Componentwise curve construction and the scope of global irreducibility

By [F1], the components $X_a$ are finitely many clopen normal integral affine finite-type threefolds. Restriction of $s$ has zero scheme $S\cap X_a$, still finite and smooth in the smooth locus; its coherent kernel restricts to $Q{\mathrm{\mid }}_{X_a}$ and has rank two on $U_a$. The closed $Z_a$ has codimension at least two there, and normality over the perfect field makes $W_a$ a nonempty smooth integral threefold. Apply the global dual construction of [F3] on each component with this particular $W_a$, obtaining the indexed family described in the Statement. In particular, in its construction a frame $g_{a,i}{\psi }_1,g_{a,i}{\psi }_2$ on $D(g_{a,i})$ has globally extendable controlling products even for the coordinate $1\mathrm{/}{g}_{a,i}$, whose products with the frame are ${\psi }_1,{\psi }_2$. Thus the chartwise lemma in [F3] applies to the actual global family, including its common-core and avoidance conditions, and gives a functional on each $Q{\mathrm{\mid }}_{U_a}$ with the stated zero curve. Sections on a finite disjoint union are tuples of sections, so these functionals assemble into $t$. Its zero scheme is precisely the disjoint union of the component zero schemes; smoothness, geometric integrality of each curve, pure codimension two and avoidance hold componentwise. When $X$ is integral this is a single curve. The fixed $Z$ was chosen before these parameters.

For the obstruction take two copies of $X_0={\mathbb{A}}_k^3$, with $P={\mathcal{O}}^3$ and point functional $(x,y,z)$ on each. Its zero set is the reduced origin, an allowed smooth point section. On $U_0=X_0\setminus \{0\}$ let $Q_0$ be its rank-two kernel. Dualizing the locally split sequence gives $0\to {\mathcal{O}}_{U_0}\stackrel{(x,y,z{)}^t}{\to }{\mathcal{O}}_{U_0}^3\to Q_0^{\vee }\to 0$. Every section of $Q_0^{\vee }$ is represented by a polynomial vector $b\in k[x,y,z{]}^3$: indeed $\mathrm{\Gamma }(U_0,\mathcal{O})=A$ since $A_x\cap A_y\cap A_z=A$ in the fraction field, and $H^1(U_0,\mathcal{O})=0$. For the latter, use the affine cover $D(x),D(y),D(z)$: its augmented localization complex is the filtered colimit of cochain Koszul complexes on $x^n,y^n,z^n$. These are regular sequences, so the Koszul cohomology below degree three is zero (by induction, tensoring the one-element complexes and using that each next element is a nonzerodivisor on the previous quotient). Filtered colimits of modules are exact. The resulting vanishing in augmented degree two is precisely $H^1(U_0,\mathcal{O})=0$. The long exact section sequence gives the claimed polynomial representative.

The zero scheme of this section on $U_0$ is defined by the $2\times 2$ minors of the matrix with rows $(x,y,z)$ and $b$: locally one of $x,y,z$ is invertible, and elementary row/column elimination identifies the two residual coordinates with those minors up to that unit. Its minor ideal $J\subset A$ is proper because every minor vanishes at the origin. By [F2] $J$ has a prime over it of height at most two. Such a prime is not the origin's maximal ideal of height three, so $V(J)\cap U_0$ is nonempty. Thus no functional $Q_0\to {\mathcal{O}}_{U_0}$ is nowhere zero. On two copies every global functional has zeros in both clopen components, and its zero locus is disconnected and cannot be irreducible. Trivial $P$ has $c_1=c_2=0$, so that additional article hypothesis does not remove this fixed-pair counterexample. This does not obstruct first reducing the global theorem to components, or assert that every possible preceding point section gives the same pair.