The Stacks project

Lemma 51.2.2. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Set $X = \mathop{\mathrm{Spec}}(A)$, $Z = V(I)$, $U = X \setminus Z$, and $j : U \to X$ the inclusion morphism. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ U$-module. Then

  1. there exists an $A$-module $M$ such that $\mathcal{F}$ is the restriction of $\widetilde{M}$ to $U$,

  2. given $M$ there is an exact sequence

    \[ 0 \to H^0_ Z(M) \to M \to H^0(U, \mathcal{F}) \to H^1_ Z(M) \to 0 \]

    and isomorphisms $H^ p(U, \mathcal{F}) = H^{p + 1}_ Z(M)$ for $p \geq 1$,

  3. we may take $M = H^0(U, \mathcal{F})$ in which case we have $H^0_ Z(M) = H^1_ Z(M) = 0$.

Proof. The existence of $M$ follows from Properties, Lemma 28.22.1 and the fact that quasi-coherent sheaves on $X$ correspond to $A$-modules (Schemes, Lemma 26.7.5). Then we look at the distinguished triangle

\[ R\Gamma _ Z(X, \widetilde{M}) \to R\Gamma (X, \widetilde{M}) \to R\Gamma (U, \widetilde{M}|_ U) \to R\Gamma _ Z(X, \widetilde{M})[1] \]

of Cohomology, Lemma 20.34.5. Since $X$ is affine we have $R\Gamma (X, \widetilde{M}) = M$ by Cohomology of Schemes, Lemma 30.2.2. By our choice of $M$ we have $\mathcal{F} = \widetilde{M}|_ U$ and hence this produces an exact sequence

\[ 0 \to H^0_ Z(X, \widetilde{M}) \to M \to H^0(U, \mathcal{F}) \to H^1_ Z(X, \widetilde{M}) \to 0 \]

and isomorphisms $H^ p(U, \mathcal{F}) = H^{p + 1}_ Z(X, \widetilde{M})$ for $p \geq 1$. By Lemma 51.2.1 we have $H^ i_ Z(M) = H^ i_ Z(X, \widetilde{M})$ for all $i$. Thus (1) and (2) do hold. Finally, setting $M' = H^0(U, \mathcal{F})$ we see that the kernel and cokernel of $M \to M'$ are $I$-power torsion. Therefore $\widetilde{M}|_ U \to \widetilde{M'}|_ U$ is an isomorphism and we can indeed use $M'$ as predicted in (3). It goes without saying that we obtain zero for both $H^0_ Z(M')$ and $H^0_ Z(M')$. $\square$

Comments (0)

Post a comment

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DWR. Beware of the difference between the letter 'O' and the digit '0'.