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 $(R' \to S \otimes _ R R', I', M \otimes _ R R')$.

## 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

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.

**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

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

$R'_{\mathfrak p'}$ 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

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 $. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)