The Stacks project

38.25 Variants of a lemma

In this section we discuss variants of Algebra, Lemmas 10.128.4 and 10.99.1. The most general version is Proposition 38.25.13; this was stated as [Lemma 4.2.2, GruRay] but the proof in loc.cit. only gives the weaker result as stated in Lemma 38.25.5. The intricate proof of Proposition 38.25.13 is due to Ofer Gabber. As we currently have no application for the proposition we encourage the reader to skip to the next section after reading the proof of Lemma 38.25.5; this lemma will be used in the next section to prove Theorem 38.26.1.

Situation 38.25.1. Let $\varphi : A \to B$ 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.

In this situation it is our goal to show that $u$ is $A$-universally injective, $N$ is of finite presentation over $B$, and $N$ is flat as an $A$-module. If this is true, we will say the lemma holds in the given situation.

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. $\square$

Lemma 38.25.3. Let $A_0$ be a local ring. If the lemma holds for every Situation 38.25.1 with $A = A_0$, with $B$ a localization of a polynomial algebra over $A$, and $N$ of finite presentation over $B$, then the lemma holds for every Situation 38.25.1 with $A = A_0$.

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

\[ \xymatrix{ 0 \ar[r] & M \otimes _ A I/I^2 \ar[r] & M/I^2M \ar[r] & M/IM \ar[r] & 0 \\ & N \otimes _ A I/I^2 \ar[r] \ar[u]^ u & N/I^2N \ar[r] \ar[u]^ u & N/IN \ar[r] \ar[u]^ u & 0 } \]

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 $I = 0$. $\square$

The following lemma discusses the special case of Situation 38.25.1 where $M$ has a $B$-module structure and $u$ is $B$-linear. This is the case most often used in practice and it is significantly easier to prove than the general case.

Lemma 38.25.5. Let $A \to B$ be a local ring homomorphism of local rings which is essentially of finite type. Let $u : N \to M$ be a $B$-module map. If $N$ is a finite $B$-module, $M$ is flat over $A$, and $\overline{u} : N/\mathfrak m_ A N \to M/\mathfrak m_ A M$ is injective, then $u$ is $A$-universally injective, $N$ is of finite presentation over $B$, and $N$ is flat over $A$.

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 15.121.3 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. $\square$

Consider the following situation

38.25.6.1
\begin{equation} \label{flat-equation-star} \begin{matrix} A \to B\text{ of finite presentation, }S \subset B \text{ a multiplicative subset, and } \\ N\text{ a finitely presented }S^{-1}B\text{-module} \end{matrix} \end{equation}

In this situation a pure spreadout is an affine open $U \subset \mathop{\mathrm{Spec}}(B)$ with $\mathop{\mathrm{Spec}}(S^{-1}B) \subset U$ and a finitely presented $\mathcal{O}(U)$-module $N'$ extending $N$ such that $N'$ is $A$-projective and $N' \to N = S^{-1}N'$ is $A$-universally injective.

In (38.25.6.1) if $A \to A_1$ is a ring map, then we can base change: take $B_1 = B \otimes _ A A_1$, let $S_1 \subset B_1$ be the image of $S$, and let $N_1 = N \otimes _ A A_1$. This works because $S_1^{-1}B_1 = S^{-1}B \otimes _ A A_1$. We will use this without further mention in the following.

Lemma 38.25.7. In (38.25.6.1) if there exists a pure spreadout, then

  1. elements of $N$ have content ideals in $A$, and

  2. if $u : N \to M$ is a morphism to a flat $A$-module $M$ such that $N/\mathfrak m N \to M/\mathfrak m M$ is injective for all maximal ideals $\mathfrak m$ of $A$, then $u$ is $A$-universally injective.

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

\[ N' \to N \xrightarrow {1/s} N \xrightarrow {u} M \]

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. $\square$

Lemma 38.25.8. In (38.25.6.1) for every $\mathfrak p \in \mathop{\mathrm{Spec}}(A)$ there is a finitely generated ideal $I \subset \mathfrak pA_\mathfrak p$ such that over $A_\mathfrak p/I$ we have a pure spreadout.

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

\[ N'/IN' = N'_0/\mathfrak m_{A_0} N'_0 \otimes _{\kappa _{A_0}} A/I \]

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. $\square$

Lemma 38.25.9. In (38.25.6.1) assume $N$ is $A$-flat, $M$ is a flat $A$-module, and $u : N \to M$ is an $A$-module map such that $u \otimes \text{id}_{\kappa (\mathfrak p)}$ is injective for all $\mathfrak p \in \mathop{\mathrm{Spec}}(A)$. Then $u$ is $A$-universally injective.

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

\[ \text{Ann}_ A(I^ n) \otimes _ R N \longrightarrow \text{Ann}_ A(I^ n) \otimes _ R M \]

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). $\square$

Lemma 38.25.10. Let $A$ be a local domain which is not a field. Let $S$ be a set of finitely generated ideals of $A$. Assume that $S$ is closed under products and such that $\bigcup _{I \in S} V(I)$ is the complement of the generic point of $\mathop{\mathrm{Spec}}(A)$. Then $\bigcap _{I \in S} I = (0)$.

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 $f \in I$. $\square$

Lemma 38.25.11. Let $A$ be a local ring. Let $I, J \subset A$ be ideals. If $J$ is finitely generated and $I \subset J^ n$ for all $n \geq 1$, then $V(I)$ contains the closed points of $\mathop{\mathrm{Spec}}(A) \setminus V(J)$.

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 $A/\mathfrak p$. $\square$

Lemma 38.25.12. Let $A$ be a local ring. Let $I \subset A$ be an ideal. Let $U \subset \mathop{\mathrm{Spec}}(A)$ be quasi-compact open. Let $M$ be an $A$-module. Assume that

  1. $M/IM$ is flat over $A/I$,

  2. $M$ is flat over $U$,

Then $M/I_2M$ is flat over $A/I_2$ where $I_2 = \mathop{\mathrm{Ker}}(I \to \Gamma (U, I/I^2))$.

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 $I_2$. $\square$

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

\[ S = \mathop{\mathrm{Spec}}(A) \supset S_0 \supset S_1 \supset \ldots \supset S_ t = \emptyset \]

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

\[ \xymatrix{ N \otimes _ A (I_0^ n/I_0^{n + 1}) \ar[r] \ar[d] & M \otimes _ A (I_0^ n/I_0^{n + 1}) \ar@{=}[d] \\ I_0^ n N /I_0^{n + 1} N \ar[r] & I_0^ n M /I_0^{n + 1} M } \]

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. $\square$


Comments (0)


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ASZ. Beware of the difference between the letter 'O' and the digit '0'.