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
$H^0_ I(M') = 0$ and $H^ p_ I(M) = H^ p_ I(M')$ and $H^ p_ I(N) = 0$ for all $p > 0$,
$H^0_ Z(M') = 0$ 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
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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)