The Stacks project

Lemma 15.102.2. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $N$ be an $A$-module annihilated by $I$. There exists an integer $n > 0$ such that $\text{Tor}^ A_ p(I^ nM, N) \to \text{Tor}^ A_ p(M, N)$ is zero for all $p \geq 0$.

Proof. By Lemma 15.101.7 we can factor $I^ nM \to M$ as $I^ nM \to M \otimes _ A^\mathbf {L} I \to M$. We claim the composition

\[ I^ nM \otimes _ A^\mathbf {L} N \to (M \otimes _ A^\mathbf {L} I) \otimes _ A^\mathbf {L} N \to M \otimes _ A^\mathbf {L} N \]

is zero. Namely, the diagram

\[ \xymatrix{ (M \otimes _ A^\mathbf {L} I) \otimes _ A^\mathbf {L} N \ar[rr] \ar[rd] & & M \otimes _ A^\mathbf {L} (I \otimes _ A^\mathbf {L} N) \ar[ld] \\ & M \otimes _ A^\mathbf {L} N } \]

commutes (details omitted) and the map $I \otimes _ A^\mathbf {L} N \to N$ is zero as $N$ is annihilated by $I$. $\square$

Comments (0)

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Your email address will not be published.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