Lemma 47.11.6. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $M$ be an $A$-module. Let $Z = V(I)$. Then $H^0_ I(M) = H^0_ Z(M)$. Let $N$ be the common value and set $M' = M/N$. Then

1. $H^0_ I(M') = 0$ and $H^ p_ I(M) = H^ p_ I(M')$ and $H^ p_ I(N) = 0$ for all $p > 0$,

2. $H^0_ Z(M') = 0$ and $H^ p_ Z(M) = H^ p_ Z(M')$ and $H^ p_ Z(N) = 0$ for all $p > 0$.

Proof. By definition $H^0_ I(M) = M[I^\infty ]$ is $I$-power torsion. By Lemma 47.9.1 we see that

$H^0_ Z(M) = \mathop{\mathrm{Ker}}(M \longrightarrow M_{f_1} \times \ldots \times M_{f_ r})$

if $I = (f_1, \ldots , f_ r)$. Thus $H^0_ I(M) \subset H^0_ Z(M)$ and conversely, if $x \in H^0_ Z(M)$, then it is annihilated by a $f_ i^{e_ i}$ for some $e_ i \geq 1$ hence annihilated by some power of $I$. This proves the first equality and moreover $N$ is $I$-power torsion. By Lemma 47.8.1 we see that $R\Gamma _ I(N) = N$. By Lemma 47.9.1 we see that $R\Gamma _ Z(N) = N$. This proves the higher vanishing of $H^ p_ I(N)$ and $H^ p_ Z(N)$ in (1) and (2). The vanishing of $H^0_ I(M')$ and $H^0_ Z(M')$ follow from the preceding remarks and the fact that $M'$ is $I$-power torsion free by More on Algebra, Lemma 15.88.4. The equality of higher cohomologies for $M$ and $M'$ follow immediately from the long exact cohomology sequence. $\square$

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