Reduction to the double conics

Proof. Let $\Pi$ be the plane defining $S$, and let $Z=C\cap\Pi$. By Lemma 103.2, $Z$ is curvilinear of type $(2,2)$ or $(4)$, respectively. It is the complete intersection of the two restricted quadrics. In type $(2,2)$ let $\lambda=0$ be the line joining the two support points; its square vanishes on both length-two parts of $Z$. In type $(4)$ let $\lambda=0$ be the tangent line to $Z$. In $\mathcal O_Z\cong\mathbb C[t]/(t^4)$ its equation belongs to $(t^2)$, so its square again vanishes on $Z$. The twisted Koszul resolution $$0\longrightarrow\mathcal O_{\Pi}(-2) \longrightarrow\mathcal O_{\Pi}^{\oplus2} \longrightarrow\mathcal I_Z(2)\longrightarrow0$$ shows that $H^0(\Pi,\mathcal I_Z(2))$ is exactly the given pencil. Thus $\lambda^2$ is one of its members.

Write the pencil as $\langle q,\lambda^2\rangle$, with $q=0$ smooth. Such a choice exists: Bertini gives smoothness away from $Z$, and curvilinearity makes the differential of every member independent of $\lambda^2$ nonzero at each support point of $Z$. In type $(2,2)$, the line meets $q=0$ in two distinct points. Suitable coordinates give $q=xz-y^2$ and $\lambda=y$. Its only singular members are $y^2$ and $xz$, the latter being the two tangent lines to $q=0$ at these points. In type $(4)$, the line is tangent to $q=0$, and we may take $q=xz-y^2$ and $\lambda=x$. Every finite member $xz-y^2+t x^2$ is smooth. The graph equation identifies the fibres on $S$ with these plane conics scheme-theoretically. In particular, there is precisely one non-reduced fibre, namely $2\ell$.

A length-$k$ part of $Z$ has local ideal $(v,u^k)$. Its $k$ successive point blow-ups resolve the pencil; their first $k-1$ exceptional curves form the $A_{k-1}$ chain contracted on $S$. All base multiplicities are one, giving $F=2h-e_1-e_2-e_3-e_4$. In type $(2,2)$ the two tangent lines have strict transforms $L_1=h-e_1-e_2$ and $L_2=h-e_3-e_4$. They are smooth, have self-intersection $-1$, intersect once away from the base scheme, and are disjoint from the exceptional roots $e_1-e_2,e_3-e_4$. In type $(4)$ the exceptional roots are $e_1-e_2,e_2-e_3,e_3-e_4$. This proves all the assertions. ◻

Proof. The Zariski decomposition of $D_a-vF$ is $$\begin{array}{c|c|c} v&P&N\\ \hline 0\le v\le1&D_a-vF&0\\ 1\le v\le a/2&(a-2v)h&(v-1)(e_1+e_2+e_3+e_4). \end{array}$$ The first positive part equals $(a-2)h+(1-v)F$, so it is nef. In the second interval $P$ is nef, $N$ is exceptional for the morphism to the plane, and $P\cdot N=0$. The pseudo-effective threshold is $a/2$. Thus the volume and the degree of the moving restriction to $F$ are $$\begin{array}{c|c|c} v&P^2&P\cdot F\\ \hline 0\le v\le1&a^2-4-4(a-2)v&2(a-2)\\ 1\le v\le a/2&(a-2v)^2&2(a-2v). \end{array}$$ Their relevant integrals are $$\int_0^{a/2}P^2\,dv=a(a-2)+\frac{(a-2)^3}{6},\qquad \int_0^{a/2}(P\cdot F)^2\,dv =4(a-2)^2+\frac23(a-2)^3.$$ The integrated values are $109/176$ and $37/44$.

The restriction of $e_1+\cdots+e_4$ to $F$ has coefficient $r=2$ at each of its two points in type $2A_1$, and coefficient $r=4$ at its single point in type $A_3$. Indeed, in the latter case $$e_1+\cdots+e_4=(e_1-e_2)+2(e_2-e_3)+3(e_3-e_4)+4e_4,$$ and $F$ meets only the last curve. The additional point moment is obtained by integrating $$2\int_1^{a/2}(P\cdot F)r(v-1)\,dv=\frac r6(a-2)^3.$$ Its integrated value is $5r/176$. This gives the asserted numbers. ◻

Proof. The Zariski decomposition of $D_a-vL_1$ is $$\begin{array}{c|c|c} v&P&N\\ \hline 0\le v\le a-2&D_a-vL_1&0\\ a-2\le v\le1&D_a-vL_1-(v-a+2)L_2&(v-a+2)L_2\\ 1\le v\le a-1&(a-v-1)(2h-e_3-e_4)& (v-a+2)L_2+(v-1)(e_1+e_2). \end{array}$$ The classes $h$, $F$ and $2h-e_3-e_4$ are nef. At the first wall the positive part is $(3-a)F+(a-2)(2h-e_3-e_4)$, and at the second wall it is $(a-2)(2h-e_3-e_4)$. This proves nefness throughout the first two intervals by convexity. The negative supports are negative definite and orthogonal to $P$. The last displayed positive part is nef, and the pseudo-effective threshold is $a-1$.

In these three intervals the volumes are, respectively, $$(a-v)^2-2(1-v)^2-2,\qquad (a-v)^2-2(1-v)^2-2+(v-a+2)^2,\qquad 2(a-v-1)^2,$$ while the degrees on $L_1$ are $a-2+v$, $2a-4$ and $2(a-v-1)$. Direct integration gives $$\int P^2\,dv=\frac{(a-2)(a^2+2a-2)}3,\qquad \int(P\cdot L_1)^2\,dv=\frac{(14-a)(a-2)^2}3.$$ The extra fixed-part point integral at $L_1\cap L_2$ is $$\frac{2(a-2)(a^2-7a+13)}3,$$ and that at $L_1\cap E$ is $4(a-2)^3/3$. For the latter coefficient, use $e_1+e_2=(e_1-e_2)+2e_2$. Applying the preceding integration in $a$ gives $73/88$, $59/88$, $7/44$ and $5/22$, respectively. Adding the appropriate fixed-part term gives the claim. ◻

Proof. All such points are smooth. Apply the surface flag inequality to the fibre through $p$, using its component if the fibre is reducible. The preceding lemmas bound the moving contribution, divided by the log discrepancy, by the larger of the integrated curve and point moments. These bounds also apply to a divisorial valuation whose centre is the flag curve itself.

Here the integrations precede the maximum. To see this directly, work at a sufficiently divisible finite level and choose, in each restricted section space, a basis simultaneously adapted to the given valuation and to the flag curve $C$. Remove the prescribed fixed divisor and the powers of $C$ from its elements. The resulting normalized basis divisor has the form $c_mC+B_m$, where $C$ is not a component of $B_m$. The calculations above bound $c_m$ by the integrated curve moment plus a term tending to zero, and bound the coefficient of $B_m|_C$ at $p$ by the integrated point moment plus a term tending to zero. Inversion of adjunction gives $$\operatorname{ord}_T(c_mC+B_m) \le c_m\operatorname{ord}_T(C) +b_m\bigl(A_S(T)-\operatorname{ord}_T(C)\bigr),$$ where $b_m$ is the latter bound. Passing to the limit gives the maximum used above. This argument uses the bases of the individual spaces; it does not require the moving parts alone to form a graded algebra.

It remains to include the fixed part $(u-1)R|_S$. The curve $R|_S$ is smooth and avoids the singularities of $S$, so $\operatorname{ord}_T(R|_S)\le A_S(T)$ for every divisor $T$ over $S$. Its contribution is therefore at most $$\frac3{22}\int_1^2(u-1)(2-u)(6-u)\,du\, A_S(T) =\frac9{88}A_S(T).$$ This contribution is zero if $p\notin R$. Moreover, $R\cap E$ is empty. Thus a smooth conic flag gives at most $83/88$ at points in $R$, at most $79/88$ at points in $E$ in type $2A_1$, and at most $21/22$ at points in $E$ in type $A_3$. At all other points it gives at most $37/44$. A component of a reducible fibre gives at most $82/88$ on $R$, at most $79/88$ on $E$, and at most $73/88$ elsewhere. Every displayed number is strictly smaller than one. ◻

Proof. Let $Z=c_X(F)$ and choose a general point $p\in Z$. Lemma 103.4 excludes surface centres and smooth quintic fibres. Since the singular quintic fibres are finite in number, $Z$ lies in one of them, say $S$. The estimates just proved exclude the complement of its double conic if $S$ has type $2A_1$ or $A_3$.

It remains to consider a fibre with one $A_1$ or $A_2$ point $o$. By Lemma 103.4, $Z$ is contained in the conic fibre $C_o$ through $o$. This fibre is a pair of distinct lines meeting at $o$. Let $T\in|B|$ be the fibre containing it. Its image in $\mathbb P^3$ is smooth: a rank-three quadric has only double lines or pairs of generator lines as singular plane sections, and in the latter case their intersection is its vertex, which cannot lie on the smooth complete intersection $C$. Here $\pi(o)\in C$. The line $\Lambda$ meets the two components at distinct smooth points, since it is disjoint from $C$ and their intersection is $\pi(o)$. Thus $T$ is a smooth sextic del Pezzo surface.

The third estimate of Lemma 103.4 confines $Z$ to the finitely many intersections of the smooth elliptic curve $E|_T$ with the six $(-1)$-curves. Consequently $Z$ is a point. The strict Abban–Zhuang inequality along $T$, whose refined threshold is at least one, now gives $$\beta_X(F)\ge\operatorname{ord}_F(T) \left(1-\frac{109}{176}\right)>0,$$ a contradiction. This proves the reduction. ◻