The Stacks project

Lemma 51.22.2. Let $0 \to K \to L \to M \to 0$ be a short exact sequence of $A$-modules such that $K$ and $L$ are annihilated by $I^ n$ and $M$ is an $(A, n, c)$-module. Then the kernel of $p^*K \to p^*L$ is scheme theoretically supported on $Y_ c$.

Proof. Let $\mathop{\mathrm{Spec}}(B) \subset X$ be an affine open. The restriction of the exact sequence over $\mathop{\mathrm{Spec}}(B)$ corresponds to the sequence of $B$-modules

\[ K \otimes _ A B \to L \otimes _ A B \to M \otimes _ A B \to 0 \]

which is isomorphismic to the sequence

\[ K \otimes _{A/I^ n} B/I^ nB \to L \otimes _{A/I^ n} B/I^ nB \to M \otimes _{A/I^ n} B/I^ nB \to 0 \]

Hence the kernel of the first map is the image of the module $\text{Tor}_1^{A/I^ n}(M, B/I^ nB)$. Recall that the exceptional divisor $Y$ is cut out by $I\mathcal{O}_ X$. Hence it suffices to show that $\text{Tor}_1^{A/I^ n}(M, B/I^ nB)$ is annihilated by $I^ c$. Since multiplication by $a \in I^ c$ on $M$ factors through a finite free $A/I^ n$-module, this is clear. $\square$

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Your email address will not be published.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 0GA8. Beware of the difference between the letter 'O' and the digit '0'.