The Stacks project

Lemma 21.22.3. Let $I$ be an ideal of a ring $A$. Let $\mathcal{C}$ be a site. Let

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

be an inverse system of sheaves of $A$-modules on $\mathcal{C}$ such that $\mathcal{F}_ n = \mathcal{F}_{n + 1}/I^ n\mathcal{F}_{n + 1}$. Let $p \geq 0$. Assume

\[ \bigoplus \nolimits _{n \geq 0} H^ p(\mathcal{C}, I^ n\mathcal{F}_{n + 1}) \]

satisfies the ascending chain condition as a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-module. Then the limit topology on $M = \mathop{\mathrm{lim}}\nolimits H^ p(\mathcal{C}, \mathcal{F}_ n)$ is the $I$-adic topology.

Proof. Set $F^ n = \mathop{\mathrm{Ker}}(M \to H^ p(\mathcal{C}, \mathcal{F}_ n))$ for $n \geq 1$ and $F^0 = M$. Observe that $I F^ n \subset F^{n + 1}$. In particular $I^ n M \subset F^ n$. Hence the $I$-adic topology is finer than the limit topology. For the converse, we will show that given $n$ there exists an $m \geq n$ such that $F^ m \subset I^ nM$1. We have injective maps

\[ F^ n/F^{n + 1} \longrightarrow H^ p(\mathcal{C}, \mathcal{F}_{n + 1}) \]

whose image is contained in the image of $H^ p(\mathcal{C}, I^ n\mathcal{F}_{n + 1}) \to H^ p(\mathcal{C}, \mathcal{F}_{n + 1})$. Denote

\[ E_ n \subset H^ p(\mathcal{C}, I^ n\mathcal{F}_{n + 1}) \]

the inverse image of $F^ n/F^{n + 1}$. Then $\bigoplus E_ n$ is a graded $\bigoplus I^ n/I^{n + 1}$-submodule of $\bigoplus H^ p(\mathcal{C}, I^ n\mathcal{F}_{n + 1})$ and $\bigoplus E_ n \to \bigoplus F^ n/F^{n + 1}$ is a homomorphism of graded modules; details omitted. By assumption $\bigoplus E_ n$ is generated by finitely many homogeneous elements over $\bigoplus I^ n/I^{n + 1}$. Since $E_ n \to F^ n/F^{n + 1}$ is surjective, we see that the same thing is true of $\bigoplus F^ n/F^{n + 1}$. Hence we can find $r$ and $c_1, \ldots , c_ r \geq 0$ and $a_ i \in F^{c_ i}$ whose images in $\bigoplus F^ n/F^{n + 1}$ generate. Set $c = \max (c_ i)$.

For $n \geq c$ we claim that $I F^ n = F^{n + 1}$. The claim shows that $F^{n + c} = I^ nF^ c \subset I^ nM$ as desired. To prove the claim suppose $a \in F^{n + 1}$. The image of $a$ in $F^{n + 1}/F^{n + 2}$ is a linear combination of our $a_ i$. Therefore $a - \sum f_ i a_ i \in F^{n + 2}$ for some $f_ i \in I^{n + 1 - c_ i}$. Since $I^{n + 1 - c_ i} = I \cdot I^{n - c_ i}$ as $n \geq c_ i$ we can write $f_ i = \sum g_{i, j} h_{i, j}$ with $g_{i, j} \in I$ and $h_{i, j}a_ i \in F^ n$. Thus we see that $F^{n + 1} = F^{n + 2} + IF^ n$. A simple induction argument gives $F^{n + 1} = F^{n + e} + IF^ n$ for all $e > 0$. It follows that $IF^ n$ is dense in $F^{n + 1}$. Choose generators $k_1, \ldots , k_ r$ of $I$ and consider the continuous map

\[ u : (F^ n)^{\oplus r} \longrightarrow F^{n + 1},\quad (x_1, \ldots , x_ r) \mapsto \sum k_ i x_ i \]

(in the limit topology). By the above the image of $(F^ m)^{\oplus r}$ under $u$ is dense in $F^{m + 1}$ for all $m \geq n$. By the open mapping lemma (More on Algebra, Lemma 15.36.5) we find that $u$ is open. Hence $u$ is surjective. Hence $IF^ n = F^{n + 1}$ for $n \geq c$. This concludes the proof. $\square$

[1] In fact, there exist a $c \geq 0$ such that $F^{n + c} \subset I^ nM$ for all $n$.

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