The Stacks project

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

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. $\square$