Lemma 52.25.1. For any coherent triple $(\mathcal{F}, \mathcal{F}_0, \alpha )$ there exists a coherent $\mathcal{O}_ X$-module $\mathcal{F}'$ such that $f : \mathcal{F}' \to \mathcal{F}'$ is injective, an isomorphism $\alpha ' : \mathcal{F}'|_ U \to \mathcal{F}$, and a map $\alpha '_0 : \mathcal{F}'/f\mathcal{F}' \to \mathcal{F}_0$ such that $\alpha \circ (\alpha ' \bmod f) = \alpha '_0|_{U_0}$.
Proof. Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction to $U$ of the coherent $\mathcal{O}_ X$-module associated to $M$, see Local Cohomology, Lemma 51.8.2. Since $\mathcal{F}$ is $f$-torsion free, we may replace $M$ by its quotient by $f$-power torsion. On the other hand, let $M_0 = \Gamma (X_0, \mathcal{F}_0)$ so that $\mathcal{F}_0$ is the coherent $\mathcal{O}_{X_0}$-module associated to the finite $A/fA$-module $M_0$. By Cohomology of Schemes, Lemma 30.10.5 there exists an $n$ such that the isomorphism $\alpha _0$ corresponds to an $A/fA$-module homomorphism $\mathfrak m^ n M/fM \to M_0$ (whose kernel and cokernel are annihilated by a power of $\mathfrak m$, but we don't need this). Thus if we take $M' = \mathfrak m^ n M$ and we let $\mathcal{F}'$ be the coherent $\mathcal{O}_ X$-module associated to $M'$, then the lemma is clear. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)