Points of type $A_2$

Proof. The tangent conic of the fibre is reduced, with two components. Its vertex $r$ is not in the scheme $Z$ of Lemma 43.3. Indeed, a point of $Z$ is the tangent direction of a line on $V$ through $p$. Every hyperplane section singular at $p$ contains that line, so its cubic part vanishes in this direction. At an $A_2$ point the cubic part is nonzero on the radical of the quadratic part, which gives a contradiction.

Thus $Z$ is contained in the two components away from their intersection. Since $Z$ is noncollinear, its lengths on these components are two and one. Let $x=0$ be the component carrying length two, and let $y=0$ be the other component. The quadrics containing $Z$ and divisible by $x$ form a pencil. It contains $xy$ and a quadric $x\ell$, where $\ell(r)\ne0$: the residual linear forms are precisely the lines through the length-one point on $y=0$. Complete $x,y$ by a linear coordinate $z$ nonzero at $r$. The two remaining normal hyperplane coordinates can therefore be chosen to have weighted leading forms $xz$ and $z^2$. For the second assertion use that some quadric of the net is nonzero at $r$, since $r\notin Z$.

The fibre has local equation $xy+\gamma z^3$ plus terms of weight greater than six, with $\gamma\ne0$. Rescaling the coordinates gives $xy-z^3$. Together with the three linear coordinates and the constant section, this proves the claimed list of leading forms.

We verify that no further leading generators have been missed. The relations in (49.1) are the five maximal Pfaffians of the skew-symmetric matrix $$\begin{pmatrix} 0&h&x&z&0\\ -h&0&P&y&z\\ -x&-P&0&R&Q\\ -z&-y&-R&0&P\\ 0&-z&-Q&-P&0 \end{pmatrix},$$ with the following signs. Deleting rows and columns $1,\ldots,5$ in order gives $$P^2-yQ+zR,\quad xP-zQ,\quad hP-z^2,\quad hQ-xz, \quad hR-xy+zP.$$ Thus its maximal Pfaffians generate precisely the displayed ideal. After inverting $h$ they give a polynomial algebra in $h^{\pm1},z,x,y$. On $h=0$ the reduced equations are $z=0$, $xy=xP=0$, and $yQ=P^2$; both components have dimension three. Thus the ideal has height three and its Pfaffian resolution is Cohen–Macaulay. No component is contained in $h=0$, so the algebra is a domain. Its ordinary Hilbert series is $(1+3t+t^2)/(1-t)^4$, the Hilbert series of $V$.

The algebra generated by the displayed leading forms embeds in the associated graded section algebra. Since both have the same ordinary Hilbert function, they are equal. This dimension argument checks all possible cancellations, not only quadratic relations.

For the Cox algebra, the same argument applies after $h=ae,R=be$. The boundary $h=R=0$ in (49.1) has dimension two; therefore the boundary $e=0$ after substitution has dimension four. The complementary open set has dimension five and is a domain. The height-three Pfaffian resolution proves Cohen–Macaulayness. The dimension-two common zero locus says exactly that $h,R$ form a regular sequence. Lemma 43.1 now identifies the substituted algebra with the full initial global section algebra and proves its primeness. ◻

Proof. The exceptional divisor is $\mathbb P(2,3,3)$. Its coarse space is $\mathbb P(1,1,2)$ with coordinates $x,y,t=z^3$, and $\Delta_G=\frac23\{t=0\}$. The discrepancy is the sum $2+3+3=8$.

In each native-degree piece take the lowest $z$-power, refining it by any fixed monomial order. The leading term of $B$ is $xy$. The thirteen distinct degree-one leading monomials are $$1,z,x,y,z^2,zx,xy,zxy,x^2y,xy^2,z^2xy,zx^2y,x^2y^2.$$ They form a basis of the degree-one space. Their exponent vectors, with coordinates ordered as $(z,x,y)$, generate the full degree-prefixed lattice: the monomials $1,z,x,y$ already suffice. Their convex hull has vertices $$\begin{gathered} (0,0,0),(0,0,1),(0,1,0),(0,1,2),(0,2,1),(0,2,2),\ (1,1,0),(1,2,1),(2,0,0),(2,1,1). \end{gathered}$$ Its volume and first moment are, respectively, $$\frac{10}{3},\qquad \left(\frac{55}{24},\frac{25}{8},\frac{65}{24}\right).$$ For an explicit description, this polytope is cut out by $\alpha,\beta,\gamma\ge0$ and $$\begin{gathered} \beta-\gamma\le1,\quad\beta\le2,\quad \alpha-2\beta+2\gamma\le2,\quad\alpha+2\gamma\le4,\ \alpha+\beta-\gamma\le2,\qquad \alpha+\beta+\gamma\le4. \end{gathered}$$ Triangulating its boundary and using the determinant formula gives the stated volume and moment. Since the volume is the leading coefficient of $N_m$, the semigroup-volume lemma shows that the actual asymptotic pivot means for the full refined spaces are $$\mu_z=\frac{11}{16},\qquad \mu_x=\frac{15}{16},\qquad \mu_y=\frac{13}{16}.$$ The native weighted degree is $2\alpha+3\beta+3\gamma$. Thus this also computes the expected vanishing of the primary extraction directly: $S_X(G)=2\mu_z+3\mu_x+3\mu_y=53/8$.

We spell out the log-canonicity step because the exceptional weighted projective space is not well formed. In a fixed finite degree, choose a one-parameter subgroup which realizes the chosen monomial order on all monomials occurring in that degree. In every native-degree piece choose a basis simultaneously adapted to this weight filtration and an arbitrary divisorial valuation over $G$. Its limit basis consists of the pivot monomials. On the coarse surface the limiting basis divisor therefore has asymptotic coefficients $$\frac{11}{48}\{t=0\} +\frac{15}{16}\{x=0\} +\frac{13}{16}\{y=0\}.$$ Together with the different $\frac23\{t=0\}$, this toric boundary is log canonical after multiplication of the basis divisor by any $c<16/15$: the three coefficient conditions are $\frac23+c\frac{11}{48}\le1$, $c\frac{15}{16}\le1$ and $c\frac{13}{16}\le1$. They can also be checked on the index-two cover at the vertex of $\mathbb P(1,1,2)$; elsewhere the toric boundary has simple normal crossings. Log-canonical-threshold semicontinuity and the simultaneous basis construction give the same bound for the original valuation. Passing to the asymptotic threshold and then letting $c$ increase to $16/15$ proves the assertion. ◻

Proof. Abban–Zhuang gives $\delta_p(X)\ge\min\{8/(53/8),16/15\}=16/15$. ◻