
## 15.19 Flattening over a closed subsets of source and base

In this section we slightly generalize the discussion in Section 15.18. We strongly suggest the reader first read and understand that section.

Situation 15.19.1. Let $R \to S$ be a ring map. Let $J \subset S$ be an ideal. Let $M$ be an $S$-module.

In this situation, given an $R$-algebra $R'$ and an ideal $I' \subset R'$ we set $S' = S \otimes _ R R'$ and $M' = M \otimes _ R R'$. We will consider the condition

15.19.1.1
$$\label{more-algebra-equation-flat-at-primes} \forall \mathfrak q' \in V(I'S' + JS') \subset \mathop{\mathrm{Spec}}(S') : M'_{\mathfrak q'}\text{ is flat over }R'.$$

Geometrically, this means that $M'$ is flat over $R'$ along the intersection of the inverse image of $V(I')$ with the inverse image of $V(J)$. Since $(R \to S, J, M)$ are fixed, condition (15.19.1.1) only depends on the pair $(R', I')$ where $R'$ is viewed as an $R$-algebra.

Lemma 15.19.2. In Situation 15.19.1 let $R' \to R''$ be an $R$-algebra map. Let $I' \subset R'$ and $I'R'' \subset I'' \subset R''$ be ideals. If (15.19.1.1) holds for $(R', I')$, then (15.19.1.1) holds for $(R'', I'')$.

Proof. Assume (15.19.1.1) holds for $(R', I')$. Let $I''S'' + JS'' \subset \mathfrak q''$ be a prime of $S''$. Let $\mathfrak q' \subset S'$ be the corresponding prime of $S'$. Then both $I'S' \subset \mathfrak q'$ and $JS' \subset \mathfrak q'$ because the corresponding conditions hold for $\mathfrak q''$. Note that $(M'')_{\mathfrak q''}$ is a localization of the base change $M'_{\mathfrak q'} \otimes _ R R''$. Hence $(M'')_{\mathfrak q''}$ is flat over $R''$ as a localization of a flat module, see Algebra, Lemmas 10.38.7 and 10.38.19. $\square$

Lemma 15.19.3. In Situation 15.19.1 let $R' \to R''$ be an $R$-algebra map. Let $I' \subset R'$ and $I'R'' \subset I'' \subset R''$ be ideals. Assume

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 (15.19.1.1) holds for $(R'', I'')$, then (15.19.1.1) holds for $(R', I')$.

Proof. Assume (15.19.1.1) holds for $(R'', I'')$. Pick a prime $I'S' + JS' \subset \mathfrak q' \subset S'$. 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 $\overline{\mathfrak q}'' \subset \kappa (\mathfrak q') \otimes _{\kappa (\mathfrak p')} \kappa (\mathfrak p'')$. This corresponds to a prime $\mathfrak q'' \subset S'' = S' \otimes _{R'} R''$ which lies over $\mathfrak q'$ and over $\mathfrak p''$. In particular we see that $I''S'' \subset \mathfrak q''$ and that $JS'' \subset \mathfrak q''$. 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 10.99.1 implies that $M'_{\mathfrak q'}$ is flat over $R'_{\mathfrak p'}$ which is what we wanted to prove. $\square$

Lemma 15.19.4. In Situation 15.19.1 assume $R \to S$ is essentially of finite presentation and $M$ is 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 (15.19.1.1) holds for $(R', I')$, then there exists a $\lambda \in \Lambda$ such that (15.19.1.1) holds for $(R_\lambda , I_\lambda )$.

Proof. We first prove the lemma in case $R \to S$ is of finite presentation and then we explain what needs to be changed in the general case. 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 10.13.2. By Algebra, Theorem 10.128.4 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 $V(I'S' + JS')$ 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 $V(I'S' + JS') \subset \bigcup D(g'_ j)$. 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 $V(I'S' + JS') \subset \bigcup D(g'_ j)$ means that $I'S' + JS' + (g'_1, \ldots , g'_ m) = S'$ which can be expressed as

$1 = \sum y_ tk_ t + \sum z_ sh_ s + \sum f_ jg'_ j$

for some $z_ s \in I'$, $y_ t \in J$, $k_ t, h_ s, f_ j \in S'$. After increasing $\lambda$ we may assume such an equation holds in $S_\lambda$. Hence we may assume that $V(I_\lambda S_\lambda + J S_\lambda ) \subset \bigcup D(g_{j, \lambda })$. By Algebra, Lemma 10.162.1 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 corresponding to points of $V(I_\lambda S_\lambda + J S_\lambda )$.

In the case that $S$ is essentially of finite presentation, we can write $S = \Sigma ^{-1}C$ where $R \to C$ is of finite presentation and $\Sigma \subset C$ is a multiplicative subset. We can also write $M = \Sigma ^{-1}N$ for some finitely presented $C$-module $N$, see Algebra, Lemma 10.125.3. At this point we introduce $C_\lambda$, $C'$, $N_\lambda$, $N'$. Then in the discussion above we obtain an open $U' \subset \mathop{\mathrm{Spec}}(C')$ over which $N'$ is flat over $R'$. The assumption that (15.19.1.1) is true means that $V(I'S' + JS')$ maps into $U'$, because for a prime $\mathfrak q' \subset S'$, corresponding to a prime $\mathfrak r' \subset C'$ we have $M'_{\mathfrak q'} = N'_{\mathfrak r'}$. Thus we can find $g'_ j \in C'$ such that $\bigcup D(g'_ j)$ contains the image of $V(I'S' + JS')$. The rest of the proof is exactly the same as before. $\square$

Lemma 15.19.5. In Situation 15.19.1. Let $I \subset R$ be an ideal. Assume

1. $R$ is a Noetherian ring,

2. $S$ is a Noetherian ring,

3. $M$ is a finite $S$-module, and

4. for each $n \geq 1$ and any prime $\mathfrak q \in V(J + IS)$ the module $(M/I^ n M)_{\mathfrak q}$ is flat over $R/I^ n$.

Then (15.19.1.1) holds for $(R, I)$, i.e., for every prime $\mathfrak q \in V(J + IS)$ the localization $M_{\mathfrak q}$ is flat over $R$.

Proof. Let $\mathfrak q \in V(J + IS)$. Then Algebra, Lemma 10.98.11 applied to $R \to S_{\mathfrak q}$ and $M_{\mathfrak q}$ implies that $M_{\mathfrak q}$ is flat over $R$. $\square$

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