The Stacks project

Lemma 15.89.2. Assume $(\varphi : R \to S, I)$ satisfies the equivalent conditions of Lemma 15.89.1. The following are equivalent

  1. for any $I$-power torsion module $M$, the natural map $M \to M \otimes _ R S$ is an isomorphism, and

  2. $R/I \to S/IS$ is an isomorphism.

Proof. The implication (1) $\Rightarrow $ (2) is immediate. Assume (2). First assume that $M$ is annihilated by $I$. In this case, $M$ is an $R/I$-module. Hence, we have an isomorphism

\[ M \otimes _ R S = M \otimes _{R/I} S/IS = M \otimes _{R/I} R/I = M \]

proving the claim. Next we prove by induction that $M \to M \otimes _ R S$ is an isomorphism for any module $M$ is annihilated by $I^ n$. Assume the induction hypothesis holds for $n$ and assume $M$ is annihilated by $I^{n + 1}$. Then we have a short exact sequence

\[ 0 \to I^ nM \to M \to M/I^ nM \to 0 \]

and as $R \to S$ is flat this gives rise to a short exact sequence

\[ 0 \to I^ nM \otimes _ R S \to M \otimes _ R S \to M/I^ nM \otimes _ R S \to 0 \]

Using that the canonical map is an isomorphism for $M' = I^ nM$ and $M'' = M/I^ nM$ (by induction hypothesis) we conclude the same thing is true for $M$. Finally, suppose that $M$ is a general $I$-power torsion module. Then $M = \bigcup M_ n$ where $M_ n$ is annihilated by $I^ n$ and we conclude using that tensor products commute with colimits. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 15.89: Formal glueing of module categories

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Your email address will not be published.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 05E9. Beware of the difference between the letter 'O' and the digit '0'.