Extracting a double conic fibre

Proof. The fibre plane is cut out by $F_1|_y$, a nonzero linear form in the four fibre coordinates of $T$. Its intersection with $e=0$ is tangent to the smooth conic $C$: restricting the double-line equation to that line gives twice the point $q$. Choose coordinates on $\Pi$ so that $C=(x^2+s z=0)$, $q=[1:0:0]$, and its tangent line is $z=0$. The fibre-plane equation is therefore a nonzero multiple of $z+c e y_0$. Replace $z$ by $z+c e y_0$. On this plane the other quadric is a square; replacing $x$ by $x+d e y_0$ and scaling its equation makes this square $x^2$. A further replacement of $s$ by $s+c' e y_0$ absorbs the remaining $e y_0 z$ term. These changes preserve $\Pi$, its conic, and the two divisor gradings.

The remaining terms of $F_1$ contain $u$ or $v$; its terms of weight two are linear in $s,t$ times $u,v$. The remaining terms of $F_2$, after $x^2+s z$, contain $e u$ or $e v$, and their weight-two terms are $e$ times a linear form in $s,t$ times $u,v$. This gives (54.1). On $y_0=1$, eliminate $z$. The first transverse terms along $\Lambda$ are $x^2+q_1(s,t)u+q_2(s,t)v$. If $q_1,q_2$ vanished simultaneously, the total space $X$ would be singular at the corresponding point. Thus they define a degree-two morphism $\mathbb P^1\to\mathbb P^1$, whose graph is $D$. At $t=0$ one has $q=sR$, so $R(1,0;u,v)$ is a nonzero linear form. The equations $t=R=0$ cut out one reduced point, and the same identity shows that it belongs to $D$. ◻

Proof. Where $q_1$ is nonzero, eliminate $u$ in the local equation. Then $\mathcal J=(x^2,v)$, and its blow-up is the transverse weighted blow-up of weights $(1,2)$. Its charts are $v=x^2r$ and $x^2=vr$. The second chart is an $A_1$ surface singularity times the smooth curve direction, so the blow-up is normal. Since $g$ is flat, blowing up the fibre ideal is base change from $\mathop{\mathrm{Bl}}_y\mathbb P^2$. Its exceptional scheme is $(2\Lambda)\times\mathbb P^1$, proving the product description and the identity $2F|_F=-f$. The formula for $L|_F$ follows from $L\Lambda=1$.

On the smooth chart, $dx\wedge dv=x^2dx\wedge dr$, giving discrepancy three. On the other chart the index cover for $F$ is $x=ab$, $v=a^2$, $r=b^2$, modulo $(a,b)\mapsto(-a,-b)$. Adjunction gives coefficient $1/2$ on its singular-locus curve; Lemma 54.1 identifies that curve with $D$. The quotient description also proves that $(X^\sharp,F)$ is plt. Finally, $-2F$ is the relatively ample tautological divisor of the blow-up. These facts verify all the extraction hypotheses. ◻

Proof. On $y_0\ne0$, eliminate $z$ and put $t=e y_0$. The remaining equation is $$y_0x^2+q_1(s,t)u+q_2(s,t)v=0.$$ It is irreducible. Indeed, it is a primitive linear polynomial in $u,v$ over $\mathbb C[s,t,y_0^{\pm1},x]$, since $q_1,q_2$ are coprime. Thus this open set is integral of dimension five.

On the boundary $y_0=0$, write $R(s,0;u,v)=s(au+bv)$ and $T(s,0;u,v)=s(cu+dv)$. Here $(a,b)\ne(0,0)$. The two equations become $$s(au+bv)=0,\qquad x^2+s z+es(cu+dv)=0.$$ On $s\ne0$ they cut out a four-dimensional set. On $s=0$ one has $x=0$, and the four variables $z,e,u,v$ remain. Therefore the boundary has dimension four. The two equations have height two, so their quotient is Cohen–Macaulay and unmixed. No component is contained in the boundary; the integral open set has only one component and is generically reduced. Unmixedness now proves that $R_0$ is a domain.

The displayed ideal is contained in the weight initial ideal of $(F_1,F_2)$. The original quotient and the displayed candidate quotient are complete intersections of bidegrees $(1,1)$ and $(2,0)$. The weight initial quotient has the same bigraded Hilbert function as the original quotient, so the candidate and initial quotients have equal Hilbert functions. Each piece is finite-dimensional, despite the negative $B$-degree of $e$, since its exponent is bounded by the $H$-degree. Equality of these Hilbert functions proves equality of the ideals.

Finally, localize at $y_0\ne0$ and at a nonzero $q_i$, and trivialize $H$ by either $s$ or $t$. The weight filtration is then the local $(1,2)$ filtration on the smooth transverse coordinates, hence $\mathop{\mathrm{ord}}_F$. Localization is injective on the initial domain. Therefore no further increase of order occurs for a nonzero global initial form. This identifies the actual divisorial filtration with (54.2). ◻

Proof. Use Lemma 54.3, set $y_0=1$, and substitute $e=t$, $z=R$, and $x^2=-q$. Every monomial becomes $x^\epsilon$ times a polynomial of bidegree $(M,r)$ in $s,t,u,v$. More explicitly, a monomial with $x$-exponent $2a+\epsilon$, $z$-exponent $c$, $e$-exponent $d$, and total $u,v$-exponent $b$ has $r=a+c+b$ and $b\le m+d$. After substitution it contains $q^aR^ct^d$. Since $q=sR+tT$, it lies in $(t,R)^k$ whenever $r>m$. The same monomial has $a+c+d\le m-\epsilon$, so $r\le2m-\epsilon$. This proves the inclusion and the vanishing assertion.

For $r\le m$, every polynomial of bidegree $(M,r)$ lifts, by using $s$, $t=e y_0$, $u,v$, and enough factors $y_0$. For $r>m$, one has $k=r-m\le M$. The ideal sheaf of $P$ is generated by $t$ of bidegree $(1,0)$ and $R$ of bidegree $(1,1)$. Its $k$th power is generated by $t^{k-c}R^c$, for $0\le c\le k$. The power resolution, after twisting, is $$0\longrightarrow\bigoplus_{c=0}^{k-1} \mathcal O_F(M-k-1,r-c-1) \longrightarrow\bigoplus_{c=0}^{k} \mathcal O_F(M-k,r-c) \longrightarrow\mathcal I_P^k(M,r)\longrightarrow0.$$ The first term has no $H^1$, since $M-k-1\ge-1$ and $r-c-1\ge0$. Consequently, every section is a sum of $t^{k-c}R^c$ times polynomials of bidegree $(M-k,r-c)$. Lift $R$ to $z$, $t$ to $e$ in these fixed factors, and $t$ to $e y_0$ in the remaining polynomial. The inequality for the available base degree is $r-c=m+(k-c)$, exactly the required equality. Thus every section lifts to a section of $mL$ of native weight $2r+\epsilon$ after multiplication by $x^\epsilon$.

At a point of $\mathbb P^1\times\mathbb P^1$, the condition of vanishing to order $k$ imposes $k(k+1)/2$ independent conditions on bidegree $(M,r)$ when $M,r\ge k-1$. One can check independence on the local monomials of total degree less than $k$. This gives (54.3). The odd factor has square $-q$, so its reflexive zero divisor is $D/2$. Its class checks the formula: $(M,r)+\epsilon(1,1/2)=(m,j/2)$, the class of $(m\omega^*L-jF)|_F$. ◻