Lemma 10.100.7. Let $R \to S$ be a ring map. Let $M$ be an $R$-module. Assume

$R$ is Artinian

$R \to S$ is injective, and

$M \otimes _ R S$ is a flat $S$-module.

Then $M$ is a flat $R$-module.

Lemma 10.100.7. Let $R \to S$ be a ring map. Let $M$ be an $R$-module. Assume

$R$ is Artinian

$R \to S$ is injective, and

$M \otimes _ R S$ is a flat $S$-module.

Then $M$ is a flat $R$-module.

**Proof.**
First proof: Let $I \subset R$ be the Jacobson radical of $R$. Then $I$ is nilpotent and $M/IM$ is flat over $R/I$ as $R/I$ is a product of fields, see Section 10.52. Hence $M$ is flat by an application of Lemma 10.100.5.

Second proof: By Lemma 10.52.6 we may write $R = \prod R_ i$ as a finite product of local Artinian rings. This induces similar product decompositions for both $R$ and $S$. Hence we reduce to the case where $R$ is local Artinian (details omitted).

Assume that $R \to S$, $M$ are as in the lemma satisfying (1), (2), and (3) and in addition that $R$ is local with maximal ideal $\mathfrak m$. Let $A$ be a set and $x_\alpha \in A$ be elements such that $\overline{x}_\alpha $ forms a basis for $M/\mathfrak mM$ over $R/\mathfrak m$. By Nakayama's Lemma 10.19.1 we see that the elements $x_\alpha $ generate $M$ as an $R$-module. Set $N = S \otimes _ R M$ and $I = \mathfrak mS$. Then $\{ 1 \otimes x_\alpha \} _{\alpha \in A}$ is a family of elements of $N$ which form a basis for $N/IN$. Moreover, since $N$ is flat over $S$ we have $\text{Tor}_1^ S(S/I, N) = 0$. Thus we conclude from Lemma 10.100.3 that $N$ is free on $\{ 1 \otimes x_\alpha \} _{\alpha \in A}$. The injectivity of $R \to S$ then guarantees that there cannot be a nontrivial relation among the $x_\alpha $ with coefficients in $R$. $\square$

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