The Stacks project

Proof. Set $N_ n = H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ and let $\delta _ n : M_ n \to N_ n$ be the boundary map on cohomology coming from the short exact sequence $0 \to I^ n\mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$. Then $\bigoplus \mathop{\mathrm{Im}}(\delta _ n) \subset \bigoplus N_ n$ is a graded submodule. Namely, if $s \in M_ n$ and $f \in I^ m$, then we have a commutative diagram

The middle vertical map is given by lifting a local section of $\mathcal{F}_{n + 1}$ to a section of $\mathcal{F}_{n + m + 1}$ and then multiplying by $f$; similarly for the other vertical arrows. We conclude that $\delta _{n + m}(fs) = f \delta _ n(s)$. By assumption we can find $s_ j \in M_{n_ j}$, $j = 1, \ldots , N$ such that $\delta _{n_ j}(s_ j)$ generate $\bigoplus \mathop{\mathrm{Im}}(\delta _ n)$ as a graded module. Let $n > c = \max (n_ j)$. Let $s \in M_ n$. Then we can find $f_ j \in I^{n - n_ j}$ such that $\delta _ n(s) = \sum f_ j \delta _{n_ j}(s_ j)$. We conclude that $\delta (s - \sum f_ j s_ j) = 0$, i.e., we can find $s' \in M_{n + 1}$ mapping to $s - \sum f_ js_ j$ in $M_ n$. It follows that

Namely, the elements $f_ js_ j$ map to zero in $M_{n - c}$. This proves the lemma. $\square$

Proof. The proof is exactly the same as the proof of Lemma 20.35.1. In fact, the result will follow from the arguments given there as soon as we show that $\bigoplus N_ n$ is a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-submodule of $\bigoplus H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ and that the boundary maps $\delta _ n : M_ n \to H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ have image contained in $N_ n$.

Suppose that $\xi \in N_ n$ and $f \in I^ k$. Choose $m \gg n + k$. Choose $\xi ' \in H^{p + 1}(X, I^ n\mathcal{F}_{m + 1})$ lifting $\xi$. We consider the diagram

constructed as in the proof of Lemma 20.35.1. We get an induced map on cohomology and we see that $f \xi ' \in H^{p + 1}(X, I^{n + k}\mathcal{F}_{m + 1})$ maps to $f \xi$. Since this is true for all $m \gg n + k$ we see that $f\xi$ is in $N_{n + k}$ as desired.

To see the boundary maps $\delta _ n$ have image contained in $N_ n$ we consider the diagrams

for $m \geq n$. Looking at the induced maps on cohomology we conclude. $\square$

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 $F^ m \subset I^ nM$³. We have injective maps

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

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

(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. $\square$

Proof. The proof is exactly the same as the proof of Lemma 20.35.3. In fact, the result will follow from the arguments given there as soon as we show that $\bigoplus N_ n$ is a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-submodule of $\bigoplus H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ and that $F^ n/F^{n + 1} \subset H^ p(X, \mathcal{F}_{n + 1})$ is contained in the image of $N_ n \to H^ p(X, \mathcal{F}_{n + 1})$. In the proof of Lemma 20.35.2 we have seen the statement on the module structure.

Let $t \in F^ n$. Choose an element $s \in H^ p(X, I^ n\mathcal{F}_{n + 1})$ which maps to the image of $t$ in $H^ p(X, \mathcal{F}_{n + 1})$. We have to show that $s$ is in $N_ n$. Now $F^ n$ is the kernel of the map from $M \to H^ p(X, \mathcal{F}_ n)$ hence for all $m \geq n$ we can map $t$ to an element $t_ m \in H^ p(X, \mathcal{F}_{m + 1})$ which maps to zero in $H^ p(X, \mathcal{F}_ n)$. Consider the cohomology sequence

coming from the short exact sequence $0 \to I^ n\mathcal{F}_{m + 1} \to \mathcal{F}_{m + 1} \to \mathcal{F}_ n \to 0$. We can choose $s_ m \in H^ p(X, I^ n\mathcal{F}_{m + 1})$ mapping to $t_ m$. Comparing the sequence above with the one for $m = n$ we see that $s_ m$ maps to $s$ up to an element in the image of $H^{p - 1}(X, \mathcal{F}_ n) \to H^ p(X, I^ n\mathcal{F}_{n + 1})$. However, this map factors through the map $H^ p(X, I^ n\mathcal{F}_{m + 1}) \to H^ p(X, I^ n\mathcal{F}_{n + 1})$ and we see that $s$ is in the image as desired. $\square$