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. $\square$

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## Comments (2)

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