The Stacks project

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 $n$. $\square$


Comments (1)

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?


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AS8. Beware of the difference between the letter 'O' and the digit '0'.