Lemma 10.99.8. Let R be a ring. Let I \subset R be an ideal. Let M be an R-module. If M/IM is flat over R/I and \text{Tor}_1^ R(R/I, M) = 0 then
M/I^ nM is flat over R/I^ n for all n \geq 1, and
for any module N which is annihilated by I^ m for some m \geq 0 we have \text{Tor}_1^ R(N, M) = 0.
In particular, if I is nilpotent, then M is flat over R.
Proof.
Assume M/IM is flat over R/I and \text{Tor}_1^ R(R/I, M) = 0. Let N be an R/I-module. Choose a short exact sequence
0 \to K \to \bigoplus \nolimits _{i \in I} R/I \to N \to 0
By the long exact sequence of \text{Tor} and the vanishing of \text{Tor}_1^ R(R/I, M) we get
0 \to \text{Tor}_1^ R(N, M) \to K \otimes _ R M \to (\bigoplus \nolimits _{i \in I} R/I) \otimes _ R M \to N \otimes _ R M \to 0
But since K, \bigoplus _{i \in I} R/I, and N are all annihilated by I we see that
\begin{align*} K \otimes _ R M & = K \otimes _{R/I} M/IM, \\ (\bigoplus \nolimits _{i \in I} R/I) \otimes _ R M & = (\bigoplus \nolimits _{i \in I} R/I) \otimes _{R/I} M/IM, \\ N \otimes _ R M & = N \otimes _{R/I} M/IM. \end{align*}
As M/IM is flat over R/I we conclude that
0 \to K \otimes _{R/I} M/IM \to (\bigoplus \nolimits _{i \in I} R/I) \otimes _{R/I} M/IM \to N \otimes _{R/} M/IM \to 0
is exact. Combining this with the above we conclude that \text{Tor}_1^ R(N, M) = 0 for any R-module N annihilated by I.
Let us prove (2) by induction on m. The case m = 1 was done in the previous paragraph. For N annihilated by I^ m for m > 1 we may choose an exact sequence 0 \to N' \to N \to N'' \to 0 with N' and N'' annihilated by I^{m - 1}. For example one can take N' = IN and N'' = N/IN. Then the exact sequence
\text{Tor}_1^ R(N', M) \to \text{Tor}_1^ R(N, M) \to \text{Tor}_1^ R(N'', M)
and induction prove the vanishing we want.
Finally, we prove (1). Given n \geq 1 we have to show that M/I^ nM is flat over R/I^ n. In other words, we have to show that the functor N \mapsto N \otimes _{R/I^ n} M/I^ nM is exact on the category of R-modules N annihilated by I^ n. However, for such N we have N \otimes _{R/I^ n} M/I^ nM = N \otimes _ R M. By the vanishing of \text{Tor}_1 in (2) we see that the functor N \mapsto N \otimes _ R M is exact on the category of N annihilated by some power of I and we conclude.
\square
Comments (3)
Comment #8472 by Et on
Comment #8730 by Kentaro Inoue on
Comment #9089 by Stacks project on