The Stacks project

Lemma 10.99.11. Let $R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $S$-module. Assume

  1. $R$ is a Noetherian ring,

  2. $S$ is a Noetherian ring,

  3. $M$ is a finite $S$-module, and

  4. for each $n \geq 1$ the module $M/I^ n M$ is flat over $R/I^ n$.

Then for every $\mathfrak q \in V(IS)$ the localization $M_{\mathfrak q}$ is flat over $R$. In particular, if $S$ is local and $IS$ is contained in its maximal ideal, then $M$ is flat over $R$.

Proof. We are going to use Lemma 10.99.10. By assumption $M/IM$ is flat over $R/I$. Hence it suffices to check that $\text{Tor}_1^ R(M, R/I)$ is zero on localization at $\mathfrak q$. By Remark 10.75.9 this Tor group is equal to $K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$. We know that the kernel of $I/I^ n \otimes _{R/I^ n} M/I^ nM \to M/I^ nM$ is zero for all $n \geq 1$. Hence an element of $K$ maps to zero in $I/I^ n \otimes _{R/I^ n} M/I^ nM$. Since

\[ I/I^ n \otimes _{R/I^ n} M/I^ nM = I/I^ n \otimes _ R M = (I \otimes _ R M)/I^{n - 1}(I \otimes _ R M) \]

we conclude that $K \subset I^{n - 1}(I \otimes _ R M)$ for all $n \geq 1$. By the Artin-Rees lemma, and more precisely Lemma 10.51.5 we conclude that $K_{\mathfrak q} = 0$, as desired. $\square$


Comments (2)

Comment #8476 by Et on

Instead of saying that there is a module map , it would be better to say that the map factors through the surjective map


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Your email address will not be published.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 0523. Beware of the difference between the letter 'O' and the digit '0'.