The Stacks project

Lemma 10.101.4. Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

  1. $M/IM$ is flat over $R/I$, and

  2. $R' \otimes _ R M$ is flat over $R'$.

Set $I_2 = \varphi ^{-1}(\varphi (I^2)R')$. Then $M/I_2M$ is flat over $R/I_2$.

Proof. We may replace $R$, $M$, and $R'$ by $R/I_2$, $M/I_2M$, and $R'/\varphi (I)^2R'$. Then $I^2 = 0$ and $\varphi $ is injective. By Lemma 10.99.8 and the fact that $I^2 = 0$ it suffices to prove that $\text{Tor}^ R_1(R/I, M) = K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is zero. Set $M' = M \otimes _ R R'$ and $I' = IR'$. By assumption the map $I' \otimes _{R'} M' \to M'$ is injective. Hence $K$ maps to zero in

\[ I' \otimes _{R'} M' = I' \otimes _ R M = I' \otimes _{R/I} M/IM. \]

Then $I \to I'$ is an injective map of $R/I$-modules. Since $M/IM$ is flat over $R/I$ the map

\[ I \otimes _{R/I} M/IM \longrightarrow I' \otimes _{R/I} M/IM \]

is injective. This implies that $K$ is zero in $I \otimes _ R M = I \otimes _{R/I} M/IM$ as desired. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 10.101: Flatness criteria over Artinian rings

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Your email address will not be published.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 051I. Beware of the difference between the letter 'O' and the digit '0'.