**Proof.**
Observe that $R\Gamma _ Y(-)$ has finite cohomological dimension by Dualizing Complexes, Lemma 47.9.1 for example. Hence we can choose a large integer $N$ such that $H^ i_ Y(M) = 0$ for all $A$-modules $M$.

Let us prove that (1) and (2) are equivalent. It is immediate that (2) implies (1). Assume (1). 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 $H^ i_ Y(F) = 0$ for $i > d$ by assumption (1). Then we see that $H^ i_ Y(M) = H^{i + 1}_ Y(N)$ for all $i > d$. Thus if we've shown the vanishing of $H^ j_ Y(N)$ for some $j > d + 1$ and all $A$-modules $N$, then we obtain the vanishing of $H^{j - 1}_ Y(M)$ for all $A$-modules $M$. By induction we find that (2) is true.

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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