$\xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] }$

be a commutative diagram in the category of rings. Let $I \subset R$ be a locally nilpotent ideal and $M$ an $S'$-module. Assume

1. $R \to S$ is of finite type,

2. $R \to S'$ is of finite presentation,

3. $M$ is a finitely presented $S'$-module,

4. $M/IM$ is flat as a $S/IS$-module, and

5. $M$ is flat as an $R$-module.

Then $M$ is a flat $S$-module and $S_\mathfrak q$ is flat and essentially of finite presentation over $R$ for every $\mathfrak q \subset S$ such that $M \otimes _ S \kappa (\mathfrak q)$ is nonzero.

Proof. If $M \otimes _ S \kappa (\mathfrak q)$ is nonzero, then $S' \otimes _ S \kappa (\mathfrak q)$ is nonzero and hence there exists a prime $\mathfrak q' \subset S'$ lying over $\mathfrak q$ (Lemma 10.17.9). Let $\mathfrak p \subset R$ be the image of $\mathfrak q$ in $\mathop{\mathrm{Spec}}(R)$. Then $I \subset \mathfrak p$ as $I$ is locally nilpotent hence $M/\mathfrak p M$ is flat over $S/\mathfrak pS$. Hence we may apply Lemma 10.128.9 to $R_\mathfrak p \to S_\mathfrak q \to S'_{\mathfrak q'}$ and $M_{\mathfrak q'}$. We conclude that $M_{\mathfrak q'}$ is flat over $S$ and $S_\mathfrak q$ is flat and essentially of finite presentation over $R$. Since $\mathfrak q'$ was an arbitrary prime of $S'$ we also see that $M$ is flat over $S$ (Lemma 10.39.18). $\square$

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