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A rank-two family avoids each prescribed codimension-two locus

PDF p. 15: Theorem6.7 second Bertini application

Statement

Fix a closed subset $Z \subset U$ of codimension at least two before choosing the functional. Set $B = Z \cup U_{s i n g}$ with reduced structure. Then $\dim B \leq 1$ , and a nonempty open $T_{a v} \subset T$ consists of parameters whose zero schemes are disjoint from $B$ . The allowed open may depend on $Z$ ; this is $\forall Z \exists t$ , not $\exists t \forall Z$ .

Facts & Assumptions

Given: The integral normal threefold and fixed $Z$ ; the globally value-generating family of Coherent dual extension supplies global Swan controlling sections.

Goal: Combine the prescribed avoidance with all later generic conditions.

[F1]

External normality and perfect-field smooth-locus inputs gives normality's $(R_{1})$ and the smooth/regular equality over the perfect field. External finite-type dimension, rational-point and etale-fiber inputs gives dimension and image-closure bounds.

[F2]

Finite-dimensional families of generated bundle functionals and Kernels of surjections of vector bundles identify zero incidence over $B$ with the total space of the rank- $(N - 2)$ value kernel.

Proof

1.1F1F2given

Codimension at least two in the integral threefold gives $\dim Z \leq 1$ . By [F1] the singular locus is closed of dimension at most one. Hence their union $B$ is a closed finite-type subset of $U$ with $\dim B \leq 1$ . The restricted evaluation surjection $O_{B}^{N} \to E ∣_{B}$ has locally free kernel of rank $N - 2$ , even if $B$ is singular. Thus its total space, the actual zero incidence $I_{B}$ , has dimension at most $1 + N - 2 = N - 1$ .

2.1F1F2step 1.1∎

The closure of the image of $I_{B}$ in $T$ has dimension at most $N - 1$ by [F1]. Its complement $T_{a v}$ is nonempty open. Its inverse image in $I_{B}$ is empty and remains empty after every field extension. Hence all its zero fibers avoid $B$ . Intersecting this open with any finitely many nonempty parameter opens retains a nonempty open since $T$ is irreducible. The construction starts with the chosen $Z$ , giving exactly the stated quantifier order.

Depends on

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