Lemma 15.21.5. Let $R \to S$ be a finite injective homomorphism of Noetherian rings. Let $M$ be an $R$-module. If $M \otimes _ R S$ is a flat $S$-module, then $M$ is a flat $R$-module.

**Proof.**
Let $M$ be an $R$-module such that $M \otimes _ R S$ is flat over $S$. By Algebra, Lemma 10.39.8 in order to prove that $M$ is flat we may replace $R$ by any faithfully flat ring extension. By Lemma 15.21.3 we can find a finite locally free ring extension $R \subset R'$ such that $S' = S \otimes _ R R' = R'[T_1, \ldots , T_ n]/J$ for some ideal $J \subset R'[T_1, \ldots , T_ n]$ which contains the elements of the form $P_ i(T_ i)$ with $P_ i(T) = \prod _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$ for some $\alpha _{ij} \in R'$. Note that $R'$ is Noetherian and that $R' \subset S'$ is a finite extension of rings. Hence we may replace $R$ by $R'$ and assume that $S$ has a presentation as in Lemma 15.21.4. Note that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective, see Algebra, Lemma 10.36.17. Thus, using Lemma 15.21.4 we conclude that $I = \bigcap J_{\underline{k}}$ is an ideal such that $V(I) = \mathop{\mathrm{Spec}}(R)$. This means that $I \subset \sqrt{(0)}$, and since $R$ is Noetherian that $I$ is nilpotent. The maps $\Phi _{\underline{k}}$ induce commutative diagrams

from which we conclude that $M/J_{\underline{k}}M$ is flat over $R/J_{\underline{k}}$. By Lemma 15.16.1 we see that $M/IM$ is flat over $R/I$. Finally, applying Algebra, Lemma 10.101.5 we conclude that $M$ is flat over $R$. $\square$

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