Lemma 20.35.3. Let $I$ be an ideal of a ring $A$. Let $X$ be a topological space. Let

$\ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1$

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

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

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

Proof. Set $F^ n = \mathop{\mathrm{Ker}}(M \to H^ p(X, \mathcal{F}_ n))$ 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

$F^ n/F^{n + 1} \longrightarrow H^ p(X, \mathcal{F}_{n + 1})$

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

$E_ n \subset H^ p(X, I^ n\mathcal{F}_{n + 1})$

the inverse image of $F^ n/F^{n + 1}$. Then $\bigoplus E_ n$ is a graded $\bigoplus I^ n/I^{n + 1}$-submodule of $\bigoplus H^ p(X, I^ n\mathcal{F}_{n + 1})$ 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 $n$.

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