The ordinary point filtration

Proof. The two summands of the differential of the defining equations are maps from the two three-dimensional factor tangent spaces to the three-dimensional equation space. Each has rank two, since each projection has a smooth one-dimensional exceptional fibre at $P$. Smoothness of $X$ says that their images span the equation space; the two rank-two images therefore intersect in a line. Choosing bases adapted to these two planes, and then factor coordinates, gives the linear parts $x_1+y_1,x_2,y_2$. Homogenizing yields the displayed normal form.

Let $M(z)=z_0M_1+z_1M_2+z_2M_3$ be the resulting matrix pencil. On the line $z_0=0$, its zeroth row and column are $(0,0,z_2,0)$ and $(0,0,z_1,0)^{\mathsf t}$, respectively. Laplace expansion gives $$\det M(0,z_1,z_2) =-z_1z_2\det(z_1N_2+z_2N_3).$$ The determinant quartic $C$ is smooth and hence irreducible, so it does not contain the line $z_0=0$. Thus the binary determinant on the right is not identically zero. If $z_1N_2+z_2N_3=0$ for some $[z_1:z_2]$, the zeroth row and all rows other than row $2$ of $M(0,z_1,z_2)$ are supported only in column $2$. The full matrix then has rank at most two. All entries of its adjugate vanish, so Jacobi’s formula makes every first derivative of $\det M$ vanish there, contradicting smoothness of $C$. Hence $N_2,N_3$ are linearly independent and both nonzero.

A nonconstant common factor of two bilinear forms is either the entire form, making them proportional, or a linear form in one factor’s variables. Proportionality is already excluded. A common $x$-linear factor writes $q_i=\ell(x)m_i(y)$, so every matrix $z_1N_2+z_2N_3$ has rank at most one. A common $y$-linear factor has the same consequence. Both contradict the nonzero binary determinant. This proves coprimality in the four variables.

For the last assertions, use regular parameters $t=x_1,a=x_3,b=y_3$ on the affine chart $x_0=y_0=1$. The equations give $$y_1=-t+O(2),\quad x_2=-Q_2+O(3),\quad y_2=-Q_3+O(3),\qquad Q_i=A_it^2+B_ita+C_itb+D_iab.$$ The linear term of the Jacobian determinant of the first blowdown is, up to sign, $C_2t+D_2a$; that of the second is, up to sign, $B_3t+D_3b$. The ramification divisor of a blowup of a smooth curve in a smooth threefold is exactly its exceptional divisor, with multiplicity one. Since both exceptional divisors are smooth, their local equations have nonzero linear terms. This proves the two asserted nonvanishing conditions. ◻

Proof. In $\mathbb P^3\times\mathbb P^3$, the locus defined by the $G_i$ on $x_0y_0\neq0$ is the graph $$y_1=-x_1,\qquad x_2=-q_2(x_1,x_3;-x_1,y_3),\qquad y_2=-q_3(x_1,x_3;-x_1,y_3).$$ It is therefore an affine three-space with coordinates $x_1,x_3,y_3$. We show that its boundary has dimension at most two.

If $x_0=0$ and $y_0\neq0$, then $G_1$ gives $x_1=0$. One must have $x_3\neq0$, since otherwise $G_2$ also forces $x_2=0$, which is impossible in projective space. The equation $G_2$ determines $x_2$, while $G_3$ imposes the nonzero linear condition $$-B_3y_1+D_3y_3=0.$$ This locus has dimension two. The case $y_0=0,x_0\neq0$ is identical, using the nonzero condition $C_2x_1+D_2x_3=0$.

On $x_0=y_0=0$, if both pairs $(x_1,x_3)$ and $(y_1,y_3)$ are nonzero, the coprime forms $q_2,q_3$ cut out a finite subscheme of $\mathbb P^1\times\mathbb P^1$. The two additional coordinates $x_2,y_2$ give fibres of dimension two, so this stratum has dimension at most two. If either pair vanishes, that projective factor is fixed at $[0:0:1:0]$ and the other factor lies in a projective plane; again the dimension is at most two.

Thus the projective zero scheme has dimension three, and its only possible three-dimensional component is the closure of the displayed affine three-space. On the affine multicone, the locus where an entire $x$- or $y$-block vanishes has dimension four, below the expected dimension five; the other boundary strata also have dimension at most four. Hence the three affine equations have height three and form a regular sequence in the polynomial ring. Their quotient is Cohen–Macaulay, so has no embedded components. Its unique minimal component is generically smooth by the graph description. The quotient is therefore reduced and irreducible, hence a domain.

The weight initial ideal contains $(G_1,G_2,G_3)$. Both the original equations and these initial equations are regular sequences of three forms of bidegree $(1,1)$. Their quotient rings consequently have the same bihomogeneous Hilbert series. A weight degeneration preserves this Hilbert series, so the inclusion of ideals must be equality.

Finally, localizing the initial domain at $x_0y_0$ is injective. The localization is $$\mathbb C[x_0^{\pm1},y_0^{\pm1},t,a,b].$$ On a fixed bihomogeneous piece, division by $x_0^my_0^n$ identifies its image with a polynomial subspace in the degree-one parameters $t,a,b$. The map to the ordinary tangent cone at $P$ sends these parameters to themselves, and is consequently injective on each bihomogeneous piece. Therefore no further cancellation can raise the vanishing order of a bihomogeneous section beyond its weight initial degree. The two filtrations agree. ◻

Proof. A common quadratic factor would make $Q_2,Q_3$, and hence $N_2,N_3$, proportional, which is excluded by Lemma 33.1. Suppose a common linear factor is $\alpha t+\beta a+\gamma b$. If $\beta\gamma\neq0$, absence of $a^2,b^2$ in the $Q_i$ forces both remaining factors to be multiples of $t$, again giving proportional quadrics. If $\beta\neq0$ and $\gamma=0$, the quotients have the form $u_it+v_ib$. Their matrices then satisfy $$N_i=\begin{pmatrix}\alpha\\\beta\end{pmatrix} \begin{pmatrix}-u_i&v_i\end{pmatrix},$$ so every pencil member has rank at most one, a contradiction. The case $\gamma\neq0,\beta=0$ is symmetric. Thus the only possible common factor is $t$, and it occurs precisely when $D_2=D_3=0$. The greatest common divisor cannot have larger degree, since the quadrics are not proportional.

In this case write the pencil coefficients as $A(z),B(z),C(z)$. Its binary determinant is $B(z)C(z)$ and is not identically zero. If the two residual forms are proportional, there are nonzero $b,c$ and a nonzero linear form $k(z)$ such that $B(z)=bk(z)$ and $C(z)=ck(z)$. If $A$ were proportional to $k$, the entire matrix $N(z)$ would vanish at the root of $k$, contradicting smoothness as in Lemma 33.1. Therefore $A,k$ are linearly independent, and the quadratic pencil has precisely the stated span. ◻

Proof. The multigraded Hilbert series is $$H(u,v,z)= \frac{\prod_{k=1}^3(1-uvz^{\rho_k})} {\prod_{i=0}^3(1-uz^{\alpha_i})(1-vz^{\beta_i})}.$$ At $z=1$ this becomes $$H_0(u,v)=\frac{(1-uv)^3}{(1-u)^4(1-v)^4}.$$ Hence $$N_m=\sum_{k=0}^3(-1)^k\binom3k \binom{m-k+3}{3}^{\!2} =\frac{(2m+1)(5m^2+5m+3)}3.$$ Here and below a binomial coefficient is zero when its nonnegative upper argument is smaller than its lower argument. Logarithmic differentiation in $z$ gives $$\left.z\frac{\partial H}{\partial z}\right|_{z=1} =H_0(u,v)\left( \frac{u\sum_i\alpha_i}{1-u} +\frac{v\sum_i\beta_i}{1-v} -\frac{uvR}{1-uv}\right).$$ The two coordinate terms have the same diagonal coefficient, by symmetry in $u,v$. Denoting that coefficient by $F_m$, and the diagonal coefficient of $H_0uv/(1-uv)$ by $G_m$, we obtain $$w_m=VF_m-RG_m.$$ Explicit coefficient extraction yields $$\begin{aligned} F_m &=\sum_{k=0}^3(-1)^k\binom3k \binom{m-k+3}{4}\binom{m-k+3}{3} =\frac{m(35m^3+30m^2+25m+6)}{24},\\ G_m &=\sum_{k=0}^2(-1)^k\binom2k \binom{m-k+2}{3}^{\!2} =\frac{m^2(5m^2+1)}6. \end{aligned}$$ This proves (33.2). Since $N_m=(10/3)m^3+O(m^2)$, division by $mN_m$ gives the stated limit. The final assertion follows from the definition of expected vanishing, once the identification with the valuation associated graded ring has been established. ◻

Proof. Lemma 33.2 identifies the ordinary grading with its valuation filtration. Lemma 36.1 below does the same for the weighted grading. Substitute their weights in Lemma 33.5. The log discrepancy of a weighted blow-up of a smooth threefold point is the sum of its three primitive weights. ◻

Proof. Lemma 33.2 identifies the valuation graded ring, including all cancellations. Dehomogenization gives the two displayed sets of generators. Distribute the $r$ copies of $t$ between the unused places in the two factors. Such a distribution exists precisely when the three inequalities in (33.3) hold. Every generator product has polynomial degree at most $4m$. Conversely, for $0\le j\le m$ all degree-$j$ monomials in $t,a,b$ satisfy those inequalities, so $W^G_{m,j}=H^0(\mathbb P^2,\mathcal O(j))$. The ray $j=m/2$ therefore contains a complete ample series. All integral degrees $0\le j\le4m$ occur, since the generator degrees in each factor are $0,1,1,2$ and the ring is a domain. These observations give the required positivity and support hypotheses. ◻