Lemma 10.99.1. 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 $R$.

## 10.99 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 10.128.

**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. $\square$

Lemma 10.99.2. 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 10.99.1.
$\square$

Lemma 10.99.3. 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 10.99.2.
$\square$

Lemma 10.99.4. 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 10.99.1 we see that $u$ is injective. On the other hand, by Nakayama's Lemma 10.20.1 the map is surjective. The lemma follows.
$\square$

Lemma 10.99.5. 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 10.99.1. 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 $F_1 \to F_0$.
$\square$

In the rest of this section we prove two versions of what is called the “*local criterion of flatness*”. Note also the interesting Lemma 10.128.1 below.

Lemma 10.99.6. 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 10.75.2.
$\square$

Lemma 10.99.7 (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 10.39.5 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. $\square$

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 10.99.6).

Lemma 10.99.8. 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

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

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

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. $\square$

Lemma 10.99.9. 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 $M \otimes _ R I/I^2 \to IM/I^2M$ is injective, then $M/I^2M$ is flat over $R/I^2$.

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 10.99.8 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 $n$. $\square$

Lemma 10.99.10 (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 10.99.8 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. $\square$

In particular, in the situation of Lemma 10.99.10, 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 10.99.11. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Assume

$R$ is a Noetherian ring,

$S$ is a Noetherian ring,

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

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 10.99.10. 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. $\square$

Lemma 10.99.12. 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

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. $\square$

Lemma 10.99.13. 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

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. $\square$

Lemma 10.99.14. Let

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

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

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

$\text{Tor}_1^ R(M, R/I) \to \text{Tor}_1^{R'}(M', R'/I')$ 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 10.39.7 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 $M$. $\square$

Please compare the lemma below to Lemma 10.101.8 (the case of a nilpotent ideal) and Lemma 10.128.8 (the case of finitely presented algebras).

Lemma 10.99.15 (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

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

The module $M$ is not zero.

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

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

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. $\square$

## 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)