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10.127. More flatness criteria

The following lemma is often used in algebraic geometry to show that a finite morphism from a normal surface to a smooth surface is flat. It is a partial converse to Lemma 10.111.9 because an injective finite local ring map certainly satisfies condition (3).

Lemma 10.127.1. Let $R \to S$ be a local homomorphism of Noetherian local rings. Assume

  1. $R$ is regular,
  2. $S$ Cohen-Macaulay,
  3. $\dim(S) = \dim(R) + \dim(S/\mathfrak m_R S)$.

Then $R \to S$ is flat.

Proof. By induction on $\dim(R)$. The case $\dim(R) = 0$ is trivial, because then $R$ is a field. Assume $\dim(R) > 0$. By (3) this implies that $\dim(S) > 0$. Let $\mathfrak q_1, \ldots, \mathfrak q_r$ be the minimal primes of $S$. Note that $\mathfrak q_i \not \supset \mathfrak m_R S$ since $$ \dim(S/\mathfrak q_i) = \dim(S) > \dim(S/\mathfrak m_R S) $$ the first equality by Lemma 10.103.3 and the inequality by (3). Thus $\mathfrak p_i = R \cap \mathfrak q_i$ is not equal to $\mathfrak m_R$. Pick $x \in \mathfrak m$, $x \not \in \mathfrak m^2$, and $x \not \in \mathfrak p_i$, see Lemma 10.14.2. Hence we see that $x$ is not contained in any of the minimal primes of $S$. Hence $x$ is a nonzerodivisor on $S$ by (2), see Lemma 10.103.2 and $S/xS$ is Cohen-Macaulay with $\dim(S/xS) = \dim(S) - 1$. By (1) and Lemma 10.105.3 the ring $R/xR$ is regular with $\dim(R/xR) = \dim(R) - 1$. By induction we see that $R/xR \to S/xS$ is flat. Hence we conclude by Lemma 10.98.10 and the remark following it. $\square$

Lemma 10.127.2. Let $R \to S$ be a homomorphism of Noetherian local rings. Assume that $R$ is a regular local ring and that a regular system of parameters maps to a regular sequence in $S$. Then $R \to S$ is flat.

Proof. Suppose that $x_1, \ldots, x_d$ are a system of parameters of $R$ which map to a regular sequence in $S$. Note that $S/(x_1, \ldots, x_d)S$ is flat over $R/(x_1, \ldots, x_d)$ as the latter is a field. Then $x_d$ is a nonzerodivisor in $S/(x_1, \ldots, x_{d - 1})S$ hence $S/(x_1, \ldots, x_{d - 1})S$ is flat over $R/(x_1, \ldots, x_{d - 1})$ by the local criterion of flatness (see Lemma 10.98.10 and remarks following). Then $x_{d - 1}$ is a nonzerodivisor in $S/(x_1, \ldots, x_{d - 2})S$ hence $S/(x_1, \ldots, x_{d - 2})S$ is flat over $R/(x_1, \ldots, x_{d - 2})$ by the local criterion of flatness (see Lemma 10.98.10 and remarks following). Continue till one reaches the conclusion that $S$ is flat over $R$. $\square$

The following lemma is the key to proving that results for finitely presented modules over finitely presented rings over a base ring follow from the corresponding results for finite modules in the Noetherian case.

Lemma 10.127.3. Let $R \to S$, $M$, $\Lambda$, $R_\lambda \to S_\lambda$, $M_\lambda$ be as in Lemma 10.126.13. Assume that $M$ is flat over $R$. Then for some $\lambda \in \Lambda$ the module $M_\lambda$ is flat over $R_\lambda$.

Proof. Pick some $\lambda \in \Lambda$ and consider $$ \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/\mathfrak m_\lambda) = \mathop{\rm Ker}(\mathfrak m_\lambda \otimes_{R_\lambda} M_\lambda \to M_\lambda). $$ See Remark 10.74.9. The right hand side shows that this is a finitely generated $S_\lambda$-module (because $S_\lambda$ is Noetherian and the modules in question are finite). Let $\xi_1, \ldots, \xi_n$ be generators. Because $M$ is flat over $R$ we have that $0 = \mathop{\rm Ker}(\mathfrak m_\lambda R \otimes_R M \to M)$. Since $\otimes$ commutes with colimits we see there exists a $\lambda' \geq \lambda$ such that each $\xi_i$ maps to zero in $\mathfrak m_{\lambda}R_{\lambda'} \otimes_{R_{\lambda'}} M_{\lambda'}$. Hence we see that $$ \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/\mathfrak m_\lambda) \longrightarrow \text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/\mathfrak m_{\lambda}R_{\lambda'}) $$ is zero. Note that $M_\lambda \otimes_{R_\lambda} R_\lambda/\mathfrak m_\lambda$ is flat over $R_\lambda/\mathfrak m_\lambda$ because this last ring is a field. Hence we may apply Lemma 10.98.14 to get that $M_{\lambda'}$ is flat over $R_{\lambda'}$. $\square$

Using the lemma above we can start to reprove the results of Section 10.98 in the non-Noetherian case.

Lemma 10.127.4. Suppose that $R \to S$ is a local homomorphism of local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Let $u : M \to N$ be a map of $S$-modules. Assume

  1. $S$ is essentially of finite presentation over $R$,
  2. $M$, $N$ are finitely presented over $S$,
  3. $N$ is flat over $R$, and
  4. $\overline{u} : M/\mathfrak mM \to N/\mathfrak mN$ is injective.

Then $u$ is injective, and $N/u(M)$ is flat over $R$.

Proof. By Lemma 10.126.13 and its proof we can find a system $R_\lambda \to S_\lambda$ of local ring maps together with maps of $S_\lambda$-modules $u_\lambda : M_\lambda \to N_\lambda$ satisfying the conclusions (1) – (6) for both $N$ and $M$ of that lemma and such that the colimit of the maps $u_\lambda$ is $u$. By Lemma 10.127.3 we may assume that $N_\lambda$ is flat over $R_\lambda$ for all sufficiently large $\lambda$. Denote $\mathfrak m_\lambda \subset R_\lambda$ the maximal ideal and $\kappa_\lambda = R_\lambda / \mathfrak m_\lambda$, resp. $\kappa = R/\mathfrak m$ the residue fields.

Consider the map $$ \Psi_\lambda : M_\lambda/\mathfrak m_\lambda M_\lambda \otimes_{\kappa_\lambda} \kappa \longrightarrow M/\mathfrak m M. $$ Since $S_\lambda/\mathfrak m_\lambda S_\lambda$ is essentially of finite type over the field $\kappa_\lambda$ we see that the tensor product $S_\lambda/\mathfrak m_\lambda S_\lambda \otimes_{\kappa_\lambda} \kappa$ is essentially of finite type over $\kappa$. Hence it is a Noetherian ring and we conclude the kernel of $\Psi_\lambda$ is finitely generated. Since $M/\mathfrak m M$ is the colimit of the system $M_\lambda/\mathfrak m_\lambda M_\lambda$ and $\kappa$ is the colimit of the fields $\kappa_\lambda$ there exists a $\lambda' > \lambda$ such that the kernel of $\Psi_\lambda$ is generated by the kernel of $$ \Psi_{\lambda, \lambda'} : M_\lambda/\mathfrak m_\lambda M_\lambda \otimes_{\kappa_\lambda} \kappa_{\lambda'} \longrightarrow M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}. $$ By construction there exists a multiplicative subset $W \subset S_\lambda \otimes_{R_\lambda} R_{\lambda'}$ such that $S_{\lambda'} = W^{-1}(S_\lambda \otimes_{R_\lambda} R_{\lambda'})$ and $$ W^{-1}(M_\lambda/\mathfrak m_\lambda M_\lambda \otimes_{\kappa_\lambda} \kappa_{\lambda'}) = M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}. $$ Now suppose that $x$ is an element of the kernel of $$ \Psi_{\lambda'} : M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'} \otimes_{\kappa_{\lambda'}} \kappa \longrightarrow M/\mathfrak m M. $$ Then for some $w \in W$ we have $wx \in M_\lambda/\mathfrak m_\lambda M_\lambda \otimes \kappa$. Hence $wx \in \mathop{\rm Ker}(\Psi_\lambda)$. Hence $wx$ is a linear combination of elements in the kernel of $\Psi_{\lambda, \lambda'}$. Hence $wx = 0$ in $M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'} \otimes_{\kappa_{\lambda'}} \kappa$, hence $x = 0$ because $w$ is invertible in $S_{\lambda'}$. We conclude that the kernel of $\Psi_{\lambda'}$ is zero for all sufficiently large $\lambda'$!

By the result of the preceding paragraph we may assume that the kernel of $\Psi_\lambda$ is zero for all $\lambda$ sufficiently large, which implies that the map $M_\lambda/\mathfrak m_\lambda M_\lambda \to M/\mathfrak m M$ is injective. Combined with $\overline{u}$ being injective this formally implies that also $\overline{u_\lambda} : M_\lambda/\mathfrak m_\lambda M_\lambda \to N_\lambda/\mathfrak m_\lambda N_\lambda$ is injective. By Lemma 10.98.1 we conclude that (for all sufficiently large $\lambda$) the map $u_\lambda$ is injective and that $N_\lambda/u_\lambda(M_\lambda)$ is flat over $R_\lambda$. The lemma follows. $\square$

Lemma 10.127.5. Suppose that $R \to S$ is a local ring homomorphism of local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Suppose

  1. $S$ is essentially of finite presentation over $R$,
  2. $S$ is flat over $R$, and
  3. $f \in S$ 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 10.127.4. $\square$

Lemma 10.127.6. Suppose that $R \to S$ is a local ring homomorphism of local rings. Denote $\mathfrak m$ the maximal ideal of $R$. Suppose

  1. $R \to S$ is essentially of finite presentation,
  2. $R \to S$ is flat, and
  3. $f_1, \ldots, f_c$ 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 10.127.5. $\square$

Here is the version of the local criterion of flatness for the case of local ring maps which are locally of finite presentation.

Lemma 10.127.7. Let $R \to S$ be a local homomorphism of local rings. Let $I \not = R$ be an ideal in $R$. Let $M$ be an $S$-module. Assume

  1. $S$ is essentially of finite presentation over $R$,
  2. $M$ is of finite presentation over $S$,
  3. $\text{Tor}_1^R(M, R/I) = 0$, and
  4. $M/IM$ is flat over $R/I$.

Then $M$ is flat over $R$.

Proof. Let $\Lambda$, $R_\lambda \to S_\lambda$, $M_\lambda$ be as in Lemma 10.126.13. Denote $I_\lambda \subset R_\lambda$ the inverse image of $I$. In this case the system $R/I \to S/IS$, $M/IM$, $R_\lambda \to S_\lambda/I_\lambda S_\lambda$, and $M_\lambda/I_\lambda M_\lambda$ satisfies the conclusions of Lemma 10.126.13 as well. Hence by Lemma 10.127.3 we may assume (after shrinking the index set $\Lambda$) that $M_\lambda/I_\lambda M_\lambda$ is flat for all $\lambda$. Pick some $\lambda$ and consider $$ \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/I_\lambda) = \mathop{\rm Ker}(I_\lambda \otimes_{R_\lambda} M_\lambda \to M_\lambda). $$ See Remark 10.74.9. The right hand side shows that this is a finitely generated $S_\lambda$-module (because $S_\lambda$ is Noetherian and the modules in question are finite). Let $\xi_1, \ldots, \xi_n$ be generators. Because $\text{Tor}^1_R(M, R/I) = 0$ and since $\otimes$ commutes with colimits we see there exists a $\lambda' \geq \lambda$ such that each $\xi_i$ maps to zero in $\text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_{\lambda'})$. The composition of the maps $$ \xymatrix{ R_{\lambda'} \otimes_{R_\lambda} \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/I_\lambda) \ar[d]^{\text{surjective by Lemma 10.98.12}} \\ \text{Tor}_1^{R_\lambda}(M_\lambda, R_{\lambda'}/I_\lambda R_{\lambda'}) \ar[d]^{\text{surjective up to localization by Lemma 10.98.13}} \\ \text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_\lambda R_{\lambda'}) \ar[d]^{\text{surjective by Lemma 10.98.12}} \\ \text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_{\lambda'}). } $$ is surjective up to a localization by the reasons indicated. The localization is necessary since $M_{\lambda'}$ is not equal to $M_\lambda \otimes_{R_\lambda} R_{\lambda'}$. Namely, it is equal to $M_\lambda \otimes_{S_\lambda} S_{\lambda'}$ and $S_{\lambda'}$ is the localization of $S_{\lambda} \otimes_{R_\lambda} R_{\lambda'}$ whence the statement up to a localization (or tensoring with $S_{\lambda'}$). Note that Lemma 10.98.12 applies to the first and third arrows because $M_\lambda/I_\lambda M_\lambda$ is flat over $R_\lambda/I_\lambda$ and because $M_{\lambda'}/I_\lambda M_{\lambda'}$ is flat over $R_{\lambda'}/I_\lambda R_{\lambda'}$ as it is a base change of the flat module $M_\lambda/I_\lambda M_\lambda$. The composition maps the generators $\xi_i$ to zero as we explained above. We finally conclude that $\text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_{\lambda'})$ is zero. This implies that $M_{\lambda'}$ is flat over $R_{\lambda'}$ by Lemma 10.98.10. $\square$

Please compare the lemma below to Lemma 10.98.15 (the case of Noetherian local rings) and Lemma 10.100.8 (the case of a nilpotent ideal in the base).

Lemma 10.127.8 (Critère de platitude par fibres). Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$ be the maximal ideal. Assume

  1. The ring maps $R \to S$ and $R \to S'$ are essentially of finite presentation.
  2. The module $M$ is of finite presentation over $S'$.
  3. The module $M$ is not zero.
  4. The module $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-module.
  5. The module $M$ is a flat $R$-module.

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

Proof. As in the proof of Lemma 10.126.11 we may first write $R = \mathop{\rm colim}\nolimits R_\lambda$ as a directed colimit of local $\mathbf{Z}$-algebras which are essentially of finite type. Denote $\mathfrak p_\lambda$ the maximal ideal of $R_\lambda$. Next, we may assume that for some $\lambda_1 \in \Lambda$ there exist $f_{j, \lambda_1} \in R_{\lambda_1}[x_1, \ldots, x_n]$ such that $$ S = \mathop{\rm colim}\nolimits_{\lambda \geq \lambda_1} S_\lambda, \text{ with } S_\lambda = (R_\lambda[x_1, \ldots, x_n]/ (f_{1, \lambda}, \ldots, f_{u, \lambda}))_{\mathfrak q_\lambda} $$ For some $\lambda_2 \in \Lambda$, $\lambda_2 \geq \lambda_1$ there exist $g_{j, \lambda_2} \in R_{\lambda_2}[x_1, \ldots, x_n, y_1, \ldots, y_m]$ with images $\overline{g}_{j, \lambda_2} \in S_{\lambda_2}[y_1, \ldots, y_m]$ such that $$ S' = \mathop{\rm colim}\nolimits_{\lambda \geq \lambda_2} S'_\lambda, \text{ with } S'_\lambda = (S_\lambda[y_1, \ldots, y_m]/ (\overline{g}_{1, \lambda}, \ldots, \overline{g}_{v, \lambda}))_{\overline{\mathfrak q}'_\lambda} $$ Note that this also implies that $$ S'_\lambda = (R_\lambda[x_1, \ldots, x_n, y_1, \ldots, y_m]/ (g_{1, \lambda}, \ldots, g_{v, \lambda}))_{\mathfrak q'_\lambda} $$ Choose a presentation $$ (S')^{\oplus s} \to (S')^{\oplus t} \to M \to 0 $$ of $M$ over $S'$. Let $A \in \text{Mat}(t \times s, S')$ be the matrix of the presentation. For some $\lambda_3 \in \Lambda$, $\lambda_3 \geq \lambda_2$ we can find a matrix $A_{\lambda_3} \in \text{Mat}(t \times s, S_{\lambda_3})$ which maps to $A$. For all $\lambda \geq \lambda_3$ we let $M_\lambda = \mathop{\rm Coker}((S'_\lambda)^{\oplus s} \xrightarrow{A_\lambda} (S'_\lambda)^{\oplus t})$.

With these choices, we have for each $\lambda_3 \leq \lambda \leq \mu$ that $S_\lambda \otimes_{R_{\lambda}} R_\mu \to S_\mu$ is a localization, $S'_\lambda \otimes_{S_{\lambda}} S_\mu \to S'_\mu$ is a localization, and the map $M_\lambda \otimes_{S'_\lambda} S'_\mu \to M_\mu$ is an isomorphism. This also implies that $S'_\lambda \otimes_{R_{\lambda}} R_\mu \to S'_\mu$ is a localization. Thus, since $M$ is flat over $R$ we see by Lemma 10.127.3 that for all $\lambda$ big enough the module $M_\lambda$ is flat over $R_\lambda$. Moreover, note that $ \mathfrak m = \mathop{\rm colim}\nolimits \mathfrak p_\lambda $, $ S/\mathfrak mS = \mathop{\rm colim}\nolimits S_\lambda/\mathfrak p_\lambda S_\lambda $, $ S'/\mathfrak mS' = \mathop{\rm colim}\nolimits S'_\lambda/\mathfrak p_\lambda S'_\lambda $, and $ M/\mathfrak mM = \mathop{\rm colim}\nolimits M_\lambda/\mathfrak p_\lambda M_\lambda $. Also, for each $\lambda_3 \leq \lambda \leq \mu$ we see (from the properties listed above) that $$ S'_\lambda/\mathfrak p_\lambda S'_\lambda \otimes_{S_{\lambda}/\mathfrak p_\lambda S_\lambda} S_\mu/\mathfrak p_\mu S_\mu \longrightarrow S'_\mu/\mathfrak p_\mu S'_\mu $$ is a localization, and the map $$ M_\lambda / \mathfrak p_\lambda M_\lambda \otimes_{S'_\lambda/\mathfrak p_\lambda S'_\lambda} S'_\mu /\mathfrak p_\mu S'_\mu \longrightarrow M_\mu/\mathfrak p_\mu M_\mu $$ is an isomorphism. Hence the system $(S_\lambda/\mathfrak p_\lambda S_\lambda \to S'_\lambda/\mathfrak p_\lambda S'_\lambda, M_\lambda/\mathfrak p_\lambda M_\lambda)$ is a system as in Lemma 10.126.13 as well. We may apply Lemma 10.127.3 again because $M/\mathfrak m M$ is assumed flat over $S/\mathfrak mS$ and we see that $M_\lambda/\mathfrak p_\lambda M_\lambda$ is flat over $S_\lambda/\mathfrak p_\lambda S_\lambda$ for all $\lambda$ big enough. Thus for $\lambda$ big enough the data $R_\lambda \to S_\lambda \to S'_\lambda, M_\lambda$ satisfies the hypotheses of Lemma 10.98.15. Pick such a $\lambda$. Then $S = S_\lambda \otimes_{R_\lambda} R$ is flat over $R$, and $M = M_\lambda \otimes_{S_\lambda} S$ is flat over $S$ (since the base change of a flat module is flat). $\square$

The following is an easy consequence of the ''critère de platitude par fibres'' Lemma 10.127.8. For more results of this kind see More on Flatness, Section 37.1.

Lemma 10.127.9. Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$ be the maximal ideal. Assume

  1. $R \to S'$ is essentially of finite presentation,
  2. $R \to S$ is essentially of finite type,
  3. $M$ is of finite presentation over $S'$,
  4. $M$ is not zero,
  5. $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-module, and
  6. $M$ is a flat $R$-module.

Then $S$ is essentially of finite presentation and flat over $R$ and $M$ is a flat $S$-module.

Proof. As $S$ is essentially of finite presentation over $R$ we can write $S = C_{\overline{\mathfrak q}}$ for some finite type $R$-algebra $C$. Write $C = R[x_1, \ldots, x_n]/I$. Denote $\mathfrak q \subset R[x_1, \ldots, x_n]$ be the prime ideal corresponding to $\overline{\mathfrak q}$. Then we see that $S = B/J$ where $B = R[x_1, \ldots, x_n]_{\mathfrak q}$ is essentially of finite presentation over $R$ and $J = IB$. We can find $f_1, \ldots, f_k \in J$ such that the images $\overline{f}_i \in B/\mathfrak mB$ generate the image $\overline{J}$ of $J$ in the Noetherian ring $B/\mathfrak mB$. Hence there exist finitely generated ideals $J' \subset J$ such that $B/J' \to B/J$ induces an isomorphism $$ (B/J') \otimes_R R/\mathfrak m \longrightarrow B/J \otimes_R R/\mathfrak m = S/\mathfrak mS. $$ For any $J'$ as above we see that Lemma 10.127.8 applies to the ring maps $$ R \longrightarrow B/J' \longrightarrow S' $$ and the module $M$. Hence we conclude that $B/J'$ is flat over $R$ for any choice $J'$ as above. Now, if $J' \subset J' \subset J$ are two finitely generated ideals as above, then we conclude that $B/J' \to B/J''$ is a surjective map between flat $R$-algebras which are essentially of finite presentation which is an isomorphism modulo $\mathfrak m$. Hence Lemma 10.127.4 implies that $B/J' = B/J''$, i.e., $J' = J''$. Clearly this means that $J$ is finitely generated, i.e., $S$ is essentially of finite presentation over $R$. Thus we may apply Lemma 10.127.8 to $R \to S \to S'$ and we win. $\square$

Lemma 10.127.10 (Critère de platitude par fibres: locally nilpotent case). Let $$ \xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] } $$ be a commutative diagram in the category of rings. Let $I \subset R$ be a locally nilpotent ideal and $M$ an $S'$-module. Assume

  1. $R \to S$ is of finite type,
  2. $R \to S'$ is of finite presentation,
  3. $M$ is a finitely presented $S'$-module,
  4. $M/IM$ is flat as a $S/IS$-module, and
  5. $M$ is flat as an $R$-module.

Then $M$ is a flat $S$-module and $S_\mathfrak q$ is flat and essentially of finite presentation over $R$ for every $\mathfrak q \subset S$ such that $M \otimes_S \kappa(\mathfrak q)$ is nonzero.

Proof. If $M \otimes_S \kappa(\mathfrak q)$ is nonzero, then $S' \otimes_S \kappa(\mathfrak q)$ is nonzero and hence there exists a prime $\mathfrak q' \subset S'$ lying over $\mathfrak q$ (Lemma 10.16.9). Let $\mathfrak p \subset R$ be the image of $\mathfrak q$ in $\mathop{\rm Spec}(R)$. Then $I \subset \mathfrak p$ as $I$ is locally nilpotent hence $M/\mathfrak p M$ is flat over $S/\mathfrak pS$. Hence we may apply Lemma 10.127.9 to $R_\mathfrak p \to S_\mathfrak q \to S'_{\mathfrak q'}$ and $M_{\mathfrak q'}$. We conclude that $M_{\mathfrak q'}$ is flat over $S$ and $S_\mathfrak q$ is flat and essentially of finite presentation over $R$. Since $\mathfrak q'$ was an arbitrary prime of $S'$ we also see that $M$ is flat over $S$ (Lemma 10.38.19). $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 31283–31863 (see updates for more information).

    \section{More flatness criteria}
    \label{section-more-flatness-criteria}
    
    \noindent
    The following lemma is often used in algebraic geometry to show that a finite
    morphism from a normal surface to a smooth surface is flat. It is a partial
    converse to
    Lemma \ref{lemma-finite-flat-over-regular-CM}
    because an injective finite local ring map certainly satisfies condition (3).
    
    \begin{lemma}
    \label{lemma-CM-over-regular-flat}
    Let $R \to S$ be a local homomorphism of Noetherian local
    rings. Assume
    \begin{enumerate}
    \item $R$ is regular,
    \item $S$ Cohen-Macaulay,
    \item $\dim(S) = \dim(R) + \dim(S/\mathfrak m_R S)$.
    \end{enumerate}
    Then $R \to S$ is flat.
    \end{lemma}
    
    \begin{proof}
    By induction on $\dim(R)$. The case $\dim(R) = 0$ is trivial, because
    then $R$ is a field. Assume $\dim(R) > 0$. By (3) this implies that
    $\dim(S) > 0$. Let $\mathfrak q_1, \ldots, \mathfrak q_r$ be the minimal
    primes of $S$. Note that $\mathfrak q_i \not \supset \mathfrak m_R S$ since
    $$
    \dim(S/\mathfrak q_i) = \dim(S) > \dim(S/\mathfrak m_R S)
    $$
    the first equality by Lemma \ref{lemma-maximal-chain-CM} and the
    inequality by (3). Thus
    $\mathfrak p_i = R \cap \mathfrak q_i$ is not equal to $\mathfrak m_R$.
    Pick $x \in \mathfrak m$, $x \not \in \mathfrak m^2$, and
    $x \not \in \mathfrak p_i$, see
    Lemma \ref{lemma-silly}.
    Hence we see that $x$ is not contained in any of the minimal
    primes of $S$. Hence $x$ is a nonzerodivisor on $S$ by (2), see
    Lemma \ref{lemma-reformulate-CM} and
    $S/xS$ is Cohen-Macaulay with $\dim(S/xS) = \dim(S) - 1$.
    By (1) and Lemma \ref{lemma-regular-ring-CM} the ring $R/xR$ is regular
    with $\dim(R/xR) = \dim(R) - 1$.
    By induction we see that $R/xR \to S/xS$ is flat. Hence we
    conclude by Lemma \ref{lemma-variant-local-criterion-flatness}
    and the remark following it.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-flat-over-regular}
    Let $R \to S$ be a homomorphism of Noetherian local rings.
    Assume that $R$ is a regular local ring and that a regular system
    of parameters maps to a regular sequence in $S$. Then $R \to S$
    is flat.
    \end{lemma}
    
    \begin{proof}
    Suppose that $x_1, \ldots, x_d$ are a system of parameters of $R$
    which map to a regular sequence in $S$. Note that
    $S/(x_1, \ldots, x_d)S$ is flat over $R/(x_1, \ldots, x_d)$
    as the latter is a field. Then $x_d$ is a nonzerodivisor in
    $S/(x_1, \ldots, x_{d - 1})S$ hence $S/(x_1, \ldots, x_{d - 1})S$
    is flat over $R/(x_1, \ldots, x_{d - 1})$ by the local criterion
    of flatness (see Lemma \ref{lemma-variant-local-criterion-flatness}
    and remarks following). Then $x_{d - 1}$ is a nonzerodivisor in
    $S/(x_1, \ldots, x_{d - 2})S$ hence $S/(x_1, \ldots, x_{d - 2})S$
    is flat over $R/(x_1, \ldots, x_{d - 2})$ by the local criterion
    of flatness (see Lemma \ref{lemma-variant-local-criterion-flatness}
    and remarks following). Continue till one reaches the conclusion
    that $S$ is flat over $R$.
    \end{proof}
    
    \noindent
    The following lemma is the key to proving that results for finitely presented
    modules over finitely presented rings over a base ring follow from the
    corresponding results for finite modules in the Noetherian case.
    
    \begin{lemma}
    \label{lemma-colimit-eventually-flat}
    Let $R \to S$, $M$, $\Lambda$, $R_\lambda \to S_\lambda$, $M_\lambda$
    be as in Lemma \ref{lemma-limit-module-essentially-finite-presentation}.
    Assume that $M$ is flat over $R$.
    Then for some $\lambda \in \Lambda$ the module
    $M_\lambda$ is flat over $R_\lambda$.
    \end{lemma}
    
    \begin{proof}
    Pick some $\lambda \in \Lambda$ and consider
    $$
    \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/\mathfrak m_\lambda)
    =
    \Ker(\mathfrak m_\lambda \otimes_{R_\lambda} M_\lambda
    \to M_\lambda).
    $$
    See Remark \ref{remark-Tor-ring-mod-ideal}. The right hand side
    shows that this is a finitely generated $S_\lambda$-module (because
    $S_\lambda$ is Noetherian and the modules in question are finite).
    Let $\xi_1, \ldots, \xi_n$ be generators.
    Because $M$ is flat over $R$ we
    have that $0 = \Ker(\mathfrak m_\lambda R \otimes_R M \to M)$.
    Since $\otimes$ commutes with colimits we see there exists
    a $\lambda' \geq \lambda$ such that each $\xi_i$ maps to
    zero in
    $\mathfrak m_{\lambda}R_{\lambda'} \otimes_{R_{\lambda'}} M_{\lambda'}$.
    Hence we see that
    $$
    \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/\mathfrak m_\lambda)
    \longrightarrow
    \text{Tor}_1^{R_{\lambda'}}(M_{\lambda'},
    R_{\lambda'}/\mathfrak m_{\lambda}R_{\lambda'})
    $$
    is zero. Note that
    $M_\lambda \otimes_{R_\lambda} R_\lambda/\mathfrak m_\lambda$
    is flat over $R_\lambda/\mathfrak m_\lambda$ because this last
    ring is a field. Hence we may apply Lemma
    \ref{lemma-another-variant-local-criterion-flatness}
    to get that $M_{\lambda'}$ is flat over $R_{\lambda'}$.
    \end{proof}
    
    \noindent
    Using the lemma above we can start to reprove the results of
    Section \ref{section-criteria-flatness}
    in the non-Noetherian case.
    
    \begin{lemma}
    \label{lemma-mod-injective-general}
    Suppose that $R \to S$ is a local homomorphism of local rings.
    Denote $\mathfrak m$ the maximal ideal of $R$.
    Let $u : M \to N$ be a map of $S$-modules.
    Assume
    \begin{enumerate}
    \item $S$ is essentially of finite presentation over $R$,
    \item $M$, $N$ are finitely presented over $S$,
    \item $N$ is flat over $R$, and
    \item $\overline{u} : M/\mathfrak mM \to N/\mathfrak mN$ is injective.
    \end{enumerate}
    Then $u$ is injective, and $N/u(M)$ is flat over $R$.
    \end{lemma}
    
    \begin{proof}
    By
    Lemma \ref{lemma-limit-module-essentially-finite-presentation}
    and its proof we can find a system $R_\lambda \to S_\lambda$ of
    local ring maps together with maps of $S_\lambda$-modules
    $u_\lambda : M_\lambda \to N_\lambda$ satisfying the conclusions
    (1) -- (6) for both $N$ and $M$ of that lemma and such that the
    colimit of the maps $u_\lambda$ is $u$. By
    Lemma \ref{lemma-colimit-eventually-flat}
    we may assume that $N_\lambda$ is flat over $R_\lambda$ for all
    sufficiently large $\lambda$. Denote $\mathfrak m_\lambda \subset R_\lambda$
    the maximal ideal and $\kappa_\lambda = R_\lambda / \mathfrak m_\lambda$,
    resp.\ $\kappa = R/\mathfrak m$ the residue fields.
    
    \medskip\noindent
    Consider the map
    $$
    \Psi_\lambda :
    M_\lambda/\mathfrak m_\lambda M_\lambda \otimes_{\kappa_\lambda} \kappa
    \longrightarrow
    M/\mathfrak m M.
    $$
    Since $S_\lambda/\mathfrak m_\lambda S_\lambda$ is essentially of finite type
    over the field $\kappa_\lambda$ we see that the tensor product
    $S_\lambda/\mathfrak m_\lambda S_\lambda \otimes_{\kappa_\lambda} \kappa$
    is essentially of finite type over $\kappa$. Hence it is a Noetherian
    ring and we conclude the kernel of $\Psi_\lambda$ is finitely generated.
    Since $M/\mathfrak m M$ is the colimit of the system
    $M_\lambda/\mathfrak m_\lambda M_\lambda$ and $\kappa$ is the colimit of
    the fields $\kappa_\lambda$ there exists a $\lambda' > \lambda$ such that
    the kernel of $\Psi_\lambda$ is generated by the kernel of
    $$
    \Psi_{\lambda, \lambda'} :
    M_\lambda/\mathfrak m_\lambda M_\lambda
    \otimes_{\kappa_\lambda}
    \kappa_{\lambda'}
    \longrightarrow
    M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}.
    $$
    By construction there exists a multiplicative subset
    $W \subset S_\lambda \otimes_{R_\lambda} R_{\lambda'}$ such that
    $S_{\lambda'} = W^{-1}(S_\lambda \otimes_{R_\lambda} R_{\lambda'})$ and
    $$
    W^{-1}(M_\lambda/\mathfrak m_\lambda M_\lambda
    \otimes_{\kappa_\lambda}
    \kappa_{\lambda'})
    =
    M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}.
    $$
    Now suppose that $x$ is an element of the kernel of
    $$
    \Psi_{\lambda'} :
    M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}
    \otimes_{\kappa_{\lambda'}} \kappa
    \longrightarrow
    M/\mathfrak m M.
    $$
    Then for some $w \in W$ we have
    $wx \in M_\lambda/\mathfrak m_\lambda M_\lambda \otimes \kappa$.
    Hence $wx \in \Ker(\Psi_\lambda)$. Hence $wx$ is a linear
    combination of elements in the kernel of $\Psi_{\lambda, \lambda'}$.
    Hence $wx = 0$ in $M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}
    \otimes_{\kappa_{\lambda'}} \kappa$, hence $x = 0$ because $w$ is invertible
    in $S_{\lambda'}$.
    We conclude that the kernel of $\Psi_{\lambda'}$ is zero for all sufficiently
    large $\lambda'$!
    
    \medskip\noindent
    By the result of the preceding paragraph we may assume that
    the kernel of $\Psi_\lambda$ is zero for all $\lambda$ sufficiently large,
    which implies that the map
    $M_\lambda/\mathfrak m_\lambda M_\lambda \to M/\mathfrak m M$
    is injective. Combined with $\overline{u}$ being injective this
    formally implies that also
    $\overline{u_\lambda} : M_\lambda/\mathfrak m_\lambda M_\lambda
    \to N_\lambda/\mathfrak m_\lambda N_\lambda$ is injective.
    By
    Lemma \ref{lemma-mod-injective}
    we conclude that (for all sufficiently large $\lambda$) the map
    $u_\lambda$ is injective and that $N_\lambda/u_\lambda(M_\lambda)$ is flat
    over $R_\lambda$.
    The lemma follows.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-grothendieck-general}
    Suppose that $R \to S$ is a local ring homomorphism of local rings.
    Denote $\mathfrak m$ the maximal ideal of $R$.
    Suppose
    \begin{enumerate}
    \item $S$ is essentially of finite presentation over $R$,
    \item $S$ is flat over $R$, and
    \item $f \in S$ is a nonzerodivisor in $S/{\mathfrak m}S$.
    \end{enumerate}
    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-general}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-grothendieck-regular-sequence-general}
    Suppose that $R \to S$ is a local ring homomorphism of local rings.
    Denote $\mathfrak m$ the maximal ideal of $R$.
    Suppose
    \begin{enumerate}
    \item $R \to S$ is essentially of finite presentation,
    \item $R \to S$ is flat, and
    \item $f_1, \ldots, f_c$ is a sequence of elements of
    $S$ such that the images $\overline{f}_1, \ldots, \overline{f}_c$
    form a regular sequence in $S/{\mathfrak m}S$.
    \end{enumerate}
    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-general}.
    \end{proof}
    
    \noindent
    Here is the version of the local criterion of flatness for the case
    of local ring maps which are locally of finite presentation.
    
    \begin{lemma}
    \label{lemma-variant-local-criterion-flatness-general}
    Let $R \to S$ be a local homomorphism of local rings.
    Let $I \not = R$ be an ideal in $R$. Let $M$ be an $S$-module. Assume
    \begin{enumerate}
    \item $S$ is essentially of finite presentation over $R$,
    \item $M$ is of finite presentation over $S$,
    \item $\text{Tor}_1^R(M, R/I) = 0$, and
    \item $M/IM$ is flat over $R/I$.
    \end{enumerate}
    Then $M$ is flat over $R$.
    \end{lemma}
    
    \begin{proof}
    Let $\Lambda$, $R_\lambda \to S_\lambda$, $M_\lambda$ be as in
    Lemma \ref{lemma-limit-module-essentially-finite-presentation}.
    Denote $I_\lambda \subset R_\lambda$ the inverse image of $I$.
    In this case the system
    $R/I \to S/IS$, $M/IM$, $R_\lambda \to S_\lambda/I_\lambda S_\lambda$,
    and $M_\lambda/I_\lambda M_\lambda$ satisfies the conclusions of
    Lemma \ref{lemma-limit-module-essentially-finite-presentation}
    as well. Hence by
    Lemma \ref{lemma-colimit-eventually-flat}
    we may assume (after shrinking the index set $\Lambda$)
    that $M_\lambda/I_\lambda M_\lambda$ is flat for all $\lambda$.
    Pick some $\lambda$ and consider
    $$
    \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/I_\lambda)
    =
    \Ker(I_\lambda \otimes_{R_\lambda} M_\lambda
    \to M_\lambda).
    $$
    See Remark \ref{remark-Tor-ring-mod-ideal}. The right hand side
    shows that this is a finitely generated $S_\lambda$-module (because
    $S_\lambda$ is Noetherian and the modules in question are finite).
    Let $\xi_1, \ldots, \xi_n$ be generators.
    Because $\text{Tor}^1_R(M, R/I) = 0$ and since $\otimes$ commutes
    with colimits we see there exists
    a $\lambda' \geq \lambda$ such that each $\xi_i$ maps to
    zero in
    $\text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_{\lambda'})$.
    The composition of the maps
    $$
    \xymatrix{
    R_{\lambda'} \otimes_{R_\lambda}
    \text{Tor}_1^{R_\lambda}(M_\lambda, R_\lambda/I_\lambda)
    \ar[d]^{\text{surjective by Lemma \ref{lemma-surjective-on-tor-one}}} \\
    \text{Tor}_1^{R_\lambda}(M_\lambda, R_{\lambda'}/I_\lambda R_{\lambda'})
    \ar[d]^{\text{surjective up to localization by
    Lemma \ref{lemma-surjective-on-tor-one-trivial}}} \\
    \text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_\lambda R_{\lambda'})
    \ar[d]^{\text{surjective by Lemma \ref{lemma-surjective-on-tor-one}}} \\
    \text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_{\lambda'}).
    }
    $$
    is surjective up to a localization by the reasons indicated.
    The localization is necessary since $M_{\lambda'}$ is not equal
    to $M_\lambda \otimes_{R_\lambda} R_{\lambda'}$. Namely, it is equal
    to $M_\lambda \otimes_{S_\lambda} S_{\lambda'}$ and $S_{\lambda'}$
    is the localization of $S_{\lambda} \otimes_{R_\lambda} R_{\lambda'}$ whence
    the statement up to a localization (or tensoring with $S_{\lambda'}$).
    Note that
    Lemma \ref{lemma-surjective-on-tor-one}
    applies to the first and third arrows because
    $M_\lambda/I_\lambda M_\lambda$ is flat over
    $R_\lambda/I_\lambda$ and because $M_{\lambda'}/I_\lambda M_{\lambda'}$
    is flat over $R_{\lambda'}/I_\lambda R_{\lambda'}$ as it is a base
    change of the flat module $M_\lambda/I_\lambda M_\lambda$.
    The composition maps the generators $\xi_i$ to zero as we explained above.
    We finally conclude that
    $\text{Tor}_1^{R_{\lambda'}}(M_{\lambda'}, R_{\lambda'}/I_{\lambda'})$
    is zero. This implies that $M_{\lambda'}$ is flat over $R_{\lambda'}$ by
    Lemma \ref{lemma-variant-local-criterion-flatness}.
    \end{proof}
    
    \noindent
    Please compare the lemma below to
    Lemma \ref{lemma-criterion-flatness-fibre-Noetherian}
    (the case of Noetherian local rings) and
    Lemma \ref{lemma-criterion-flatness-fibre-nilpotent}
    (the case of a nilpotent ideal in the base).
    
    \begin{lemma}[Crit\`ere de platitude par fibres]
    \label{lemma-criterion-flatness-fibre}
    Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring
    homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$
    be the maximal ideal. Assume
    \begin{enumerate}
    \item The ring maps $R \to S$ and $R \to S'$ are essentially
    of finite presentation.
    \item The module $M$ is of finite presentation over $S'$.
    \item The module $M$ is not zero.
    \item The module $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-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}
    As in the proof of Lemma \ref{lemma-limit-essentially-finite-presentation}
    we may first write $R = \colim R_\lambda$ as a directed colimit
    of local $\mathbf{Z}$-algebras which are essentially of finite type.
    Denote $\mathfrak p_\lambda$ the maximal ideal of $R_\lambda$.
    Next, we may assume that for some $\lambda_1 \in \Lambda$ there
    exist $f_{j, \lambda_1} \in R_{\lambda_1}[x_1, \ldots, x_n]$
    such that
    $$
    S =
    \colim_{\lambda \geq \lambda_1} S_\lambda, \text{ with }
    S_\lambda =
    (R_\lambda[x_1, \ldots, x_n]/
    (f_{1, \lambda}, \ldots, f_{u, \lambda}))_{\mathfrak q_\lambda}
    $$
    For some $\lambda_2 \in \Lambda$,
    $\lambda_2 \geq \lambda_1$ there exist
    $g_{j, \lambda_2} \in R_{\lambda_2}[x_1, \ldots, x_n, y_1, \ldots, y_m]$
    with images
    $\overline{g}_{j, \lambda_2} \in S_{\lambda_2}[y_1, \ldots, y_m]$
    such that
    $$
    S' =
    \colim_{\lambda \geq \lambda_2} S'_\lambda, \text{ with }
    S'_\lambda =
    (S_\lambda[y_1, \ldots, y_m]/
    (\overline{g}_{1, \lambda}, \ldots,
    \overline{g}_{v, \lambda}))_{\overline{\mathfrak q}'_\lambda}
    $$
    Note that this also implies that
    $$
    S'_\lambda =
    (R_\lambda[x_1, \ldots, x_n, y_1, \ldots, y_m]/
    (g_{1, \lambda}, \ldots, g_{v, \lambda}))_{\mathfrak q'_\lambda}
    $$
    Choose a presentation
    $$
    (S')^{\oplus s} \to (S')^{\oplus t} \to M \to 0
    $$
    of $M$ over $S'$. Let $A \in \text{Mat}(t \times s, S')$ be
    the matrix of the presentation. For some $\lambda_3 \in \Lambda$,
    $\lambda_3 \geq \lambda_2$
    we can find a matrix $A_{\lambda_3} \in \text{Mat}(t \times s, S_{\lambda_3})$
    which maps to $A$. For all $\lambda \geq \lambda_3$ we let
    $M_\lambda = \Coker((S'_\lambda)^{\oplus s} \xrightarrow{A_\lambda}
    (S'_\lambda)^{\oplus t})$.
    
    \medskip\noindent
    With these choices, we have for each $\lambda_3 \leq \lambda \leq \mu$
    that $S_\lambda \otimes_{R_{\lambda}} R_\mu \to S_\mu$ is a localization,
    $S'_\lambda \otimes_{S_{\lambda}} S_\mu \to S'_\mu$ is a localization, and
    the map $M_\lambda \otimes_{S'_\lambda} S'_\mu \to M_\mu$ is an
    isomorphism. This also implies that
    $S'_\lambda \otimes_{R_{\lambda}} R_\mu \to S'_\mu$ is a localization.
    Thus, since $M$ is flat over $R$ we see by
    Lemma \ref{lemma-colimit-eventually-flat} that
    for all $\lambda$ big enough the module $M_\lambda$ is
    flat over $R_\lambda$.
    Moreover, note that
    $
    \mathfrak m = \colim \mathfrak p_\lambda
    $,
    $
    S/\mathfrak mS = \colim S_\lambda/\mathfrak p_\lambda S_\lambda
    $,
    $
    S'/\mathfrak mS' = \colim S'_\lambda/\mathfrak p_\lambda S'_\lambda
    $,
    and
    $
    M/\mathfrak mM = \colim M_\lambda/\mathfrak p_\lambda M_\lambda
    $. Also, for each $\lambda_3 \leq \lambda \leq \mu$ we see (from the
    properties listed above) that
    $$
    S'_\lambda/\mathfrak p_\lambda S'_\lambda
    \otimes_{S_{\lambda}/\mathfrak p_\lambda S_\lambda}
    S_\mu/\mathfrak p_\mu S_\mu
    \longrightarrow
    S'_\mu/\mathfrak p_\mu S'_\mu
    $$
    is a localization, and the map
    $$
    M_\lambda / \mathfrak p_\lambda M_\lambda
    \otimes_{S'_\lambda/\mathfrak p_\lambda S'_\lambda}
    S'_\mu /\mathfrak p_\mu S'_\mu
    \longrightarrow
    M_\mu/\mathfrak p_\mu M_\mu
    $$
    is an isomorphism. Hence the system
    $(S_\lambda/\mathfrak p_\lambda S_\lambda \to
    S'_\lambda/\mathfrak p_\lambda S'_\lambda,
    M_\lambda/\mathfrak p_\lambda M_\lambda)$
    is a system as in
    Lemma \ref{lemma-limit-module-essentially-finite-presentation} as well.
    We may apply Lemma \ref{lemma-colimit-eventually-flat} again because
    $M/\mathfrak m M$ is assumed flat over $S/\mathfrak mS$ and we see that
    $M_\lambda/\mathfrak p_\lambda M_\lambda$ is flat over
    $S_\lambda/\mathfrak p_\lambda S_\lambda$ for all $\lambda$ big enough.
    Thus for $\lambda$ big enough the data
    $R_\lambda \to S_\lambda \to S'_\lambda, M_\lambda$ satisfies
    the hypotheses of Lemma \ref{lemma-criterion-flatness-fibre-Noetherian}.
    Pick such a $\lambda$. Then $S = S_\lambda \otimes_{R_\lambda} R$
    is flat over $R$, and $M = M_\lambda \otimes_{S_\lambda} S$
    is flat over $S$ (since the base change of a flat module is flat).
    \end{proof}
    
    \noindent
    The following is an easy consequence of the ``crit\`ere de platitude par
    fibres'' Lemma \ref{lemma-criterion-flatness-fibre}. For more results of
    this kind see More on Flatness, Section \ref{flat-section-introduction}.
    
    \begin{lemma}
    \label{lemma-criterion-flatness-fibre-fp-over-ft}
    Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring
    homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$
    be the maximal ideal. Assume
    \begin{enumerate}
    \item $R \to S'$ is essentially of finite presentation,
    \item $R \to S$ is essentially of finite type,
    \item $M$ is of finite presentation over $S'$,
    \item $M$ is not zero,
    \item $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-module, and
    \item $M$ is a flat $R$-module.
    \end{enumerate}
    Then $S$ is essentially of finite presentation and flat over $R$
    and $M$ is a flat $S$-module.
    \end{lemma}
    
    \begin{proof}
    As $S$ is essentially of finite presentation over $R$ we can write
    $S = C_{\overline{\mathfrak q}}$ for some finite type $R$-algebra $C$.
    Write $C = R[x_1, \ldots, x_n]/I$. Denote
    $\mathfrak q \subset R[x_1, \ldots, x_n]$ be the prime ideal corresponding
    to $\overline{\mathfrak q}$. Then we see that $S = B/J$ where
    $B = R[x_1, \ldots, x_n]_{\mathfrak q}$ is essentially of finite presentation
    over $R$ and $J = IB$. We can find $f_1, \ldots, f_k \in J$ such that
    the images $\overline{f}_i \in B/\mathfrak mB$
    generate the image $\overline{J}$ of $J$ in the Noetherian ring
    $B/\mathfrak mB$. Hence there exist finitely generated ideals
    $J' \subset J$ such that $B/J' \to B/J$ induces an isomorphism
    $$
    (B/J') \otimes_R R/\mathfrak m \longrightarrow
    B/J \otimes_R R/\mathfrak m = S/\mathfrak mS.
    $$
    For any $J'$ as above we see that
    Lemma \ref{lemma-criterion-flatness-fibre}
    applies to the ring maps
    $$
    R \longrightarrow B/J' \longrightarrow S'
    $$
    and the module $M$. Hence we conclude that $B/J'$ is flat over $R$
    for any choice $J'$ as above. Now, if $J' \subset J' \subset J$ are
    two finitely generated ideals as above, then we conclude that
    $B/J' \to B/J''$ is a surjective map between flat $R$-algebras
    which are essentially of finite presentation which is an isomorphism
    modulo $\mathfrak m$. Hence
    Lemma \ref{lemma-mod-injective-general}
    implies that $B/J' = B/J''$, i.e., $J' = J''$. Clearly this means that
    $J$ is finitely generated, i.e., $S$ is essentially of finite presentation
    over $R$. Thus we may apply
    Lemma \ref{lemma-criterion-flatness-fibre}
    to $R \to S \to S'$ and we win.
    \end{proof}
    
    \begin{lemma}[Crit\`ere de platitude par fibres: locally nilpotent case]
    \label{lemma-criterion-flatness-fibre-locally-nilpotent}
    Let
    $$
    \xymatrix{
    S \ar[rr] & & S' \\
    & R \ar[lu] \ar[ru]
    }
    $$
    be a commutative diagram in the category of rings.
    Let $I \subset R$ be a locally nilpotent ideal and
    $M$ an $S'$-module. Assume
    \begin{enumerate}
    \item $R \to S$ is of finite type,
    \item $R \to S'$ is of finite presentation,
    \item $M$ is a finitely presented $S'$-module,
    \item $M/IM$ is flat as a $S/IS$-module, and
    \item $M$ is flat as an $R$-module.
    \end{enumerate}
    Then $M$ is a flat $S$-module and $S_\mathfrak q$ is flat
    and essentially of finite presentation over $R$
    for every $\mathfrak q \subset S$ such that
    $M \otimes_S \kappa(\mathfrak q)$ is nonzero.
    \end{lemma}
    
    \begin{proof}
    If $M \otimes_S \kappa(\mathfrak q)$ is nonzero, then
    $S' \otimes_S \kappa(\mathfrak q)$ is nonzero and hence
    there exists a prime $\mathfrak q' \subset S'$ lying over
    $\mathfrak q$ (Lemma \ref{lemma-in-image}). Let
    $\mathfrak p \subset R$ be the image of $\mathfrak q$ in $\Spec(R)$.
    Then $I \subset \mathfrak p$ as $I$ is locally nilpotent
    hence $M/\mathfrak p M$ is flat over $S/\mathfrak pS$.
    Hence we may apply Lemma \ref{lemma-criterion-flatness-fibre-fp-over-ft}
    to $R_\mathfrak p \to S_\mathfrak q \to S'_{\mathfrak q'}$
    and $M_{\mathfrak q'}$. We conclude that $M_{\mathfrak q'}$
    is flat over $S$ and $S_\mathfrak q$ is flat and essentially
    of finite presentation over $R$.
    Since $\mathfrak q'$ was an arbitrary prime of $S'$ we also
    see that $M$ is flat over $S$ (Lemma \ref{lemma-flat-localization}).
    \end{proof}

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