## 15.18 Flattening over a closed subset of the base

Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. In the following we will consider the following condition

15.18.0.1
$$\label{more-algebra-equation-flat-at-primes-over} \forall \mathfrak q \in V(IS) \subset \mathop{\mathrm{Spec}}(S) : M_{\mathfrak q}\text{ is flat over }R.$$

Geometrically, this means that $M$ is flat over $R$ along the inverse image of $V(I)$ in $\mathop{\mathrm{Spec}}(S)$. If $R$ and $S$ are Noetherian rings and $M$ is a finite $S$-module, then (15.18.0.1) is equivalent to the condition that $M/I^ nM$ is flat over $R/I^ n$ for all $n \geq 1$, see Algebra, Lemma 10.99.11.

Lemma 15.18.1. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Let $R \to R'$ be a ring map and $IR' \subset I' \subset R'$ an ideal. If (15.18.0.1) holds for $(R \to S, I, M)$, then (15.18.0.1) holds for $(R' \to S \otimes _ R R', I', M \otimes _ R R')$.

Proof. Assume (15.18.0.1) holds for $(R \to S, I \subset R, M)$. Let $I'(S \otimes _ R R') \subset \mathfrak q'$ be a prime of $S \otimes _ R R'$. Let $\mathfrak q \subset S$ be the corresponding prime of $S$. Then $IS \subset \mathfrak q$. Note that $(M \otimes _ R R')_{\mathfrak q'}$ is a localization of the base change $M_{\mathfrak q} \otimes _ R R'$. Hence $(M \otimes _ R R')_{\mathfrak q'}$ is flat over $R'$ as a localization of a flat module, see Algebra, Lemmas 10.39.7 and 10.39.18. $\square$

Lemma 15.18.2. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Let $R \to R'$ be a ring map and $IR' \subset I' \subset R'$ an ideal such that

1. the map $V(I') \to V(I)$ induced by $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is surjective, and

2. $R'_{\mathfrak p'}$ is flat over $R$ for all primes $\mathfrak p' \in V(I')$.

If (15.18.0.1) holds for $(R' \to S \otimes _ R R', I', M \otimes _ R R')$, then (15.18.0.1) holds for $(R \to S, I, M)$.

Proof. Assume (15.18.0.1) holds for $(R' \to S \otimes _ R R', IR', M \otimes _ R R')$. Pick a prime $IS \subset \mathfrak q \subset S$. Let $I \subset \mathfrak p \subset R$ be the corresponding prime of $R$. By assumption there exists a prime $\mathfrak p' \in V(I')$ of $R'$ lying over $\mathfrak p$ and $R_{\mathfrak p} \to R'_{\mathfrak p'}$ is flat. Choose a prime $\overline{\mathfrak q}' \subset \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p')$ which corresponds to a prime $\mathfrak q' \subset S \otimes _ R R'$ which lies over $\mathfrak q$ and over $\mathfrak p'$. Note that $(S \otimes _ R R')_{\mathfrak q'}$ is a localization of $S_{\mathfrak q} \otimes _{R_{\mathfrak p}} R'_{\mathfrak p'}$. By assumption the module $(M \otimes _ R R')_{\mathfrak q'}$ is flat over $R'_{\mathfrak p'}$. Hence Algebra, Lemma 10.100.1 implies that $M_{\mathfrak q}$ is flat over $R_{\mathfrak p}$ which is what we wanted to prove. $\square$

Lemma 15.18.3. Let $R \to S$ be a ring map of finite presentation. Let $M$ be an $S$-module of finite presentation. Let $R' = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } R_\lambda$ be a directed colimit of $R$-algebras. Let $I_\lambda \subset R_\lambda$ be ideals such that $I_\lambda R_\mu \subset I_\mu$ for all $\mu \geq \lambda$ and set $I' = \mathop{\mathrm{colim}}\nolimits _\lambda I_\lambda$. If (15.18.0.1) holds for $(R' \to S \otimes _ R R', I', M \otimes _ R R')$, then there exists a $\lambda \in \Lambda$ such that (15.18.0.1) holds for $(R_\lambda \to S \otimes _ R R_\lambda , I_\lambda , M \otimes _ R R_\lambda )$.

Proof. We are going to write $S_\lambda = S \otimes _ R R_\lambda$, $S' = S \otimes _ R R'$, $M_\lambda = M \otimes _ R R_\lambda$, and $M' = M \otimes _ R R'$. The base change $S'$ is of finite presentation over $R'$ and $M'$ is of finite presentation over $S'$ and similarly for the versions with subscript $\lambda$, see Algebra, Lemma 10.14.2. By Algebra, Theorem 10.129.4 the set

$U' = \{ \mathfrak q' \in \mathop{\mathrm{Spec}}(S') \mid M'_{\mathfrak q'}\text{ is flat over }R'\}$

is open in $\mathop{\mathrm{Spec}}(S')$. Note that $V(I'S')$ is a quasi-compact space which is contained in $U'$ by assumption. Hence there exist finitely many $g'_ j \in S'$, $j = 1, \ldots , m$ such that $D(g'_ j) \subset U'$ and such that $V(I'S') \subset \bigcup D(g'_ j)$. Note that in particular $(M')_{g'_ j}$ is a flat module over $R'$.

We are going to pick increasingly large elements $\lambda \in \Lambda$. First we pick it large enough so that we can find $g_{j, \lambda } \in S_{\lambda }$ mapping to $g'_ j$. The inclusion $V(I'S') \subset \bigcup D(g'_ j)$ means that $I'S' + (g'_1, \ldots , g'_ m) = S'$ which can be expressed as $1 = \sum z_ sh_ s + \sum f_ jg'_ j$ for some $z_ s \in I'$, $h_ s, f_ j \in S'$. After increasing $\lambda$ we may assume such an equation holds in $S_\lambda$. Hence we may assume that $V(I_\lambda S_\lambda ) \subset \bigcup D(g_{j, \lambda })$. By Algebra, Lemma 10.168.1 we see that for some sufficiently large $\lambda$ the modules $(M_\lambda )_{g_{j, \lambda }}$ are flat over $R_\lambda$. In particular the module $M_\lambda$ is flat over $R_\lambda$ at all the primes lying over the ideal $I_\lambda$. $\square$

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