The weighted extraction at a repeated residual direction

Proof. The one-sided nonvanishing conditions give $C_2\neq0$ and $B_3\neq0$; proportionality therefore gives $B_2C_2\lambda\neq0$. Write $B(z)=B_2(z_1+\lambda z_2)$ and $C(z)=C_2(z_1+\lambda z_2)$. Their common root $r=[0:-\lambda:1]$ is distinct from the two coordinate endpoints. The matrix $N(r)$ has only its upper-left entry possibly nonzero. That entry is nonzero, since otherwise the full matrix $M(r)$ would have rank at most two, contradicting smoothness of the determinant quartic. In particular $A_3-\lambda A_2\neq0$.

Let $\eta$ denote the coefficient of $x_3y_3$ in $F_1$. We prove $\eta\neq0$ explicitly. At $r$, the zeroth row of $M(r)$ forces a right kernel vector to have second coordinate $y_2=0$; its first row and the nonzero upper-left entry of $N(r)$ then force $y_1=0$. Its right kernel vector is consequently of the form $(u,0,0,1)$, after scaling. Similarly its left kernel vector is of the form $(v,0,0,1)$. The zeroth row and column of $M_1$ are supported only in coordinate $1$, while $(M_1)_{00}=0$. Therefore $$(v,0,0,1)M_1(u,0,0,1)^{\mathsf t}=\eta.$$ The rank of $M(r)$ is three, so Jacobi’s formula says $\partial\det M/\partial z_0(r)$ is a nonzero scalar times $\eta$. The restriction of $\det M$ to $z_0=0$ has a double root at $r$, so its other two first derivatives vanish at $r$. Smoothness of the quartic forces its $z_0$-derivative to be nonzero; hence $\eta\neq0$.

For completeness, the required normalization uses only rescalings. Set $B=B_2,C=C_2$. Choose nonzero scalars $l,m,k,n$ with $$m/l=-B/C,\qquad k=\eta lm,\qquad n=Blk.$$ Replace the old coordinates $x_3,y_3,x_1,y_1,x_2,y_2$ by $lx_3,my_3,kx_1,ky_1,nx_2,\lambda n y_2$, respectively, and divide $F_1,F_2,F_3$ by $k,n,\lambda n$. Their lowest weighted terms are exactly those displayed. All remaining terms in $F_1$ have weight at least three. The $x_1y_1$ terms in $F_2,F_3$ have weight four, and their other omitted terms contain $x_2$ or $y_2$ and another nonconstant coordinate, hence also have weight at least four.

Let $G_1,G_2,G_3$ denote the three displayed initial forms. Their locus on $x_0y_0\neq0$ is the asserted graph, hence an affine three-space. We check that its projective boundary has dimension at most two. If $x_0=0,y_0\neq0$, then $x_3\neq0$: otherwise $G_1,G_2$ force $x_1=x_2=0$. The first two equations determine $x_1,x_2$, and $G_3$ becomes the nonzero condition $$-y_0y_1+y_3^2=0$$ after multiplication by an invertible factor. This stratum has dimension two. The opposite mixed stratum is analogous. If $x_0=y_0=0$, the equations reduce to $$x_3y_3=0,\qquad x_3y_1+x_1y_3=0.$$ If $x_3=y_3=0$, the locus is $\mathbb P^1\times\mathbb P^1$. If $x_3=0,y_3\neq0$, then $x_1=0$ and the first factor is fixed; the remaining locus lies in a projective plane. The final case is symmetric. All these strata have dimension at most two.

The affine-multicone argument of Lemma 33.2 now applies verbatim: the three equations have height three, the quotient is a complete intersection, and no component is supported on the boundary or on the dimension-four irrelevant loci. The unique component is generically smooth, so the quotient is a domain. Equality of the bihomogeneous complete-intersection Hilbert series proves that these three forms generate the whole weight initial ideal.

The formal implicit-function equations in the regular parameters $t,a,b$ give $$y_1=-t-ab+O_w(3),\qquad x_2=Q+O_w(4),\qquad y_2=Q+O_w(4).$$ Localizing the initial domain at $x_0y_0$ gives the Laurent polynomial ring in $x_0,y_0$ over $\mathbb C[t,a,b]$, with weights $(2,1,1)$ on $t,a,b$. On any fixed bihomogeneous piece, dehomogenization is injective and its map to the monomial associated graded ring is the identity on $t,a,b$. This proves the asserted equality of filtrations. ◻

Proof. This is the toric weighted blowup of a smooth three-dimensional regular local scheme. The primitive exceptional ray is $(2,1,1)$, so its log discrepancy is the sum of its weights, namely four. The two charts indexed by the weight-one coordinates are smooth. The remaining chart is the quotient of coordinates $(s,u,v)$, with $t=s^2,a=su,b=sv$, by $$(s,u,v)\longmapsto(-s,-u,-v).$$ Its exceptional divisor is the image of $s=0$. The cover is unramified in codimension one both on the threefold and on that divisor. The smooth pair with boundary $s=0$ is plt; the quotient pair is therefore plt and adjunction has zero different, since there is no divisorial ramification on the divisor. The stated ray subdivides the only determinant-two cone into determinant-one cones, giving the indicated smooth resolution. ◻

Proof. The following thirteen sections of $W_1$ have distinct leading terms: $$1,b,a,ab,t,tb,aY,bQ,aQ,abQ,tY,tQ,Q^2.$$ Here $abQ=-(tQ+YQ)$ belongs to $W_1$ by (36.1). Their exponent vectors, in coordinate order $(t,a,b)$, are (36.5)$$\label{f212:eq:weighted-degree-one-pivots} \begin{aligned} &(0,0,0),(0,0,1),(0,1,0),(0,1,1),\\ &(1,0,0),(1,0,1),(1,1,0),(1,1,1),\\ &(1,2,0),(1,2,1),(2,0,0),(2,1,0),(2,2,0). \end{aligned}$$ Since $N_1=13$, these sections form a basis. Three more leading terms, in diagonal degrees two, two and three, respectively, are obtained from the identities (36.6)$$\begin{aligned} P_1&=b^2Q+abQ-a^2bY=b^3(t-a^2), &\operatorname{in}(P_1)&=tb^3,\label{f212:eq:weighted-P1}\end{aligned}$$

(36.7)$$\begin{aligned} P_2&=-b^2Q^2-abQ^2-aY^2Q-taYQ=-P_1Q, &\operatorname{in}(P_2)&=t^2ab^3.\label{f212:eq:weighted-P2} \end{aligned}$$ Moreover, (36.8)$$\begin{aligned} P_3={}&b^2Q^4+2YQ^4+2Y^2bQ^3-2aY^2Q^3\\ &\quad-tY^3Q^2-2t^2Y^2Q^2-t^3YQ^2\\ ={}&b^3Q^2(bt^2-a^3t-a^4b), &\operatorname{in}(P_3)&=t^4a^2b^4. \label{f212:eq:weighted-P3} \end{aligned}$$ Each summand on the first sides of these identities satisfies (36.4) with $m=2,2,3$, respectively. This proves membership in the asserted diagonal degree; membership is not inferred merely from the factored expressions.

Let $P$ be the convex hull of the thirteen vectors in (36.5), together with $$\left(\frac12,0,\frac32\right),\qquad \left(1,\frac12,\frac32\right),\qquad \left(\frac43,\frac23,\frac43\right).$$ These are the vectors of the additional leading monomials divided by their diagonal degrees. The vertices of $P$ are the following eleven points: $$\begin{array}{c|rrrrrrrrrrr} &O&A&B&C&D&E&F&G&H&I&J\\\hline t&0&0&0&0&1&1&2&2&1/2&1&4/3\\ a&0&0&1&1&2&2&0&2&0&1/2&2/3\\ b&0&1&0&1&0&1&0&0&3/2&3/2&4/3 \end{array}$$ An exact facet description, also fixing the body independently of its construction, is (36.9)$$\label{f212:eq:weighted-body-facets} \begin{gathered} t,a,b\ge0,\qquad b-t\le1,\quad a-t\le1,\quad a\le2,\\ a+3b-t\le4,\qquad t-a+b\le2,\qquad t+a+3b\le6,\\ 2t+b\le4,\qquad 2t+a+2b\le6. \end{gathered}$$ Here is a rational integration certificate. Triangulate the boundary facets not contained in a plane through $O$ as in the table and cone each triangle to $O$. These tetrahedra have disjoint interiors and cover $P$; this follows directly from (36.9). $$\begin{array}{c|cc|c} \text{triangle}&\mathop{\mathrm{vol}}[O,\text{triangle}]& \text{triangle}&\mathop{\mathrm{vol}}[O,\text{triangle}]\\\hline ACH&1/12&JGF&8/9\\ EGD&1/3&EJG&2/3\\ IEJ&1/6&EBD&1/6\\ EBC&1/6&IHF&1/4\\ IJF&1/9&IEC&1/3\\ ICH&1/6&& \end{array}$$ For a triangle with vertices $U,V,Z$, its tetrahedron has volume $|\det(U,V,Z)|/6$ and first moment $\mathop{\mathrm{vol}}[O,U,V,Z](U+V+Z)/4$. Thus summation of the displayed rational entries gives (36.10)$$\label{f212:eq:weighted-body-integrals} \mathop{\mathrm{vol}}(P)=\frac{10}{3},\qquad \int_P(t,a,b)\,dV =\left(\frac{37}{12},\frac{811}{288},\frac{533}{288}\right).$$

Consider the affine semigroup generated by the sixteen certified leading monomials, retaining diagonal degree as its first coordinate. Its group is the full lattice: degree one contains the constant monomial and the three coordinate monomials $t,a,b$. The degree-$m$ part of this semigroup consequently has cardinality $$\mathop{\mathrm{vol}}(P)m^3+O(m^2),$$ and its three exponent sums have leading terms $m^4\int_P(t,a,b)\,dV$. These are the usual lattice-point asymptotics for an affine semigroup of full lattice; equivalently, one compares with its saturation using a conductor element. Every such monomial is a leading monomial of an actual refined section, by multiplicativity. On the other hand, (36.3) gives $N_m=(10/3)m^3+O(m^2)$. The full initial series therefore has only $O(m^2)$ further monomials. All their exponents are at most $6m$, so they change each exponent sum by only $O(m^3)$. It follows from (36.10) that the full refined series has the asserted asymptotic means. This comparison does not require the displayed generators to form a complete SAGBI basis. ◻

Proof. Lemma 34.1 and the computed means give $$\delta(G;W)\ge\frac1{\max\{37/40,811/960,533/960\}}=\frac{40}{37}.$$ The quotient-chart proof of Lemma 36.2 verifies plt extraction and zero different. The bounded support and ample-series conditions were checked directly from (36.4). Corollary 33.6 gives $A_X(G)/S_X(G)=16/13$. Proposition 2.7 now gives the stated local bound. ◻