The Stacks project

31.36 Modifications

In this section we will collect results of the type: after a modification such and such are true. We will later see that a modification can be dominated by a blowup (More on Flatness, Lemma 38.31.4).

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. $\square$

Lemma 31.36.2. Let $S$ be a scheme. Let $X$, $Y$ be schemes over $S$. Assume $X$ is Noetherian and $Y$ is proper over $S$. Given an $S$-rational map $f : U \to Y$ from $X$ to $Y$ there exists a morphism $p : X' \to X$ and an $S$-morphism $f' : X' \to Y$ such that

  1. $p$ is proper and $p^{-1}(U) \to U$ is an isomorphism,

  2. $f'|_{p^{-1}(U)}$ is equal to $f \circ p|_{p^{-1}(U)}$.

Proof. Denote $j : U \to X$ the inclusion morphism. Let $X' \subset Y \times _ S X$ be the scheme theoretic image of $(f, j) : U \to Y \times _ S X$ (Morphisms, Definition 29.6.2). The projection $g : Y \times _ S X \to X$ is proper (Morphisms, Lemma 29.41.5). The composition $p : X' \to X$ of $X' \to Y \times _ S X$ and $g$ is proper (Morphisms, Lemmas 29.41.6 and 29.41.4). Since $g$ is separated and $U \subset X$ is retrocompact (as $X$ is Noetherian) we conclude that $p^{-1}(U) \to U$ is an isomorphism by Morphisms, Lemma 29.6.8. On the other hand, the composition $f' : X' \to Y$ of $X' \to Y \times _ S X$ and the projection $Y \times _ S X \to Y$ agrees with $f$ on $p^{-1}(U)$. $\square$

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