Lemma 47.17.1. Let $A$ be a Noetherian ring. Let $\mathfrak p$ be a minimal prime of $A$. Then $H^ i(\omega _ A^\bullet )_\mathfrak p$ is nonzero for exactly one $i$.

## 47.17 Dualizing complexes and dimension functions

Our results in the local setting have the following consequence: a Noetherian ring which has a dualizing complex is a universally catenary ring of finite dimension.

**Proof.**
The complex $\omega _ A^\bullet \otimes _ A A_\mathfrak p$ is a dualizing complex for $A_\mathfrak p$ (Lemma 47.15.6). The dimension of $A_\mathfrak p$ is zero as $\mathfrak p$ is minimal. Hence the result follows from Lemma 47.16.8.
$\square$

Let $A$ be a Noetherian ring and let $\omega _ A^\bullet $ be a dualizing complex. Lemma 47.15.12 allows us to define a function

by mapping $\mathfrak p$ to the integer of Lemma 47.15.12 for the dualizing complex $(\omega _ A^\bullet )_\mathfrak p$ over $A_\mathfrak p$ (Lemma 47.15.6) and the residue field $\kappa (\mathfrak p)$. To be precise, we define $\delta (\mathfrak p)$ to be the unique integer such that

is a normalized dualizing complex over the Noetherian local ring $A_\mathfrak p$.

Lemma 47.17.2. Let $A$ be a Noetherian ring and let $\omega _ A^\bullet $ be a dualizing complex. Let $A \to B$ be a surjective ring map and let $\omega _ B^\bullet = R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ be the dualizing complex for $B$ of Lemma 47.15.9. Then we have

**Proof.**
This follows from the definition of the functions and Lemma 47.16.2.
$\square$

Lemma 47.17.3. Let $A$ be a Noetherian ring and let $\omega _ A^\bullet $ be a dualizing complex. The function $\delta = \delta _{\omega _ A^\bullet }$ defined above is a dimension function (Topology, Definition 5.20.1).

**Proof.**
Let $\mathfrak p \subset \mathfrak q$ be an immediate specialization. We have to show that $\delta (\mathfrak p) = \delta (\mathfrak q) + 1$. We may replace $A$ by $A/\mathfrak p$, the complex $\omega _ A^\bullet $ by $\omega _{A/\mathfrak p}^\bullet = R\mathop{\mathrm{Hom}}\nolimits (A/\mathfrak p, \omega _ A^\bullet )$, the prime $\mathfrak p$ by $(0)$, and the prime $\mathfrak q$ by $\mathfrak q/\mathfrak p$, see Lemma 47.17.2. Thus we may assume that $A$ is a domain, $\mathfrak p = (0)$, and $\mathfrak q$ is a prime ideal of height $1$.

Then $H^ i(\omega _ A^\bullet )_{(0)}$ is nonzero for exactly one $i$, say $i_0$, by Lemma 47.17.1. In fact $i_0 = -\delta ((0))$ because $(\omega _ A^\bullet )_{(0)}[-\delta ((0))]$ is a normalized dualizing complex over the field $A_{(0)}$.

On the other hand $(\omega _ A^\bullet )_\mathfrak q[-\delta (\mathfrak q)]$ is a normalized dualizing complex for $A_\mathfrak q$. By Lemma 47.16.11 we see that

is nonzero only for $e = -\dim (A_\mathfrak q) = -1$. We conclude

as desired. $\square$

Lemma 47.17.4. Let $A$ be a Noetherian ring which has a dualizing complex. Then $A$ is universally catenary of finite dimension.

**Proof.**
Because $\mathop{\mathrm{Spec}}(A)$ has a dimension function by Lemma 47.17.3 it is catenary, see Topology, Lemma 5.20.2. Hence $A$ is catenary, see Algebra, Lemma 10.105.2. It follows from Proposition 47.15.11 that $A$ is universally catenary.

Because any dualizing complex $\omega _ A^\bullet $ is in $D^ b_{\textit{Coh}}(A)$ the values of the function $\delta _{\omega _ A^\bullet }$ in minimal primes are bounded by Lemma 47.17.1. On the other hand, for a maximal ideal $\mathfrak m$ with residue field $\kappa $ the integer $i = -\delta (\mathfrak m)$ is the unique integer such that $\mathop{\mathrm{Ext}}\nolimits _ A^ i(\kappa , \omega _ A^\bullet )$ is nonzero (Lemma 47.15.12). Since $\omega _ A^\bullet $ has finite injective dimension these values are bounded too. Since the dimension of $A$ is the maximal value of $\delta (\mathfrak p) - \delta (\mathfrak m)$ where $\mathfrak p \subset \mathfrak m$ are a pair consisting of a minimal prime and a maximal prime we find that the dimension of $\mathop{\mathrm{Spec}}(A)$ is bounded. $\square$

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 10.157.1.

**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

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. $\square$

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