The Stacks project

Proof. Applying Algebra, Lemma 10.99.1 we see that $u$ is injective and that $N/u(M)$ is flat over $A$. Then $u$ is $A$-universally injective (Algebra, Lemma 10.39.12) and $N$ is $A$-flat (Algebra, Lemma 10.39.13). Since $B$ is Noetherian in this case we see that $N$ is of finite presentation. $\square$

Proof. Let $A \to B$, $u : N \to M$ be as in Situation 38.25.1. Write $B = C/I$ where $C$ is the localization of a polynomial algebra over $A$ at a prime. If we can show that $N$ is finitely presented as a $C$-module, then a fortiori this shows that $N$ is finitely presented as a $B$-module (see Algebra, Lemma 10.6.4). Hence we may assume that $B$ is the localization of a polynomial algebra. Next, write $N = B^{\oplus n}/K$ for some submodule $K \subset B^{\oplus n}$. Since $B/\mathfrak m_ AB$ is Noetherian (as it is essentially of finite type over a field), there exist finitely many elements $k_1, \ldots , k_ s \in K$ such that for $K' = \sum Bk_ i$ and $N' = B^{\oplus n}/K'$ the canonical surjection $N' \to N$ induces an isomorphism $N'/\mathfrak m_ AN' \cong N/\mathfrak m_ AN$. Now, if the lemma holds for the composition $u' : N' \to M$, then $u'$ is injective, hence $N' = N$ and $u' = u$. Thus the lemma holds for the original situation. $\square$

Proof. It suffices to prove this when $B$ is essentially of finite presentation over $A$ and $N$ is of finite presentation over $B$, see Lemma 38.25.3. Let us temporarily make the additional assumption that $N$ is flat over $A$. Then $N$ is a filtered colimit $N = \mathop{\mathrm{colim}}\nolimits _ i F_ i$ of free $A$-modules $F_ i$ such that the transition maps $u_{ii'} : F_ i \to F_{i'}$ are injective modulo $\mathfrak m_ A$, see Lemma 38.19.5. Each of the compositions $u_ i : F_ i \to M$ is $A$-universally injective by Lemma 38.7.5 wherefore $u = \mathop{\mathrm{colim}}\nolimits u_ i$ is $A$-universally injective as desired.

Assume $A$ is a henselian local ring, $B$ is essentially of finite presentation over $A$, $N$ of finite presentation over $B$. By Theorem 38.24.1 there exists a finitely generated ideal $I \subset A$ such that $N/IN$ is flat over $A/I$ and such that $N/I^2N$ is not flat over $A/I^2$ unless $I = 0$. The result of the previous paragraph shows that the lemma holds for $u \bmod I : N/IN \to M/IM$ over $A/I$. Consider the commutative diagram

whose rows are exact by right exactness of $\otimes $ and the fact that $M$ is flat over $A$. Note that the left vertical arrow is the map $N/IN \otimes _{A/I} I/I^2 \to M/IM \otimes _{A/I} I/I^2$, hence is injective. A diagram chase shows that the lower left arrow is injective, i.e., $\text{Tor}^1_{A/I^2}(I/I^2, M/I^2) = 0$ see Algebra, Remark 10.75.9. Hence $N/I^2N$ is flat over $A/I^2$ by Algebra, Lemma 10.99.8 a contradiction unless $I = 0$. $\square$

Proof. Let $A \to A^ h$ be the henselization of $A$. Let $B'$ be the localization of $B \otimes _ A A^ h$ at the maximal ideal $\mathfrak m_ B \otimes A^ h + B \otimes \mathfrak m_{A^ h}$. Since $B \to B'$ is flat (hence faithfully flat, see Algebra, Lemma 10.39.17), we may replace $A \to B$ with $A^ h \to B'$, the module $M$ by $M \otimes _ B B'$, the module $N$ by $N \otimes _ B B'$, and $u$ by $u \otimes \text{id}_{B'}$, see Algebra, Lemmas 10.83.2 and 10.39.9. Thus we may assume that $A$ is a henselian local ring. In this case our lemma follows from the more general Lemma 38.25.4. $\square$

Proof. Recall that an $A$-module is flat if and only if it is torsion free, see More on Algebra, Lemma 15.22.10. Let $T \subset N$ be the $A$-torsion. Then $u(T) = 0$ and $N/T$ is $A$-flat. Hence $N/T$ is finitely presented over $B$, see More on Algebra, Lemma 15.25.6. Thus $T$ is a finite $B$-module, see Algebra, Lemma 10.5.3. Since $N/T$ is $A$-flat we see that $T/\mathfrak m_ A T \subset N/\mathfrak m_ A N$, see Algebra, Lemma 10.39.12. As $\overline{u}$ is injective but $u(T) = 0$, we conclude that $T/\mathfrak m_ A T = 0$. Hence $T = 0$ by Nakayama's lemma, see Algebra, Lemma 10.20.1. At this point we have proved two out of the three assertions ($N$ is $A$-flat and of finite presentation over $B$) and what is left is to show that $u$ is universally injective.

By Algebra, Theorem 10.82.3 it suffices to show that $N \otimes _ A Q \to M \otimes _ A Q$ is injective for every finitely presented $A$-module $Q$. By More on Algebra, Lemma 15.124.3 we may assume $Q = A/(a)$ with $a \in \mathfrak m_ A$ nonzero. Thus it suffices to show that $N/aN \to M/aM$ is injective. Let $x \in N$ with $u(x) \in aM$. By Lemma 38.19.6 we know that $x$ has a content ideal $I \subset A$. Since $I$ is finitely generated (More on Algebra, Lemma 15.24.2) and $A$ is a valuation ring, we have $I = (b)$ for some $b$ (by Algebra, Lemma 10.50.15). By More on Algebra, Lemma 15.24.3 the element $u(x)$ has content ideal $I$ as well. Since $u(x) \in aM$ we see that $(b) \subset (a)$ by More on Algebra, Definition 15.24.1. Since $x \in bN$ we conclude $x \in aN$ as desired. $\square$

Proof. Choose $U$, $N'$ as in the definition of a pure spreadout. Any element $x' \in N'$ has a content ideal in $A$ because $N'$ is $A$-projective (this can easily be seen directly, but it also follows from More on Algebra, Lemma 15.24.4 and Algebra, Example 10.91.1). Since $N' \to N$ is $A$-universally injective, we see that the image $x \in N$ of any $x' \in N'$ has a content ideal in $A$ (it is the same as the content ideal of $x'$). For a general $x \in N$ we choose $s \in S$ such that $s x$ is in the image of $N' \to N$ and we use that $x$ and $sx$ have the same content ideal.

Let $u : N \to M$ be as in (2). To show that $u$ is $A$-universally injective, we may replace $A$ by a localization at a maximal ideal (small detail omitted). Assume $A$ is local with maximal ideal $\mathfrak m$. Pick $s \in S$ and consider the composition

Each of these maps is injective modulo $\mathfrak m$, hence the composition is $A$-universally injective by Lemma 38.7.5. Since $N = \mathop{\mathrm{colim}}\nolimits _{s \in S} (1/s)N'$ we conclude that $u$ is $A$-inversally injective as a colimit of universally injective maps. $\square$

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. $\square$

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). $\square$

Proof. Since $\mathfrak m_ A \subset A$ is not the generic point of $\mathop{\mathrm{Spec}}(A)$ we see that $I \subset \mathfrak m_ A$ for at least one $I \in S$. Hence $\bigcap _{I \in S} I \subset \mathfrak m_ A$. Let $f \in \mathfrak m_ A$ be nonzero. Then $V(f) \subset \bigcup _{I \in S} V(I)$. Since the constructible topology on $V(f)$ is quasi-compact (Topology, Lemma 5.23.2 and Algebra, Lemma 10.26.2) we find that $V(f) \subset V(I_1) \cup \ldots \cup V(I_ n)$ for some $I_ j \in S$. Because $I_1 \ldots I_ n \in S$ we see that $V(f) \subset V(I)$ for some $I$. As $I$ is finitely generated this implies that $I^ m \subset (f)$ for some $m$ and since $S$ is closed under products we see that $I \subset (f^2)$ for some $I \in S$. Then it is not possible to have $f \in I$. $\square$

Proof. Let $\mathfrak p \subset A$ be a closed point of $\mathop{\mathrm{Spec}}(A) \setminus V(J)$. We want to show that $I \subset \mathfrak p$. If not, then some $f \in I$ maps to a nonzero element of $A/\mathfrak p$. Note that $V(J) \cap \mathop{\mathrm{Spec}}(A/\mathfrak p)$ is the set of non-generic points. Hence by Lemma 38.25.10 applied to the collection of ideals $J^ nA/\mathfrak p$ we conclude that the image of $f$ is zero in $A/\mathfrak p$. $\square$

Proof. It suffices to show that $M \otimes _ A I/I_2 \to IM/I_2M$ is injective, see Algebra, Lemma 10.99.9. This is true over $U$ by assumption (2). Thus it suffices to show that $M \otimes _ A I/I_2$ injects into its sections over $U$. We have $M \otimes _ A I/I_2 = M/IM \otimes _ A I/I_2$ and $M/IM$ is a filtered colimit of finite free $A/I$-modules (Algebra, Theorem 10.81.4). Hence it suffices to show that $I/I_2$ injects into its sections over $U$, which follows from the construction of $I_2$. $\square$

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. $\square$