Lemma 10.99.10 (Variant of the local criterion). Let $R \to S$ be a local homomorphism of Noetherian local rings. Let $I \not= R$ be an ideal in $R$. Let $M$ be a finite $S$-module. If $\text{Tor}_1^ R(M, R/I) = 0$ and $M/IM$ is flat over $R/I$, then $M$ is flat over $R$.

**Proof.**
First proof: By Lemma 10.99.8 we see that $\text{Tor}_1^ R(\kappa , M)$ is zero where $\kappa $ is the residue field of $R$. Hence we see that $M$ is flat over $R$ by Lemma 10.99.7.

Second proof: Let $\mathfrak m$ be the maximal ideal of $R$. We will show that $\mathfrak m \otimes _ R M \to M$ is injective, and then apply Lemma 10.99.7. Suppose that $\sum f_ i \otimes x_ i \in \mathfrak m \otimes _ R M$ and that $\sum f_ i x_ i = 0$ in $M$. By the equational criterion for flatness Lemma 10.39.11 applied to $M/IM$ over $R/I$ we see there exist $\overline{a}_{ij} \in R/I$ and $\overline{y}_ j \in M/IM$ such that $x_ i \bmod IM = \sum _ j \overline{a}_{ij} \overline{y}_ j $ and $0 = \sum _ i (f_ i \bmod I) \overline{a}_{ij}$. Let $a_{ij} \in R$ be a lift of $\overline{a}_{ij}$ and similarly let $y_ j \in M$ be a lift of $\overline{y}_ j$. Then we see that

Since $x_ i - \sum a_{ij} y_ j \in IM$ and $\sum f_ i a_{ij} \in I$ we see that there exists an element in $I \otimes _ R M$ which maps to our given element $\sum f_ i \otimes x_ i$ in $\mathfrak m \otimes _ R M$. But $I \otimes _ R M \to M$ is injective by assumption (see Remark 10.75.9) and we win. $\square$

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