The Stacks project

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 10.39.7 and 10.39.18. $\square$

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 10.100.1 implies that $M'_{\mathfrak q'}$ is flat over $R'_{\mathfrak p'}$ which is what we wanted to prove. $\square$

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

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

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. $\square$

Proof. Let $\mathfrak q \in V(J + IS)$. Then Algebra, Lemma 10.99.11 applied to $R \to S_{\mathfrak q}$ and $M_{\mathfrak q}$ implies that $M_{\mathfrak q}$ is flat over $R$. $\square$