The Stacks project

52.3 Formal sections, II

We suggest looking at Cohomology, Sections 20.36 and 20.39 first.

Lemma 52.3.1. Let $X$ be a scheme. Let $f \in \Gamma (X, \mathcal{O}_ X)$. Let

\[ \ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1 \]

be an inverse system of quasi-coherent $\mathcal{O}_ X$-modules. The following are equivalent

  1. for all $n \geq 1$ the map $f : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ to give a short exact sequence $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$,

  2. for all $n \geq 1$ the map $f^ n : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_1$ to give a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$

  3. there exists an $\mathcal{O}_ X$-module $\mathcal{G}$ which is $f$-divisible such that $\mathcal{F}_ n = \mathcal{G}[f^ n]$.

  4. there exists an $\mathcal{O}_ X$-module $\mathcal{F}$ which is $f$-torsion free such that $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$.

Proof. The equivalence of (1), (2), (3) and the implication (4) $\Rightarrow $ (1) are proven in Cohomology, Lemma 20.36.1. Assume (1) holds. Set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. By Lemma 52.2.1 part (4) we have $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$. Let $U \subset X$ be open and $s = (s_ n) \in \mathcal{F}(U) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U)$. Choose $n \geq 1$. If $fs = 0$, then $s_{n + 1}$ is in the kernel of $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ by condition (1). Hence $s_ n = 0$. Since $n$ was arbitrary, we see $s = 0$. Thus $\mathcal{F}$ is $f$-torsion free. $\square$

reference

Lemma 52.3.2. Let $A$ be a ring and $f \in A$. Let $X$ be a scheme over $A$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume that $\mathcal{F}[f^ n] = \mathop{\mathrm{Ker}}(f^ n : \mathcal{F} \to \mathcal{F})$ stabilizes. Then

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

where the right hand side indicates the derived completion with respect to the ideal $(f) \subset A$. Consequently, for $p \in \mathbf{Z}$ we obtain a commutative diagram

\[ \xymatrix{ & 0 & 0 \\ 0 \ar[r] & \widehat{H^ p(X, \mathcal{F})} \ar[r] \ar[u] & \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}/f^ n\mathcal{F}) \ar[r] \ar[u] & T_ f(H^{p + 1}(X, \mathcal{F})) \ar[r] & 0 \\ 0 \ar[r] & H^0(H^ p(X, \mathcal{F})^\wedge ) \ar[r] \ar[u] & H^ p(X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F}) \ar[r] \ar[u] & T_ f(H^{p + 1}(X, \mathcal{F})) \ar[r] \ar@{=}[u] & 0 \\ & R^1\mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F})[f^ n] \ar[u] \ar[r]^\cong & R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}/f^ n\mathcal{F}) \ar[u] \\ & 0 \ar[u] & 0 \ar[u] } \]

with exact rows and columns where $\widehat{H^ p(X, \mathcal{F})} = \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F})/f^ n H^ p(X, \mathcal{F})$ is the usual $f$-adic completion and $T_ f(-)$ denotes the $f$-adic Tate module as in More on Algebra, Example 15.93.5.

Proof. By Lemma 52.2.1 we have $\mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F} = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n \mathcal{F}$. Everything else follows from Cohomology, Example 20.39.3. $\square$


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