## 52.23 Algebraization of coherent formal modules, VI

In this section we add a few more easier to prove cases.

Proposition 52.23.1. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume

1. there exist $f_1, \ldots , f_ d \in I$ such that for $y \in U \cap Y$ the ideal $I\mathcal{O}_{X, y}$ is generated by $f_1, \ldots , f_ d$ and $f_1, \ldots , f_ d$ form a $\mathcal{F}_ y^\wedge$-regular sequence,

2. $H^0(U, \mathcal{F}_1)$ and $H^1(U, \mathcal{F}_1)$ 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. For every $n$ we have a short exact sequence

$0 \to I^ n\mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$

Since $f_1, \ldots , f_ d$ forms a regular sequence (and hence quasi-regular, see Algebra, Lemma 10.69.2) on each of the “stalks” $\mathcal{F}_ y^\wedge$ and since we have $I\mathcal{F}_ n = (f_1, \ldots , f_ d)\mathcal{F}_ n$ for all $n$, we find that

$I^ n\mathcal{F}_{n + 1} = \bigoplus \nolimits _{e_1 + \ldots + e_ d = n} \mathcal{F}_1 \cdot f_1^{e_1} \ldots f_ d^{e_ d}$

by checking on stalks. Using the assumption of finiteness of $H^0(U, \mathcal{F}_1)$ and induction, we first conclude that $M_ n = H^0(U, \mathcal{F}_ n)$ is a finite $A$-module for all $n$. In this way we see that condition (c) of Lemma 52.16.10 holds. We also see that

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

is a finite graded $R = \bigoplus I^ n/I^{n +1}$-module. By Lemma 52.2.1 we conclude that condition (a) of Lemma 52.16.10 is satisfied. Finally, condition (b) of Lemma 52.16.10 is satisfied because $\bigoplus H^0(U, I^ n\mathcal{F}_{n + 1})$ is a finite graded $R$-module and we can apply Lemma 52.2.3. $\square$

Remark 52.23.2. In the situation of Proposition 52.23.1 if we assume $A$ has a dualizing complex, then the condition that $H^0(U, \mathcal{F}_1)$ and $H^1(U, \mathcal{F}_1)$ are finite is equivalent to

$\text{depth}(\mathcal{F}_{1, y}) + \dim (\mathcal{O}_{\overline{\{ y\} }, z}) > 2$

for all $y \in U \cap Y$ and $z \in Z \cap \overline{\{ y\} }$. See Local Cohomology, Lemma 51.12.1. This holds for example if $\mathcal{F}_1$ is a finite locally free $\mathcal{O}_{U \cap Y}$-module, $Y$ is $(S_2)$, and $\text{codim}(Z', Y') \geq 3$ for every pair of irreducible components $Y'$ of $Y$, $Z'$ of $Z$ with $Z' \subset Y'$.

Proposition 52.23.3. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume there is Noetherian local ring $(R, \mathfrak m)$ and a ring map $R \to A$ such that

1. $I = \mathfrak m A$,

2. for $y \in U \cap Y$ the stalk $\mathcal{F}_ y^\wedge$ is $R$-flat,

3. $H^0(U, \mathcal{F}_1)$ and $H^1(U, \mathcal{F}_1)$ 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. The proof is exactly the same as the proof of Proposition 52.23.1. Namely, if $\kappa = R/\mathfrak m$ then for $n \geq 0$ there is an isomorphism

$I^ n \mathcal{F}_{n + 1} \cong \mathcal{F}_1 \otimes _\kappa \mathfrak m^ n/\mathfrak m^{n + 1}$

and the right hand side is a finite direct sum of copies of $\mathcal{F}_1$. This can be checked by looking at stalks. Everything else is exactly the same. $\square$

Remark 52.23.4. Proposition 52.23.3 is a local version of [Theorem 2.10 (i), Baranovsky]. It is straightforward to deduce the global results from the local one; we will sketch the argument. Namely, suppose $(R, \mathfrak m)$ is a complete Noetherian local ring and $X \to \mathop{\mathrm{Spec}}(R)$ is a proper morphism. For $n \geq 1$ set $X_ n = X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R/\mathfrak m^ n)$. Let $Z \subset X_1$ be a closed subset of the special fibre. Set $U = X \setminus Z$ and denote $j : U \to X$ the inclusion morphism. Suppose given an object

$(\mathcal{F}_ n) \text{ of } \textit{Coh}(U, \mathfrak m\mathcal{O}_ U)$

which is flat over $R$ in the sense that $\mathcal{F}_ n$ is flat over $R/\mathfrak m^ n$ for all $n$. Assume that $j_*\mathcal{F}_1$ and $R^1j_*\mathcal{F}_1$ are coherent modules. Then affine locally on $X$ we get a canonical extension of $(\mathcal{F}_ n)$ by Proposition 52.23.3 and formation of this extension commutes with localization (by Lemma 52.16.11). Thus we get a canonical global object $(\mathcal{G}_ n)$ of $\textit{Coh}(X, \mathfrak m\mathcal{O}_ X)$ whose restriction of $U$ is $(\mathcal{F}_ n)$. By Grothendieck's existence theorem (Cohomology of Schemes, Proposition 30.25.4) we see there exists a coherent $\mathcal{O}_ X$-module $\mathcal{G}$ whose completion is $(\mathcal{G}_ n)$. In this way we see that $(\mathcal{F}_ n)$ is algebraizable, i.e., it is the completion of a coherent $\mathcal{O}_ U$-module.

We add that the coherence of $j_*\mathcal{F}_1$ and $R^1j_*\mathcal{F}_1$ is a condition on the special fibre. Namely, if we denote $j_1 : U_1 \to X_1$ the special fibre of $j : U \to X$, then we can think of $\mathcal{F}_1$ as a coherent sheaf on $U_1$ and we have $j_*\mathcal{F}_1 = j_{1, *}\mathcal{F}_1$ and $R^1j_*\mathcal{F}_1 = R^1j_{1, *}\mathcal{F}_1$. Hence for example if $X_1$ is $(S_2)$ and irreducible, we have $\dim (X_1) - \dim (Z) \geq 3$, and $\mathcal{F}_1$ is a locally free $\mathcal{O}_{U_1}$-module, then $j_{1, *}\mathcal{F}_1$ and $R^1j_{1, *}\mathcal{F}_1$ are coherent modules.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).