The Stacks project

Lemma 47.16.5. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Let $M$ be a finite $A$-module and let $d = \dim (\text{Supp}(M))$. Then

  1. if $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, \omega _ A^\bullet )$ is nonzero, then $i \in \{ -d, \ldots , 0\} $,

  2. the dimension of the support of $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, \omega _ A^\bullet )$ is at most $-i$,

  3. $\text{depth}(M)$ is the smallest integer $\delta \geq 0$ such that $\mathop{\mathrm{Ext}}\nolimits ^{-\delta }_ A(M, \omega _ A^\bullet ) \not= 0$.

Proof. We prove this by induction on $d$. If $d = 0$, this follows from Lemma 47.16.4 and Matlis duality (Proposition 47.7.8) which guarantees that $\mathop{\mathrm{Hom}}\nolimits _ A(M, E)$ is nonzero if $M$ is nonzero.

Assume the result holds for modules with support of dimension $< d$ and that $M$ has depth $> 0$. Choose an $f \in \mathfrak m$ which is a nonzerodivisor on $M$ and consider the short exact sequence

\[ 0 \to M \to M \to M/fM \to 0 \]

Since $\dim (\text{Supp}(M/fM)) = d - 1$ (Algebra, Lemma 10.63.10) we may apply the induction hypothesis. Writing $E^ i = \mathop{\mathrm{Ext}}\nolimits ^ i_ A(M, \omega _ A^\bullet )$ and $F^ i = \mathop{\mathrm{Ext}}\nolimits ^ i_ A(M/fM, \omega _ A^\bullet )$ we obtain a long exact sequence

\[ \ldots \to F^ i \to E^ i \xrightarrow {f} E^ i \to F^{i + 1} \to \ldots \]

By induction $E^ i/fE^ i = 0$ for $i + 1 \not\in \{ -\dim (\text{Supp}(M/fM)), \ldots , -\text{depth}(M/fM)\} $. By Nakayama's lemma (Algebra, Lemma 10.20.1) and Algebra, Lemma 10.72.7 we conclude $E^ i = 0$ for $i \not\in \{ -\dim (\text{Supp}(M)), \ldots , -\text{depth}(M)\} $. Moreover, in the boundary case $i = - \text{depth}(M)$ we deduce that $E^ i$ is nonzero as $F^{i + 1}$ is nonzero by induction. Since $E^ i/fE^ i \subset F^{i + 1}$ we get

\[ \dim (\text{Supp}(F^{i + 1})) \geq \dim (\text{Supp}(E^ i/fE^ i)) \geq \dim (\text{Supp}(E^ i)) - 1 \]

(see lemma used above) we also obtain the dimension estimate (2).

If $M$ has depth $0$ and $d > 0$ we let $N = M[\mathfrak m^\infty ]$ and set $M' = M/N$ (compare with Lemma 47.11.6). Then $M'$ has depth $> 0$ and $\dim (\text{Supp}(M')) = d$. Thus we know the result for $M'$ and since $R\mathop{\mathrm{Hom}}\nolimits _ A(N, \omega _ A^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _ A(N, E)$ (Lemma 47.16.4) the long exact cohomology sequence of $\mathop{\mathrm{Ext}}\nolimits $'s implies the result for $M$. $\square$


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 0A7U. Beware of the difference between the letter 'O' and the digit '0'.