
Lemma 49.2.2. Let $I$ be an ideal of a ring $A$. Let $X$ be a scheme over $\mathop{\mathrm{Spec}}(A)$. Let

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

be an inverse system of $\mathcal{O}_ X$-modules such that $\mathcal{F}_ n = \mathcal{F}_{n + 1}/I^ n\mathcal{F}_{n + 1}$. Given $n$ define

$H^1_ n = \bigcap \nolimits _{m \geq n} \mathop{\mathrm{Im}}\left( H^1(X, I^ n\mathcal{F}_{m + 1}) \to H^1(X, I^ n\mathcal{F}_{n + 1}) \right)$

If $\bigoplus H^1_ n$ satisfies the ascending chain condition as a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-module, then the inverse system $M_ n = \Gamma (X, \mathcal{F}_ n)$ satisfies the Mittag-Leffler condition.

Proof. The proof is exactly the same as the proof of Lemma 49.2.1. In fact, the result will follow from the arguments given there as soon as we show that $\bigoplus H^1_ n$ is a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-submodule of $\bigoplus H^1(X, I^ n\mathcal{F}_{n + 1})$ and that the boundary maps $\delta _ n$ have image contained in $H^1_ n$.

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

$\xymatrix{ 0 \ar[r] & I^ n\mathcal{F}_{m + 1} \ar[d]_ f \ar[r] & \mathcal{F}_{m + 1} \ar[d]_ f \ar[r] & \mathcal{F}_ n \ar[d]_ f \ar[r] & 0 \\ 0 \ar[r] & I^{n + k}\mathcal{F}_{m + 1} \ar[r] & \mathcal{F}_{m + 1} \ar[r] & \mathcal{F}_{n + k} \ar[r] & 0 }$

constructed as in the proof of Lemma 49.2.1. We get an induced map on cohomology and we see that $f \xi ' \in H^1(X, I^{n + k}\mathcal{F}_{m + 1})$ maps to $f \xi$. Since this is true for all $m \gg n + k$ we see that $f\xi$ is in $H^1_{n + k}$ as desired.

To see the boundary maps $\delta _ n$ have image contained in $H^1_ n$ we consider the diagrams

$\xymatrix{ 0 \ar[r] & I^ n\mathcal{F}_{m + 1} \ar[d] \ar[r] & \mathcal{F}_{m + 1} \ar[d] \ar[r] & \mathcal{F}_ n \ar[d] \ar[r] & 0 \\ 0 \ar[r] & I^ n\mathcal{F}_{n + 1} \ar[r] & \mathcal{F}_{n + 1} \ar[r] & \mathcal{F}_ n \ar[r] & 0 }$

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

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