# The Stacks Project

## Tag 00R3

### 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 31256–31836 (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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