The Stacks project

Proof. We may replace $A$ by $A_\mathfrak p$. Thus we may assume $A$ is local and $\mathfrak p$ is the maximal ideal $\mathfrak m$ of $A$. We may write $N = S^{-1}N'$ for some finitely presented $B$-module $N'$ by clearing denominators in a presentation of $N$ over $S^{-1}B$. Since $B/\mathfrak m B$ is Noetherian, the kernel $K$ of $N'/\mathfrak m N' \to N/\mathfrak m N$ is finitely generated. Thus we can pick $s \in S$ such that $K$ is annihilated by $s$. After replacing $B$ by $B_ s$ which is allowed as it just means passing to an affine open subscheme of $\mathop{\mathrm{Spec}}(B)$, we find that the elements of $S$ are injective on $N'/\mathfrak m N'$. At this point we choose a local subring $A_0 \subset A$ essentially of finite type over $\mathbf{Z}$, a finite type ring map $A_0 \to B_0$ such that $B = A \otimes _{A_0} B_0$, and a finite $B_0$-module $N'_0$ such that $N' = B \otimes _{B_0} N'_0 = A \otimes _{A_0} N'_0$. We claim that $I = \mathfrak m_{A_0} A$ works. Namely, we have

which is free over $A/I$. Multiplication by the elements of $S$ is injective after dividing out by the maximal ideal, hence $N'/IN' \to N/IN$ is universally injective for example by Lemma 38.7.6. $\square$