The Stacks project

Proof. First we claim that $u_ n : N/{\mathfrak m}^ nN \to M/{\mathfrak m}^ nM$ is injective for all $n \geq 1$. We proceed by induction, the base case is that $\overline{u} = u_1$ is injective. By our assumption that $M$ is flat over $R$ we have a short exact sequence $0 \to M \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} \to M/{\mathfrak m}^{n + 1}M \to M/{\mathfrak m}^ n M \to 0$. Also, $M \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} = M/{\mathfrak m}M \otimes _{R/{\mathfrak m}} {\mathfrak m}^ n/{\mathfrak m}^{n + 1}$. We have a similar exact sequence $N \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} \to N/{\mathfrak m}^{n + 1}N \to N/{\mathfrak m}^ n N \to 0$ for $N$ except we do not have the zero on the left. We also have $N \otimes _ R {\mathfrak m}^ n/{\mathfrak m}^{n + 1} = N/{\mathfrak m}N \otimes _{R/{\mathfrak m}} {\mathfrak m}^ n/{\mathfrak m}^{n + 1}$. Thus the map $u_{n + 1}$ is injective as both $u_ n$ and the map $\overline{u} \otimes \text{id}_{{\mathfrak m}^ n/{\mathfrak m}^{n + 1}}$ are.

By Krull's intersection theorem (Lemma 10.51.4) applied to $N$ over the ring $S$ and the ideal $\mathfrak mS$ we have $\bigcap \mathfrak m^ nN = 0$. Thus the injectivity of $u_ n$ for all $n$ implies $u$ is injective.

To show that $M/u(N)$ is flat over $R$, it suffices to show that $\text{Tor}_1^ R(M/u(N), R/I) = 0$ for every ideal $I \subset R$, see Lemma 10.75.8. From the short exact sequence

and the flatness of $M$ we obtain an exact sequence of Tors

See Lemma 10.75.2. Thus it suffices to show that $N/IN$ injects into $M/IM$. Note that $R/I \to S/IS$ is a local homomorphism of Noetherian local rings, $N/IN \to M/IM$ is a map of $R/I$-modules, $N/IN$ is finite over $S/IS$, and $M/IM$ is flat over $R/I$ and $u \bmod I : N/IN \to M/IM$ is injective modulo $\mathfrak m$. Thus we may apply the first part of the proof to $u \bmod I$ and we conclude. $\square$

Proof. Follows directly from Lemma 10.99.1. $\square$

Proof. Induction and Lemma 10.99.2. $\square$

Proof. Let $\overline{x}_1, \ldots , \overline{x}_ n$ be a basis for the free module $M/\mathfrak mM$. Choose $x_1, \ldots , x_ n \in M$ with $x_ i$ mapping to $\overline{x}_ i$. Let $u : S^{\oplus n} \to M$ be the map which maps the $i$th standard basis vector to $x_ i$. By Lemma 10.99.1 we see that $u$ is injective. On the other hand, by Nakayama's Lemma 10.20.1 the map is surjective. The lemma follows. $\square$

Proof. By induction on $e$. If $e = 1$, then this is exactly Lemma 10.99.1. If $e > 1$, we see by Lemma 10.99.1 that $F_ e \to F_{e-1}$ is injective and that $C = \mathop{\mathrm{Coker}}(F_ e \to F_{e-1})$ is a finite $S$-module flat over $R$. Hence we can apply the induction hypothesis to the complex $0 \to C \to F_{e-2} \to \ldots \to F_0$. We deduce that $C \to F_{e-2}$ is injective and the exactness of the complex follows, as well as the flatness of the cokernel of $F_1 \to F_0$. $\square$

Proof. By descending induction on the length of $N$. If the length of $N$ is $1$, then $N \cong \kappa $ and we are done. If the length of $N$ is more than $1$, then we can fit $N$ into a short exact sequence $0 \to N' \to N \to N'' \to 0$ where $N'$, $N''$ are finite length $R$-modules of smaller length. The vanishing of $\text{Tor}_1^ R(N, M)$ follows from the vanishing of $\text{Tor}_1^ R(N', M)$ and $\text{Tor}_1^ R(N'', M)$ (induction hypothesis) and the long exact sequence of Tor groups, see Lemma 10.75.2. $\square$

Proof. Let $I \subset R$ be an ideal. By Lemma 10.39.5 it suffices to show that $I \otimes _ R M \to M$ is injective. By Remark 10.75.9 we see that this kernel is equal to $\text{Tor}_1^ R(M, R/I)$. By Lemma 10.99.6 we see that $J \otimes _ R M \to M$ is injective for all ideals of finite colength.

Choose $n >> 0$ and consider the following short exact sequence

This is a sub sequence of the short exact sequence $0 \to R \to R^{\oplus 2} \to R \to 0$. Thus we get the diagram

Note that $I + \mathfrak m^ n$ and $\mathfrak m^ n$ are ideals of finite colength. Thus a diagram chase shows that $\mathop{\mathrm{Ker}}((I \cap \mathfrak m^ n)\otimes _ R M \to M) \to \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is surjective. We conclude in particular that $K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is contained in the image of $(I \cap \mathfrak m^ n) \otimes _ R M$ in $I \otimes _ R M$. By Artin-Rees, Lemma 10.51.2 we see that $K$ is contained in $\mathfrak m^{n-c}(I \otimes _ R M)$ for some $c > 0$ and all $n >> 0$. Since $I \otimes _ R M$ is a finite $S$-module (!) and since $S$ is Noetherian, we see that this implies $K = 0$. Namely, the above implies $K$ maps to zero in the $\mathfrak mS$-adic completion of $I \otimes _ R M$. But the map from $S$ to its $\mathfrak mS$-adic completion is faithfully flat by Lemma 10.97.3. Hence $K = 0$, as desired. $\square$

Proof. Assume $M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$. Let $N$ be an $R/I$-module. Choose a short exact sequence

By the long exact sequence of $\text{Tor}$ and the vanishing of $\text{Tor}_1^ R(R/I, M)$ we get

But since $K$, $\bigoplus _{i \in I} R/I$, and $N$ are all annihilated by $I$ we see that

As $M/IM$ is flat over $R/I$ we conclude that

is exact. Combining this with the above we conclude that $\text{Tor}_1^ R(N, M) = 0$ for any $R$-module $N$ annihilated by $I$.

Let us prove (2) by induction on $m$. The case $m = 1$ was done in the previous paragraph. For $N$ annihilated by $I^ m$ for $m > 1$ we may choose an exact sequence $0 \to N' \to N \to N'' \to 0$ with $N'$ and $N''$ annihilated by $I^{m - 1}$. For example one can take $N' = IN$ and $N'' = N/IN$. Then the exact sequence

and induction prove the vanishing we want.

Finally, we prove (1). Given $n \geq 1$ we have to show that $M/I^ nM$ is flat over $R/I^ n$. In other words, we have to show that the functor $N \mapsto N \otimes _{R/I^ n} M/I^ nM$ is exact on the category of $R$-modules $N$ annihilated by $I^ n$. However, for such $N$ we have $N \otimes _{R/I^ n} M/I^ nM = N \otimes _ R M$. By the vanishing of $\text{Tor}_1$ in (2) we see that the functor $N \mapsto N \otimes _ R M$ is exact on the category of $N$ annihilated by some power of $I$ and we conclude. $\square$

Proof. The first statement is a consequence of Lemma 10.99.8 applied with $R$ replaced by $R/I^2$ and $M$ replaced by $M/I^2M$ using that

see Remark 10.75.9. The second statement follows in the same manner using induction on $n$ to show that $M/I^{n + 1}M$ is flat over $R/I^{n + 1}$ for $n = 1, \ldots , k$. Here we use that

for every $n$. $\square$

Proof. First proof: By Lemma 10.99.8 we see that $\text{Tor}_1^ R(\kappa , M)$ is zero where $\kappa $ is the residue field of $R$. Hence we see that $M$ is flat over $R$ by Lemma 10.99.7.

Second proof: Let $\mathfrak m$ be the maximal ideal of $R$. We will show that $\mathfrak m \otimes _ R M \to M$ is injective, and then apply Lemma 10.99.7. Suppose that $\sum f_ i \otimes x_ i \in \mathfrak m \otimes _ R M$ and that $\sum f_ i x_ i = 0$ in $M$. By the equational criterion for flatness Lemma 10.39.11 applied to $M/IM$ over $R/I$ we see there exist $\overline{a}_{ij} \in R/I$ and $\overline{y}_ j \in M/IM$ such that $x_ i \bmod IM = \sum _ j \overline{a}_{ij} \overline{y}_ j $ and $0 = \sum _ i (f_ i \bmod I) \overline{a}_{ij}$. Let $a_{ij} \in R$ be a lift of $\overline{a}_{ij}$ and similarly let $y_ j \in M$ be a lift of $\overline{y}_ j$. Then we see that

Since $x_ i - \sum a_{ij} y_ j \in IM$ and $\sum f_ i a_{ij} \in I$ we see that there exists an element in $I \otimes _ R M$ which maps to our given element $\sum f_ i \otimes x_ i$ in $\mathfrak m \otimes _ R M$. But $I \otimes _ R M \to M$ is injective by assumption (see Remark 10.75.9) and we win. $\square$

Proof. We are going to use Lemma 10.99.10. By assumption $M/IM$ is flat over $R/I$. Hence it suffices to check that $\text{Tor}_1^ R(M, R/I)$ is zero on localization at $\mathfrak q$. By Remark 10.75.9 this Tor group is equal to $K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$. We know that the kernel of $I/I^ n \otimes _{R/I^ n} M/I^ nM \to M/I^ nM$ is zero for all $n \geq 1$. Hence an element of $K$ maps to zero in $I/I^ n \otimes _{R/I^ n} M/I^ nM$. Since

we conclude that $K \subset I^{n - 1}(I \otimes _ R M)$ for all $n \geq 1$. By the Artin-Rees lemma, and more precisely Lemma 10.51.5 we conclude that $K_{\mathfrak q} = 0$, as desired. $\square$

Proof. Let $F_\bullet $ be a free resolution of $M$ over $R$. The complex $F_2 \otimes _ R R' \to F_1\otimes _ R R' \to F_0 \otimes _ R R'$ computes $\text{Tor}_1^ R(M, R')$. The complex $F_2 \otimes _ R R'' \to F_1\otimes _ R R'' \to F_0 \otimes _ R R''$ computes $\text{Tor}_1^ R(M, R'')$. Note that $F_ i \otimes _ R R' \otimes _{R'} R'' = F_ i \otimes _ R R''$. Let $K' = \mathop{\mathrm{Ker}}(F_1\otimes _ R R' \to F_0 \otimes _ R R')$ and similarly $K'' = \mathop{\mathrm{Ker}}(F_1\otimes _ R R'' \to F_0 \otimes _ R R'')$. Thus we have an exact sequence

By the assumption that $M \otimes _ R R'$ is flat over $R'$, the sequence

is still exact. This means that $K' \otimes _{R'} R'' \to K''$ is surjective. Since $\text{Tor}_1^ R(M, R')$ is a quotient of $K'$ and $\text{Tor}_1^ R(M, R'')$ is a quotient of $K''$ we win. $\square$

Proof. Let $F_2 \to F_1 \to F_0 \to M \to 0$ be a free resolution of $M$ over $R$. Set $F_ i' = F_ i \otimes _ R R'$. The sequence $F_2' \to F_1' \to F_0' \to M' \to 0$ may no longer be exact at $F_1'$. A free resolution of $M'$ over $R'$ therefore looks like

for a suitable free module $F_2''$ over $R'$. Next, note that $F_ i \otimes _ R R'/I' = F_ i' / IF_ i' = F_ i'/I'F_ i'$. So the complex $F_2'/I'F_2' \to F_1'/I'F_1' \to F_0'/I'F_0'$ computes $\text{Tor}_1^ R(M, R'/I')$. On the other hand $F_ i' \otimes _{R'} R'/I' = F_ i'/I'F_ i'$ and similarly for $F_2''$. Thus the complex $F_2'/I'F_2' \oplus F_2''/I'F_2'' \to F_1'/I'F_1' \to F_0'/I'F_0'$ computes $\text{Tor}_1^{R'}(M', R'/I')$. Since the vertical map on complexes

clearly induces a surjection on cohomology we win. $\square$

Proof. Since $S'$ is a localization of $S \otimes _ R R'$ we see that $M'$ is a localization of $M \otimes _ R R'$. Note that by Lemma 10.39.7 the module $M/IM \otimes _{R/I} R'/I' = M \otimes _ R R' /I'(M \otimes _ R R')$ is flat over $R'/I'$. Hence also $M'/I'M'$ is flat over $R'/I'$ as the localization of a flat module is flat. By Lemma 10.99.10 it suffices to show that $\text{Tor}_1^{R'}(M', R'/I')$ is zero. Since $M'$ is a localization of $M \otimes _ R R'$, the last assumption implies that it suffices to show that $\text{Tor}_1^ R(M, R/I) \otimes _ R R' \to \text{Tor}_1^{R'}(M \otimes _ R R', R'/I')$ is surjective.

By Lemma 10.99.13 we see that $\text{Tor}_1^ R(M, R'/I') \to \text{Tor}_1^{R'}(M \otimes _ R R', R'/I')$ is surjective. So now it suffices to show that $\text{Tor}_1^ R(M, R/I) \otimes _ R R' \to \text{Tor}_1^ R(M, R'/I')$ is surjective. This follows from Lemma 10.99.12 by looking at the ring maps $R \to R/I \to R'/I'$ and the module $M$. $\square$

Proof. Set $I = \mathfrak mS \subset S$. Then we see that $M/IM$ is a flat $S/I$-module because of (3). Since $\mathfrak m \otimes _ R S' \to I \otimes _ S S'$ is surjective we see that also $\mathfrak m \otimes _ R M \to I \otimes _ S M$ is surjective. Consider

As $M$ is flat over $R$ the composition is injective and so both arrows are injective. In particular $\text{Tor}_1^ S(S/I, M) = 0$ see Remark 10.75.9. By Lemma 10.99.10 we conclude that $M$ is flat over $S$. Note that since $M/\mathfrak m_{S'}M$ is not zero by Nakayama's Lemma 10.20.1 we see that actually $M$ is faithfully flat over $S$ by Lemma 10.39.15 (since it forces $M/\mathfrak m_ SM \not= 0$).

Consider the exact sequence $0 \to \mathfrak m \to R \to \kappa \to 0$. This gives an exact sequence $0 \to \text{Tor}_1^ R(\kappa , S) \to \mathfrak m \otimes _ R S \to I \to 0$. Since $M$ is flat over $S$ this gives an exact sequence $0 \to \text{Tor}_1^ R(\kappa , S)\otimes _ S M \to \mathfrak m \otimes _ R M \to I \otimes _ S M \to 0$. By the above this implies that $\text{Tor}_1^ R(\kappa , S)\otimes _ S M = 0$. Since $M$ is faithfully flat over $S$ this implies that $\text{Tor}_1^ R(\kappa , S) = 0$ and we conclude that $S$ is flat over $R$ by Lemma 10.99.7. $\square$