The quintic del Pezzo pencil

Proof. The codimension-three Pfaffian resolution of the homogeneous coordinate ring of $V_5$ is $$0\longrightarrow P(-5)\longrightarrow P(-3)^5 \longrightarrow P(-2)^5\longrightarrow P \longrightarrow\mathcal A\longrightarrow0, \qquad P=\mathbb C[x_0,\ldots,x_6].$$ It gives projective normality and $H^1(V,\mathcal O_V(dH))=H^2(V,\mathcal O_V(dH))=0$ for every integer $d$. In particular $\mathcal A$ is Cohen–Macaulay of dimension four. The affine cone over $C$ has dimension two, so $f,g$ form a regular sequence in $\mathcal A$.

After inverting $e$, the proposed graph ring is $\mathcal A[e,e^{-1}]$, of dimension five. Its boundary $e=0$ is $\mathcal A/(f,g)[a,b]$, of dimension four. Thus the two graph relations have height two in the Cohen–Macaulay ring $\mathcal A[a,b,e]$. They form a regular sequence, and their quotient is Cohen–Macaulay of pure dimension five. No minimal component is contained in $e=0$, and a Cohen–Macaulay ring has no embedded associated primes. Hence $e$ is a nonzerodivisor. Localization is therefore injective and proves the domain assertion and the stated extended Rees presentation.

For $m\ge1$, the powers of the ideal of the complete intersection $C=(f=g=0)$ have the resolution $$0\longrightarrow\mathcal O_V(-(m+1)H)^m \longrightarrow\mathcal O_V(-mH)^{m+1} \longrightarrow\mathcal I_C^m\longrightarrow0.$$ Twisting by $2mH$ and using $H^1(V,(m-1)H)=0$ shows that $H^0(V,\mathcal I_C^m(2mH))$ is generated by $f^{m-i}g^iH^0(V,mH)$, $0\le i\le m$. A graph monomial of degree $m(2H-E)$ whose total exponent in $a,b$ is $m+r$ has $e$-exponent $r$ and remaining $H$-degree $m-r$. Its numerator after division by $e^m$ lies in $(f,g)^{m+r}\mathcal A_{m-r}\subset(f,g)^m\mathcal A_m$. Conversely, the terms with $r=0$ give every generator just listed. This identifies the entire graph component with that global section space. The blow-up identity $H^0(X,2mH-mE)=H^0(V,\mathcal I_C^m(2mH))$ proves the assertion; $m=0$ is immediate.

Apply the same dimension and localization argument to $\mathcal A_0,f_0,g_0$. It proves that $\mathcal R_0$ is a Cohen–Macaulay domain. Its bigraded Hilbert series is obtained from that of $\mathcal A_0[a,b,e]$ by multiplying by the two graph-relation factors of degree $H$. It is consequently equal to the bigraded Hilbert series of $\mathcal R$. The graph initial equations give a surjection $\mathcal R_0\to\mathop{\mathrm{gr}}\mathcal R$. Every fixed divisor-degree component is finite-dimensional: in degree $dH-kE$, the total exponent of $a,b$ lies between $\max\{k,0\}$ and $d$, and the exponent of $e$ is then determined. Weight degeneration preserves the dimension of every such component. The surjection is therefore an isomorphism, which proves that no further initial equations occur. ◻

Proof. A fibre is isomorphic to its image $S\in|H|$ on $V$, since blowing up the Cartier divisor $C\subset S$ does not change $S$. Since $\operatorname{Pic}(V)=\mathbb Z[H]$ and $H$ is ample, $S$ is reduced and irreducible. The differentials of two generators of the pencil are independent along $C$, so $S$ is smooth there. A positive-dimensional singular locus would meet the ample curve $C\subset S$. Thus $S$ has isolated singularities and is normal by Serre’s criterion. Adjunction gives $-K_S=H|_S$ and $K_S^2=5$.

By the classification of normal Gorenstein del Pezzo surfaces [26, Theorem 2.2], either $S$ has Du Val singularities or it is an elliptic cone. The vertex of an elliptic cone of degree five has embedding dimension five, whereas a Cartier divisor on a smooth threefold has embedding dimension at most three. The second possibility is therefore excluded. The exceptional curves on the minimal resolution form a root subsystem of $A_4$, giving the stated list. ◻

Proof. Translate the coordinates in (43.2) to a representative. Subtracting constants and linear terms from $x_2,x_3,x_6$ gives the following three nets, in local coordinates $U,V,W$: $$\begin{array}{c|ccc} \dim\operatorname{Aut}(V)p& q_1&q_2&q_3\\\hline 1&U^2+VW&V^2&UV\\ 2&U^2+VW&UV+V^2&UV-U^2-2VW\\ 3&U^2+VW&UW+V^2&UV-W^2. \end{array}$$ Their common zero schemes have the stated lengths and partitions, as one checks on the three affine charts of $\mathbb P^2$. Each ideal is generated by these three quadrics, with two independent linear syzygies. This is the Hilbert–Burch resolution of a noncollinear length-three scheme. ◻

Proof. We use the chart (43.2) and translate to the three orbit representatives. Let $N_1,N_2,N_3$ be the parts of $x_2,x_3,x_6$ left after subtracting their constants and linear terms. A hyperplane section singular at the representative has equation $aN_1+bN_2+cN_3=0$.

At $(0,0,0)$ these polynomials are $$U^2+VW,\quad U^3+UVW+V^2,\quad UV-U^2W-VW^2.$$ The determinant of their quadratic combination is $-a^3/4$. Thus $a\ne0$ gives $A_1$. If $a=0,c\ne0$, normalize $c=1$. The quadratic form $UV+bV^2$ has rank two. Eliminating its two nonsingular directions leaves $-W^5+O(W^6)$: the critical solutions start with $U=W^2+O(W^3)$ and $V=2W^3+O(W^4)$. Hence the singularity is $A_4$. If $a=c=0$, the section is singular along $U=V=0$, which is impossible for a fibre.

At $(0,0,1)$ the three polynomials are $$U^2+VW,\quad U^3+UVW+UV+V^2,\quad UV-U^2-U^2W-2VW-VW^2.$$ The quadratic determinant is $-(a-c)(a-2c)^2/4$. Outside these two lines it gives $A_1$. On $a=c\ne0$, set $c=1$; the radical is $(U,V,W)=(t,0,(b+1)t)$, and the cubic restricted to it is $-t^3$. Thus this case is $A_2$. On $a=2c\ne0$, again set $c=1$. The equation is $$U^2+(b+1)UV+bV^2+bU^3+bUVW-U^2W-VW^2.$$ For $b\ne1$, eliminating $U,V$ leaves $W^4/(b-1)^2+O(W^5)$, since their critical solutions begin with $U=(b+1)W^2/(b-1)^2$ and $V=-2W^2/(b-1)^2$. This gives $A_3$. At $b=1$ the quadratic part has rank one, incompatible with the list in Lemma 43.2. Finally $a=c=0$ gives a section singular along $U=V=0$.

At $(0,1,0)$ the polynomials are $$U^2+VW,\quad U^3+UVW+UW+V^2,\quad UV-U^2W-VW^2-W^2.$$ The quadratic determinant is $-(a+b-c)(a^2-ab+ac+b^2+bc+c^2)/4$. Its three linear factors are permuted cyclically by $(U,V,W)\mapsto(\omega^2U,V,\omega W)$, where $\omega^3=1$ and $\omega\ne1$. This is an actual automorphism fixing the representative: in (43.1) it is induced by the diagonal torus with weights $(x_0,x_5,x_1,x_4,x_2,x_6,x_3)=(0,1,2,3,4,5,6)$. It therefore suffices to consider $c=a+b$. Put $D=a^2+ab+b^2$. A radical vector is $r=(-a-2b,2a+b,a-b)$, and the cubic part takes value $-3D^2$ on $r$. If $D\ne0$, the singularity is $A_2$. If $D=0$, the gradient of the cubic part at $r$ is $D(2a+7b,-a+b,-5a-4b)=0$; the quadratic gradient also vanishes. The section is then singular along the affine line $tr$, and cannot be a fibre. All remaining hyperplane sections have nondegenerate quadratic part and give $A_1$. ◻

Proof. The normal bundle has degree $(-K_V)\ell-2=0$, and is a subbundle of $N_{\ell/\mathbb P^6}=\mathcal O_\ell(1)^{\oplus5}$. Its splitting is therefore one of the two displayed types. The Hilbert scheme of lines is $\mathbb P(\operatorname{Sym}^2\mathbb C^2)$ with its natural $\operatorname{PGL}_2$ action; see [27, Section 4.2]. It has two orbits, the conic of double binary forms and its complement. The line $u=v=0$ is the tangent line to the closed orbit at $w=0$. In the chart at infinity put $t=1/w$, $a=tu$, and $b=tuv-u^2-v/t$. The normal transition has linear part $(a,b)=(tu,-v/t)$, giving the splitting $\mathcal O(1)\oplus\mathcal O(-1)$. The general line is ordinary: the evaluation map from the universal family is generically finite and separable, so its differential is surjective at a general point, which excludes the second normal bundle. Thus the closed line orbit consists precisely of lines of the second kind. Finally, the stabilizer of the closed-orbit point acts on its tangent line as the full triangular subgroup; restricting the binary sextic representation to the span of $x^6,x^5y$ gives exactly the stated fractional-linear transformations. ◻

Proof. In a fixed native-degree piece, filter first by $T$. On each associated graded piece use the filtration induced by the spaces of sections vanishing along $B$. For every integer $q$, taking the $T$-associated graded preserves the dimension of the subspace with $B$-order at least $q$. Thus the total induced $B$-weight is the original $B$-weight. Choose on each $T$-graded piece a basis compatible with this induced filtration and the order at $p$. Lift each vector with its indicated $T$- and $B$-orders. These lifts form a basis compatible with both divisor filtrations, and their $T$-initial forms compute the point invariant. After removing the $T$ factor, a lift with $B$-order $q$ still has the factor $(B|_T)^q$. Its zeros lie away from $p$. Subtract their degree from the total degree of the restricted divisor, sum over the bases, and divide by $mN_m$. Passing to the limit gives the displayed bound. Before restriction the same bases give effective divisors after removal of both factors, proving the assertion about numerical classes. Only two filtrations are imposed on each already restricted graded piece; no simultaneous-basis assertion for three unrelated filtrations is used. ◻

Proof. For the first assertion apply inversion of adjunction to $(G,\Delta+(1-d)T)$ and retain the original coefficient $d$ when comparing discrepancies. The basis-divisor proof of Proposition 2.7 then gives the stated first ratio and the usual curve term. For the second assertion use the same finite-level proof with restricted coefficient tending to one. Keeping, rather than dropping, the coefficient of $G$ gives the displayed inequality in the limit. Its last factor is positive because the centre on the extracted model is contained in $G$. ◻