The Stacks project

10.98 Criteria for flatness

In this section we prove some important technical lemmas in the Noetherian case. We will (partially) generalize these to the non-Noetherian case in Section 10.127.

Lemma 10.98.1. Suppose that $R \to S$ is a local homomorphism of Noetherian local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Let $M$ be a flat $R$-module and $N$ a finite $S$-module. Let $u : N \to M$ be a map of $R$-modules. If $\overline{u} : N/\mathfrak m N \to M/\mathfrak m M$ is injective then $u$ is injective. In this case $M/u(N)$ is flat over $R$.

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 10.50.4) 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 $I \otimes _ R M/u(N) \to M/u(N)$ is injective for every ideal $I \subset R$, see Lemma 10.38.5. Consider the diagram

\[ \begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & N/IN & \to & M/IM & \to & M/(IM + u(N)) & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & N & \to & M & \to & M/u(N) & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & N \otimes _ R I & \to & M \otimes _ R I & \to & M/u(N)\otimes _ R I & \to & 0 \end{matrix} \]

The arrow $M \otimes _ R I \to M$ is injective. By the snake lemma (Lemma 10.4.1) we see that it suffices to prove 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. $\square$

Lemma 10.98.2. Suppose that $R \to S$ is a flat and local ring homomorphism of Noetherian local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Suppose $f \in S$ is a nonzerodivisor in $S/{\mathfrak m}S$. Then $S/fS$ is flat over $R$, and $f$ is a nonzerodivisor in $S$.

Proof. Follows directly from Lemma 10.98.1. $\square$

Lemma 10.98.3. Suppose that $R \to S$ is a flat and local ring homomorphism of Noetherian local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Suppose $f_1, \ldots , f_ c$ is a sequence of elements of $S$ such that the images $\overline{f}_1, \ldots , \overline{f}_ c$ form a regular sequence in $S/{\mathfrak m}S$. Then $f_1, \ldots , f_ c$ is a regular sequence in $S$ and each of the quotients $S/(f_1, \ldots , f_ i)$ is flat over $R$.

Proof. Induction and Lemma 10.98.2. $\square$

Lemma 10.98.4. Let $R \to S$ be a local homomorphism of Noetherian local rings. Let $\mathfrak m$ be the maximal ideal of $R$. Let $M$ be a finite $S$-module. Suppose that (a) $M/\mathfrak mM$ is a free $S/\mathfrak mS$-module, and (b) $M$ is flat over $R$. Then $M$ is free and $S$ is flat over $R$.

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 10.98.1 we see that $u$ is injective. On the other hand, by Nakayama's Lemma 10.19.1 the map is surjective. The lemma follows. $\square$

Lemma 10.98.5. Let $R \to S$ be a local homomorphism of local Noetherian rings. Let $\mathfrak m$ be the maximal ideal of $R$. Let $0 \to F_ e \to F_{e-1} \to \ldots \to F_0$ be a finite complex of finite $S$-modules. Assume that each $F_ i$ is $R$-flat, and that the complex $0 \to F_ e/\mathfrak m F_ e \to F_{e-1}/\mathfrak m F_{e-1} \to \ldots \to F_0 / \mathfrak m F_0$ is exact. Then $0 \to F_ e \to F_{e-1} \to \ldots \to F_0$ is exact, and moreover the module $\mathop{\mathrm{Coker}}(F_1 \to F_0)$ is $R$-flat.

Proof. By induction on $e$. If $e = 1$, then this is exactly Lemma 10.98.1. If $e > 1$, we see by Lemma 10.98.1 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 $F_1 \to F_0$. $\square$

In the rest of this section we prove two versions of what is called the “local criterion of flatness”. Note also the interesting Lemma 10.127.1 below.

Lemma 10.98.6. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and residue field $\kappa = R/\mathfrak m$. Let $M$ be an $R$-module. If $\text{Tor}_1^ R(\kappa , M) = 0$, then for every finite length $R$-module $N$ we have $\text{Tor}_1^ R(N, M) = 0$.

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 10.74.2. $\square$

Lemma 10.98.7 (Local criterion for flatness). Let $R \to S$ be a local homomorphism of local Noetherian rings. Let $\mathfrak m$ be the maximal ideal of $R$, and let $\kappa = R/\mathfrak m$. Let $M$ be a finite $S$-module. If $\text{Tor}_1^ R(\kappa , M) = 0$, then $M$ is flat over $R$.

Proof. Let $I \subset R$ be an ideal. By Lemma 10.38.5 it suffices to show that $I \otimes _ R M \to M$ is injective. By Remark 10.74.9 we see that this kernel is equal to $\text{Tor}_1^ R(M, R/I)$. By Lemma 10.98.6 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

\[ 0 \to I \cap \mathfrak m^ n \to I \oplus \mathfrak m^ n \to I + \mathfrak m^ n \to 0 \]

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

\[ \xymatrix{ (I\cap \mathfrak m^ n) \otimes _ R M \ar[r] \ar[d] & I \otimes _ R M \oplus \mathfrak m^ n \otimes _ R M \ar[r] \ar[d] & (I + \mathfrak m^ n) \otimes _ R M \ar[d] \\ M \ar[r] & M \oplus M \ar[r] & M } \]

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 10.50.2 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 10.96.3. Hence $K = 0$, as desired. $\square$

In the following we often encounter the conditions “$M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$”. The following lemma gives some consequences of these conditions (it is a generalization of Lemma 10.98.6).

Lemma 10.98.8. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. If $M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$ then

  1. $M/I^ nM$ is flat over $R/I^ n$ for all $n \geq 1$, and

  2. for any module $N$ which is annihilated by $I^ m$ for some $m \geq 0$ we have $\text{Tor}_1^ R(N, M) = 0$.

In particular, if $I$ is nilpotent, then $M$ is flat over $R$.

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

\[ 0 \to K \to \bigoplus \nolimits _{i \in I} R/I \to N \to 0 \]

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

\[ 0 \to \text{Tor}_1^ R(N, M) \to K \otimes _ R M \to (\bigoplus \nolimits _{i \in I} R/I) \otimes _ R M \to N \otimes _ R M \to 0 \]

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

\begin{align*} K \otimes _ R M & = K \otimes _{R/I} M/IM, \\ (\bigoplus \nolimits _{i \in I} R/I) \otimes _ R M & = (\bigoplus \nolimits _{i \in I} R/I) \otimes _{R/I} M/IM, \\ N \otimes _ R M & = N \otimes _{R/I} M/IM. \end{align*}

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

\[ 0 \to K \otimes _{R/I} M/IM \to (\bigoplus \nolimits _{i \in I} R/I) \otimes _{R/I} M/IM \to N \otimes _{R/} M/IM \to 0 \]

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

In particular, if we apply this to the module $I/I^2$, then we conclude that the sequence

\[ 0 \to I^2 \otimes _ R M \to I \otimes _ R M \to I/I^2 \otimes _ R M \to 0 \]

is short exact. This implies that $I^2 \otimes _ R M \to M$ is injective and it implies that $I/I^2 \otimes _{R/I} M/IM = IM/I^2M$.

Let us prove that $M/I^2M$ is flat over $R/I^2$. Let $I^2 \subset J$ be an ideal. We have to show that $J/I^2 \otimes _{R/I^2} M/I^2M \to M/I^2M$ is injective, see Lemma 10.38.5. As $M/IM$ is flat over $R/I$ we know that the map $(I + J)/I \otimes _{R/I} M/IM \to M/IM$ is injective. The sequence

\[ (I \cap J)/I^2 \otimes _{R/I^2} M/I^2M \to J/I^2 \otimes _{R/I^2} M/I^2M \to (I + J)/I \otimes _{R/I} M/IM \to 0 \]

is exact, as you get it by tensoring the exact sequence $0 \to (I \cap J) \to J \to (I + J)/I \to 0$ by $M/I^2M$. Hence suffices to prove the injectivity of the map $(I \cap J)/I^2 \otimes _{R/I} M/IM \to IM/I^2M$. However, the map $(I \cap J)/I^2 \to I/I^2$ is injective and as $M/IM$ is flat over $R/I$ the map $(I \cap J)/I^2 \otimes _{R/I} M/IM \to I/I^2 \otimes _{R/I} M/IM$ is injective. Since we have previously seen that $I/I^2 \otimes _{R/I} M/IM = IM/I^2M$ we obtain the desired injectivity.

Hence we have proven that the assumptions imply: (a) $\text{Tor}_1^ R(N, M) = 0$ for all $N$ annihilated by $I$, (b) $I^2 \otimes _ R M \to M$ is injective, and (c) $M/I^2M$ is flat over $R/I^2$. Thus we can continue by induction to get the same results for $I^ n$ for all $n \geq 1$. $\square$

Lemma 10.98.9. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module.

  1. If $M/IM$ is flat over $R/I$ and $M \otimes _ R I/I^2 \to IM/I^2M$ is injective, then $M/I^2M$ is flat over $R/I^2$.

  2. If $M/IM$ is flat over $R/I$ and $M \otimes _ R I^ n/I^{n + 1} \to I^ nM/I^{n + 1}M$ is injective for $n = 1, \ldots , k$, then $M/I^{k + 1}M$ is flat over $R/I^{k + 1}$.

Proof. The first statement is a consequence of Lemma 10.98.8 applied with $R$ replaced by $R/I^2$ and $M$ replaced by $M/I^2M$ using that

\[ \text{Tor}_1^{R/I^2}(M/I^2M, R/I) = \mathop{\mathrm{Ker}}(M \otimes _ R I/I^2 \to IM/I^2M), \]

see Remark 10.74.9. 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

\[ \text{Tor}_1^{R/I^{n + 1}}(M/I^{n + 1}M, R/I) = \mathop{\mathrm{Ker}}(M \otimes _ R I^ n/I^{n + 1} \to I^ nM/I^{n + 1}M) \]

for every $n$. $\square$

Lemma 10.98.10 (Variant of the local criterion). Let $R \to S$ be a local homomorphism of Noetherian local rings. Let $I \not= R$ be an ideal in $R$. Let $M$ be a finite $S$-module. If $\text{Tor}_1^ R(M, R/I) = 0$ and $M/IM$ is flat over $R/I$, then $M$ is flat over $R$.

Proof. First proof: By Lemma 10.98.8 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 10.98.7.

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 10.98.7. 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 10.38.11 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

\begin{eqnarray*} \sum f_ i \otimes x_ i & = & \sum f_ i \otimes x_ i + \sum f_ ia_{ij} \otimes y_ j - \sum f_ i \otimes a_{ij} y_ j \\ & = & \sum f_ i \otimes (x_ i - \sum a_{ij} y_ j) + \sum (\sum f_ i a_{ij}) \otimes y_ j \end{eqnarray*}

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 10.74.9) and we win. $\square$

In particular, in the situation of Lemma 10.98.10, suppose that $I = (x)$ is generated by a single element $x$ which is a nonzerodivisor in $R$. Then $\text{Tor}_1^ R(M, R/(x)) = (0)$ if and only if $x$ is a nonzerodivisor on $M$.

Lemma 10.98.11. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Assume

  1. $R$ is a Noetherian ring,

  2. $S$ is a Noetherian ring,

  3. $M$ is a finite $S$-module, and

  4. for each $n \geq 1$ the module $M/I^ n M$ is flat over $R/I^ n$.

Then for every $\mathfrak q \in V(IS)$ the localization $M_{\mathfrak q}$ is flat over $R$. In particular, if $S$ is local and $IS$ is contained in its maximal ideal, then $M$ is flat over $R$.

Proof. We are going to use Lemma 10.98.10. 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 10.74.9 this Tor group is equal to $K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$. We know for each $n \geq 1$ that the kernel $\mathop{\mathrm{Ker}}(I/I^ n \otimes _{R/I^ n} M/I^ nM \to M/I^ nM)$ is zero. Since there is a module map $I/I^ n \otimes _{R/I^ n} M/I^ nM \to (I \otimes _ R M)/I^{n - 1}(I \otimes _ R M)$ we conclude that $K \subset I^{n - 1}(I \otimes _ R M)$ for each $n$. By the Artin-Rees lemma, and more precisely Lemma 10.50.5 we conclude that $K_{\mathfrak q} = 0$, as desired. $\square$

Lemma 10.98.12. Let $R \to R' \to R''$ be ring maps. Let $M$ be an $R$-module. Suppose that $M \otimes _ R R'$ is flat over $R'$. Then the natural map $\text{Tor}_1^ R(M, R') \otimes _{R'} R'' \to \text{Tor}_1^ R(M, R'')$ is onto.

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

\[ 0 \to K' \to F_1\otimes _ R R' \to F_0 \otimes _ R R' \to M \otimes _ R R' \to 0. \]

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

\[ K' \otimes _{R'} R'' \to F_1 \otimes _ R R'' \to F_0 \otimes _ R R'' \to M \otimes _ R R'' \to 0 \]

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. $\square$

Lemma 10.98.13. Let $R \to R'$ be a ring map. Let $I \subset R$ be an ideal and $I' = IR'$. Let $M$ be an $R$-module and set $M' = M \otimes _ R R'$. The natural map $\text{Tor}_1^ R(R'/I', M) \to \text{Tor}_1^{R'}(R'/I', M')$ is surjective.

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

\[ F_2' \oplus F_2'' \to F_1' \to F_0' \to M' \to 0 \]

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

\[ \xymatrix{ F_2'/I'F_2' \ar[r] \ar[d] & F_1'/I'F_1' \ar[r] \ar[d] & F_0'/I'F_0' \ar[d] \\ F_2'/I'F_2' \oplus F_2''/I'F_2'' \ar[r] & F_1'/I'F_1' \ar[r] & F_0'/I'F_0' } \]

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

Lemma 10.98.14. Let

\[ \xymatrix{ S \ar[r] & S' \\ R \ar[r] \ar[u] & R' \ar[u] } \]

be a commutative diagram of local homomorphisms of local Noetherian rings. Let $I \subset R$ be a proper ideal. Let $M$ be a finite $S$-module. Denote $I' = IR'$ and $M' = M \otimes _ S S'$. Assume that

  1. $S'$ is a localization of the tensor product $S \otimes _ R R'$,

  2. $M/IM$ is flat over $R/I$,

  3. $\text{Tor}_1^ R(M, R/I) \to \text{Tor}_1^{R'}(M', R'/I')$ is zero.

Then $M'$ is flat over $R'$.

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 10.38.7 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 10.98.10 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 10.98.13 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 10.98.12 by looking at the ring maps $R \to R/I \to R'/I'$ and the module $M$. $\square$

Please compare the lemma below to Lemma 10.100.8 (the case of a nilpotent ideal) and Lemma 10.127.8 (the case of finitely presented algebras).

Lemma 10.98.15 (Critère de platitude par fibres; Noetherian case). Let $R$, $S$, $S'$ be Noetherian local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $\mathfrak m \subset R$ be the maximal ideal. Let $M$ be an $S'$-module. Assume

  1. The module $M$ is finite over $S'$.

  2. The module $M$ is not zero.

  3. The module $M/\mathfrak m M$ is a flat $S/\mathfrak m S$-module.

  4. The module $M$ is a flat $R$-module.

Then $S$ is flat over $R$ and $M$ is a flat $S$-module.

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

\[ \mathfrak m \otimes _ R M \to I \otimes _ S M \to M. \]

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 10.74.9. By Lemma 10.98.10 we conclude that $M$ is flat over $S$. Note that since $M/\mathfrak m_{S'}M$ is not zero by Nakayama's Lemma 10.19.1 we see that actually $M$ is faithfully flat over $S$ by Lemma 10.38.15 (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 10.98.7. $\square$


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