Lemma 15.89.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$, $N$ be finite $A$-modules with $N$ annihilated by $I$. For each $p > 0$ there exists an $n$ such that the map $\mathop{\mathrm{Ext}}\nolimits _ A^ p(M, N) \to \text{Ext}_ A^ p(I^ nM, N)$ is zero.

**Proof.**
The result is clear for $p = 0$ (with $n = 1$). Choose a short exact sequence $0 \to K \to A^{\oplus t} \to M \to 0$. For $n$ pick a short exact sequence $0 \to L \to A^{\oplus s} \to I^ nM \to 0$. It is clear that we can construct a map of short exact sequences

such that $A^{\oplus s} \to A^{\oplus t}$ has image in $(I^ n)^{\oplus t}$. By Artin-Rees (Algebra, Lemma 10.50.2) we see that $L \to K$ has image contained in $I^{n - c}K$ if $n \geq c$. At this point the exact sequence

and the corresponding sequence for $\mathop{\mathrm{Ext}}\nolimits ^1_ A(I^ nM, N)$ show that the lemma holds for $p = 1$ with $n = c + 1$. Moreover, we see that the result for $p - 1$ and the module $K$ implies the result for $p$ and the module $M$ by the commutativity of the diagram

for $p > 1$. Some details omitted. $\square$

## 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Your email address will not be published.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.

## Comments (0)