The Stacks project

Lemma 52.12.1. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $\mathcal{F}$ be a coherent module on $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume

  1. $A$ is $I$-adically complete and has a dualizing complex,

  2. if $x \in \text{Ass}(\mathcal{F})$, $x \not\in V(I)$, $\overline{\{ x\} } \cap V(I) \not\subset V(\mathfrak a)$ and $z \in \overline{\{ x\} } \cap V(\mathfrak a)$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$,

  3. one of the following holds:

    1. the restriction of $\mathcal{F}$ to $U \setminus V(I)$ is $(S_1)$

    2. the dimension of $V(\mathfrak a)$ is at most $2$1.

Then we obtain an isomorphism

\[ \mathop{\mathrm{colim}}\nolimits H^0(V, \mathcal{F}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F}) \]

where the colimit is over opens $V \subset U$ containing $U \cap V(I)$.

Proof. Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction to $U$ of the coherent module associated to $M$, see Local Cohomology, Lemma 51.8.2. Set $d = \text{cd}(A, I)$. Let $\mathfrak p$ be a prime of $A$ not contained in $V(I)$ and let $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$. Then either $\mathfrak p$ is not an associated prime of $M$ and hence $\text{depth}(M_\mathfrak p) \geq 1$ or we have $\dim ((A/\mathfrak p)_\mathfrak q) > d + 1$ by (2). Thus the hypotheses of Lemma 52.8.5 are satisfied for $s = 1$ and $d$; here we use condition (3). Thus we find there exists an ideal $J_0 \subset \mathfrak a$ with $V(J_0) \cap V(I) = V(\mathfrak a)$ such that for any $J \subset J_0$ with $V(J) \cap V(I) = V(\mathfrak a)$ the maps

\[ H^ i_ J(M) \longrightarrow H^ i(R\Gamma _\mathfrak a(M)^\wedge ) \]

are isomorphisms for $i = 0, 1$. Consider the morphisms of exact triangles

\[ \xymatrix{ R\Gamma _ J(M) \ar[d] \ar[r] & M \ar[r] \ar[d] & R\Gamma (V, \mathcal{F}) \ar[d] \ar[r] & R\Gamma _ J(M)[1] \ar[d] \\ R\Gamma _ J(M)^\wedge \ar[r] & M \ar[r] & R\Gamma (V, \mathcal{F})^\wedge \ar[r] & R\Gamma _ J(M)^\wedge [1] \\ R\Gamma _\mathfrak a(M)^\wedge \ar[r] \ar[u] & M \ar[r] \ar[u] & R\Gamma (U, \mathcal{F})^\wedge \ar[r] \ar[u] & R\Gamma _\mathfrak a(M)^\wedge [1] \ar[u] } \]

where $V = \mathop{\mathrm{Spec}}(A) \setminus V(J)$. Recall that $R\Gamma _\mathfrak a(M)^\wedge \to R\Gamma _ J(M)^\wedge $ is an isomorphism (because $\mathfrak a$, $\mathfrak a + I$, and $J + I$ cut out the same closed subscheme, for example see proof of Lemma 52.8.5). Hence $R\Gamma (U, \mathcal{F})^\wedge = R\Gamma (V, \mathcal{F})^\wedge $. This produces a commutative diagram

\[ \xymatrix{ 0 \ar[r] & H^0_ J(M) \ar[r] \ar[d] & M \ar[r] \ar[d] \ar[r] & \Gamma (V, \mathcal{F}) \ar[d] \ar[r] & H^1_ J(M) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & H^0(R\Gamma _ J(M)^\wedge ) \ar[r] & M \ar[r] & H^0(R\Gamma (V, \mathcal{F})^\wedge ) \ar[r] & H^1(R\Gamma _ J(M)^\wedge ) \ar[r] & 0 \\ 0 \ar[r] & H^0(R\Gamma _\mathfrak a(M)^\wedge ) \ar[r] \ar[u] & M \ar[r] \ar[u] & H^0(R\Gamma (U, \mathcal{F})^\wedge ) \ar[r] \ar[u] & H^1(R\Gamma _\mathfrak a(M)^\wedge ) \ar[r] \ar[u] & 0 } \]

with exact rows and isomorphisms for the lower vertical arrows. Hence we obtain an isomorphism $\Gamma (V, \mathcal{F}) \to H^0(R\Gamma (U, \mathcal{F})^\wedge )$. By Lemmas 52.6.20 and 52.7.2 we have

\[ R\Gamma (U, \mathcal{F})^\wedge = R\Gamma (U, \mathcal{F}^\wedge ) = R\Gamma (U, R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F}) \]

and we find $H^0(R\Gamma (U, \mathcal{F})^\wedge ) = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ by Cohomology, Lemma 20.36.1. $\square$

[1] In the sense that the difference of the maximal and minimal values on $V(\mathfrak a)$ of a dimension function on $\mathop{\mathrm{Spec}}(A)$ is at most $2$.

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 0EIF. Beware of the difference between the letter 'O' and the digit '0'.