Proposition 51.16.6. Let $A$ be a Noetherian local ring with completion $A^\wedge $. Let $I \subset A$ be an ideal such that

for every minimal prime $\mathfrak p \subset A^\wedge $ of dimension $\dim (A)$. Then $\text{cd}(A, I) < \dim (A)$.

[Theorem 3.1, CD]

Proposition 51.16.6. Let $A$ be a Noetherian local ring with completion $A^\wedge $. Let $I \subset A$ be an ideal such that

\[ \dim V(IA^\wedge + \mathfrak p) \geq 1 \]

for every minimal prime $\mathfrak p \subset A^\wedge $ of dimension $\dim (A)$. Then $\text{cd}(A, I) < \dim (A)$.

**Proof.**
Since $A \to A^\wedge $ is faithfully flat we have $H^ d_{V(I)}(A) \otimes _ A A^\wedge = H^ d_{V(IA^\wedge )}(A^\wedge )$ by Dualizing Complexes, Lemma 47.9.3. Thus we may assume $A$ is complete.

Assume $A$ is complete. Let $\mathfrak p_1, \ldots , \mathfrak p_ n \subset A$ be the minimal primes of dimension $d$. Consider the complete local ring $A_ i = A/\mathfrak p_ i$. We have $H^ d_{V(I)}(A_ i) = H^ d_{V(IA_ i)}(A_ i)$ by Dualizing Complexes, Lemma 47.9.2. By Lemma 51.16.2 it suffices to prove the lemma for $(A_ i, IA_ i)$. Thus we may assume $A$ is a complete local domain.

Assume $A$ is a complete local domain. We can choose a prime ideal $\mathfrak p \supset I$ with $\dim (A/\mathfrak p) = 1$. By Lemma 51.16.1 it suffices to prove the lemma for $\mathfrak p$.

By Lemma 51.16.4 it suffices to show that $H^0_{V(\mathfrak p)}(\omega _ A^\bullet ) = 0$. Recall that

\[ H^0_{V(\mathfrak p)}(\omega _ A^\bullet ) = \mathop{\mathrm{colim}}\nolimits \text{Ext}^0_ A(A/\mathfrak p^ n, \omega _ A^\bullet ) \]

By Lemma 51.16.5 we see that the colimit is the same as

\[ \mathop{\mathrm{colim}}\nolimits \text{Ext}^0_ A(A/\mathfrak p^{(n)}, \omega _ A^\bullet ) \]

Since $\text{depth}(A/\mathfrak p^{(n)}) = 1$ we see that these ext groups are zero by Lemma 51.9.4 as desired. $\square$

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