Lemma 51.16.4. Let $A$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Let $I \subset A$ be an ideal. If $H^0_{V(I)}(\omega _ A^\bullet ) = 0$, then $\text{cd}(A, I) < \dim (A)$.

**Proof.**
Set $d = \dim (A)$. Let $\mathfrak p_1, \ldots , \mathfrak p_ n \subset A$ be the minimal primes of dimension $d$. Recall that the finite $A$-module $H^{-i}(\omega _ A^\bullet )$ is nonzero only for $i \in \{ 0, \ldots , d\} $ and that the support of $H^{-i}(\omega _ A^\bullet )$ has dimension $\leq i$, see Lemma 51.9.4. Set $\omega _ A = H^{-d}(\omega _ A^\bullet )$. By prime avoidence (Algebra, Lemma 10.14.2) we can find $f \in A$, $f \not\in \mathfrak p_ i$ which annihilates $H^{-i}(\omega _ A^\bullet )$ for $i < d$. Consider the distinguished triangle

See Derived Categories, Remark 13.12.4. By Derived Categories, Lemma 13.12.5 we see that $f^ d$ induces the zero endomorphism of $\tau _{\geq -d + 1}\omega _ A^\bullet $. Using the axioms of a triangulated category, we find a map

whose composition with $\omega _ A[d] \to \omega _ A^\bullet $ is multiplication by $f^ d$ on $\omega _ A[d]$. Thus we conclude that $f^ d$ annihilates $H^ d_{V(I)}(\omega _ A)$. By Lemma 51.16.3 we conlude $H^ d_{V(I)}(\omega _ A) = 0$. Then we conclude by Lemma 51.16.2 and the fact that $(\omega _ A)_{\mathfrak p_ i}$ is nonzero (see for example Dualizing Complexes, Lemma 47.16.11). $\square$

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