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. $\square$
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