Lemma 51.2.6. Let $I \subset I' \subset A$ be finitely generated ideals of a Noetherian ring $A$. Let $M$ be an $A$-module. Let $i \geq 0$ be an integer. Consider the map

$\Psi : H^ i_{V(I')}(M) \to H^ i_{V(I)}(M)$

The following are true:

1. if $H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for all $\mathfrak p \in V(I) \setminus V(I')$, then $\Psi$ is surjective,

2. if $H^{i - 1}_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for all $\mathfrak p \in V(I) \setminus V(I')$, then $\Psi$ is injective,

3. if $H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = H^{i - 1}_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ for all $\mathfrak p \in V(I) \setminus V(I')$, then $\Psi$ is an isomorphism.

Proof. Proof of (1). Let $\xi \in H^ i_{V(I)}(M)$. Since $A$ is Noetherian, there exists a largest ideal $I \subset I'' \subset I'$ such that $\xi$ is the image of some $\xi '' \in H^ i_{V(I'')}(M)$. If $V(I'') = V(I')$, then we are done. If not, choose a generic point $\mathfrak p \in V(I'')$ not in $V(I')$. Then we have $H^ i_{V(I'')}(M)_\mathfrak p = H^ i_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ by assumption. By Lemma 51.2.5 we can increase $I''$ which contradicts maximality.

Proof of (2). Let $\xi ' \in H^ i_{V(I')}(M)$ be in the kernel of $\Psi$. Since $A$ is Noetherian, there exists a largest ideal $I \subset I'' \subset I'$ such that $\xi '$ maps to zero in $H^ i_{V(I'')}(M)$. If $V(I'') = V(I')$, then we are done. If not, then choose a generic point $\mathfrak p \in V(I'')$ not in $V(I')$. Then we have $H^{i - 1}_{V(I'')}(M)_\mathfrak p = H^{i - 1}_{\mathfrak pA_\mathfrak p}(M_\mathfrak p) = 0$ by assumption. By Lemma 51.2.5 we can increase $I''$ which contradicts maximality.

Part (3) is formal from parts (1) and (2). $\square$

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