Geometry and the local estimates

Proof. The cubic equations generate the ideal sheaf of $C$. At a point of $C$, their differentials span its two-dimensional conormal space. The cubics singular at that point form one point of the parameter plane. Varying the point of $C$ gives a subset of dimension at most one. Bertini applies away from $C$, so a general cubic in the net is smooth.

Adjunction on $S_0$ gives $C^2=2g(C)-2+A\cdot C=15$. Thus $D=C-2A$ satisfies $A\cdot D=1$ and $D^2=-1$. Riemann–Roch gives $\chi(\mathcal O_{S_0}(D))=1$, while $h^2(D)=h^0(-A-D)=0$ by ampleness of $A$. Hence $D$ is effective. Its $A$-degree is one, so it is a line. Two distinct lines have nonnegative intersection and cannot both represent a class of square $-1$, proving uniqueness.

The blow-up of the Cartier divisor $C\subset S_0$ is $S_0$ itself. The restriction formulas follow. The planes through $\ell$ cut out $\ell$ plus a residual conic, giving the base point free pencil $|A-\ell|$. ◻

Proof. We include the algebra behind this familiar determinantal description. On a smooth cubic from Lemma 10.3, the hypersurface sequence reduces the vanishing of $H^1(\mathcal I_C(n))$ to that of $H^1(S_0,tA-\ell)$ for every integer $t$. For $t\ge0$, use $0\to\mathcal O(tA-\ell)\to\mathcal O(tA)\to\mathcal O_\ell(t)\to0$; the restriction of degree-$t$ polynomials to a line is surjective and $H^1(tA)=0$. For $t<0$, Serre duality reduces the assertion to $H^1(kA+\ell)=0$, $k=-t-1\ge0$, and the sequence with quotient $\mathcal O_\ell(k-1)$ gives this vanishing. Thus $C$ is arithmetically Cohen–Macaulay.

For completeness, the three given cubic equations generate the ideal sheaf, so their kernel $\mathcal K$ is locally free of rank two. The preceding vanishing gives $H^2(\mathcal K(n))=0$ for every $n$. Since $\det\mathcal K=\mathcal O(-9)$, rank-two duality and Serre duality give $H^1(\mathcal K(n))=0$ as well. Horrocks’ splitting criterion [19] therefore writes $\mathcal K=\mathcal O(-a)\oplus\mathcal O(-b)$. There is no constant relation among the three cubics, and $a+b=9$, so $\{a,b\}=\{4,5\}$. The Hilbert–Burch theorem [16] now gives the displayed matrix and its minors.

If the linear column had rank two, coordinates would make it $(x,y,0)$. Its minors would be $yq_2,-xq_2,xq_1-yq_0$. Unless $q_2=0$, irreducibility and nonplanarity of $C$ would force $C\subset(q_2=0)$. But $C$ lies on no quadric: on $S_0$, such a quadric would give an effective divisor $2A-C\sim-\ell$. If $q_2=0$, the minors do not have codimension two. Rank at most one similarly forces a planar component or the wrong codimension. Thus the linear forms are independent. They vanish at a point $p_0$, which belongs to the determinantal curve. If also $q(p_0)=0$, all the minors have zero differential there, contradicting smoothness of $C$.

Twisting the resolution by $4$ identifies the space of linear syzygies with $H^0(\mathcal O_{\mathbb P^3}\oplus\mathcal O_{\mathbb P^3}(-1))=\mathbb C$, so its nonzero element is unique up to a scalar. To check independence of the cubic generators, regard $f$ and $l$ as columns, so that the relation is $l^{\mathsf t}f=0$. If $f'=Mf$ for $M\in\operatorname{GL}_3(\mathbb C)$, its linear syzygy is a nonzero scalar multiple of $M^{-\mathsf t}l$. Invertible recombination and multiplication by a nonzero scalar preserve the span of the three coefficients. Hence $U_C$ and its common zero $p_0$ are intrinsic to $C$. In particular, their construction does not involve the choice of a cubic surface through $C$. ◻

Proof. Off $C$, the two equations specify the point $y=l\times q$. Along $C$, their coefficient matrix has rank one: rank zero would make the differentials of all its minors vanish, contradicting smoothness of $C$. Near any point of $C$, a unit matrix entry allows row and column operations reducing the incidence equations to the usual equation for the blow-up of a two-generated ideal defining a smooth codimension-two centre. Thus (10.1) is the graph of the cubic net, including over $C$. The restriction formulas follow.

For each $y$, the linear form $l_y=\sum l_i y_i$ is nonzero because the $l_i$ are independent. Its zero plane is denoted by $\Pi_y$. The quadratic $q_y=\sum q_i y_i$ does not vanish identically on $\Pi_y$. Otherwise, after an invertible linear recombination of the three rows, $l_y=l_0$ and $q_y=q_0=l_0r$ for a linear form $r$. On $\Pi_y$, two minors then vanish identically and the remaining one is a cubic. Their common zero scheme contains a plane curve, contradicting the fact that $C$ is irreducible and nonplanar. Hence the second equation cuts out a nonzero conic on every plane fibre. It is a relative effective Cartier divisor of degree two in the plane bundle, and its equation remains a nonzerodivisor on every fibre. This proves flatness. ◻

Proof. Write a cubic of the net as $S_a=(a\cdot f=0)$, where $a\in\mathbb P^2$ is a coefficient vector. The condition that it contains the conic over $y$ is $a\cdot y=0$, a line in the parameter plane. For $c\in C\setminus\{p_0\}$, the relation $q(c)\parallel l(c)$ and the rank-two Jacobian of $C$ show that the unique cubic singular at $c$ has coefficient $a=l(c)$. At $p_0$ its coefficient is $q(p_0)$. This defines the extension of the projection $c\mapsto l(c)$. Its image is not a line, since otherwise $C$ would lie in a plane through $p_0$. Hence only finitely many members of our parameter line are singular along $C$.

On $X$, Bertini makes a general member of the pencil smooth away from its base fibre. Along a smooth point of that fibre, $\varphi$ is smooth. If the fibre is a union of two lines, it has one node; locally the conic equation has a nonzero derivative in a base direction there, since $X$ is smooth. Choosing $M$ transverse to that direction makes its inverse image smooth at the node. Thus a general $S$ is smooth, and the preceding paragraph ensures that its cubic image is smooth along $C$ too.

For $a=[a_0:a_1:a_2]$, let $\ell_a\subset\mathbb P^3$ be the line defined by $l_i(x)a_j-l_j(x)a_i=0$ for $0\le i<j\le2$. Independence of the $l_i$ shows that this is a line through $p_0$. Its equations imply $a\cdot(l\times q)=0$, so it lies in $S_a$. If $S_a$ is smooth, this line cannot meet $C$ away from $p_0$, for such an intersection would give $a=l(c)$ and make $S_a$ singular at $c$. At $p_0$ it is transverse to $C$: the tangent direction of $C$ in the $l$-coordinates is $q(p_0)$, and $a\not\parallel q(p_0)$. Its intersection with $C$ is therefore one. It is the line $\ell=\ell_{S_a}$ of Lemma 10.3, since any other line has intersection at least two with $2A+\ell$. Distinct coefficient vectors $a\in\mathbb P^2$ give distinct lines $\ell_a$: projection from $p_0$ sends $\ell_a\setminus\{p_0\}$ to $a$. Thus the line depends on the chosen cubic $S_a$, whereas their common point $p_0$ depends only on $C$.

If $P\notin E$ and $\pi(P)\ne p_0$, at most one coefficient $a$, namely $a=l(\pi(P))$, makes $\ell$ pass through $P$. Avoid that coefficient. If $P\in E$ lies above $c\ne p_0$, the line $\ell$ avoids $P$ automatically, as it meets $C$ only at $p_0$. ◻

Proof. A nonreduced plane conic is a double line. If the rank-one locus of the conic equation contained a curve in the base, work at its general point and write the equation as $x^2+u q(y,z)+$ higher terms in the transverse base parameter $u$. Differentiation in the direction tangent to the rank-one locus vanishes along $x=u=0$. The binary quadratic $q$ has a zero on the double line. At that zero the other base derivative vanishes as well, making the total space singular. This contradicts smoothness of $X$. The rank-one locus is closed and hence is finite on the projective surface $\mathbb P^2$. ◻