The full refinement at a reduced-fibre node on $E$

Proof. The seven $H$-sections on this chart are $$1,\ y,\ -A-by,\ Ab+(b^2-A)y,\ zA,\ zb,\ z,$$ and the three $B$-sections are $1,b,A$. Among their products take the sixteen expressions obtained from the displayed table by replacing $q,Q,g$ by $-A-by,A,Ab+(b^2-A)y$, respectively. Their initial forms are exactly the table. It remains to prove generation; generation of the original ring alone would not imply generation of its initial algebra.

Inside the same projective bundle $P$ consider instead $$X_0=(q_0+fzy_1+hCz^2=0).$$ On $C=y_0=1$ its equation is exactly $A=Q$, so this chart is $\mathbb A^3_{b,y,z}$. Its generic conic is nonsingular: its discriminant is $C(\kappa AC-hB^2)$, a nonzero polynomial. The quadratic $q_0$ is nonzero on every base fibre, including the branch conic, where it has rank one. Thus this hypersurface has no vertical component and is integral. We do not assume that $X_0$ is smooth or K-semistable.

Here is the needed section calculation. For $r\ge0$, the Steiner resolution of $G$ gives the symmetric Koszul resolution $$0\longrightarrow \mathcal O(-2)\otimes\operatorname{Sym}^{r-2}\mathbb C^4 \longrightarrow \mathcal O(-1)^{\oplus2}\otimes\operatorname{Sym}^{r-1}\mathbb C^4 \longrightarrow \mathcal O\otimes\operatorname{Sym}^r\mathbb C^4 \longrightarrow\operatorname{Sym}^rG\longrightarrow0,$$ where terms with negative symmetric exponent are omitted. After twisting by any $d\ge0$, this proves vanishing of higher cohomology and surjectivity on sections from the third term. Consequently $$H^0(P,mH+mB)=\bigoplus_{a=0}^m z^a H^0(\mathbb P^2,\operatorname{Sym}^{m-a}G(m+a))$$ is generated by degree-one sections: a monomial of base degree $m+a$ factors into $a$ quadratic and $m-a$ linear factors. The ideal of $X_0$ in this degree is $(m-2)H+(m+1)B$. For $m\ge2$ its pushforward has summands $\operatorname{Sym}^{m-2-a}G(m+1+a)$ and has no higher cohomology. For $m=0,1$ all fibre cohomology vanishes because the relative degrees are $-2,-1$. Restriction is therefore surjective in every degree. In particular the section ring of $(X_0,L)$ is generated in degree one and has the same Hilbert function as $(X,L)$: $$N_m=\frac{(2m+1)(13m^2+13m+6)}6.$$

Restriction to the displayed dense affine chart embeds this section ring in $\mathbb C[b,y,z,u]$, where $u$ records the anticanonical degree. Its sixteen degree-one generators are precisely $u$ times the table. Their independence also follows directly, degree by degree, from the polynomials in the table. The algebra they generate is contained in the actual initial algebra of $X$. In degree $m$ the former already has dimension $N_m$, while the latter has dimension $\dim R_m=N_m$. The inclusion is consequently an equality in every degree. ◻

Proof. At each marked point, use regular coordinates whose second coordinate vanishes on the specified tangent. Direct substitution in Lemma 80.2 shows that the weight-$(1,2)$ order of every degree-one generator is at least $j-1$. Its ordinary order at $p_c$ is at least $j-2$. These inequalities multiply, so all products satisfy the conditions defining $D_{m,j}$. The two successive blowups express the first condition as ordinary order at least $\lceil r/2\rceil$ and weighted order at least $r$.

When $h\ne0$, every generator of degree five is divisible by $Q$, and every generator has degree at most five. Thus products of anticanonical degree $m$ and point degree $j$ are divisible by $Q^{\max\{j-4m,0\}}$. When $h=0$, the table gives the stronger divisibilities $y^{\max\{j-3m,0\}}$ and $(b+y-z)^{\max\{j-4m,0\}}$. This proves inclusion in the asserted spaces. These are algebraic fixed-component conditions; in particular the first coefficient in the second case must not be replaced by the smaller coefficient in the Zariski decomposition of $D_{m,j}$ alone.

We verify equality by dimensions. For reference the negative curves on $T$ in the first case are $$\begin{gathered} r_i=e_i-f_i,\quad f_i,\quad t_i=\ell-e_i-f_i,\quad l_{ik}=\ell-e_i-e_k\quad(i\ne k),\\ Q_i=2\ell-\sum_{k=a,d,c}(e_k+f_k)+f_i. \end{gathered}$$ There are fifteen, with $l_{ik}=l_{ki}$. In the second case they are the twelve curves $$r_i,\ f_i\ (i=a,d,c),\qquad L_1,L_2,t_c,l_{ad},l_{dc},Q_a.$$ Here $Q_a$ is the smooth conic $y^2-yz+bz=0$. These are weak del Pezzo surfaces of degree three. Here is a direct check of nefness of the anticanonical class. In the first case, for each $i$, the effective divisor $Q_i+(\ell-e_i)+r_i$ is anticanonical, where the middle term is a general line through $p_i$. These divisors have no common curve component. In the second case use the same divisors for $i=a,d$, interpreting $Q_d=L_1+l_{dc}+r_c$. Again they have no common component. Thus $-K_T$ is nef, and its square is three. To check the negative-curve lists, the three points are noncollinear. In the first case none of the three specified tangents joins two marked points, and the three five-point conics are smooth. There is no conic through all six points. In the second case the only additional root is $L_1$, meeting $r_c$; the other roots are $r_a,r_d$. A line or conic gives all remaining negative curves listed above. Equivalently one may enumerate the plane classes with square $-1,-2$ and anticanonical degree $1,0$; the proximity inequalities give exactly these lists. There are no higher-degree candidates: Cauchy’s inequality applied to $\sum m_i=3d-1$, $\sum m_i^2=d^2+1$ in the $(-1)$ case, and to $\sum m_i=3d$, $\sum m_i^2=d^2+2$ in the $(-2)$ case, forces $d\le2$ in both cases. Adjunction and nefness of $-K_T$ exclude more negative irreducible curves.

We now prove nefness in every degree using six fixed classes. Put $P=-K_T$, $U=2\ell-e_a-e_d-e_c$, and put $L=0$ in the first case and $L=L_1$ in the second case. Define $$\begin{aligned} A_2&=8\ell-3(e_a+f_a+e_d+f_d)-4e_c-2f_c-2L,\\ A_1&=4\ell-2(e_a+e_d)-f_a-f_d-2e_c-f_c-L,\\ A_0&=3\ell-e_a-f_a-e_d-f_d-2e_c-f_c-L. \end{aligned}$$ The classes $\ell,P,U$ are nef: the first is a pullback of a line, the second was treated above, and the third is the pullback of the base-point-free conic system through the three proper points. For $A_2,A_1,A_0$, intersection with each curve in the relevant negative-curve list is nonnegative. Their squares are $8,1,0$, respectively. Their line degrees are $8,4,3$ in the first case and $6,3,2$ in the second. They therefore lie in the closure of the positive cone, in its component containing $\ell$. Every irreducible curve of nonnegative square lies in the same closed component; the Hodge index theorem gives nonnegative intersection with these curves as well. Thus all six classes are nef.

Write $M_{m,j}$ for the proposed moving divisor. The following identities express it as a nonnegative integral combination of these classes and also account for every parity: $$\begin{array}{c|c|l} \text{range}&\text{indices}&M_{m,j}\\ \hline 0\le j\le m&&j\ell\\ m\le j\le3m&j-m=2k+\epsilon& (m-k-\epsilon)\ell+kP+\epsilon U\\ 3m\le j\le4m&j-3m=q=2k+\epsilon& (m-q)P+kA_2+\epsilon A_1\\ 4m\le j\le5m&q=j-4m,\quad m-q=2k+\epsilon& qA_0+kA_2+\epsilon A_1, \end{array}$$ where $\epsilon\in\{0,1\}$. In the second row $2k+\epsilon\le2m$ implies $m-k-\epsilon\ge0$; all other coefficients are visibly nonnegative. The identities follow by substituting the coefficients of $D_{m,j}$ and its stated fixed part. In particular, in the last two rows the subtraction of $L$ is retained when $h=0$. This proves nefness for all $m,j$, without extrapolating from finitely many degrees. For a nef integral divisor $M$ on either surface, Kawamata–Viehweg vanishing applied to $M-K_T$ gives $$h^0(T,M)=1+\frac{M^2-M\cdot K_T}{2}.$$ Let $M_{m,j}$ denote the moving divisor in either case. Its Riemann–Roch value is the same in the two cases. Summing the displayed quadratic polynomial gives the identities $$\begin{aligned} \sum_{j=0}^{5m}h^0(T,M_{m,j})&= \frac{(2m+1)(13m^2+13m+6)}6=N_m,\\ \sum_{j=0}^{5m}j h^0(T,M_{m,j})&= \frac{143m^4+208m^3+133m^2+32m}{12}. \end{aligned}$$ For an explicit way to perform the sum, take $j\le m$ first, where the summand is $\binom{j+2}{2}$. In the remaining terms write $j=m+2k+\epsilon$, $\epsilon=0,1$, and split at $j=3m,4m$. Substitute $a=k+\epsilon$, $b_0=k$ in $D_{m,j}$ and use $K_T=-3\ell+\sum(e_i+f_i)$. The sums of $1,k,k^2,k^3$ give the two identities, for both parities of $m$. This is an all-degree polynomial identity.

The inclusions already proved have total dimension at most the left side of the first identity. Their actual total dimension is $N_m$ by Lemma 80.2. Every inclusion is therefore an equality. Generation in anticanonical degree one also proves vanishing above $5m$. ◻

Proof. Put $t=j/m$. The asymptotic moving divisors of Lemma 80.3, in the case $h\ne0$, are $$\begin{array}{c|l} 0\le t\le1&P_t=t\ell\\ 1\le t\le3&P_t=t\ell-\frac{t-1}{2}\sum_i(e_i+f_i)\\ 3\le t\le4&P_t=t\ell-\frac{t-1}{2}(e_a+f_a+e_d+f_d) -(t-2)e_c-f_c\\ 4\le t\le5&P_t=(8-t)\ell-\frac{7-t}{2}(e_a+f_a+e_d+f_d)-2e_c-f_c. \end{array}$$ If $h=0$, replace the third row by $P_t-(t-3)L_1$ and the fourth by $P_t-L_1$, leaving the first two rows unchanged. In either case the squares are, respectively, $$t^2,\qquad-\tfrac12t^2+3t-\tfrac32,\qquad -t^2+6t-6,\qquad10-2t.$$ The normalization is $\frac3{26}\int_0^5P_t^2\,dt=1$. The exact section spaces give $\operatorname{vol}(L-tJ)=3\int_t^5P_u^2\,du$, hence the displayed function. Finally $$S_X(J)=\frac1{26}\int_0^5\operatorname{vol}(L-tJ)\,dt =\frac3{26}\int_0^5tP_t^2\,dt=\frac{11}{4}.$$ The weighted Hilbert identity in the preceding lemma is an independent finite-degree computation of the same limit. ◻