Proposition 52.28.5. In the situation above let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(P, \mathcal{I})$. Assume for all $q \in Q$ and for all primes $\mathfrak p \in \mathcal{O}_{P, q}^\wedge $, $\mathfrak p \not\in V(\mathcal{I}_ q^\wedge )$ we have

\[ \text{depth}((\mathcal{F}_ q^\wedge )_\mathfrak p) + \dim (\mathcal{O}_{P, q}^\wedge /\mathfrak p) + \dim (\overline{\{ q\} }) > 2 \]

Then $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ P$-module.

**Proof.**
By Cohomology of Schemes, Lemma 30.23.6 to prove the lemma, we may replace $(\mathcal{F}_ n)$ by an object differing from it by $\mathcal{I}$-torsion (see below for more precision). Let $T' = \{ q \in Q \mid \dim (\overline{\{ q\} }) = 0\} $ and $T = \{ q \in Q \mid \dim (\overline{\{ q\} }) \leq 1\} $. The assumption in the proposition is exactly that $Q \subset P$, $(\mathcal{F}_ n)$, and $T' \subset T \subset Q$ satisfy the conditions of Lemma 52.21.2 with $d = 1$; besides trivial manipulations of inequalities, use that $V(\mathfrak p) \cap V(\mathcal{I}^\wedge _ y) = \{ \mathfrak m^\wedge _ y\} \Leftrightarrow \dim (\mathcal{O}_{P, q}^\wedge /\mathfrak p) = 1$ as $\mathcal{I}_ y^\wedge $ is generated by $1$ element. Combining these two remarks, we may replace $(\mathcal{F}_ n)$ by the object $(\mathcal{H}_ n)$ of $\textit{Coh}(P, \mathcal{I})$ found in Lemma 52.21.2. Thus we may and do assume $(\mathcal{F}_ n)$ is pro-isomorphic to an inverse system $(\mathcal{F}_ n'')$ of coherent $\mathcal{O}_ P$-modules such that $\text{depth}(\mathcal{F}''_{n, q}) + \dim (\overline{\{ q\} }) \geq 2$ for all $q \in Q$.

We will use More on Morphisms, Lemma 37.51.1 and we will use the notation used and results found More on Morphisms, Section 37.51 without further mention; this proof will not make sense without at least understanding the statement of the lemma. Observe that in our case $A = \bigoplus _{m \geq 0} \Gamma (P, \mathcal{L}^{\otimes m})$ is a finite type $k$-algebra all of whose graded parts are finite dimensional $k$-vector spaces, see Cohomology of Schemes, Lemma 30.16.1.

By Cohomology of Schemes, Lemma 30.23.9 the pull back by $\pi |_ U : U \to P$ is an object $(\pi |_ U^*\mathcal{F}_ n)$ of $\textit{Coh}(U, f\mathcal{O}_ U)$ which is pro-isomorphic to the inverse system $(\pi |_ U^*\mathcal{F}_ n'')$ of coherent $\mathcal{O}_ U$-modules. We claim

\[ \text{depth}(\pi |_ U^*\mathcal{F}''_{n, y}) + \delta _ Z^ Y(y) \geq 3 \]

for all $y \in U \cap Y$. Since all the points of $Z$ are closed, we see that $\delta _ Z^ Y(y) \geq \dim (\overline{\{ y\} })$ for all $y \in U \cap Y$, see Lemma 52.18.1. Let $q \in Q$ be the image of $y$. Since the morphism $\pi : U \to P$ is smooth of relative dimension $1$ we see that either $y$ is a closed point of a fibre of $\pi $ or a generic point. Thus we see that

\[ \text{depth}(\pi ^*\mathcal{F}''_{n, y}) + \delta _ Z^ Y(y) \geq \text{depth}(\pi ^*\mathcal{F}''_{n, y}) + \dim (\overline{\{ y\} }) = \text{depth}(\mathcal{F}''_{n, q}) + \dim (\overline{\{ q\} }) + 1 \]

because either the depth goes up by $1$ or the dimension. This proves the claim.

By Lemma 52.22.1 we conclude that $(\pi |_ U^*\mathcal{F}_ n)$ canonically extends to $X$. Observe that

\[ M_ n = \Gamma (U, \pi |_ U^*\mathcal{F}_ n) = \bigoplus \nolimits _{m \in \mathbf{Z}} \Gamma (P, \mathcal{F}_ n \otimes _{\mathcal{O}_ P} \mathcal{L}^{\otimes m}) \]

is canonically a graded $A$-module, see More on Morphisms, Equation (37.51.0.2). By Properties, Lemma 28.18.2 we have $\pi |_ U^*\mathcal{F}_ n = \widetilde{M_ n}|_ U$. Thus we may apply Lemma 52.28.4 to find a finite graded $A$-module $N$ such that $(M_ n)$ and $(N/I^ nN)$ are pro-isomorphic in the category of graded $A$-modules modulo $A_+$-torsion modules. Let $\mathcal{F}$ be the coherent $\mathcal{O}_ P$-module associated to $N$, see Cohomology of Schemes, Proposition 30.15.3. The same proposition tells us that $(\mathcal{F}/\mathcal{I}^ n\mathcal{F})$ is pro-isomorphic to $(\mathcal{F}_ n)$. Since both are objects of $\textit{Coh}(P, \mathcal{I})$ we win by Lemma 52.15.3.
$\square$

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