The Stacks project

Lemma 31.36.1. Let $X$ be an integral scheme. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. There exists a modification $f : X' \to X$ such that $f^*\mathcal{E}$ has a filtration whose successive quotients are invertible $\mathcal{O}_{X'}$-modules.

Proof. We prove this by induction on the rank $r$ of $\mathcal{E}$. If $r = 1$ or $r = 0$ the lemma is obvious. Assume $r > 1$. Let $P = \mathbf{P}(\mathcal{E})$ with structure morphism $\pi : P \to X$, see Constructions, Section 27.21. Then $\pi $ is proper (Lemma 31.30.4). There is a canonical surjection

\[ \pi ^*\mathcal{E} \to \mathcal{O}_ P(1) \]

whose kernel is finite locally free of rank $r - 1$. Choose a nonempty open subscheme $U \subset X$ such that $\mathcal{E}|_ U \cong \mathcal{O}_ U^{\oplus r}$. Then $P_ U = \pi ^{-1}(U)$ is isomorphic to $\mathbf{P}^{r - 1}_ U$. In particular, there exists a section $s : U \to P_ U$ of $\pi $. Let $X' \subset P$ be the scheme theoretic image of the morphism $U \to P_ U \to P$. Then $X'$ is integral (Morphisms, Lemma 29.6.7), the morphism $f = \pi |_{X'} : X' \to X$ is proper (Morphisms, Lemmas 29.41.6 and 29.41.4), and $f^{-1}(U) \to U$ is an isomorphism. Hence $f$ is a modification (Morphisms, Definition 29.51.11). By construction the pullback $f^*\mathcal{E}$ has a two step filtration whose quotient is invertible because it is equal to $\mathcal{O}_ P(1)|_{X'}$ and whose sub $\mathcal{E}'$ is locally free of rank $r - 1$. By induction we can find a modification $g : X'' \to X'$ such that $g^*\mathcal{E}'$ has a filtration as in the statement of the lemma. Thus $f \circ g : X'' \to X$ is the required modification. $\square$

Comments (0)

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AYP. Beware of the difference between the letter 'O' and the digit '0'.