The distinguished surface and a double conic

Proof. On $\{u_0=0\}$ the equation is $y_0q_0(u_1,u_2)+y_1q_1(u_1,u_2)=0$. If the binary quadratics $q_0,q_1$ had a common factor, this equation would be reducible, contrary to the hypothesis on $D$. They are therefore relatively prime. Consequently, the curve defined by this equation in $\mathbb P^1_y\times\mathbb P^1_{u_1,u_2}$ is the graph of the morphism $[u_1:u_2]\mapsto[-q_1:q_0]$, of degree two. Denote this graph by $R$. Thus $D=\mathbb P^1_x\times R$. On this product, $A|_D=\mathcal O(1,0)$, $B|_D=\mathcal O(0,2)$, and $(\xi+A+B)|_D=\mathcal O(0,1)$, whence $\xi|_D=\mathcal O(-1,-1)$. A degree-two morphism between smooth projective lines in characteristic zero has two ramification points, with distinct images.

Suppose that the conic over a base point is twice a line $\ell$. The line $\ell$ cannot be $\{u_0=0\}$: otherwise the corresponding member of the pencil $\langle q_0,q_1\rangle$ would vanish identically. Its restriction to $u_0=0$ is therefore a nonzero square. This is exactly the ramification condition for $R\to\mathbb P^1_y$, and the unique zero of the square is $D\cap\ell$. ◻

Proof. The two singular members of the base-point-free pencil of binary quadratics defining $D$ are distinct squares. Hence we may arrange $q=y_0u_2^2+y_1u_1^2$. Choose the first branch value to be $y_1=0$, and move the first coordinate of the double conic to $x_1=0$. Since the conic there has rank one, its equation is a nonzero multiple of $(u_2+\lambda u_0)^2$ on $x_0=y_0=1$. The homogeneous change $z=u_2+\lambda x_0y_0u_0$ is permitted by the classes of these coordinates. Rescaling gives the stated coefficient of $z^2$.

The quadratic terms in $u_1,u_2$ are independent of $x$. Thus the coefficient of $a$ in the initial equation is divisible by $s$, whereas the coefficient of $b$ restricts to $r^2$ on $s=0$. Homogeneity gives exactly the displayed form of $F_0$. At a point of $\ell$, both fibre derivatives vanish because the fibre is a double line. The two remaining derivatives are $f$ and $g$. Smoothness of $X$ therefore implies that $f,g$ have no common zero. In particular $f$ is not identically zero.

Write $g_0=r^2+e x_0y_0u_0r+h x_0^2y_0^2u_0^2$. Consider $F_0$ as a polynomial of degree one in $y_1$. Its leading coefficient is $g_0$. A common divisor of this coefficient and its constant term would divide $y_0z^2$; since $g_0$ is independent of $z$ and is not divisible by $y_0$, no such divisor exists. Gauss’s lemma now proves irreducibility. Finally, the conditions obtained above only assert that $f,g$ are nonzero and relatively prime. They do not assert that their individual roots are distinct. ◻

Proof. The fibre ideal is $(a,b)$. Wherever $g\ne0$, the equation allows us to eliminate $b$, and $(a,b)=(a,z^2)$. More precisely, write the equation as $z^2+aF+bG_1=0$ with $G_1$ a unit on this open set; then the two ideals coincide in the local ring of $X$. The analogous statement holds wherever $f\ne0$. These two open sets cover $\ell$. The normalization of the blowup of $(a,z^2)$ is the weighted blowup of the regular parameters $(z,a)$ with weights $(1,2)$. Its log discrepancy is $1+2=3$.

The weighted exceptional equation is $z^2+af+bg=0$, with $z$ of weight one and $a,b$ of weight two. Its second Veronese eliminates $z^2$ and leaves the two base coordinates $a,b$. Since $a,b$ are coordinates on the fixed base surface, their transition functions along $\ell$ are trivial. Hence its coarse surface is the stated product. The weight-two chart is an index-two quotient chart; the index-two locus on $G$ is $z=0$. Adjunction on the index-one cover therefore gives different $\Gamma/2$. The same quotient description proves that the extraction is purely log terminal. The coprimality of $f,g$ identifies $\Gamma$ with the graph of a morphism, proving its smoothness.

For the last assertion, the normal sequence of $\ell\subset X\subset Q$ is $$0\longrightarrow N_{\ell/X}\longrightarrow \mathcal O(1)\oplus\mathcal O\oplus\mathcal O \xrightarrow{(0,f,g)}\mathcal O(2)\longrightarrow0.$$ The map is surjective by coprimality, and the kernel of its last two summands is $\mathcal O(-2)$. This proves the splitting. ◻

Proof. The cubic fibre is $a=0$. At a point of $\ell$, its derivative in the $b$ direction is $g$, so its singular points on $\ell$ are precisely the zeros of $g$. If $v$ is a local parameter on $\ell$ at a simple zero, the quadratic part of its local equation in $(z,b,v)$ is $z^2+\kappa bv+\lambda bz+\nu b^2$, with $\kappa\ne0$. It is nondegenerate, so the singularity is $A_1$. At a double zero, the quadratic part has no term involving $v$ and is degenerate; the singularity is not $A_1$. These assertions concern the surface only along $\ell$, as required by the cited generality condition. If $g$ is a square, replace $r$ by $r+\lambda s$ to obtain $g=r^2$. The form of $f$ is preserved by this change, and coprimality gives $d\ne0$. ◻