Lemma 51.4.1. Let $I \subset A$ be a finitely generated ideal of a ring $A$. Set $Y = V(I) \subset X = \mathop{\mathrm{Spec}}(A)$. Let $d \geq -1$ be an integer. The following are equivalent

1. $H^ i_ Y(A) = 0$ for $i > d$,

2. $H^ i_ Y(M) = 0$ for $i > d$ for every $A$-module $M$, and

3. if $d = -1$, then $Y = \emptyset$, if $d = 0$, then $Y$ is open and closed in $X$, and if $d > 0$ then $H^ i(X \setminus Y, \mathcal{F}) = 0$ for $i \geq d$ for every quasi-coherent $\mathcal{O}_{X \setminus Y}$-module $\mathcal{F}$.

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

$H^1_ Y(A) = \mathop{\mathrm{Coker}}(A \to \Gamma (X \setminus Y, \mathcal{O}_{X \setminus Y})) \quad \text{and}\quad H^1_ Y(I) = \mathop{\mathrm{Coker}}(I \to \Gamma (X \setminus Y, \mathcal{O}_{X \setminus Y}))$

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). $\square$

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