The Stacks project

Proof. We may assume that $B$ is the localization of a finitely presented $A$-algebra $B_0$ and that $N$ is the localization of a finitely presented $B_0$-module $M_0$, see Lemma 38.25.3. By More on Morphisms, Lemma 37.54.1 there exists a “generic flatness stratification” for $\widetilde{M_0}$ on $\mathop{\mathrm{Spec}}(B_0)$ over $\mathop{\mathrm{Spec}}(A)$. Translating back to $N$ we find a sequence of closed subschemes

with $S_ i \subset S$ cut out by a finitely generated ideal of $A$ such that the pullback of $\widetilde{N}$ to $\mathop{\mathrm{Spec}}(B) \times _ S (S_ i \setminus S_{i + 1})$ is flat over $S_ i \setminus S_{i + 1}$. We will prove the proposition by induction on $t$ (the base case $t = 1$ will be proved in parallel with the other steps). Let $\mathop{\mathrm{Spec}}(A/J_ i)$ be the scheme theoretic closure of $S_ i \setminus S_{i + 1}$.

Claim 1. $N/J_ iN$ is flat over $A/J_ i$. This is immediate for $i = t - 1$ and follows from the induction hypothesis for $i > 0$. Thus we may assume $t > 1$, $S_{t - 1} \not= \emptyset $, and $J_0 = 0$ and we have to prove that $N$ is flat. Let $J \subset A$ be the ideal defining $S_1$. By induction on $t$ again, we also have flatness modulo powers of $J$. Let $A^ h$ be the henselization of $A$ and let $B'$ be the localization of $B \otimes _ A A^ h$ at the maximal ideal $\mathfrak m_ B \otimes A^ h + B \otimes \mathfrak m_{A^ h}$. Then $B \to B'$ is faithfully flat. Set $N' = N \otimes _ B B'$. Note that $N'$ is $A^ h$-flat if and only if $N$ is $A$-flat. By Theorem 38.24.1 there is a smallest ideal $I \subset A^ h$ such that $N'/IN'$ is flat over $A^ h/I$, and $I$ is finitely generated. By the above $I \subset J^ nA^ h$ for all $n \geq 1$. Let $S_ i^ h \subset \mathop{\mathrm{Spec}}(A^ h)$ be the inverse image of $S_ i \subset \mathop{\mathrm{Spec}}(A)$. By Lemma 38.25.11 we see that $V(I)$ contains the closed points of $U = \mathop{\mathrm{Spec}}(A^ h) - S_1^ h$. By construction $N'$ is $A^ h$-flat over $U$. By Lemma 38.25.12 we see that $N'/I_2N'$ is flat over $A/I_2$, where $I_2 = \mathop{\mathrm{Ker}}(I \to \Gamma (U, I/I^2))$. Hence $I = I_2$ by minimality of $I$. This implies that $I = I^2$ locally on $U$, i.e., we have $I\mathcal{O}_{U, u} = (0)$ or $I\mathcal{O}_{U, u} = (1)$ for all $u \in U$. Since $V(I)$ contains the closed points of $U$ we see that $I = 0$ on $U$. Since $U \subset \mathop{\mathrm{Spec}}(A^ h)$ is scheme theoretically dense (because replaced $A$ by $A/J_0$ in the beginning of this paragraph), we see that $I = 0$. Thus $N'$ is $A^ h$-flat and hence Claim 1 holds.

We return to the situation as laid out before Claim 1. With $A^ h$ the henselization of $A$, with $B'$ the localization of $B \otimes _ A A^ h$ at the maximal ideal $\mathfrak m_ B \otimes A^ h + B \otimes \mathfrak m_{A^ h}$, and with $N' = N \otimes _ B B'$ we now see that the flattening ideal $I \subset A^ h$ of Theorem 38.24.1 is nilpotent. If $nil(A^ h)$ denotes the ideal of nilpotent elements, then $nil(A^ h) = nil(A) A^ h$ (More on Algebra, Lemma 15.45.5). Hence there exists a finitely generated nilpotent ideal $I_0 \subset A$ such that $N/I_0N$ is flat over $A/I_0$.

Claim 2. For every prime ideal $\mathfrak p \subset A$ the map $\kappa (\mathfrak p) \otimes _ A N \to \kappa (\mathfrak p) \otimes _ A M$ is injective. We say $\mathfrak p$ is bad it this is false. Suppose that $C$ is a nonempty chain of bad primes and set $\mathfrak p^* = \bigcup _{\mathfrak p \in C} \mathfrak p$. By Lemma 38.25.8 there is a finitely generated ideal $\mathfrak a \subset \mathfrak p^*A_{\mathfrak p^*}$ such that there is a pure spreadout over $V(\mathfrak a)$. If $\mathfrak p^*$ were good, then it would follow from Lemma 38.25.7 that the points of $V(\mathfrak a)$ are good. However, since $\mathfrak a$ is finitely generated and since $\mathfrak p^*A_{\mathfrak p^*} = \bigcup _{\mathfrak p \in C}A_{\mathfrak p^*}$ we see that $V(\mathfrak a)$ contains a $\mathfrak p \in C$, contradiction. Hence $\mathfrak p^*$ is bad. By Zorn's lemma, if there exists a bad prime, there exists a maximal one, say $\mathfrak p$. In other words, we may assume every $\mathfrak p' \supset \mathfrak p$, $\mathfrak p' \not= \mathfrak p$ is good. In this case we see that for every $f \in A$, $f \not\in \mathfrak p$ the map $u \otimes \text{id}_{A/(\mathfrak p + f)}$ is universally injective, see Lemma 38.25.9. Thus it suffices to show that $N/\mathfrak p N$ is separated for the topology defined by the submodules $f(N/\mathfrak pN)$. Since $B \to B'$ is faithfully flat, it is enough to prove the same for the module $N'/\mathfrak p N'$. By Lemma 38.19.5 and More on Algebra, Lemma 15.24.4 elements of $N'/\mathfrak pN'$ have content ideals in $A^ h/\mathfrak pA^ h$. Thus it suffices to show that $\bigcap _{f \in A, f \not\in \mathfrak p} f(A^ h/\mathfrak p A^ h) = 0$. Then it suffices to show the same for $A^ h/\mathfrak q A^ h$ for every prime $\mathfrak q \subset A^ h$ minimal over $\mathfrak p A^ h$. Because $A \to A^ h$ is the henselization, every $\mathfrak q$ contracts to $\mathfrak p$ and every $\mathfrak q' \supset \mathfrak q$, $\mathfrak q' \not= \mathfrak q$ contracts to a prime $\mathfrak p'$ which strictly contains $\mathfrak p$. Thus we get the vanishing of the intersections from Lemma 38.25.10.

At this point we can put everything together. Namely, using Claim 1 and Claim 2 we see that $N/I_0 N \to M/I_0M$ is $A/I_0$-universally injective by Lemma 38.25.9. Then the diagrams

show that the left vertical arrows are injective. Hence by Algebra, Lemma 10.99.9 we see that $N$ is flat. In a similar way the universal injectivity of $u$ can be reduced (even without proving flatness of $N$ first) to the one modulo $I_0$. This finishes the proof. $\square$