Lemma 10.99.9. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module.

1. If $M/IM$ is flat over $R/I$ and $M \otimes _ R I/I^2 \to IM/I^2M$ is injective, then $M/I^2M$ is flat over $R/I^2$.

2. If $M/IM$ is flat over $R/I$ and $M \otimes _ R I^ n/I^{n + 1} \to I^ nM/I^{n + 1}M$ is injective for $n = 1, \ldots , k$, then $M/I^{k + 1}M$ is flat over $R/I^{k + 1}$.

Proof. The first statement is a consequence of Lemma 10.99.8 applied with $R$ replaced by $R/I^2$ and $M$ replaced by $M/I^2M$ using that

$\text{Tor}_1^{R/I^2}(M/I^2M, R/I) = \mathop{\mathrm{Ker}}(M \otimes _ R I/I^2 \to IM/I^2M),$

see Remark 10.75.9. The second statement follows in the same manner using induction on $n$ to show that $M/I^{n + 1}M$ is flat over $R/I^{n + 1}$ for $n = 1, \ldots , k$. Here we use that

$\text{Tor}_1^{R/I^{n + 1}}(M/I^{n + 1}M, R/I) = \mathop{\mathrm{Ker}}(M \otimes _ R I^ n/I^{n + 1} \to I^ nM/I^{n + 1}M)$

for every $n$. $\square$

Comment #8473 by Et on

Where is the induction here specifically done? Wouldn't it be enough just to replace R by R/I^k+1, M by M/I^k+1 and I by I/I^k+1 and apply the precceding lemma?

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).