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

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

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

$\text{depth}(M)$ 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 $M$.
$\square$

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