The Stacks project

Proof. Write $M_ n = H^ p(X, I^ n\mathcal{F})$ for $n \geq 1$ and $M_0 = H^ p(X, \mathcal{F})$ so that we have maps $\ldots \to M_3 \to M_2 \to M_1 \to M_0$. Setting $B = \bigoplus _{n \geq 0} I^ n$, then $M = \bigoplus _{n \geq 0} M_ n$ is a finite graded $B$-module, see Lemma 30.20.1. Observe that the products $B_ n \otimes M_ m \to M_{m + n}$, $a \otimes m \mapsto a \cdot m$ are compatible with the maps in our inverse system in the sense that the diagrams

commute for $n, m' \geq 0$ and $m \geq m'$.

Proof of (1). Choose $d_1, \ldots , d_ t \geq 0$ and $x_ i \in M_{d_ i}$ such that $M$ is generated by $x_1, \ldots , x_ t$ over $B$. For any $c \geq \max \{ d_ i\}$ we conclude that $B_{n - c} \cdot M_ c = M_ n$ for $n \geq c$ and we conclude (1) is true.

Proof of (2). Let $c$ be as in the proof of (1). Let $n + m \geq c$. We have $M_{n + m} = B_{n + m - c} \cdot M_ c$. If $c > n$ then we use $M_ c \to M_ n$ and the compatibility of products with transition maps pointed out above to conclude that the image of $M_{n + m} \to M_ n$ is contained in $I^{n + m - c}M_ n$. If $c \leq n$, then we write $M_{n + m} = B_ m \cdot B_{n - c} \cdot M_ c = B_ m \cdot M_ n$ to see that the image is contained in $I^ m M_ n$. This proves (2).

Let $K_ n \subset M_ n$ be the kernel of the map $M_ n \to M_0$. The compatibility of products with transition maps pointed out above shows that $K = \bigoplus K_ n \subset M$ is a graded $B$-submodule. As $B$ is Noetherian and $M$ is a finitely generated graded $B$-module, this shows that $K$ is a finitely generated graded $B$-module. Choose $d'_1, \ldots , d'_{t'} \geq 0$ and $y_ i \in K_{d'_ i}$ such that $K$ is generated by $y_1, \ldots , y_{t'}$ over $B$. Set $c = \max (d'_ i, d'_ j)$. Since $y_ i \in \mathop{\mathrm{Ker}}(M_{d'_ i} \to M_0)$ we see that $B_ n \cdot y_ i \subset \mathop{\mathrm{Ker}}(M_{n + d'_ i} \to M_ n)$. In this way we see that $K_ n = \mathop{\mathrm{Ker}}(M_ n \to M_{n - c})$ for $n \geq c$. This proves (3).

Consider the following commutative solid diagram

Since the kernel of the surjective arrow $I^ n \otimes _ A M_0 \to I^ nM_0$ maps into $K_ n$ by the above we obtain the dotted arrow and the composition $I^ nM_0 \to M_{n - c} \to M_0$ is the canonical map. Then clearly the composition $I^ nM_0 \to M_{n - c} \to I^{n - 2c}M_0$ is the canonical map for $n \geq 2c$. Consider the composition $M_ n \to I^{n - c}M_0 \to M_{n - 2c}$. The first map sends an element of the form $a \cdot m$ with $a \in I^{n - c}$ and $m \in M_ c$ to $a m'$ where $m'$ is the image of $m$ in $M_0$. Then the second map sends this to $a \cdot m'$ in $M_{n - 2c}$ and we see (4) is true.

Part (5) is an immediate consequence of (4) and the definition of morphisms of pro-objects. $\square$