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 10.104.2. 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. $\square$

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