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). $\square$