Situation 52.16.1. Here $A$ is a Noetherian ring and $I \subset \mathfrak a \subset A$ are ideals. We set $X = \mathop{\mathrm{Spec}}(A)$, $Y = V(I) = \mathop{\mathrm{Spec}}(A/I)$, and $Z = V(\mathfrak a) = \mathop{\mathrm{Spec}}(A/\mathfrak a)$. Furthermore $U = X \setminus Z$.

## 52.16 Algebraization of coherent formal modules, I

The essential surjectivity of the completion functor (see below) was studied systematically in [SGA2], [MRaynaud-book], and [MRaynaud-paper]. We work in the following affine situation.

In this section we try to find conditions that guarantee an object of $\textit{Coh}(U, I\mathcal{O}_ U)$ is in the image of the completion functor $\textit{Coh}(\mathcal{O}_ U) \to \textit{Coh}(U, I\mathcal{O}_ U)$. See Cohomology of Schemes, Section 30.23 and Section 52.15.

Lemma 52.16.2. In Situation 52.16.1. Consider an inverse system $(M_ n)$ of $A$-modules such that

$M_ n$ is a finite $A$-module,

$M_ n$ is annihilated by $I^ n$,

the kernel and cokernel of $M_{n + 1}/I^ nM_{n + 1} \to M_ n$ are $\mathfrak a$-power torsion.

Then $(\widetilde{M}_ n|_ U)$ is in $\textit{Coh}(U, I\mathcal{O}_ U)$. Conversely, every object of $\textit{Coh}(U, I\mathcal{O}_ U)$ arises in this manner.

**Proof.**
We omit the verification that $(\widetilde{M}_ n|_ U)$ is in $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. By Local Cohomology, Lemma 51.8.2 we see that $\mathcal{F}_ n = \widetilde{M_ n}$ for some finite $A/I^ n$-module $M_ n$. After dividing $M_ n$ by $H^0_\mathfrak a(M_ n)$ we may assume $M_ n \subset H^0(U, \mathcal{F}_ n)$, see Dualizing Complexes, Lemma 47.11.6 and the already referenced lemma. After replacing inductively $M_{n + 1}$ by the inverse image of $M_ n$ under the map $M_{n + 1} \to H^0(U, \mathcal{F}_{n + 1}) \to H^0(U, \mathcal{F}_ n)$, we may assume $M_{n + 1}$ maps into $M_ n$. This gives a inverse system $(M_ n)$ satisfying (1) and (2) such that $\mathcal{F}_ n = \widetilde{M_ n}$. To see that (3) holds, use that $M_{n + 1}/I^ nM_{n + 1} \to M_ n$ is a map of finite $A$-modules which induces an isomorphism after applying $\widetilde{\ }$ and restriction to $U$ (here we use the first referenced lemma one more time).
$\square$

In Situation 52.16.1 we can study the completion functor Cohomology of Schemes, Equation (30.23.3.1)

If $A$ is $I$-adically complete, then this functor is fully faithful on suitable subcategories by our earlier work on algebraization of formal sections, see Section 52.15 and Lemma 52.19.6 for some sample results. Next, let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Still assuming $A$ is $I$-adically complete, we can ask: When is $(\mathcal{F}_ n)$ in the essential image of the completion functor displayed above?

Lemma 52.16.3. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Consider the following conditions:

$(\mathcal{F}_ n)$ is in the essential image of the functor (52.16.2.1),

$(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module,

$(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ V$-module for $U \cap Y \subset V \subset U$ open,

$(\mathcal{F}_ n)$ is the completion of the restriction to $U$ of a coherent $\mathcal{O}_ X$-module,

$(\mathcal{F}_ n)$ is the restriction to $U$ of the completion of a coherent $\mathcal{O}_ X$-module,

there exists an object $(\mathcal{G}_ n)$ of $\textit{Coh}(X, I\mathcal{O}_ X)$ whose restriction to $U$ is $(\mathcal{F}_ n)$.

Then conditions (1), (2), (3), (4), and (5) are equivalent and imply (6). If $A$ is $I$-adically complete then condition (6) implies the others.

**Proof.**
Parts (1) and (2) are equivalent, because the completion of a coherent $\mathcal{O}_ U$-module $\mathcal{F}$ is by definition the image of $\mathcal{F}$ under the functor (52.16.2.1). If $V \subset U$ is an open subscheme containing $U \cap Y$, then we have

since the category of coherent $\mathcal{O}_ V$-modules supported on $V \cap Y$ is the same as the category of coherent $\mathcal{O}_ U$-modules supported on $U \cap Y$. Thus the completion of a coherent $\mathcal{O}_ V$-module is an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Having said this the equivalence of (2), (3), (4), and (5) holds because the functors $\textit{Coh}(\mathcal{O}_ X) \to \textit{Coh}(\mathcal{O}_ U) \to \textit{Coh}(\mathcal{O}_ V)$ are essentially surjective. See Properties, Lemma 28.22.5.

It is always the case that (5) implies (6). Assume $A$ is $I$-adically complete. Then any object of $\textit{Coh}(X, I\mathcal{O}_ X)$ corresponds to a finite $A$-module by Cohomology of Schemes, Lemma 30.23.1. Thus we see that (6) implies (5) in this case. $\square$

Example 52.16.4. Let $k$ be a field. Let $A = k[x, y][[t]]$ with $I = (t)$ and $\mathfrak a = (x, y, t)$. Let us use notation as in Situation 52.16.1. Observe that $U \cap Y = (D(x) \cap Y) \cup (D(y) \cap Y)$ is an affine open covering. For $n \geq 1$ consider the invertible module $\mathcal{L}_ n$ of $\mathcal{O}_ U/t^ n\mathcal{O}_ U$ given by glueing $A_ x/t^ nA_ x$ and $A_ y/t^ nA_ y$ via the invertible element of $A_{xy}/t^ nA_{xy}$ which is the image of any power series of the form

with $a_ n \in k[x, y]$ and $\varphi (n) \in \mathbf{N}$. Then $(\mathcal{L}_ n)$ is an invertible object of $\textit{Coh}(U, I\mathcal{O}_ U)$ which is not the completion of a coherent $\mathcal{O}_ U$-module $\mathcal{L}$. We only sketch the argument and we omit most of the details. Let $y \in U \cap Y$. Then the completion of the stalk $\mathcal{L}_ y$ would be an invertible module hence $\mathcal{L}_ y$ is invertible. Thus there would exist an open $V \subset U$ containing $U \cap Y$ such that $\mathcal{L}|_ V$ is invertible. By Divisors, Lemma 31.28.3 we find an invertible $A$-module $M$ with $\widetilde{M}|_ V \cong \mathcal{L}|_ V$. However the ring $A$ is a UFD hence we see $M \cong A$ which would imply $\mathcal{L}_ n \cong \mathcal{O}_ U/I^ n\mathcal{O}_ U$. Since $\mathcal{L}_2 \not\cong \mathcal{O}_ U/I^2\mathcal{O}_ U$ by construction we get a contradiction as desired.

Note that if we take $a_ n = 0$ for $n \geq 2$, then we see that $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{L}_ n)$ is nonzero: in this case we the function $x$ on $D(x)$ and the function $x + t/y$ on $D(y)$ glue. On the other hand, if we take $a_ n = 1$ and $\varphi (n) = 2^ n$ or even $\varphi (n) = n^2$ then the reader can show that $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{L}_ n)$ is zero; this gives another proof that $(\mathcal{L}_ n)$ is not algebraizable in this case.

If in Situation 52.16.1 the ring $A$ is not $I$-adically complete, then Lemma 52.16.3 suggests the correct thing is to ask whether $(\mathcal{F}_ n)$ is in the essential image of the restriction functor

However, we can no longer say that this means $(\mathcal{F}_ n)$ is algebraizable. Thus we introduce the following terminology.

Definition 52.16.5. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. We say *$(\mathcal{F}_ n)$ extends to $X$* if there exists an object $(\mathcal{G}_ n)$ of $\textit{Coh}(X, I\mathcal{O}_ X)$ whose restriction to $U$ is isomorphic to $(\mathcal{F}_ n)$.

This notion is equivalent to being algebraizable over the completion.

Lemma 52.16.6. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $A', I', \mathfrak a'$ be the $I$-adic completions of $A, I, \mathfrak a$. Set $X' = \mathop{\mathrm{Spec}}(A')$ and $U' = X' \setminus V(\mathfrak a')$. The following are equivalent

$(\mathcal{F}_ n)$ extends to $X$, and

the pullback of $(\mathcal{F}_ n)$ to $U'$ is the completion of a coherent $\mathcal{O}_{U'}$-module.

**Proof.**
Recall that $A \to A'$ is a flat ring map which induces an isomorphism $A/I \to A'/I'$. See Algebra, Lemmas 10.97.2 and 10.97.4. Thus $X' \to X$ is a flat morphism inducing an isomorphism $Y' \to Y$. Thus $U' \to U$ is a flat morphism which induces an isomorphism $U' \cap Y' \to U \cap Y$. This implies that in the commutative diagram

the vertical functors are equivalences. See Cohomology of Schemes, Lemma 30.23.10. The lemma follows formally from this and the results of Lemma 52.16.3. $\square$

In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. To figure out if $(\mathcal{F}_ n)$ extends to $X$ it makes sense to look at the $A$-module

Observe that $M$ has a limit topology which is (a priori) coarser than the $I$-adic topology since $M \to H^0(U, \mathcal{F}_ n)$ annihilates $I^ nM$. There are canonical maps

One could hope that $\widetilde{M}$ restricts to a coherent module on $U$ and that $(\mathcal{F}_ n)$ is the completion of this module. This is naive because this has almost no chance of being true if $A$ is not complete. But even if $A$ is $I$-adically complete this notion is very difficult to work with. A less naive approach is to consider the following requirement.

Definition 52.16.7. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. We say *$(\mathcal{F}_ n)$ canonically extends to $X$* if the the inverse system

in $\mathit{QCoh}(\mathcal{O}_ X)$ is pro-isomorphic to an object $(\mathcal{G}_ n)$ of $\textit{Coh}(X, I\mathcal{O}_ X)$.

We will see in Lemma 52.16.8 that the condition in Definition 52.16.7 is stronger than the condition of Definition 52.16.5.

Lemma 52.16.8. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. If $(\mathcal{F}_ n)$ canonically extends to $X$, then

$(\widetilde{H^0(U, \mathcal{F}_ n)})$ is pro-isomorphic to an object $(\mathcal{G}_ n)$ of $\textit{Coh}(X, I \mathcal{O}_ X)$ unique up to unique isomorphism,

the restriction of $(\mathcal{G}_ n)$ to $U$ is isomorphic to $(\mathcal{F}_ n)$, i.e., $(\mathcal{F}_ n)$ extends to $X$,

the inverse system $\{ H^0(U, \mathcal{F}_ n)\} $ satisfies the Mittag-Leffler condition, and

the module $M$ in (52.16.6.1) is finite over the $I$-adic completion of $A$ and the limit topology on $M$ is the $I$-adic topology.

**Proof.**
The existence of $(\mathcal{G}_ n)$ in (1) follows from Definition 52.16.7. The uniqueness of $(\mathcal{G}_ n)$ in (1) follows from Lemma 52.15.3. Write $\mathcal{G}_ n = \widetilde{M_ n}$. Then $\{ M_ n\} $ is an inverse system of finite $A$-modules with $M_ n = M_{n + 1}/I^ n M_{n + 1}$. By Definition 52.16.7 the inverse system $\{ H^0(U, \mathcal{F}_ n)\} $ is pro-isomorphic to $\{ M_ n\} $. Hence we see that the inverse system $\{ H^0(U, \mathcal{F}_ n)\} $ satisfies the Mittag-Leffler condition and that $M = \mathop{\mathrm{lim}}\nolimits M_ n$ (as topological modules). Thus the properties of $M$ in (4) follow from Algebra, Lemmas 10.98.2, 10.96.12, and 10.96.3. Since $U$ is quasi-affine the canonical maps

are isomorphisms (Properties, Lemma 28.18.2). We conclude that $(\mathcal{G}_ n|_ U)$ and $(\mathcal{F}_ n)$ are pro-isomorphic and hence isomorphic by Lemma 52.15.3. $\square$

Lemma 52.16.9. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $A \to A'$ be a flat ring map. Set $X' = \mathop{\mathrm{Spec}}(A')$, let $U' \subset X'$ be the inverse image of $U$, and denote $g : U' \to U$ the induced morphism. Set $(\mathcal{F}'_ n) = (g^*\mathcal{F}_ n)$, see Cohomology of Schemes, Lemma 30.23.9. If $(\mathcal{F}_ n)$ canonically extends to $X$, then $(\mathcal{F}'_ n)$ canonically extends to $X'$. Moreover, the extension found in Lemma 52.16.8 for $(\mathcal{F}_ n)$ pulls back to the extension for $(\mathcal{F}'_ n)$.

**Proof.**
Let $f : X' \to X$ be the induced morphism. We have $H^0(U', \mathcal{F}'_ n) = H^0(U, \mathcal{F}_ n) \otimes _ A A'$ by flat base change, see Cohomology of Schemes, Lemma 30.5.2. Thus if $(\mathcal{G}_ n)$ in $\textit{Coh}(X, I\mathcal{O}_ X)$ is pro-isomorphic to $(\widetilde{H^0(U, \mathcal{F}_ n)})$, then $(f^*\mathcal{G}_ n)$ is pro-isomorphic to

This finishes the proof. $\square$

Lemma 52.16.10. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $M$ be as in (52.16.6.1). Assume

the inverse system $H^0(U, \mathcal{F}_ n)$ has Mittag-Leffler,

the limit topology on $M$ agrees with the $I$-adic topology, and

the image of $M \to H^0(U, \mathcal{F}_ n)$ is a finite $A$-module for all $n$.

Then $(\mathcal{F}_ n)$ extends canonically to $X$. In particular, if $A$ is $I$-adically complete, then $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module.

**Proof.**
Since $H^0(U, \mathcal{F}_ n)$ has the Mittag-Leffler condition and since the limit topology on $M$ is the $I$-adic topology we see that $\{ M/I^ nM\} $ and $\{ H^0(U, \mathcal{F}_ n)\} $ are pro-isomorphic inverse systems of $A$-modules. Thus if we set

then we see that to verify the condition in Definition 52.16.7 it suffices to show that $M$ is a finite module over the $I$-adic completion of $A$. This follows from the fact that $M/I^ n M$ is finite by condition (c) and the above and Algebra, Lemma 10.96.12. $\square$

The following is in some sense the most straightforward possible application Lemma 52.16.10 above.

Lemma 52.16.11. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume

$I = (f)$ is a principal ideal for a nonzerodivisor $f \in \mathfrak a$,

$\mathcal{F}_ n$ is a finite locally free $\mathcal{O}_ U/f^ n\mathcal{O}_ U$-module,

$H^1_\mathfrak a(A/fA)$ and $H^2_\mathfrak a(A/fA)$ are finite $A$-modules.

Then $(\mathcal{F}_ n)$ extends canonically to $X$. In particular, if $A$ is complete, then $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module.

**Proof.**
We will prove this by verifying hypotheses (a), (b), and (c) of Lemma 52.16.10.

Since $\mathcal{F}_ n$ is locally free over $\mathcal{O}_ U/f^ n\mathcal{O}_ U$ we see that we have short exact sequences $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$ for all $n$. Thus condition (b) holds by Cohomology, Lemma 20.36.2.

As $f$ is a nonzerodivisor we obtain short exact sequences

and we have corresponding short exact sequences $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$. We will use Local Cohomology, Lemma 51.8.2 without further mention. Our assumptions imply that $H^0(U, \mathcal{O}_ U/f\mathcal{O}_ U)$ and $H^1(U, \mathcal{O}_ U/f\mathcal{O}_ U)$ are finite $A$-modules. Hence the same thing is true for $\mathcal{F}_1$, see Local Cohomology, Lemma 51.12.2. Using induction and the short exact sequences we find that $H^0(U, \mathcal{F}_ n)$ are finite $A$-modules for all $n$. In this way we see hypothesis (c) is satisfied.

Finally, as $H^1(U, \mathcal{F}_1)$ is a finite $A$-module we can apply Cohomology, Lemma 20.36.4 to see hypothesis (a) holds. $\square$

Remark 52.16.12. In Lemma 52.16.11 if $A$ is universally catenary with Cohen-Macaulay formal fibres (for example if $A$ has a dualizing complex), then the condition that $H^1_\mathfrak a(A/fA)$ and $H^2_\mathfrak a(A/fA)$ are finite $A$-modules, is equivalent with

for all $\mathfrak p \in V(f) \setminus V(\mathfrak a)$ and $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$ by Local Cohomology, Theorem 51.11.6.

For example, if $A/fA$ is $(S_2)$ and if every irreducible component of $Z = V(\mathfrak a)$ has codimension $\geq 3$ in $Y = \mathop{\mathrm{Spec}}(A/fA)$, then we get the finiteness of $H^1_\mathfrak a(A/fA)$ and $H^2_\mathfrak a(A/fA)$. This should be contrasted with the slightly weaker conditions found in Lemma 52.20.1 (see also Remark 52.20.2).

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