Lemma 47.17.5. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Let $d = \dim (A)$ and $\omega _ A = H^{-d}(\omega _ A^\bullet )$. Then

the support of $\omega _ A$ is the union of the irreducible components of $\mathop{\mathrm{Spec}}(A)$ of dimension $d$,

$\omega _ A$ satisfies $(S_2)$, see Algebra, Definition 10.155.1.

**Proof.**
We will use Lemma 47.16.5 without further mention. By Lemma 47.16.11 the support of $\omega _ A$ contains the irreducible components of dimension $d$. Let $\mathfrak p \subset A$ be a prime. By Lemma 47.17.3 the complex $(\omega _ A^\bullet )_{\mathfrak p}[-\dim (A/\mathfrak p)]$ is a normalized dualizing complex for $A_\mathfrak p$. Hence if $\dim (A/\mathfrak p) + \dim (A_\mathfrak p) < d$, then $(\omega _ A)_\mathfrak p = 0$. This proves the support of $\omega _ A$ is the union of the irreducible components of dimension $d$, because the complement of this union is exactly the primes $\mathfrak p$ of $A$ for which $\dim (A/\mathfrak p) + \dim (A_\mathfrak p) < d$ as $A$ is catenary (Lemma 47.17.4). On the other hand, if $\dim (A/\mathfrak p) + \dim (A_\mathfrak p) = d$, then

\[ (\omega _ A)_\mathfrak p = H^{-\dim (A_\mathfrak p)}\left( (\omega _ A^\bullet )_{\mathfrak p}[-\dim (A/\mathfrak p)] \right) \]

Hence in order to prove $\omega _ A$ has $(S_2)$ it suffices to show that the depth of $\omega _ A$ is at least $\min (\dim (A), 2)$. We prove this by induction on $\dim (A)$. The case $\dim (A) = 0$ is trivial.

Assume $\text{depth}(A) > 0$. Choose a nonzerodivisor $f \in \mathfrak m$ and set $B = A/fA$. Then $\dim (B) = \dim (A) - 1$ and we may apply the induction hypothesis to $B$. By Lemma 47.16.10 we see that multiplication by $f$ is injective on $\omega _ A$ and we get $\omega _ A/f\omega _ A \subset \omega _ B$. This proves the depth of $\omega _ A$ is at least $1$. If $\dim (A) > 1$, then $\dim (B) > 0$ and $\omega _ B$ has depth $ > 0$. Hence $\omega _ A$ has depth $> 1$ and we conclude in this case.

Assume $\dim (A) > 0$ and $\text{depth}(A) = 0$. Let $I = A[\mathfrak m^\infty ]$ and set $B = A/I$. Then $B$ has depth $\geq 1$ and $\omega _ A = \omega _ B$ by Lemma 47.16.9. Since we proved the result for $\omega _ B$ above the proof is done.
$\square$

## Comments (3)

Comment #3402 by Manoj Kummini on

Comment #3403 by Manoj Kummini on

Comment #3465 by Johan on