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 10.14.2. 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 $V(I'S' + JS')$ 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 10.168.1 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 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 10.126.3. 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. $\square$

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