Lemma 51.4.7. Let $I \subset A$ be a finitely generated ideal of a ring $A$. If $M$ is a finite $A$-module, then $H^ i_{V(I)}(M) = 0$ for $i > \dim (\text{Supp}(M))$. In particular, we have $\text{cd}(A, I) \leq \dim (A)$.

**Proof.**
We first prove the second statement. Recall that $\dim (A)$ denotes the Krull dimension. By Lemma 51.4.6 we may assume $A$ is local. If $V(I) = \emptyset $, then the result is true. If $V(I) \not= \emptyset $, then $\dim (\mathop{\mathrm{Spec}}(A) \setminus V(I)) < \dim (A)$ because the closed point is missing. Observe that $U = \mathop{\mathrm{Spec}}(A) \setminus V(I)$ is a quasi-compact open of the spectral space $\mathop{\mathrm{Spec}}(A)$, hence a spectral space itself. See Algebra, Lemma 10.26.2 and Topology, Lemma 5.23.5. Thus Cohomology, Proposition 20.22.4 implies $H^ i(U, \mathcal{F}) = 0$ for $i \geq \dim (A)$ which implies what we want by Lemma 51.4.1. In the Noetherian case the reader may use Grothendieck's Cohomology, Proposition 20.20.7.

We will deduce the first statement from the second. Let $\mathfrak a$ be the annihilator of the finite $A$-module $M$. Set $B = A/\mathfrak a$. Recall that $\mathop{\mathrm{Spec}}(B) = \text{Supp}(M)$, see Algebra, Lemma 10.40.5. Set $J = IB$. Then $M$ is a $B$-module and $H^ i_{V(I)}(M) = H^ i_{V(J)}(M)$, see Dualizing Complexes, Lemma 47.9.2. Since $\text{cd}(B, J) \leq \dim (B) = \dim (\text{Supp}(M))$ by the first part we conclude. $\square$

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