The Stacks project

Slight generalization of [Lemme 1, Beauville-Laszlo].

Lemma 15.88.8. Let $R \to R'$ be a ring map. Let $I \subset R$ be an ideal such that $R/I^ n \to R'/I^ nR'$ is an isomorphism for $n > 0$. For any $I$-power torsion $R$-module $M$ the map $M \to M \otimes _ R R'$ is an isomorphism. For example, if $I$ is finitely generated and $R^\wedge $ is the completion of $R$ with respect to $I$, then we have $M \cong M \otimes _ R R^\wedge $.

Proof. If $M$ is annihilated by $I^ n$, then

\[ M \otimes _ R R' \cong M \otimes _{R/I^ n} R'/I^ n R' \cong M \otimes _{R/I^ n} R/I^ n \cong M. \]

If $M$ is $I$-power torsion, then $M = \bigcup M[I^ n]$. Since tensor products commute with direct limits (Algebra, Lemma 10.12.9), we obtain the desired isomorphism. The last statement is a special case of the first statement by Algebra, Lemma 10.96.3. $\square$

Comments (4)

Comment #8329 by Rankeya on

Lemma 0BNK works when you replace f by a finitely generated ideal I and take M to be an I-power torsion R-module, right? Can it be stated in this generality?

The consequence also continues to hold because of Tag 05GG.

Comment #8333 by on

Indeed, this is true. We could say that the first part works if is killed by a power of and is an isomorphism modulo any power of without assuming is finitely generated. The conclusion about the -adic completion only (as far as I know) works when is finitely generated by the reference you pointed out. I will make these changes the next time I go through all the comments and I will move this lemma into Section 15.88 because it is more suitable.

Comment #8339 by Rankeya on

Thank you, this will be very helpful! I needed the fact that if is the injective hull of the residue field of a noetherian local ring (or for that matter any Artinian -module), then , but could not find a suitable reference for the last isomorphism.

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Your email address will not be published.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



View Lemma 15.88.8 as pdf