## Tag `00MD`

## 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 $$ \begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & N/IN & \to & M/IM & \to & M/(IN + u(N)) & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & N & \to & M & \to & M/u(N) & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & N \otimes_R I & \to & M \otimes_R I & \to & M/u(N)\otimes_R I & \to & 0 \end{matrix} $$ 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 $\mathop{\rm Coker}(F_1 \to F_0)$ 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 $C = \mathop{\rm Coker}(F_e \to F_{e-1})$ 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 $\mathop{\rm Ker}((I \cap \mathfrak m^n)\otimes_R M \to M) \to \mathop{\rm Ker}(I \otimes_R M \to M)$ is surjective. We conclude in particular that $K = \mathop{\rm Ker}(I \otimes_R M \to M)$ 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

- $M/I^nM$ 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 $$ 0 \to K \to \bigoplus\nolimits_{i \in I} R/I \to N \to 0 $$ 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.

- 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 10.98.8 applied with $R$ replaced by $R/I^2$ and $M$ replaced by $M/I^2M$ using that $$ \text{Tor}_1^{R/I^2}(M/I^2M, R/I) = \mathop{\rm Ker}(M \otimes_R I/I^2 \to IM/I^2M), $$ 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 $$ \text{Tor}_1^{R/I^{n + 1}}(M/I^{n + 1}M, R/I) = \mathop{\rm Ker}(M \otimes_R I^n/I^{n + 1} \to I^nM/I^{n + 1}M) $$ 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

- $R$ 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 10.98.10. 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 $K = \mathop{\rm Ker}(I \otimes_R M \to M)$. We know for each $n \geq 1$ that the kernel $\mathop{\rm Ker}(I/I^n \otimes_{R/I^n} M/I^nM \to M/I^nM)$ 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 $K' = \mathop{\rm Ker}(F_1\otimes_R R' \to F_0 \otimes_R R')$ and similarly $K'' = \mathop{\rm Ker}(F_1\otimes_R R'' \to F_0 \otimes_R R'')$. 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

- $S'$ 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 10.38.7 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

- 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 $$ \mathfrak m \otimes_R M \to I \otimes_S M \to M. $$ 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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\section{Criteria for flatness}
\label{section-criteria-flatness}
\noindent
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 \ref{section-more-flatness-criteria}.
\begin{lemma}
\label{lemma-mod-injective}
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 $R$.
\end{lemma}
\begin{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.
\medskip\noindent
By Krull's intersection theorem
(Lemma \ref{lemma-intersect-powers-ideal-module-zero})
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.
\medskip\noindent
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 \ref{lemma-flat}. Consider the diagram
$$
\begin{matrix}
& & 0 & & 0 & & 0 & & \\
& & \uparrow & & \uparrow & & \uparrow & & \\
& & N/IN & \to & M/IM & \to & M/(IN + u(N)) & \to & 0 \\
& & \uparrow & & \uparrow & & \uparrow & & \\
0 & \to & N & \to & M & \to & M/u(N) & \to & 0 \\
& & \uparrow & & \uparrow & & \uparrow & & \\
& & N \otimes_R I & \to & M \otimes_R I & \to & M/u(N)\otimes_R I & \to & 0
\end{matrix}
$$
The arrow $M \otimes_R I \to M$ is injective. By the snake lemma
(Lemma \ref{lemma-snake}) 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.
\end{proof}
\begin{lemma}
\label{lemma-grothendieck}
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 $S$.
\end{lemma}
\begin{proof}
Follows directly from Lemma \ref{lemma-mod-injective}.
\end{proof}
\begin{lemma}
\label{lemma-grothendieck-regular-sequence}
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 $R$.
\end{lemma}
\begin{proof}
Induction and Lemma \ref{lemma-grothendieck}.
\end{proof}
\begin{lemma}
\label{lemma-free-fibre-flat-free}
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 $R$.
\end{lemma}
\begin{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 \ref{lemma-mod-injective}
we see that $u$ is injective. On the other hand, by
Nakayama's Lemma \ref{lemma-NAK} the map is surjective. The
lemma follows.
\end{proof}
\begin{lemma}
\label{lemma-complex-exact-mod}
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
$\Coker(F_1 \to F_0)$ is $R$-flat.
\end{lemma}
\begin{proof}
By induction on $e$. If $e = 1$, then this is exactly
Lemma \ref{lemma-mod-injective}. If $e > 1$, we see
by Lemma \ref{lemma-mod-injective} that $F_e \to F_{e-1}$
is injective and that $C = \Coker(F_e \to F_{e-1})$
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 $F_1 \to F_0$.
\end{proof}
\noindent
In the rest of this section we prove two versions of what is called the
``{\it local criterion of flatness}''. Note also the interesting
Lemma \ref{lemma-CM-over-regular-flat} below.
\begin{lemma}
\label{lemma-prepare-local-criterion-flatness}
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
$\text{Tor}_1^R(N, M) = 0$.
\end{lemma}
\begin{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
\ref{lemma-long-exact-sequence-tor}.
\end{proof}
\begin{lemma}[Local criterion for flatness]
\label{lemma-local-criterion-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 $R$.
\end{lemma}
\begin{proof}
Let $I \subset R$ be an ideal. By Lemma \ref{lemma-flat} it suffices
to show that $I \otimes_R M \to M$ is injective. By Remark
\ref{remark-Tor-ring-mod-ideal} we see that this kernel is
equal to $\text{Tor}_1^R(M, R/I)$. By
Lemma \ref{lemma-prepare-local-criterion-flatness}
we see that $J \otimes_R M \to M$ is injective for all ideals
of finite colength.
\medskip\noindent
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
$\Ker((I \cap \mathfrak m^n)\otimes_R M \to M)
\to \Ker(I \otimes_R M \to M)$
is surjective. We conclude in particular that
$K = \Ker(I \otimes_R M \to M)$ is contained
in the image of $(I \cap \mathfrak m^n) \otimes_R M$
in $I \otimes_R M$. By Artin-Rees, Lemma \ref{lemma-Artin-Rees}
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 \ref{lemma-completion-faithfully-flat}.
Hence $K = 0$, as desired.
\end{proof}
\noindent
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 \ref{lemma-prepare-local-criterion-flatness}).
\begin{lemma}
\label{lemma-what-does-it-mean}
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
\begin{enumerate}
\item $M/I^nM$ is flat over $R/I^n$ for all $n \geq 1$, and
\item for any module $N$ which is annihilated by $I^m$ for some $m \geq 0$
we have $\text{Tor}_1^R(N, M) = 0$.
\end{enumerate}
In particular, if $I$ is nilpotent, then $M$ is flat over $R$.
\end{lemma}
\begin{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 $I$.
\medskip\noindent
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 $I/I^2 \otimes_{R/I} M/IM = IM/I^2M$.
\medskip\noindent
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 \ref{lemma-flat}.
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.
\medskip\noindent
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 $n \geq 1$.
\end{proof}
\begin{lemma}
\label{lemma-what-does-it-mean-again}
Let $R$ be a ring. Let $I \subset R$ be an ideal.
Let $M$ be an $R$-module.
\begin{enumerate}
\item 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 $R/I^2$.
\item 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 $R/I^{k + 1}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The first statement is a consequence of
Lemma \ref{lemma-what-does-it-mean} applied with $R$ replaced by $R/I^2$
and $M$ replaced by $M/I^2M$ using that
$$
\text{Tor}_1^{R/I^2}(M/I^2M, R/I) =
\Ker(M \otimes_R I/I^2 \to IM/I^2M),
$$
see Remark \ref{remark-Tor-ring-mod-ideal}.
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
$$
\text{Tor}_1^{R/I^{n + 1}}(M/I^{n + 1}M, R/I) =
\Ker(M \otimes_R I^n/I^{n + 1} \to I^nM/I^{n + 1}M)
$$
for every $n$.
\end{proof}
\begin{lemma}[Variant of the local criterion]
\label{lemma-variant-local-criterion-flatness}
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 $R$.
\end{lemma}
\begin{proof}
First proof: By
Lemma \ref{lemma-what-does-it-mean}
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 \ref{lemma-local-criterion-flatness}.
\medskip\noindent
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 \ref{lemma-local-criterion-flatness}.
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 \ref{lemma-flat-eq} 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 \ref{remark-Tor-ring-mod-ideal}) and we win.
\end{proof}
\noindent
In particular, in the situation of
Lemma \ref{lemma-variant-local-criterion-flatness}, 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 $M$.
\begin{lemma}
\label{lemma-flat-module-powers}
Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal.
Let $M$ be an $S$-module. Assume
\begin{enumerate}
\item $R$ is a Noetherian ring,
\item $S$ is a Noetherian ring,
\item $M$ is a finite $S$-module, and
\item for each $n \geq 1$ the module $M/I^n M$ is flat over
$R/I^n$.
\end{enumerate}
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 $R$.
\end{lemma}
\begin{proof}
We are going to use
Lemma \ref{lemma-variant-local-criterion-flatness}.
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 \ref{remark-Tor-ring-mod-ideal}
this Tor group is equal to $K = \Ker(I \otimes_R M \to M)$.
We know for each $n \geq 1$ that the kernel
$\Ker(I/I^n \otimes_{R/I^n} M/I^nM \to M/I^nM)$ 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 \ref{lemma-intersection-powers-ideal-module}
we conclude that $K_{\mathfrak q} = 0$, as desired.
\end{proof}
\begin{lemma}
\label{lemma-surjective-on-tor-one}
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.
\end{lemma}
\begin{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
$K' = \Ker(F_1\otimes_R R' \to F_0 \otimes_R R')$ and
similarly $K'' = \Ker(F_1\otimes_R R'' \to F_0 \otimes_R R'')$.
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.
\end{proof}
\begin{lemma}
\label{lemma-surjective-on-tor-one-trivial}
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.
\end{lemma}
\begin{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.
\end{proof}
\begin{lemma}
\label{lemma-another-variant-local-criterion-flatness}
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
\begin{enumerate}
\item $S'$ is a localization of the tensor product
$S \otimes_R R'$,
\item $M/IM$ is flat over $R/I$,
\item $\text{Tor}_1^R(M, R/I) \to \text{Tor}_1^{R'}(M', R'/I')$
is zero.
\end{enumerate}
Then $M'$ is flat over $R'$.
\end{lemma}
\begin{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 \ref{lemma-flat-base-change} 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 \ref{lemma-variant-local-criterion-flatness}
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.
\medskip\noindent
By Lemma \ref{lemma-surjective-on-tor-one-trivial} 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 \ref{lemma-surjective-on-tor-one}
by looking at the ring maps $R \to R/I \to R'/I'$ and the module $M$.
\end{proof}
\noindent
Please compare the lemma below to
Lemma \ref{lemma-criterion-flatness-fibre-nilpotent}
(the case of a nilpotent ideal) and
Lemma \ref{lemma-criterion-flatness-fibre}
(the case of finitely presented algebras).
\begin{lemma}[Crit\`ere de platitude par fibres; Noetherian case]
\label{lemma-criterion-flatness-fibre-Noetherian}
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
\begin{enumerate}
\item The module $M$ is finite over $S'$.
\item The module $M$ is not zero.
\item The module $M/\mathfrak m M$
is a flat $S/\mathfrak m S$-module.
\item The module $M$ is a flat $R$-module.
\end{enumerate}
Then $S$ is flat over $R$ and $M$ is a flat $S$-module.
\end{lemma}
\begin{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 \ref{remark-Tor-ring-mod-ideal}. By
Lemma \ref{lemma-variant-local-criterion-flatness} we conclude
that $M$ is flat over $S$. Note that since $M/\mathfrak m_{S'}M$
is not zero by Nakayama's Lemma \ref{lemma-NAK}
we see that actually $M$ is faithfully flat over $S$ by
Lemma \ref{lemma-ff} (since it forces $M/\mathfrak m_SM \not = 0$).
\medskip\noindent
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 \ref{lemma-local-criterion-flatness}.
\end{proof}
```

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