The Stacks project

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

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

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