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 10.39.7 and 10.39.18.
$\square$

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