Proposition 52.28.1. In the situation above assume there exists an integer $\sigma $ such that for all points $p \in P \setminus Q$ we have

Then the map

is an isomorphism for $0 \leq i < \sigma $.

Proposition 52.28.1. In the situation above assume there exists an integer $\sigma $ such that for all points $p \in P \setminus Q$ we have

\[ \text{depth}(\mathcal{F}_ p) + \dim (\overline{\{ p\} }) > \sigma \]

Then the map

\[ H^ i(P, \mathcal{F}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^ i(Q_ n, \mathcal{F}_ n) \]

is an isomorphism for $0 \leq i < \sigma $.

**Proof.**
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.

We may and do think of $s$ as an element $f \in A_1 \subset A$, i.e., a homogeneous element of degree $1$ of $A$. Denote $Y = V(f) \subset X$ the closed subscheme defined by $f$. Then $U \cap Y = (\pi |_ U)^{-1}(Q)$ scheme theoretically. Recall the notation $\mathcal{F}_ U = \pi ^*\mathcal{F}|_ U = (\pi |_ U)^*\mathcal{F}$. This is a coherent $\mathcal{O}_ U$-module. Choose a finite $A$-module $M$ such that $\mathcal{F}_ U = \widetilde{M}|_ U$ (for existence see Local Cohomology, Lemma 51.8.2). We claim that $H^ i_ Z(M)$ is annihilated by a power of $f$ for $i \leq \sigma + 1$.

To prove the claim we will apply Local Cohomology, Proposition 51.10.1. Translating into geometry we see that it suffices to prove for $u \in U$, $u \not\in Y$ and $z \in \overline{\{ u\} } \cap Z$ that

\[ \text{depth}(\mathcal{F}_{U, u}) + \dim (\mathcal{O}_{\overline{\{ u\} }, z}) > \sigma + 1 \]

This requires only a small amount of thought.

Observe that $Z = \mathop{\mathrm{Spec}}(A_0)$ is a finite set of closed points of $X$ because $A_0$ is a finite dimensional $k$-algebra. (The reader who would like $Z$ to be a singleton can replace the finite $k$-algebra $A_0$ by $k$; it won't affect anything else in the proof.)

The morphism $\pi : L \to P$ and its restriction $\pi |_ U : U \to P$ are smooth of relative dimension $1$. Let $u \in U$, $u \not\in Y$ and $z \in \overline{\{ u\} } \cap Z$. Let $p = \pi (u) \in P \setminus Q$ be its image. Then either $u$ is a generic point of the fibre of $\pi $ over $p$ or a closed point of the fibre. If $u$ is a generic point of the fibre, then $\text{depth}(\mathcal{F}_{U, u}) = \text{depth}(\mathcal{F}_ p)$ and $\dim (\overline{\{ u\} }) = \dim (\overline{\{ p\} }) + 1$. If $u$ is a closed point of the fibre, then $\text{depth}(\mathcal{F}_{U, u}) = \text{depth}(\mathcal{F}_ p) + 1$ and $\dim (\overline{\{ u\} }) = \dim (\overline{\{ p\} })$. In both cases we have $\dim (\overline{\{ u\} }) = \dim (\mathcal{O}_{\overline{\{ u\} }, z})$ because every point of $Z$ is closed. Thus the desired inequality follows from the assumption in the statement of the lemma.

Let $A'$ be the $f$-adic completion of $A$. So $A \to A'$ is flat by Algebra, Lemma 10.97.2. Denote $U' \subset X' = \mathop{\mathrm{Spec}}(A')$ the inverse image of $U$ and similarly for $Y'$ and $Z'$. Let $\mathcal{F}'$ on $U'$ be the pullback of $\mathcal{F}_ U$ and let $M' = M \otimes _ A A'$. By flat base change for local cohomology (Local Cohomology, Lemma 51.5.7) we have

\[ H^ i_{Z'}(M') = H^ i_ Z(M) \otimes _ A A' \]

and we find that for $i \leq \sigma + 1$ these are annihilated by a power of $f$. Consider the diagram

\[ \xymatrix{ & H^ i(U, \mathcal{F}_ U) \ar[ld] \ar[d] \ar[r] & \mathop{\mathrm{lim}}\nolimits H^ i(U, \mathcal{F}_ U/f^ n\mathcal{F}_ U) \ar@{=}[d] \\ H^ i(U, \mathcal{F}_ U) \otimes _ A A' \ar@{=}[r] & H^ i(U', \mathcal{F}') \ar[r] & \mathop{\mathrm{lim}}\nolimits H^ i(U', \mathcal{F}'/f^ n\mathcal{F}') } \]

The lower horizontal arrow is an isomorphism for $i < \sigma $ by Lemma 52.13.2 and the torsion property we just proved. The horizontal equal sign is flat base change (Cohomology of Schemes, Lemma 30.5.2) and the vertical equal sign is because $U \cap Y$ and $U' \cap Y'$ as well as their $n$th infinitesimal neighbourhoods are mapped isomorphically onto each other (as we are completing with respect to $f$).

Applying More on Morphisms, Equation (37.51.0.2) we have compatible direct sum decompositions

\[ \mathop{\mathrm{lim}}\nolimits H^ i(U, \mathcal{F}_ U/f^ n\mathcal{F}_ U) = \mathop{\mathrm{lim}}\nolimits \left( \bigoplus \nolimits _{m \in \mathbf{Z}} H^ i(Q_ n, \mathcal{F}_ n \otimes \mathcal{L}^{\otimes m}) \right) \]

and

\[ H^ i(U, \mathcal{F}_ U) = \bigoplus \nolimits _{m \in \mathbf{Z}} H^ i(P, \mathcal{F} \otimes \mathcal{L}^{\otimes m}) \]

Thus we conclude by Algebra, Lemma 10.98.4. $\square$

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