82.14 it suffices to check that $N/IN \to M/IM$ is injective for every ideal $I \subset A$. After replacing $A$ by $A/I$ we see that it suffices to prove that $u$ is injective.

Proof that $u$ is injective. Let $x \in N$ be a nonzero element of the kernel of $u$. Then there exists a weakly associated prime $\mathfrak p$ of the module $Ax$, see Algebra, Lemma 10.66.5. Replacing $A$ by $A_\mathfrak p$ we may assume $A$ is local and we find a nonzero element $y \in Ax$ whose annihilator has radical equal to $\mathfrak m_ A$, see Algebra, Lemma 10.66.2. Thus $\text{Supp}(y) \subset \mathop{\mathrm{Spec}}(S^{-1}B)$ is nonempty and contained in the closed fibre of $\mathop{\mathrm{Spec}}(S^{-1}B) \to \mathop{\mathrm{Spec}}(A)$. Let $I \subset \mathfrak m_ A$ be a finitely generated ideal so that we have a pure spreadout over $A/I$, see Lemma 38.25.8. Then $I^ n y = 0$ for some $n$. Now $y \in \text{Ann}_ M(I^ n) = \text{Ann}_ A(I^ n) \otimes _ R N$ by flatness. Thus, to get the desired contradiction, it suffices to show that

is injective. Since $N$ and $M$ are flat and since $\text{Ann}_ A(I^ n)$ is annihilated by $I^ n$, it suffices to show that $Q \otimes _ A N \to Q \otimes _ A M$ is injective for every $A$-module $Q$ annihilated by $I$. This holds by our choice of $I$ and Lemma 38.25.7 part (2). $\square$