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
\begin{eqnarray*} \sum f_ i \otimes x_ i & = & \sum f_ i \otimes x_ i + \sum f_ ia_{ij} \otimes y_ j - \sum f_ i \otimes a_{ij} y_ j \\ & = & \sum f_ i \otimes (x_ i - \sum a_{ij} y_ j) + \sum (\sum f_ i a_{ij}) \otimes y_ j \end{eqnarray*}
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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