## 52.2 Formal sections, I

Let $A$ be a ring and $I \subset A$ an ideal. Let $X$ be a scheme over $\mathop{\mathrm{Spec}}(A)$. In this section we prove some general facts on inverse systems of $\mathcal{O}_ X$-modules $\{ \mathcal{F}_ n\}$ such that $\mathcal{F}_ n = \mathcal{F}_{n + 1} / I^ n \mathcal{F}_{n + 1}$. In particular, we prove two lemmas on the behaviour of the inverse system $\{ H^0(X, \mathcal{F}_ n)\}$. These results have generalizations to higher cohomology groups which we will add here if we need them.

Lemma 52.2.1. 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 $\mathcal{F}_ n = \mathcal{F}_{n + 1}/I^ n\mathcal{F}_{n + 1}$. Assume

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

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

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

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

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

$\mathop{\mathrm{Im}}(M_{n + 1} \to M_{n - c}) = \mathop{\mathrm{Im}}(M_ n \to M_{n - c})$

This proves the lemma. $\square$

Lemma 52.2.2. 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 52.2.1. 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 52.2.1. 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. $\square$

Lemma 52.2.3. Let $I$ be a finitely generated 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 $\mathcal{F}_ n = \mathcal{F}_{n + 1}/I^ n\mathcal{F}_{n + 1}$. Assume

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

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

Proof. Set $F^ n = \mathop{\mathrm{Ker}}(M \to H^0(X, \mathcal{F}_ n))$ for $n \geq 1$ and $F^0 = M$. Observe that $I F^ n \subset F^{n + 1}$. In particular $I^ n M \subset F^ n$ and we are trying to show that given $n$ there exists an $m \geq n$ such that $F^ m \subset I^ nM$. We have an injective map of graded modules

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

By assumption the left hand side is generated by finitely many homogeneous elements. 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 $I F^ n = F^{n + 1}$. Namely, 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. $\square$

Lemma 52.2.4. Let $X$ be a scheme. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. Let

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

be an inverse system of quasi-coherent $\mathcal{O}_ X$-modules such that $\mathcal{F}_ n = \mathcal{F}_{n + 1}/\mathcal{I}^ n\mathcal{F}_{n + 1}$. Set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Then

1. $\mathcal{F} = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$,

2. for any affine open $U \subset X$ we have $H^ p(U, \mathcal{F}) = 0$ for $p > 0$, and

3. for each $p$ there is a short exact sequence $0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}_ n) \to H^ p(X, \mathcal{F}) \to \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}_ n) \to 0$.

If moreover $\mathcal{I}$ is of finite type, then

1. $\mathcal{F}_ n = \mathcal{F}/\mathcal{I}^ n\mathcal{F}$, and

2. $\mathcal{I}^ n \mathcal{F} = \mathop{\mathrm{lim}}\nolimits _{m \geq n} \mathcal{I}^ n\mathcal{F}_ m$.

Proof. Parts (1), (2), and (3) are general facts about inverse systems of quasi-coherent modules with surjective transition maps, see Derived Categories of Schemes, Lemma 36.3.2 and Cohomology, Lemma 20.35.1. Next, assume $\mathcal{I}$ is of finite type. Let $U \subset X$ be affine open. Say $U = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{I}|_ U$ corresponds to $I \subset A$. Observe that $I$ is a finitely generated ideal. By the equivalence of categories between quasi-coherent $\mathcal{O}_ U$-modules and $A$-modules (Schemes, Lemma 26.7.5) we find that $M_ n = \mathcal{F}_ n(U)$ is an inverse system of $A$-modules with $M_ n = M_{n + 1}/I^ nM_{n + 1}$. Thus

$M = \mathcal{F}(U) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U) = \mathop{\mathrm{lim}}\nolimits M_ n$

is an $I$-adically complete module with $M/I^ nM = M_ n$ by Algebra, Lemma 10.98.2. This proves (4). Part (5) translates into the statement that $\mathop{\mathrm{lim}}\nolimits _{m \geq n} I^ nM/I^ mM = I^ nM$. Since $I^ mM = I^{m - n} \cdot I^ nM$ this is just the statement that $I^ mM$ is $I$-adically complete. This follows from Algebra, Lemma 10.96.3 and the fact that $M$ is complete. $\square$

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