Proof. Choose a smooth point \(x\) of \(Z\) where both normal-basis conditions hold and which lies on no other irreducible component of the common zero set. Flowing from \(Z\) along the fields indexed by \(B\) gives local analytic coordinates \((z,b)\), where \(z\) parametrizes \(Z\) and \(b=(b_1,\ldots,b_k)\) consists of transverse flow parameters. Commutativity implies that the chosen fields are the derivatives in the \(b\)-directions. Thus (3) says that in the expansion \[f_\ell(z,b)=\sum_{\alpha\in\mathbb Z_{\ge 0}^k} f_{\ell,\alpha}(z)b^\alpha,\] every coefficient function of \(B\)-weight less than \(\varepsilon N\) vanishes identically on the chosen neighborhood in \(Z\).

Let \(S\) be the affine coordinate slice through \(x\) obtained by fixing the coordinates outside \(A\). It is transverse to \(Z\), and the zero set of the restricted equations is isolated at \(x\). The flow parameters \(b\) give regular analytic coordinates on \(S\); the coordinates \(z\) become analytic functions \(z(b)\). Substitution cannot generate a missing term of smaller \(B\)-weight. Indeed, each surviving summand already contains a monomial \(b^\alpha\) of weight at least \(\varepsilon N\), and its analytic coefficient has only nonnegative powers of \(b\). Consequently the ideal of the restricted equations is contained in the monomial ideal of weight at least \(\varepsilon N\). Its quotient therefore has length at least \[ \#\left\{\alpha\in\mathbb Z_{\ge 0}^k: \sum_{b\in B}\kappa_b\alpha_b<\varepsilon N\right\} \ge \frac{(\varepsilon N)^k}{2k!\prod_{b\in B}\kappa_b} \tag{5}\] In fact the factor \(1/2\) can be omitted: the simplex \(\{x\in\mathbb R_{\ge0}^k:\sum_{b\in B}\kappa_bx_b<\varepsilon N\}\) is covered by the disjoint half-open unit cubes \(\alpha+[0,1)^k\) indexed by the admissible multi-indices \(\alpha=\lfloor x\rfloor\). Its volume is \((\varepsilon N)^k/(k!\prod_{b\in B}\kappa_b)\). Thus the displayed bound holds for every \(N>0\). Analytic local coordinates preserve finite length, so no radius or coefficient bound for the flow map is needed.

In the regular local ring of the \(k\)-dimensional slice, the ideal generated by the restricted equations is primary to the maximal ideal. Over the infinite field \(\mathbb C\), \(k\) sufficiently general constant linear combinations of these equations still have an isolated common zero at \(x\): successive generic combinations avoid the finitely many positive-dimensional components that remain. Their ideal is contained in the original ideal. Their quotient length is therefore an upper bound for the length in (5).

Each of the \(k\) combinations has weighted degree at most \(N\) in the variables indexed by \(A\). We claim that their local intersection length is at most \[ \frac{N^k}{\prod_{a\in A}\rho_a}. \tag{6}\] Choose an integer \(M_0\) clearing the denominators of the \(\rho_a\), and make the substitutions \[x_a=\lambda_a+z_a^{M_0\rho_a}\qquad(a\in A).\] Choose the constants \(\lambda_a\) so that no preimage of the given point is ramified. There are \(\prod_{a\in A}(M_0\rho_a)\) such preimages, each with the same local intersection length. The substituted polynomials have ordinary total degree at most \(M_0N\). The ordinary Bézout bound for isolated complete-intersection zeros, with their local multiplicities, bounds the sum of these lengths by \((M_0N)^k\); see (Mondal 2021, Corollary VIII.3, p. 144). The multiplicities there are local quotient lengths, unchanged by completion (Mondal 2021, sec. IV.4.2). Dividing gives (6). This application concerns isolated zeros only and permits other components elsewhere.

Combining (5) and (6) proves the result. Neither the polynomials nor the possibly varying component enters the constant in (4). ◻

Proof. There is nothing to prove if the curve misses all the points. Suppose instead that (12) fails. For every sufficiently large auxiliary integer \(N\), linear algebra gives a nonzero polynomial \(F_N\) of \(W\)-degree at most \(N\) whose expansions at all the points have \(V\)-order at least \((1+3\sigma)N\). Indeed, the number of its coefficients has leading term \[\frac{N^{m+1}}{(m+1)!\prod_{a=0}^m w_a},\] whereas the number of conditions has leading term \[\frac{K(1+3\sigma)^{m+1}N^{m+1}} {(m+1)!\prod_{a=0}^m v_a}.\] Their comparison is the first inequality in (7). This argument requires no independence among the conditions.

On the open set \(Y\ne0\) consider the commuting frame \[ D_0=Y\partial_Y+\sum_{i=1}^m\partial_{X_i},\qquad D_i=\partial_{X_i}\quad(1\le i\le m). \tag{13}\] It preserves polynomial degree at most \(N\). In the formal coordinates of (2), \[D_0=(1+t)\partial_t,\qquad D_i=\partial_{u_i}.\] Thus differentiation by \(D_a\) decreases weighted order by at most \(v_a\). Every \(D^\gamma F_N\) with \(\sum_av_a\gamma_a\le\sigma N\) retains order at least \((1+2\sigma)N\) at each point. On a branch \(P\), its order is consequently at least \((1+2\sigma)N h_P\).

A polynomial of \(W\)-degree at most \(N\) restricts to a rational function on the complete model of \(C\) with total pole degree at most \(N\deg_W C\). Since we assumed that (12) fails, the just obtained total zero order exceeds this pole bound. The ordinary divisor equality for a nonzero rational function therefore forces \[ D^\gamma F_N\big|_C=0 \qquad\left(\sum_av_a\gamma_a\le\sigma N\right). \tag{14}\] All the assertions about branch orders hold for formal power series: each monomial has order at least its weight times \(h_P\), and cancellation can only increase the order.

We have converted excess contact into vanishing of many derivatives on \(C\). We next find a component on which that vanishing persists, as in the derivative-ideal method of zero estimates (Philippon 1986, sec. 5), and compare all its normal bases. Put \(\delta=\sigma N/L_0\). For \(0\le r\le L_0\), let \(V_r\) be the common zero set in \(Y\ne0\) of the derivatives of cost at most \(r\delta\). Let \(C^\circ\) be the part of \(C\) in this affine open set. The \(V_r\) form a decreasing chain, and (14) puts \(C^\circ\) in \(V_{L_0}\). Choose an irreducible component of \(V_{L_0}\) containing \(C^\circ\), and extend it backwards to nested irreducible components of the earlier \(V_r\). Every selected component has dimension between \(1\) and \(m\): it contains a curve, and it lies in the proper zero set of \(F_N\). Since there are \(L_0+1=m+3\) levels, two adjacent selected components coincide. Denote this persistent component by \(Z\), and write \(k=\operatorname{codim}Z\).

Take the derivatives of cost at most \(r\delta\) as the equations of the earlier of these two levels. Any further derivative of cost less than \(\delta\) vanishes along \(Z\), because the fields commute and \(Z\) is also a component at the next level. Lemma 4, with \(\varepsilon=\sigma/L_0\), gives \[ \prod_{a\in A}w_a\le C_*\prod_{b\in B}v_b \tag{15}\] for every coordinate normal basis \(A\) and every frame normal basis \(B\). The equations and \(Z\) may vary with \(N\); the uniformity proved in Lemma 4 is exactly what permits this use.

We show that \(Y\) is constant on \(Z\), working on the common smooth open subset where all the normal-basis conditions hold. Suppose first that some positive coordinate direction is nontangent to \(Z\), and choose the largest such index \(i\). Every larger positive index occurs in neither kind of normal basis, since \(D_j=\partial_{X_j}\) for \(j>0\). Some coordinate normal basis \(A\) contains \(i\). If a frame normal basis \(B\) omitted \(i\), the largest positive index where these sets differ would be \(i\). Then (9) would contradict (15). Thus every frame normal basis contains \(D_i\).

The other frame vectors consequently fail to span the normal quotient. Their ambient span is the hyperplane \[\ker(dX_i-dY/Y).\] For an ambient hyperplane \(Q\) and a tangent space \(T\), the image of \(Q\) in the quotient by \(T\) fails to be surjective precisely when \(T\subseteq Q\). We obtain \[ dX_i=dY/Y\quad\text{on }Z. \tag{16}\] This identity forces \(Y\) to be constant on the algebraic variety \(Z\). Indeed, if \(Y\) were nonconstant, choose a smooth point where \(dY\ne0\) and intersect with general algebraic hyperplanes through that point until a curve remains on which \(Y\) is nonconstant. On the smooth complete model of this curve, (16) is an identity of meromorphic differentials. A nonconstant meromorphic function \(Y\) has a zero or pole, where \(dY/Y\) has nonzero residue \(\operatorname{ord}_P Y\). An exact differential \(dX_i\) has zero residue at every point. This is a contradiction. It follows also that \(X_i\) is constant, since \(dX_i=0\) in characteristic zero.

If instead every positive coordinate direction is tangent, the \(m\) independent vectors \(\partial_{X_i}\) all lie in the tangent space of \(Z\). Properness forces that tangent space to be their span, and \(dY=0\) on \(Z\) again. We conclude in all cases that \(Y\) is constant on \(Z\). Since \(C^\circ\subseteq Z\) meets a prescribed point with \(Y=1\), we have \(Y=1\) on \(C^\circ\), and hence on \(C\) by density.

It remains to rule out a violating curve in that fiber. We work with \(C^\circ\) in the affine fiber \(Y=1\), while computing orders and degrees on the complete model of \(C\) as before. In this fiber, the degree weights are \(w_i\), the order weights are \(v_i=w_i/\theta\), and \(h_P=\min_i\operatorname{ord}_P(X_i-c_{ji})/v_i\). Choose a new auxiliary polynomial in these \(m\) variables. The ratio of the number of vanishing conditions to the number of coefficients now tends to \(K(1+3\sigma)^m\theta^m<1\), by the second inequality in (7). Thus, for sufficiently large \(N\), there is a nonzero polynomial \(G_N(X_1,\ldots,X_m)\) of weighted degree at most \(N\) and order at least \((1+3\sigma)N\) at every center.

An ordinary partial derivative of cost at most \(\sigma N\) still has order at least \((1+2\sigma)N\) at every center and degree at most \(N\). The same zero–pole comparison on \(C\) therefore gives \[\partial_X^\gamma G_N\big|_C=0 \qquad\left(\sum_{i=1}^m v_i\gamma_i\le\sigma N\right).\] If \(m=1\), this is already impossible: a nonzero polynomial on the affine line cannot vanish on a curve. Suppose henceforth that \(m\ge2\). For \(0\le r\le L_0\), let \(U_r\) be the common zero set in this fiber of the partial derivatives of \(G_N\) of cost at most \(r\delta\), with \(\delta=\sigma N/L_0\) as before. The curve \(C^\circ\) lies in \(U_{L_0}\). Choose nested irreducible components containing \(C^\circ\) backwards through this chain. Their dimensions lie between \(1\) and \(m-1\), since \(G_N\) is nonzero, so two adjacent components coincide in a component \(Z'\). Its codimension in the fiber satisfies \(1\le k\le m-1\).

For the equations at the earlier of these two levels, every further partial derivative of cost less than \(\delta\) vanishes on \(Z'\). Lemma 4, again with \(\varepsilon=\sigma/L_0\), therefore gives (15) for \(Z'\), with the same \(C_*\) and with normal-basis index sets contained in \(\{1,\ldots,m\}\).

Here the coordinate and frame bases are the same. They must have a unique index set: two different basis sets could be ordered so that the largest differing index belongs to \(A\), contradicting (9). The codimension is positive, so this unique normal basis has some index \(i\). The other coordinate directions fail to span the normal quotient; hence the tangent space is contained in \(\ker dX_i\), and \(X_i\) is constant on \(Z'\). Coordinatewise distinctness shows that \(Z'\) meets at most one prescribed point. Every prescribed point of \(C\) belongs to \(C^\circ\subseteq Z'\), so the same holds for \(C\).

Let that point be \(a_j\). Choose a coordinate \(X_l\) nonconstant on \(C\). The nonzero rational function \(X_l-c_{jl}\) has total pole degree at most \(w_l\deg_W C\), and at each branch through \(a_j\) it has order at least \((w_l/\theta)h_P\). Therefore \[\deg_W C\ge\theta^{-1}\sum_{P\mapsto a_j}h_P >(1+\sigma)\sum_{P\mapsto a_j}h_P,\] by the last inequality in (7). This final contradiction proves (12). ◻

Proof. Let \(L_i=\operatorname{lcm}(1,\ldots,T_i-1)\), interpreted as \(1\) if \(T_i=1\), and put \[D_H=\prod_{i=1}^m L_i^{\lfloor H/w_i\rfloor}.\] Scale a column \(Y^hX^\alpha\) by \(\prod_iq_i^{\alpha_i}\) and a row \((j,s,\beta)\) by \(\prod_iq_i^{-\beta_i}\). The scaled entry is zero unless \(\alpha\ge\beta\) componentwise; otherwise it is \[\prod_i\binom{\alpha_i}{\beta_i} [t^s](1+t)^h \prod_i\bigl(q_i(jr_i+G_i(t))\bigr)^{\alpha_i-\beta_i}.\] Here \(q_ijr_i=2\mathrm ijp_i\in\mathbb Z[\mathrm i]\). Since \(G_i\) has coefficients in \(L_i^{-1}\mathbb Z\), the denominator of this entry divides \(\prod_iL_i^{\alpha_i-\beta_i}\), which divides \(D_H\). Thus multiplying every entry of the scaled matrix by \(D_H\) gives a matrix over \(\mathbb Z[\mathrm i]\) with nonzero determinant. A nonzero Gaussian integer has modulus at least one, so \[ \begin{aligned} \log|\Delta_H| &\ge -M\log D_H -\sum_{\text{columns }\alpha}\sum_i\alpha_i\log q_i\\ &\quad+\sum_{\text{rows }\beta}\sum_i\beta_i\log q_i. \end{aligned} \tag{31}\] For each selected column, \(\sum_i\alpha_i\log q_i\le w\alpha\le H\). Also \(0\le w_i-\log q_i<1\), and hence \[\sum_{\text{rows }\beta}\sum_i\beta_i\log q_i \ge MH\bar b-\sum_{\text{rows }\beta}\sum_i\beta_i \ge MH\bar b-\frac{MH\theta}{w_*}.\]

For completeness, \(\log\operatorname{lcm}(1,\ldots,n)\le\Lambda n\). To see this, the factorial valuation formula shows that each prime power in \((\ell,2\ell]\) contributes its logarithmic prime weight to \(\log\binom{2\ell}{\ell}\); all the other contributions to the latter are nonnegative. Thus the increase of the log least common multiple across this interval is at most \(2\ell\log2\). Summing over dyadic intervals up to the smallest power of two above \(n\) gives at most \(4n\log2\). Therefore \[\frac{\log D_H}{H} \le\Lambda\sum_i\frac{T_i}{w_i} \le\frac{\Lambda F_0m}{v_0}+\Lambda\sum_i\frac1{w_i}.\] Substitution in (31) proves (30). ◻

Proof. Put \[\epsilon_i=j(r_i-\omega),\qquad \tau_i(t)=G_i(t)-\log(1+t).\] Since \(e^{j\omega}=1\), the substitutions in (26) can be written as \(Y=e^z\), \(X_i=z+u_i+\epsilon_i+\tau_i(t)\), where \(z=j\omega+\log(1+t)\). Expanding first in the variables \(u_i\), then in the constants \(\epsilon_i\) and tails \(\tau_i\), gives the exact row vector identity \[\begin{align*} \text{row}_{j,s,\beta} ={}&\sum_{\substack{a\ge\beta\\wa\le H}} \binom a\beta \sum_{0\le d\le a-\beta}\binom{a-\beta}{d} \epsilon^{a-\beta-d} \sum_{k=0}^s \left([t^k]\prod_i\tau_i(t)^{d_i}\right) \\[-2pt] &\hspace{35mm}\cdot \bigl([t^{s-k}]f_{a,P}(j\omega+\log(1+t))\bigr)_P. \tag{36}\end{align*}\] All scalar factors in this identity are independent of the column \(P\). This is therefore an identity to which determinant multilinearity applies directly.

The product of tails vanishes to order at least \(\sum_i d_iT_i\). Consequently a nonzero coefficient in (36) satisfies \[ \sum_i d_iw_i\le\frac{v_0}{F_0}\sum_i d_iT_i \le\frac{v_0s}{F_0}<\frac{H}{F_0}. \tag{37}\] On \(|t|=1/2\), \[|\tau_i(t)| \le\sum_{k\ge T_i}\frac{2^{-k}}k\le1.\] Cauchy’s coefficient estimate thus bounds the tail coefficient in (36) by \(2^k\le2^s\). From (24), \[|\epsilon_i|\le2Kq_i^{-\nu} \le2K e^\nu e^{-\nu w_i}.\] Moreover \(\binom a\beta\binom{a-\beta}{d}\le4^{|a|}\) and \(|a|\le H/w_*\). Combining these inequalities and using (37) gives \[\log|\xi| \le-\nu w(a-\beta)+\frac{\nu H}{F_0} +\frac{H\log2}{v_0} +\frac{H(\log4+\log(2K)+\nu)}{w_*}\] for each nonzero scalar. A zero scalar obeys the required bound as well. Finally, each coordinate of \(a\) and \(d\) is at most \(H\) because \(w_i\ge1\), while \(0\le k\le H/v_0\). This proves the term count and the Lemma. ◻

Proof. Put \(R=100K\). Equation (33) gives, on \(|z|\le R\), \[|f_{a,P}(z)|\le \exp\left\{H\left(\frac R{w_0} +\frac{\log(2R)}{w_*}\right)\right\} =:\mathcal D_H.\] Expand \[f_{a,P}(z)=\sum_{d\ge0}c_{a,d,P}(z/R)^d.\] Cauchy’s estimate gives \(|c_{a,d,P}|\le\mathcal D_H\). A row of \(T\) is a test of order \(\ell\le s\) at some center \(j\). Since \[|j\omega+\log(1+t)| \le2\pi(K-1)+\log2<50K=R/2 \qquad(|t|=1/2),\] its value on \((z/R)^d\) has modulus at most \(2^\ell2^{-d}\).

Expand the determinant in these power-series rows. The expansion is absolutely convergent: a coefficient determinant has modulus at most \(M!\mathcal D_H^M\), and the sum of the product bounds \(2^{-d}\) over all degrees is finite. If two rows have the same pair \((a,d)\), their coefficient vectors \((c_{a,d,P})_P\) agree, and that summand vanishes. Thus every nonzero summand has distinct degrees within each group \(a\), and so \[\sum_{\text{rows}}d\ge\sum_a\binom{n_a}{2}.\] Use half of the geometric decay for this lower bound and sum the other half freely over all nonnegative degrees. Since \(\sum\ell\le MH/v_0\), this gives \[|\det T| \le M!\mathcal D_H^M2^{MH/v_0} 2^{-\frac12\sum_a\binom{n_a}{2}} (1-2^{-1/2})^{-M}.\] Because \(\sum_a n_a=M\), the asserted inequality follows with, for example, \[\rho_H=\frac{\log M+\frac14\log2 -\log(1-2^{-1/2})}{H}.\] Equation (28) shows that \(\rho_H\to0\) with the fixed parameters stipulated above. ◻

Proof. Let \[D_A(H)=\#\{a\in\mathbb Z_{\ge0}^m:wa\le AH\}, \qquad L_H=\frac{\eta^2M}{H D_A(H)}.\] The simplex count, now in dimension \(m\), and (28) show that \[D_A(H)\sim\frac{(AH)^m}{m!\prod_iw_i}, \qquad L_H\longrightarrow L.\] Expand \(\Delta_H\) by the row identities in Lemma 7. For a choice of terms with indices \(a_\rho\), the product of their scalar factors has modulus at most \[\exp\left\{-\nu\sum_\rho wa_\rho +\nu MH\bar b+MH E_{\rm tr}\right\}.\] There are at most \(Q_H^M\) term choices. Apply Lemma 8 to each one.

Suppose first that at least \(\eta M\) of its indices obey \(wa\le AH\). There are at most \(D_A(H)\) such indices, so Cauchy–Schwarz gives \[\sum_a n_a^2\ge\frac{(\eta M)^2}{D_A(H)}=MH L_H.\] The scalar main exponent \(-\nu\sum_\rho w(a_\rho-\beta_\rho)\) is nonpositive, because \(a_\rho\ge\beta_\rho\). Discarding it leaves an upper bound \(-cL_H+E_{\rm an}+\rho_H\) after division by \(MH\).

In the other case, more than \((1-\eta)M\) indices have \(wa>AH\). Hence \(\sum_\rho wa_\rho\ge MH A(1-\eta)\). Discarding the nonpositive collision term leaves the upper bound \(-\nu(A(1-\eta)-\bar b)+E_{\rm an}+\rho_H\). Taking absolute values before summing the determinant expansion gives \[ \begin{aligned} \frac{\log|\Delta_H|}{MH} &\le E_{\rm an}+\varepsilon_H+ \max\{-cL_H,-\nu(A(1-\eta)-\bar b)\},\\ \varepsilon_H&=\rho_H+\frac{\log Q_H}{H}\longrightarrow0. \end{aligned} \tag{43}\] All bounds here are uniform over the row-term choices.

For the second alternative, the gap between the two main exponents is at least \(g\), since \(\bar b\le\theta\) gives \[\nu(A(1-\eta)-\bar b)-(1-\bar b) =\nu A(1-\eta)-1-(\nu-1)\bar b\ge g.\] Under (42), both quantities inside the maximum in (43), after adding \(E_{\rm an}+\varepsilon_H\), are strictly less than \(-(1-\bar b)-E_{\rm ar}\) for all sufficiently large \(H\). For the first quantity, use \(L_H\to L\) and \(1-\bar b\le1\); for the second, use the displayed uniform gap. This contradicts Lemma 6. ◻