Proposition 38.25.13. Let $A \to B$ be a local ring homomorphism of local rings which is essentially of finite type. Let $M$ be a flat $A$-module, $N$ a finite $B$-module and $u : N \to M$ an $A$-module map such that $\overline{u} : N/\mathfrak m_ AN \to M/\mathfrak m_ AM$ is injective. Then $u$ is $A$-universally injective, $N$ is of finite presentation over $B$, and $N$ is flat over $A$.

**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 Lemma 38.21.4 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)