The Stacks project

Proof. Observe that $R\Gamma _ Y(-)$ has finite cohomological dimension by Dualizing Complexes, Lemma 47.9.1 for example. Hence there exists an integer $i_0$ such that $H^ i_ Y(M) = 0$ for all $A$-modules $M$ and $i \geq i_0$.

Let us prove that (1) and (2) are equivalent. It is immediate that (2) implies (1). Assume (1). By descending induction on $i > d$ we will show that $H^ i_ Y(M) = 0$ for all $A$-modules $M$. For $i \geq i_0$ we have seen this above. To do the induction step, let $i_0 > i > d$. Choose any $A$-module $M$ and fit it into a short exact sequence $0 \to N \to F \to M \to 0$ where $F$ is a free $A$-module. Since $R\Gamma _ Y$ is a right adjoint, we see that $H^ i_ Y(-)$ commutes with direct sums. Hence $H^ i_ Y(F) = 0$ as $i > d$ by assumption (1). Then we see that $H^ i_ Y(M) = H^{i + 1}_ Y(N) = 0$ as desired.

Assume $d = -1$ and (2) holds. Then $0 = H^0_ Y(A/I) = A/I \Rightarrow A = I \Rightarrow Y = \emptyset $. Thus (3) holds. We omit the proof of the converse.

Assume $d = 0$ and (2) holds. Set $J = H^0_ I(A) = \{ x \in A \mid I^ nx = 0 \text{ for some }n > 0\} $. Then

and the kernel of the first map is equal to $J$. See Lemma 51.2.2. We conclude from (2) that $I(A/J) = A/J$. Thus we may pick $f \in I$ mapping to $1$ in $A/J$. Then $1 - f \in J$ so $I^ n(1 - f) = 0$ for some $n > 0$. Hence $f^ n = f^{n + 1}$. Then $e = f^ n \in I$ is an idempotent. Consider the complementary idempotent $e' = 1 - f^ n \in J$. For any element $g \in I$ we have $g^ m e' = 0$ for some $m > 0$. Thus $I$ is contained in the radical of ideal $(e) \subset I$. This means $Y = V(I) = V(e)$ is open and closed in $X$ as predicted in (3). Conversely, if $Y = V(I)$ is open and closed, then the functor $H^0_ Y(-)$ is exact and has vanshing higher derived functors.

If $d > 0$, then we see immediately from Lemma 51.2.2 that (2) is equivalent to (3). $\square$

Proof. The explicit description for $R\Gamma _ Y(-)$ given in Dualizing Complexes, Lemma 47.9.1 shows that (1) is true. We can deduce (2) from (1) using the fact that $R\Gamma _ Z$ depends only on the closed subset $Z$ and not on the choice of the finitely generated ideal $I \subset A$ with $V(I) = Z$. This follows either from the construction of local cohomology in Dualizing Complexes, Section 47.9 combined with More on Algebra, Lemma 15.88.6 or it follows from Lemma 51.2.1. To see (3) we use Lemma 51.4.1 and the vanishing result of Cohomology of Schemes, Lemma 30.4.2. $\square$

Proof. Use the definition and Dualizing Complexes, Lemma 47.9.6. $\square$

Proof. Use the definition and Dualizing Complexes, Lemma 47.9.3. $\square$

Proof. Let $Y = V(I)$ and $Y' = V(I_\mathfrak p) \subset \mathop{\mathrm{Spec}}(A_\mathfrak p)$. Recall that $R\Gamma _ Y(A) \otimes _ A A_\mathfrak p = R\Gamma _{Y'}(A_\mathfrak p)$ by Dualizing Complexes, Lemma 47.9.3. Thus we conclude by Algebra, Lemma 10.23.1. $\square$

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. $\square$

Proof. This follows from Lemma 51.4.1 and Cohomology of Schemes, Lemma 30.3.1. $\square$

Proof. By one of the characterizations of dimension, there exists an ideal of definition for $A$ generated by $d$ elements, see Algebra, Proposition 10.60.9. Hence $\text{cd}(A, \mathfrak m) \leq d$ by Lemma 51.4.3. Thus $H^ d_\mathfrak m(A)$ is nonzero if and only if $\text{cd}(A, \mathfrak m) = d$ if and only if $\text{cd}(A, \mathfrak m) \geq d$.

Let $A \to A^\wedge $ be the map from $A$ to its completion. Observe that $A^\wedge $ is a Noetherian local ring of the same dimension as $A$ with maximal ideal $\mathfrak m A^\wedge $. See Algebra, Lemmas 10.97.6, 10.97.4, and 10.97.3 and More on Algebra, Lemma 15.43.1. By Lemma 51.4.5 it suffices to prove the lemma for $A^\wedge $.

By the previous paragraph we may assume that $A$ is a complete local ring. Then $A$ has a normalized dualizing complex $\omega _ A^\bullet $ (Dualizing Complexes, Lemma 47.22.4). The local duality theorem (in the form of Dualizing Complexes, Lemma 47.18.4) tells us $H^ d_\mathfrak m(A)$ is Matlis dual to $\text{Ext}^{-d}(A, \omega _ A^\bullet ) = H^{-d}(\omega _ A^\bullet )$ which is nonzero for example by Dualizing Complexes, Lemma 47.16.11. $\square$

Proof. By Lemma 51.4.5 we have $\text{cd}(A, I) \geq \text{cd}(A/\mathfrak p, I(A/\mathfrak p))$. Since $V(I) \cap V(\mathfrak p) = \{ \mathfrak m\} $ we have $\text{cd}(A/\mathfrak p, I(A/\mathfrak p)) = \text{cd}(A/\mathfrak p, \mathfrak m/\mathfrak p)$. By Lemma 51.4.9 this is equal to $\dim (A/\mathfrak p)$. $\square$

Proof. Set $U = X \setminus V(I)$. Denote $j : U \to X'$ the canonical open immersion, see Divisors, Section 31.32. Since the exceptional divisor is an effective Cartier divisor (Divisors, Lemma 31.32.4) we see that $j$ is affine, see Divisors, Lemma 31.13.3. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ U$-module. Then $R^ pj_*\mathcal{F} = 0$ for $p > 0$, see Cohomology of Schemes, Lemma 30.2.3. On the other hand, we have $R^ qb_*(j_*\mathcal{F}) = 0$ for $q \geq d$ by Limits, Lemma 32.19.2. Thus by the Leray spectral sequence (Cohomology, Lemma 20.13.8) we conclude that $R^ n(b \circ j)_*\mathcal{F} = 0$ for $n \geq d$. Thus $H^ n(U, \mathcal{F}) = 0$ for $n \geq d$ (by Cohomology, Lemma 20.13.6). This means that $\text{cd}(A, I) \leq d$ by definition. $\square$