The Quasi-Riemann Hypothesis: A Zero-Free Half-Plane \(\Re s\gt 7/8\)

Proof. By Theorem 1.1 and the functional equation, the nontrivial zeros of every primitive Dirichlet L-function lie in 1/8\le\Re s\le7/8. They therefore lie in the strictly larger strip 1/16<\Re s<15/16. This supplies the weak-GRH hypothesis of Bhargava, Ivanyos, Mittal, and Saxena [2, Conjecture 6.3 and Theorem 6.7] with \epsilon=7/16. Their bound for the least quadratic nonresidue gives the asserted inequality, for example with A=32; no optimization is needed here. The assertion for each \delta>0 follows because every fixed power of \log p is O_\delta(p^\delta).

For the algorithm, first handle a=0 and test a nonzero a by Euler’s criterion. If a is a square, scan 2,3,\ldots until a quadratic nonresidue is found, testing each candidate by its Legendre symbol. The bound just proved makes this a polynomial-time deterministic search; its stopping rule does not require knowing C. Use the resulting nonresidue in the Tonelli–Shanks algorithm [10, Section 2.9, Algorithm 3 and Lemma 2.9.5]. All remaining steps are deterministic and polynomial in \log p. Writing p-1=2^e u with u odd only requires removing factors of two, not factoring u.□

Proof. Put

Because C has slope one, the triangle inequality gives

Moreover \epsilon_*\le\Delta_0-\omega<\Delta_0, so \beta_*-\epsilon_*>\sigma_0.

We also need control as Z\downarrow0. By Equation (2.2), H_\eta is bounded on \Re s>\sigma_0. The reciprocal Euler product is absolutely and uniformly bounded on \Re s\ge2. On every fixed strip 2\le\Re s\le B, the Gaussian in Equation (2.3) is O_B(e^{-(\Im s)^2}). Cauchy’s theorem on rectangles therefore moves the contour to any fixed B>2, with horizontal integrals tending to zero. Hence

Since B is arbitrary, the signal has arbitrarily rapid power decay at zero.

This bound and Equation (2.6) show that

converges locally uniformly on \Re s>\beta_*-\epsilon_*. On each compact subset, choose B larger than all occurring real parts for the integral near zero, and use Equation (2.6) near infinity. Thus F_\eta is holomorphic on that half-plane.

We identify this Mellin transform without moving a contour across a zero. Set

The function A_\eta(2+it) is continuous and integrable in t. Writing Z=e^u, Equation (2.3) becomes

The left side is integrable in u by the two endpoint estimates. Ordinary Fourier inversion therefore gives F_\eta(2+it)=A_\eta(2+it) for every real t. Both sides are holomorphic on \Re s>1, so the identity theorem gives

Equation (2.2) implies |H_\eta(s)|\ge1/2 on \Re s>\sigma_0. Consequently

is a holomorphic continuation of 1/L_F^{\mathcal S}(s,\eta). The number \epsilon_* was chosen independently of \eta. By the definition of \beta_*, some target has a zero \rho with \Re\rho>\beta_*-\epsilon_*. Its reciprocal has a pole at \rho, and the deleted factors are nonzero there. This contradicts the continuation.□

Proof. Let \xi^6=a. Since F contains all sixth roots of unity, all roots of X^6-a lie in F(\xi), and \sigma\mapsto\sigma(\xi)/\xi embeds its Galois group into \mu_6. The extension is therefore abelian. A prime outside 6a is unramified; for example this follows from the discriminant of X^6-a, which is supported on 6a. At such a prime, arithmetic Frobenius is characterized on the residue field by x\mapsto x^{q_{\mathfrak p}}. Consequently its quotient on \xi reduces to a^{(q_{\mathfrak p}-1)/6}. Reduction is injective on \mu_6 outside 6, proving the displayed identity. Multiplicativity of the Artin map gives the assertion for ideals. Artin reciprocity makes this map factor through a ray group with modulus supported on the primes dividing 6a; these are the power-residue and ray-group assertions in [22, Chapter VIII, (5.3) and (5.5)].

For the last statement, an ideal A outside S is coprime to every \pi_{\mathfrak p} and to every unit. Its symbol therefore has no zero there, and the power of each \chi_A(\pi_{\mathfrak p})\in\mu_6 depends only on v_{\mathfrak p}\pmod6. There are at most 6^{|S|+1} possible displayed numerators after this reduction. Apply the first assertion to each and take a common multiple of their ray moduli. Their prime supports are all contained in S because S contains the primes above 6.□

Proof. Let k=\mathcal O/(p) and let \psi(x)=e(x/p) be its nontrivial additive character. For a multiplicative character A of k^\times, extended by zero, put

If A is nonprincipal, the change of variables x=ty gives

The inner sum is P-1 for t=1 and -1 otherwise, and \sum_{t\in k^\times}A(t)=0. Conjugation also gives \tau(A)\tau(\overline A)=A(-1)P. When AB is nonprincipal, grouping the product \tau(A)\tau(B) by x+y gives

The group with x+y=0 vanishes because AB is nonprincipal; for a nonzero sum, division by x+y gives the displayed Jacobi factor. In particular |J(A,B)|=\sqrt P if A,B,AB are all nonprincipal.

The number of solutions of 4x(1-x)=y is 1+H^3(1-y). Summing H(y) times this identity and using \sum_yH(y)=0 gives J(H,H^3)=H(4)J(H,H). Write \tau_j=\tau(H^j). The Gauss–Jacobi identity and \tau_2\tau_4=P now give

Here H^2(-1)=1 because -1 is a cube.

It remains to fix the cubic Jacobi sum, including its unit. Put m=(P-1)/3. The definition of H^2 gives, in k,

Indeed the polynomial has degree 2m<P-1, and the sum over k of each monomial of degree less than P-1 is zero in k (the constant case is P=0 in k). The Jacobi sum belongs to \mathcal O and has absolute value \sqrt P.

For x\ne0,1, choose j_x\in\{0,1,2\} such that H^2(x(1-x))=\omega^{j_x}. The products of x and of 1-x over these x are both -1, so \prod_{x\ne0,1}x(1-x)=1. Hence \sum_{x\ne0,1}j_x\equiv0\pmod3. Since

we obtain J(H^2,H^2)\equiv P-2\equiv-1\pmod3. Divisibility by p and equality of norms imply J(H^2,H^2)=up for a unit u. As p is primary and the six units have distinct residues modulo 3, the congruence forces u=-1. Finally, \tau_2^3=J(H^2,H^2)\tau_2\tau_4=-pP. Dividing the two Gauss identities by the appropriate powers of \sqrt P proves the lemma.□

Proof. Use the Fourier transform with kernel e(-zy) and measure d\mu. For \varepsilon>0 apply Poisson summation on \mathcal O to f_\varepsilon(z)=e(z^2/c)e^{-\pi\varepsilon|z|^2}. Direct Gaussian integration gives, with r_\varepsilon=1+3q_c\varepsilon^2/16,

To check the normalization, rotate z by half the argument of c. The quadratic matrix of the resulting real two-variable Gaussian is

Its eigenvalues are conjugates with positive real part. The positive square root of the determinant, the factor 2/\sqrt3 in d\mu, and completion of the square give the formula above.

Put I_\varepsilon=\int_{\mathbb C}e^{-\pi\varepsilon|z|^2}d\mu(z) =2/(\sqrt3\varepsilon). For a fixed ideal (b), a fixed class v\bmod b, and a>0, Gaussian Poisson summation on b\mathcal O gives

The zero dual vector gives the limit and every nonzero dual vector is exponentially small. The phase e(z^2/c) is periodic modulo c. Consequently the normalized left side of Poisson’s identity tends to \Gamma(c)/|c|.

On the right side one may replace r_\varepsilon by 1. For fixed c, the total error from the phase is at most

because the sum is O_c(\varepsilon^{-2}). The amplitude error is O_c(\varepsilon), as is the damping error, by the same lattice estimate. All three are o(I_\varepsilon). The phase e(-cy^2/4) is periodic modulo 2\mathcal O. In the preceding Gaussian mean take b=2 and a=q_c/4; each of its four classes has normalized mean 1/q_c. The normalized right side therefore tends to (2|c|)^{-1}\sum_{y\bmod2}e(-cy^2/4), which proves Equation (4.3).

The representatives 0,1,\omega,\omega^2 modulo 2 give the asserted four-term expression. Multiplication by an odd v permutes these classes, so the same formula gives \Gamma(cv^2)=\Gamma(c). The group of units modulo 4 has order 12. Squaring a lift modulo 4 depends only on its class modulo 2, and its three possible squares are 1,\omega,\omega^2. The four representatives in the statement are distinct modulo this subgroup; also \lambda^2=-3\equiv1\pmod4. Evaluating the four-term expression on them gives the table. Evaluating \Gamma(ab)/(\Gamma(a)\Gamma(b)) on the two generators -1,\lambda gives the displayed exponent, proving the last claims.□

Proof. At a good prime, counting square roots in the residue field gives

The additive sum without the character is zero. Hence \Gamma(p)=\gamma_3(p), and replacing x^2/p by ux^2/p for a unit u multiplies the sum by \chi_p(u)^3. For squarefree c, representing a residue as \sum_{p\mid c}(c/p)x_p gives, in both sums,

The cross terms in the square are integral in the additive character. Thus \Gamma(c)=\gamma_3(c) for squarefree c. In particular, for distinct good primes p,q,

We use cubic reciprocity in the following precise form: for coprime primary a,b one has (a/b)_3=(b/a)_3; see [8, Equation (1.4)]. Thus the quotient \chi_b(a)/\chi_a(b) is a sign. At p,q its cube is \chi_q(p)^3\chi_p(q)^3, so this sign equals \mathfrak r(p,q). Multiplicativity in both arguments and the bicharacter property extend the equality to every coprime primary pair. The definition by \mathfrak r on noncoprime pairs involves no division of zero symbols.

The element -2 is primary and -1 is a cube. Cubic reciprocity therefore gives

This is a character of c\bmod2. Together with the dependence of \Gamma on c\bmod4, this proves that G is a function on a fixed ray group (for example the ray group modulo 12 suffices here). Indeed, if two primary generators represent the same ray class modulo 12, their quotient differs from an element congruent to one modulo 12 by a unit; reduction modulo 3 forces that unit to be one. The generators therefore have the same residue modulo 4. Its multiplicative refinement is the definition of \mathfrak r. Since \Gamma(v^2)=1 and \chi_v(4) has order dividing three, G(v^2)=\overline{\chi_v(4)}^{\,2}=\chi_v(4). The table also gives

For the last equality use \chi_p(-1)=(-1)^{(q_p-1)/6} and the equality of these parities. Now \Gamma(p^3)=\Gamma(p), and \overline{\chi_p(4)}^{\,3}=1, proving G(p^3)=\gamma_3(p). Finally \mathcal R(p,p)=1/\Gamma(p)^2=\Gamma(p)^2, as this square is a sign.

For a symmetric sign-valued bicharacter its diagonal is multiplicative:

The prime identity consequently gives \mathcal R(n,n)=\chi_n(-1) for every primary n outside S. On the square class (-1)^e\lambda^f, the table gives both this diagonal and \mathfrak r(-1,n) the value (-1)^f. This use of -1 is in the residue square-class group; the ideal (-1) is the identity in an ideal ray class group and is not being substituted there.

For coprime squarefree primary a,b outside S, the Chinese remainder theorem gives

One obtains this by representing a residue as bx+ay with x\bmod a and y\bmod b. Cubing the factor for j=2 gives one. The product of the factors for j=1 and j=2 is (\chi_a(b)\chi_b(a))^3=\mathcal R(a,b). The prime identities from Lemma 4.2, followed by Equation (4.5), therefore prove Equation (4.7) by induction on the number of prime factors.□

Proof. For an integer m with 2m>J+d, integration by parts gives

Its absolute value is bounded by the volume of \Omega times C_{m,d}p_{2m}(w). The weighted integral is finite because 2m>J+d; one can choose 2m\le J+d+2. Fourier inversion gives the monomial formula. For row vectors B(\boldsymbol t) in \ell^2, the precise inequality used there is

In one dimension, the fundamental theorem of calculus, averaging a base point over the enlarged interval, and Cauchy–Schwarz give \sup_I|F|^2\ll\int_{I+[-1,1]}(|F|^2+|F'|^2). Applying this successively in each coordinate proves Equation (4.9). It can be summed over rows before the derivative integrals, since all terms are nonnegative. A logarithmic scale range of length O(\log Z) needs O(\log Z) unit intervals; a height range [-T_1,T_1] needs O(1+T_1) intervals. Derivatives with respect to a logarithmic scale insert yW'(y), together with the constant derivative of a central normalization. Derivatives of a normalized twist y^{it} insert powers of \log y. These are again annular profiles with finite seminorm bounds. For example, if a fixed-parameter squared row bound is O((1+|t|)^H) and there are d_t height parameters, the covering and integration cost at most a fixed multiple of (1+T_1)^{H+d_t}.

On |\boldsymbol t|>T_1 one has (1+|\boldsymbol t|)^{-N}\le T_1^{-N}, proving Equation (4.10). The Mellin shift is immediate from its definition. To prove Equation (4.11), put F_\sigma(u)=e^{\sigma u}W(e^u). For |T|\ge1, integration by parts gives

Taking absolute values gives the bound on a compact real strip. For |T|<1, the absolute integral gives it after increasing the constant. The two stated classes of W have the required endpoint decay: compact support suffices in the first case, while e^{\sigma u} at -\infty and Schwartz decay at +\infty suffice in the second.

For Equation (4.12), use

Move (1+|T|)^N to the left, bound the weighted a(v) pointwise, and integrate the remaining polynomial weight against the Gaussian in T+v. The remaining w integral is the one displayed. This proves the joint bound and also the asserted join estimate. For the integrated assertion, both the map (t_1,t_2,t_3)\mapsto(t_1+t_2,t_2,t_3) and its inverse have fixed operator norm. A fixed polynomial weight in the original variables is therefore bounded by a fixed polynomial weight in the transformed ones. On a complement of a box one inserts the additional inverse power of that weight and integrates the Gaussian and the two Mellin factors separately. The appended block triangular map has the same property because all of its coefficients and those of its inverse are fixed.

A kernel depending on a common product, such as y_1y_2, gives the same norm power on both variables in the monomial formula; this preserves an equal-product-scale difference. A nonsmooth arithmetic mask has no such derivative bound and must instead be resolved before this lemma is applied.□

Proof. Such a partition is obtained by normalizing the integer translates of a nonnegative compactly supported smooth function positive on [-1/2,1/2]. Let \operatorname{supp}\chi\subset[-C,C]. Each logarithmic derivative of W_{\mathrm G}(e^k x) is a polynomial in k+\log x times the same Gaussian. The product rule therefore gives

This is summable after multiplication by e^{A|k|} for every fixed A. The first assertion follows, and the separation norms follow from Lemma 4.5. In the profile’s logarithmic Fourier transform, rescaling an annulus inserts only the unit phase e^{-ikt}, so it does not change these norms; any real normalization of a surrounding polynomial remains explicit. On the indicated tail, the logarithm of the last bound after multiplication by Z^Be^{A|k|} is at most -c\kappa^2(\log Z)^2+O(\log Z) for some fixed c>0; summing the Gaussian tail proves the stated estimate for every M. Finally, set u=\log y and complete the square in its Gaussian integral to obtain \mathcal MW_{\mathrm G}(s)=e^{s^2}. The same Gaussian bounds every e^{\sigma u}(d/du)^jW_{\mathrm G}(e^u) in L^1(\mathbb R), uniformly for \sigma in a compact interval.□

Proof. Differentiating \widehat f inserts x^\beta, and multiplication by a monomial in \xi differentiates x^\beta f before Fourier transformation. The L^1 bound for the Fourier transform, the product rule, and the bound of (1+|\xi|)^A by a finite sum of monomials of degrees at most \lceil A\rceil give the first assertion. For a radial function, r\partial_r corresponds to (\xi\cdot\partial_\xi)/2. Expanding its jth power gives finitely many polynomial multiples of Fourier derivatives, so the first assertion gives the radial one. A fixed linear change of coordinates changes only its constants.

For the Mellin assertion, differentiation under the absolutely convergent integral inserts (-s)^j. On the fixed strip this is bounded by a constant times (1+|t|)^j. Write y=e^u and integrate the Fourier transform of e^{-\sigma u}W(e^u) by parts more than h+j+1 times. The product rule bounds the resulting L^1 derivatives by the stated weighted logarithmic derivatives, because \sigma ranges over a fixed set. For a pure twist the Mellin integrand is \mathcal MW(-\sigma-it+i\omega). Substitute v=t-\omega and use (1+|v+\omega|)^{h+j}\le(1+|v|)^{h+j}(1+|\omega|)^{h+j}. Finally the product rule for a fixed annular cutoff and the chain rule for Ry give the last statement.□

Proof. A finite-order character has trivial infinite type, since \mathbb C^\times is connected. The primitive Hecke functional equation in this case is

This is the functional equation for primitive characters of trivial infinite type stated in [11, Equation (1.1)]; a nonprincipal L(s,\psi) is entire. Absolute Euler convergence bounds L(11/10+it,\psi) uniformly in Q,t. Applying the functional equation and Stirling’s formula on \Re s=-1/10 gives

The estimate for bounded t follows by continuity of the gamma quotient on that line, so the constant is uniform there as well.

For completeness, the growth hypothesis needed to use the strip principle can be obtained directly from a lattice theta integral. Let \mathfrak f=(f) be the conductor and define the periodic residue character \phi(a)=\psi((a)) when (a,\mathfrak f)=1, with value zero otherwise. For the principal conductor (1) put \phi(a)=1 for all a, including a=0. Since the ideal class group is trivial, conductor one has no nonprincipal character. Periodicity modulo \mathfrak f follows from the ray character property; \phi is trivial on the six units because (u)=(1) for a unit. Thus, for \Re s>1,

The factor six counts the generators of each ideal. If \widehat\phi(h)=Q^{-1}\sum_{a\bmod f}\phi(a)e(-ha/f), finite Fourier inversion and Gaussian Poisson summation give

Changing the generator f only reindexes the finite sum. For each fixed conductor, the nonzero dual vectors have a positive minimum length. With A_\phi=2\widehat\phi(0)/\sqrt3, this proves

for some c_\phi>0. Splitting the Mellin integral at one consequently gives

Both integrals are entire and bounded in height on each bounded real strip, with constants that may depend on the fixed conductor. Stirling’s formula for 1/\Gamma(s) therefore gives at most exponential height growth for L(s,\psi) on that strip. In the principal case the same conclusion holds after multiplication by (s-1)/(s+1). Only this qualitative growth, for each fixed conductor, is used in the strip principle.

To see explicitly that the fixed-conductor growth constants do not enter the uniform strip bound, put

It is holomorphic in the closed strip and is bounded by one absolute constant C on both vertical boundary lines, by the bounds already proved. For \delta>0, apply the maximum principle on the rectangle of height T to H_Q(s)\exp(\delta(s-1/2)^2). On the vertical sides the exponential has modulus at most e^{9\delta/25}, because |\Re s-1/2|\le3/5. On the horizontal sides its modulus is at most e^{\delta(9/25-T^2)}, which tends to zero faster than the qualitative fixed-Q exponential growth as T\to\infty. First let T\to\infty for this fixed Q,\delta, and then let \delta\downarrow0. The result is |H_Q(s)|\le C throughout the strip, with the same constant for every Q. Multiplication by Q^{3/5}|s+2|^2 proves Equation (4.14).

For the principal function, the factor (s-1)/(s+1) removes its only pole in the strip and is bounded on the boundary. Its functional equation gives the same boundary bounds with Q=1. This equation also follows from the preceding Poisson formula with \phi=1: after writing v=2t/\sqrt3, the theta relation is \Theta_\phi(2t/\sqrt3)=t^{-1}\Theta_\phi(2/(\sqrt3t)), whose split Mellin integral is invariant under s\mapsto1-s. Apply the same damped-rectangle argument to (s-1)\zeta_F(s)/((s+1)(s+2)^2).□

Proof. Write R_j=2-a-je and \mathcal C=2Q(3+|t|)^2. All the disks used here lie in \Re s>1/2, where the principal regularizer is holomorphic. On the zero-free disk choose a holomorphic logarithm g=\log\mathcal L whose value near the center is the Euler logarithm, together with the logarithm of the regularizing factor in the principal case. The value g(2+it) is uniformly bounded. Lemma 4.8 and Euler convergence to the right of 11/10 give

The possible heights differ from t by at most 3/2, which is included in \mathcal C. Borel–Carathéodory on radii R_3,R_4, whose difference is e, now gives |g|\ll_e\log\mathcal C on the disk of radius R_4.

On the disk of fixed radius r_0=49/100, one has \Re s\ge151/100>1. The absolutely convergent Euler logarithm is uniformly bounded there. For the principal function the logarithm of (s-1)/(s+1) is also bounded on this disk, using its branch in \Re s>1. Since r_0<R_6<R_4, Hadamard’s three-circles theorem applied to g gives

For fixed e, continuity on the compact interval 1/2\le a\le1 makes \theta uniformly smaller than one. For every \epsilon>0, \exp(C_e(\log\mathcal C)^\theta)\ll_{e,\epsilon}\mathcal C^\epsilon. Applying this to both e^g and e^{-g} proves Equation (4.15). Cauchy’s estimate between radii R_6 and R_8 gives the stated bound for g'=\mathcal L'/\mathcal L.

For the global assertion first suppose \beta_*\le b\le1. Choose 0<e<\min(10^{-3},v/8) and put a=b. The disk of radius R_2 is contained in \Re s>b+2e>\beta_*, so it is zero-free by the definition of \beta_* and absolute Euler convergence beyond one. The disk of radius R_8 covers the points b+v\le\Re s\le2 at its center’s height; Euler convergence covers \Re s\ge2. The preceding constants are uniform in b. If b>1, Euler convergence on \Re s\ge1+v suffices. Finally

which proves the principal reciprocal assertion.□

Proof. Every factor is nonzero on \Re s>0. For \Re s\ge\sigma_0, both its absolute value and the absolute value of its inverse are at most (1-q_p^{-\sigma_0})^{-1}. For sufficiently large q_p, depending on \sigma_0,\epsilon, -\log(1-q_p^{-\sigma_0})\le\epsilon\log q_p. The finitely many smaller primes contribute a fixed constant. This proves the first estimate. For any real \sigma,

The last inequality follows by separating the finitely many primes with q_p<2^{1/\epsilon}. Finally, logarithmic differentiation gives

If q_R\le Z^B, choose the exponent in the first estimate smaller than the desired exponent of Z divided by the bounded value of B; the case B=0 is immediate. This proves the stated interpretation for deleted factors.□

Proof. We first express the character masks through finitely many translated theta values. We then transform their rational cusps and compute the resulting automorphy multipliers and local Fourier factors; the final Mellin comparison gives the reflection identity and the bounds for its transformed test.

The fixed theta input. Let \Gamma=\mathrm{SL}_2(\mathcal O), \Gamma_1(3)=\{g\in\Gamma:g\equiv I\pmod3\}, and \Gamma_2=\langle\mathrm{SL}_2(\mathbb Z),\Gamma_1(3)\rangle. The unconditional formulas of [8, Equations (5.4), (5.7), (5.9), and (5.12)–(5.17)] give a smooth cubic theta function \theta on \mathbb H^3=\mathbb C\times\mathbb R_{>0} with

Here \kappa is one on \mathrm{SL}_2(\mathbb Z) and is one on an element of \Gamma_1(3) with lower left entry zero. The subscript 3 denotes the ordinary cubic residue symbol; for a good prime p\notin S, one has (x/p)_3=\chi_p(x)^2. These automorphy and coefficient formulas have no hypothesis about zeros of L-functions.

Put

If v\in\mathbb Z+3\mathcal O, then \mathsf L(v) is a product of an element of \mathrm{SL}_2(\mathbb Z) and a lower translation in \Gamma_1(3) with multiplier one. The analogous assertion holds for upper translations. Therefore

because \lambda-\omega^2=2+3\omega and -\lambda-\omega=-1-3\omega. If u_0=s+t\omega with 0\le s,t<3, and E=\bigl(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\bigr), then

and hence \theta\bigl(\bigl(\begin{smallmatrix}u_0&-1\\1&0\end{smallmatrix}\bigr)w\bigr) =\Theta_t(w). These are the only cusp representatives needed below.

For any of them, define d_H by

Here C_H(v) is independent of z. The three functions \Theta_0,\Theta_1,\Theta_2 are respectively F_1,F_{19},F_{10} in [8]. Its Appendix A, rows 1,19,10, expresses their coefficients in terms of \tau,\tau_1,\tau_2 in its Equations (5.7), (5.13), and (5.14). In their notation the conjugate expansion uses \overline{d_j(-\mu)}, as in their Equation (5.16). Those formulas give the support inclusion in Equation (5.2). For completeness, the two possible magnitudes from \tau at \mu=u\lambda^k n b^3 are

and \tau_1,\tau_2 have magnitude 9|b| and k=-4. Here the Gauss sums use only the cubic symbol, including at primes in S. To make their domain explicit, for every primary squarefree n define, as in [8, Equation (1.7)],

The ordinary cubic symbol is defined at every prime other than \lambda and is extended by zero on nonunits and multiplicatively in the denominator. Every primary n is prime to \lambda. At each prime divisor of n the cubic character is nontrivial, and \check e(v/n)=e(\lambda v/n) is primitive because \lambda is a unit there. The finite-field Gauss identity and the Chinese remainder theorem therefore give |\widetilde g_3(n)|=1 for every such squarefree n. This includes the primary prime -2 of norm 4; no sextic character with that denominator is used. Also, whenever (t,n)=1, the change of variable y=tv gives

This covers the twists t=\lambda^2,\omega\lambda^2,\omega^2\lambda^2 in the other source coefficient families. Thus the unnormalized Gauss sums occurring in all three cusp families have magnitude |n|. Consequently each coefficient magnitude displayed above is at most 27\,3^{k/6}|b|, including k=-4. The formulas require only that n be squarefree and that n,b be primary; they do not require (n,b)=1. Prime valuations of nb^3 also show that its representation by such n,b, when it exists, is unique. In particular the same bound is valid when the variables share primes or contain primes excluded from the primal sums.

At infinity the condition x=\lambda^3\mu\equiv1\pmod3 isolates exactly \mu=\lambda^{-3}nb^3 with n,b primary and n squarefree. The infinity support has \lambda^3\mu\in\mathcal O; every other ramified exponent in Equation (5.7) of [8] makes x divisible by \lambda, and the negative sign in the remaining pair gives x\equiv-1\pmod3. At the isolated index the coefficient in \bar\theta is

This is the unmasked source formula, valid for every primary squarefree n and every primary b, with no coprimality condition between them. Only when (n,S)=1 may it be expressed in the global sextic notation: substituting y=\lambda v in \gamma_2(n) gives

The cubic supplementary laws in [8, Equation (1.5)] make (\lambda/x)_3 periodic on all primary x modulo 9. On primary elements prime to S it equals \chi_x(\lambda)^2. A character in \mathcal X is periodic on primary elements prime to \mathfrak E modulo its conductor, and the zero condition (x,\mathfrak E)\ne1 is periodic modulo \mathfrak E. Thus a sufficiently divisible L with prime support S works for every member of the fixed finite family. This verifies its choice before the moving local primes are selected.

Finite Fourier inversion with the zero extensions. For j\in\{0,\ldots,5\} define

The character pairing furnished by e is nondegenerate on \mathcal O/(p). Changing variables in a Gauss sum, and using the orthogonality of a nontrivial multiplicative character, gives

Here and below h=0 means h\equiv0\pmod p. Each nontrivial power of \chi_p is primitive modulo the prime p, so the usual finite-field Gauss-sum identity gives |\tau_{p,m}^{\pm}|=1.

Fourier inversion on \mathcal O/(L) and on each \mathcal O/(p) now gives

This uses every additive frequency, including nonunit h_0 and zero local frequencies. We do not invoke the twisted formula in [8, Lemma 5.2 and Corollary 5.1], whose unit-Fourier-support hypothesis need not hold for these masks. The theta function with this multiplier on its infinity coefficient x=\lambda^3\mu\in\mathcal O is the finite sum

For an isolated infinity index x=nb^3, the piecewise definition of \phi first discards every term for which (nb,\mathfrak E)\ne1. This is a mask on the whole completed index: (nb^3,\mathfrak E)=1 if and only if both n and b are \mathfrak E-units. Only on that remaining domain do we use complete multiplicativity, including \chi_b(\lambda)^6=1, to write

It remains true when n,b share a moving prime, since the corresponding local powers retain their zero values, including for j_p=0. For the discarded terms the completed coefficient is defined to be zero directly; no value of \chi_n(\lambda) is evaluated there. Combining Equations (5.9) and (5.10) on the retained domain now identifies the direct Mellin series with the product T. The dual coefficient functions d_H keep the full source support in Equation (5.2); no S-mask is placed on them.

Fix h_0 and declare p active precisely when h_p\ne0. Equation (5.11) forces all primes with j_p\ne0 to be active. Choose once for each h_0\bmod L a representative and a reduced expression \lambda^2h_0/L=a_F/c_F. Multiplying numerator and denominator by one unit, normalize the numerator to be 1\pmod3 if \lambda\mid c_F and the denominator to be 1\pmod3 otherwise. Since every active prime is primary, this unit depends only on h_0. For r=\prod_{p\ {\rm active}}p, Equation (5.12) then gives

At a prime dividing c_F, the numerator is congruent to a_Fr and is a unit. At p\mid r, it is congruent to \lambda^2c_F(r/p)h_p and is again a unit. Thus this fraction is reduced, even when h_0 is a nonunit modulo L. The empty product r=1 is allowed. Replacing any lift by another changes z_h by an element of \lambda^2\mathcal O=3\mathcal O, a period of \theta.

Choice of the cusp matrix and its multiplier. The next construction determines the automorphy multiplier and reflected additive phase for each reduced cusp fraction just obtained. Set M=\lambda^{12}L^4. For every active p, the Chinese remainder theorem permits the chosen nonzero class h_p\bmod p to be lifted with h_p\equiv0\pmod{M^2}. Restrict r to one class modulo M^2. Equation (5.13) shows that a is then fixed modulo Mc_F; in fact its moving terms are divisible by M^2c_F, and a_Fr is fixed modulo M^2, which is divisible by Mc_F because c_F divides L up to a unit. It also shows a\equiv1\pmod3 if \lambda\mid c, and c\equiv1\pmod3 otherwise.

Choose \delta' by the following compatible local congruences:

The first line means the indicated prime-power parts of Mc_F. All inverses exist because a/c is reduced. Put b=(a\delta'-1)/c and g=\bigl(\begin{smallmatrix}a&b\\c&\delta'\end{smallmatrix}\bigr). Then g\in\Gamma, and division of a\delta'-1 by c gives

Consequently a,b,c,\delta' have fixed residues at the needed powers of every prime dividing L, independent of the classes h_p. Whenever a zero entry would occur in a residue-symbol calculation, one may first translate z_h by \lambda^2M^2\mathcal O and then add a multiple of the combined congruence modulus to \delta'. These changes preserve all displayed congruences and avoid the finitely many values making an entry zero. They do not change the translated theta function.

There are three cases, distinguished by v_\lambda(c). If 3\mid c, put H=I. Then G=g\in\Gamma_1(3), and cubic reciprocity for the primary a and the primary primes of r gives

If v_\lambda(c)=1, take the unique u_0\in\{\lambda,-\lambda\} with u_0\equiv c\pmod3 and put H=\mathsf L(u_0); these are the two possibilities because \mathcal O/(\lambda)\simeq\mathbb F_3. The congruences above show that G=gH^{-1}\in\Gamma_1(3). Write A=a-u_0b. Both A and a are primary, and they are coprime: (a,b)=1 and (a,u_0)=1. The determinant equation gives

It follows, by cubic reciprocity for a,A, that

For the second equality, reduce A modulo a and use (A/a)_3=(-u_0b/a)_3=(u_0/a)_3(c/a)_3^{-1}; the cubic symbol of -1 is one. Finally suppose (\lambda,c)=1. Choose u_0=s+t\omega\equiv a\pmod3 with 0\le s,t<3 and put H=\bigl(\begin{smallmatrix}u_0&-1\\1&0\end{smallmatrix}\bigr). Here c\equiv1, \delta'\equiv0, and b\equiv-1\pmod3, so

Both -b and c are primary. Moreover -bc=1-a\delta'\equiv1\pmod9. Factor \delta'=u\lambda^k\delta_0 with \delta_0 primary. The supplementary laws make (u\lambda^k/(1-a\delta'))_3=1; cubic reciprocity makes (\delta_0/(1-a\delta'))_3=1 because the numerator after reciprocity is 1\pmod{\delta_0}. Thus

The last inverse uses a\delta'\equiv1\pmod c.

In all three cases we have proved

The factor \kappa_F is constant as the active h_p vary in the fixed sector. If 3\mid c, it is (c_F/a)_3: the unit and \lambda-power factors are determined by a\bmod9, and reciprocity determines the remaining fixed-prime factors from the residue of a at the primes of c_F. In the middle case it is (-u_0/A)_3((c_F/u_0)/a)_3, determined in the same way by the fixed residues of A=a-u_0b and a. In the last case it is (a/c_F)_3, which is determined directly by the fixed denominator c_F and a\bmod c_F. The displayed congruences for a,b,c,\delta' therefore determine \kappa_F using only h_0 and r\bmod M^2. They also determine the chosen one of the three functions \Theta_b.

The reflected additive phase. Write x=\lambda^4\mu and D_F=\lambda^3c_F. At each p\mid r, Equation (5.13) gives a\equiv\sigma_ph_p\pmod p. Hence \delta'\equiv\sigma_p^{-1}h_p^{-1}\pmod p. Additive Chinese remaindering for the coprime factors of D_Fr gives the exact identity

Here \delta'_F is the class of \delta' modulo D_F, and r^{-1} in the formula for \vartheta is taken modulo D_F. For example, the local numerator at p is -\delta'(D_Fr/p)^{-1}=\epsilon_ph_p^{-1}; this verifies the signs and the powers of \lambda. The chosen congruences fix \delta'_F, since D_F divides the corresponding local moduli, and the sector fixes r^{-1}\bmod D_F. A common multiple of the finitely many D_F is therefore a fixed modulus for all the characters \vartheta.

Conjugating Equation (5.15) contributes \bar\kappa_F\prod_{p\mid r}\chi_p(\sigma_p)^{-2}\chi_p(h_p)^{-2}. For j\ne0, multiplication by Equation (5.11) leaves the local sum

Upon putting y=h^{-1}, the sum is a Gauss sum of exponent j+2. If j\ne4, its value is q_p^{1/2}\tau_{p,j+2}^{+}\chi_p(\epsilon_px)^{-j-2}, also at p\mid x because both sides then vanish. If j=4, it is the Ramanujan sum

For an active j=0, the same substitution gives

These are exactly \omega_{p,j}B_p(x); an inactive zero exponent contributes 1-q_p^{-1}. This proves every local factor, including its zero extension. The choices of h_0 and the active subset give at most q_L2^{\#\{p:j_p=0\}} terms. Splitting the j_p=4 factors multiplies this by at most 2^{\#\{p:j_p=4\}}, proving the stated counts.

The denominator and the fixed-sector data constructed before these local sums depend on the local prime set only through h_0 and the active radical r; the remaining active-frequency dependence is exactly the local dependence just summed. We use the same representatives and normalized fractions for each h_0, and the same fixed-sector choices of cusp function, fixed additive character, and \kappa_F whenever those data recur. An absent prime and an inactive zero-exponent prime have the same active radical; their other active \sigma_p,\epsilon_p are therefore also identical. The inactive coefficient 1-q_p^{-1} is the only change. This proves the simultaneous branch compatibility in the statement.

The marked reflection in Section 14.2 will use the following dependence on pairs of active primes. Since

there is a scalar \xi_{p,j} of absolute value one, depending on p,j but on no other active prime such that

For j=4 this uses 2j+2\equiv-2\pmod6; for the other exponents it follows immediately by inserting the expression for \epsilon_p. Thus the contribution of another active prime to the phase at p has exponent exactly 2j+2 modulo six. In particular an active exponent j=1 has column coupling \chi_p(x)^{-3}=\chi_p(x)^3.

Mellin normalization and continuation. Let

The Bessel identity used in [8, Lemma 5.3] is

On \Re s>1 the coefficient sums converge absolutely, so we may differentiate and integrate their expansions term by term. Since \partial_{\bar z}\check e(\mu z)=2\pi i\bar\mu\check e(\mu z), the identity with w=2s+1 gives

First use the whole-index mask to restrict to (nb,\mathfrak E)=1 as above, then insert Equations (5.9) and (5.10). Using q_{\lambda^{-3}}=27^{-1} and \bar\alpha(\lambda^{-3})=-i gives

For one translated term z_h=a/c, use the matrix g=GH constructed above. The coordinate action of g^{-1} on hyperbolic space, from [8, Equation (5.1)], gives

At z=0, the derivative of its first coordinate with respect to \bar z is -(cv)^{-2}; the derivatives of its conjugate coordinate and of its vertical coordinate with respect to \bar z are zero. The chain rule and \bar\theta(gw)=\bar\kappa(G)\bar\theta(Hw) therefore give

In particular the derivative removes C_H(v) from every cusp expansion in Equation (5.7). Substituting that expansion and then v\mapsto1/(q_cv) in the integral gives, for \Re s<0, the contribution

To verify the scalar directly, before applying the Bessel integral the factor is -2\pi i c^{-2}q_c^{2-2s} and the remaining integral is \int_0^\infty u^{2-2s}K_{1/3}(4\pi|\mu|u)\,du. The Bessel identity with w=3-2s, together with c^{-2}=\bar\alpha(c)^2/q_c, gives exactly the display.

We justify both the continuation and the contour operations here. The coefficient bound in Equation (5.2) implies absolute convergence of the reflected series when \Re(1-s)>1: the separate majorants are

Each converges in that region. The fixed support has a positive lower bound for |\mu|. The exponential decay of K_{1/3} and of all its derivatives, the coefficient bound, and the direct expansion imply exponential decay at v\to\infty for every logarithmic derivative of each translated derivative. The chain-rule formula just obtained implies exponential decay at v\to0 as well, with constants depending on the fixed translated term. Repeated integration by parts in \log v now shows that J is entire and decreases faster than every power of |\Im s| on each fixed strip.

Equation (5.18), with reciprocal gamma functions on its right, continues T to an entire function. Its direct series is bounded on any line to the right of 1. Equation (5.19) and Stirling’s formula give a polynomial bound on any line to the left of 0; the exponential parts of the two gamma products cancel. On a strip between such lines, the rapid bound for J and the reciprocal gamma factors first give |T(\sigma+it)|\ll (1+|t|)^C e^{\pi|t|} for some C. Divide T by a sufficiently large power of B+s, with B chosen so that B+s has no zero on the strip, and multiply by e^{\varepsilon s^2}. The horizontal sides of a growing rectangle tend to zero for each \varepsilon>0. The maximum principle, followed by \varepsilon\to0, transfers the polynomial bounds on the two vertical sides to the strip. This proves the stated polynomial strip growth.

Shift the integral on the left of Equation (5.1) from \Re s=a to \Re s=1/2-\sigma<0 with \sigma>1/2. The entire continuation, polynomial strip bound, and rapid Mellin decay justify the shift: Equation (4.11) applied to the weighted logarithmic derivatives of V makes the horizontal joins tend to zero. Inserting Equation (5.19) divided by Equation (5.18), and putting t=1/2-s, gives the gamma quotient R(t) and the factor

Indeed (2\pi)^{4s-2}27^{-s}3^{-5/2} =3^{-4}K^{-t} and q_c^{1-2s}q_\mu^{s-1} =q_\mu^{-1/2}(q_\mu/q_c^2)^{-t}. The reflected coefficient series is absolutely convergent on this line, so it may be interchanged with the integral. Summing its finite local factors using Equation (5.16) and the computed Gauss sums proves Equation (5.1) and Equation (5.3). Since |\phi|\le1, we have |\widehat\phi(h_0)|\le1; every other factor in the scalar has absolute value at most one. This proves |\zeta|\le1/81.

Uniform estimates for the transformed test. The first negative pole of the numerator of R(t) is -5/6. The reciprocal denominator gamma functions are entire. Hence the integrand defining V^\sharp is holomorphic for \Re t>-5/6. On -1/4\le\sigma\le A, Stirling’s formula gives

The estimate is uniform also for bounded u, since this closed strip has no numerator pole. Equation (4.11) makes \widehat V rapidly decreasing pointwise on every vertical strip. Rectangular contour shifts are therefore valid within -1/4\le\Re t\le A, including between any two nonnegative lines. Applying (x\partial_x)^j contributes (-t)^j. The three lines -1/4,0,A bound the resulting integral respectively by constant multiples of

Combining them proves Equation (5.4). The rapid large-x bound, with A arbitrarily large, also makes every dual sum in Equation (5.1) absolutely convergent when V^\sharp is represented on the zero line. Thus the shift of the kernel contour does not require a termwise shift of a conditionally convergent Dirichlet series.

For Equation (5.5), set f_Y(v)=\chi(e^v)V^\sharp(Ye^v). This is supported on a fixed compact interval. Integrating its Fourier transform by parts J+2 times and using the trivial bound for |u|\le1 gives

Leibniz’s rule and Equation (5.4) bound each norm by the right side of Equation (5.5), because e^v stays in a fixed compact subinterval of (0,\infty). Finally, for |u|\ge1, B+2 integrations by parts give

For |u|\le1 use the corresponding undifferentiated integral. On -A\le\eta\le1/4, x^\eta\le1+x^{-A}+x^{1/4}, proving Equation (5.6). The identity \widehat{x^{iu_0}V}(\eta+iu)=\widehat V(\eta+i(u+u_0)) and 1+|u|\le(1+|u+u_0|)(1+|u_0|) give the asserted norm-twist bound. In particular the Gaussian test V(x)=(2\sqrt\pi)^{-1}\exp(- (\log x)^2/4), whose Mellin transform is e^{t^2}, satisfies the hypotheses of this Proposition directly.□

Proof. If (A,m_{\rm good})=1, multiplicativity and the symmetric reciprocity factor in Lemma 4.4 give Equation (5.20). If a good prime is shared, both sides are zero: the left side is the zero extension of \chi_A(m), and its local factor on the right is zero even for a six-divisible exponent. This proves the identity on the entire stated domain.

On this domain \chi_A(u m_S) depends only on u and the S-valuations modulo six, because every numerator factor is a unit modulo A. For each of the finitely many representatives d=u\prod_{\mathfrak l\in S}\pi_{\mathfrak l}^{e_{\mathfrak l}}, 0\le e_{\mathfrak l}<6, Lemma 4.1 identifies A\mapsto\chi_A(d) with a finite-ray character whose conductor is supported at primes over 6d, all in S. Once the good ray class is fixed, A\mapsto\mathcal R(A,m_{\rm good}) is one of a fixed finite family of characters supported at the primes over 2,3. Their products with the finite family of base characters give the asserted fixed family. Choosing a common multiple of its conductors and the S-mask gives the common \mathfrak E and then L. Additional good local residue-symbol factors in a completed coefficient are placed in \mathcal P and combined at a shared prime with the same zero convention; they do not alter this fixed modulus. The argument requires m\ne0 and makes no local-prime assertion for the row m=0.□

Proof. On the support of W, the monomial h is bounded above and below by positive constants depending only on \Omega and the exponents. Each operator y_i\partial_{y_i} acting on V^\sharp(Yh(\boldsymbol y)) is a_i times the Euler derivative of V^\sharp at Yh(\boldsymbol y). It produces no additional power of Y. Leibniz’s rule and Equation (5.4) prove Equation (5.22). Applying Lemma 4.5 with j=J+d+2 gives Equation (5.23).□

Proof. The shell 2^jU<a q_\mu\le2^{j+1}U contains O(1+2^jU/a) points of either fixed lattice. Its contribution is at most C_A(1+2^jU/a)(2^jU)^{-A}. Summing the two geometric series gives O_A(C_A(U^{-A}+a^{-1}U^{1-A})), which implies the display because U\ge1. The last assertion follows from Equation (5.4).□

Proof. The factorization b=gt^2 permits g and t to share primes. Since nb=c^2mht^2, the zero convention gives the exact identity

Indeed a square has sixth power under the displayed quadratic character; this is one on units and zero at a collision. The identity therefore also holds when n or g shares a prime with t.

For the chosen ranges there are O(T) possible t. Hilbert-space Cauchy gives

Fix t in the positive sum. The condition (k,t)=1 is now a fixed row restriction. For the remaining condition use the full identity

For each k the number of possible r is at most d_{\mathcal O}(k). Rowwise Cauchy followed by summation over k costs (KNB)^\epsilon and gives a positive sum over fixed squarefree good r. Write k=rk' and c=rc'. Squarefreeness gives the fixed row restriction (k',r)=1 and the coefficient restriction (c',r)=1; terms with (r,t)>1 vanish, and otherwise (k',t)=1 is another fixed row restriction. There is no remaining (k',c') condition in an individual r-summand: the full collision mask was already expanded. A fixed ray sector for k becomes a fixed sector for k' once r is fixed.

The factor \chi_r(mh)^3 is a bounded column factor. Split the squarefree parts of m,h supported on S from their good parts: m=m_Sm_{\rm g} and h=h_Sh_{\rm g}. There are only finitely many choices of m_S,h_S. Their symbols are bounded row factors, while j=m_{\rm g}h_{\rm g} is squarefree and good, of norm O(NG/C^2). For fixed r,t,m_S,h_S, group the remaining columns by j. If C_j is the resulting coefficient, the quadratic large sieve gives

The outer sum may retain all its fixed restrictions. On a nonempty range, both quotient lengths in the last display are bounded below by a fixed positive constant; replacing them by their maxima with one changes only the comparison constant.

For each j, the number of factorizations into m_{\rm g},h_{\rm g} is divisor-bounded. There are O(C/q_r) possible c' in its range. Cauchy over these two choices therefore gives

The bounded column factor from \chi_r has been discarded only in this positive sum. The fixed bad-part symbols were row factors of unit modulus and were removed from the outer modulus before applying the sieve. The ideals m,h here are the original ideals, reconstructed from their fixed S-parts; the S-part of c' is untouched. Combining the displays proves Equation (5.27).

Suppose now |\beta|\le1. Ideal counting gives

For q_r\asymp r_0, its product with the preceding factor C/q_r is O(NG/r_0^2). There are O(r_0) possible r in this range and O(T) possible t. Since B\asymp GT^2, their contribution to Equation (5.27) is at most

Here r_0,C,T are bounded below on a nonempty range and G\ll B. The number of dyadic choices for C,G,T,r_0 is logarithmic in the norm ranges. Hilbert-space triangle over the C,G,T blocks, followed by a rescaling of \epsilon, proves Equation (5.28).□

Proof. For the fixed h_0 and the fixed class of R modulo M_{\rm ref}^2, the class of F_{\rm act}R is fixed. Proposition 5.1 therefore gives one common d and one common additive character \vartheta for the whole row set. Write x_F=u\lambda^{k+4}N_FB_F^3, so that \lambda^4\mu=x_Fnb^3. The residual local factors are exactly

Both equalities include zeros; no unit symbol has been divided. In particular, any collision with x_F is a restriction depending only on R. Every remaining local factor is fixed and depends only on the dual index. The scalar \zeta_R of the reflection depends on R but not on \mu.

Define on the specified dual support

The coefficient bound gives |\delta(n,b)|\le1. Its denominator is nonzero, and the definition remains valid when the numerator is zero. Using q_\mu=3^kq_{N_Fn}q_{B_Fb}^3 gives the exact factorization

A small Ramanujan factor or an active zero-mask factor contributes q_p^{-1/2}. A divisibility factor contributes q_p^{1/2}; for a prime assigned to N_F it cancels q_p^{-1/2} from the squarefree denominator, and for a prime assigned to B_F it leaves q_p^{-1/2} after extraction. When a prime assigned to N_F also divides b, its entire occurrence in b remains in q_b^{-1} and in \delta(n,b). These facts give exactly the common power Z^{-\nu_0}y_n^{-1/2}y_b^{-1}.

Absorb \delta, the signs of small Ramanujan terms, the bounded fixed local characters, and the fixed dual indicators into \beta(n,b). This function is bounded by one and is independent of R. Put \rho(R)=81\zeta_R\chi_R(x_F)^3; it is bounded by one because |\zeta_R|\le1/81. The factor 27/81=1/3 now proves Equation (5.32). Multiplicativity of the norm proves Equation (5.33). It does not use any coprimality between N_F and b.

By the definitions of the log-norms,

The finite set of possible c_F depends only on L. Consequently \mathfrak m_A(Y)\ll_L Z^{-(T_0-v-3\ell_b-e_\lambda)_+/4}. Apply Lemma 5.3 in dimension three and Equation (5.24) to the structural identity. The single density is common to all R because W uses the row and both dual norms as coordinates. At each Fourier mode the norm powers have absolute value one and can be absorbed into the bounded row scalar and the common dual coefficient. Only now enlarge the positive row norm from \mathcal I_H to squarefree good R with q_R\ll Z^H, retaining any fixed restrictions. No kernel estimate is used on the added rows.

Equation (5.28), with K=Z^H, N_{\rm col}=Z^v, and B_{\rm col}=Z^{\ell_b}, bounds the separated squared norm by

The squared common coefficient in the structural identity has exponent -v-2\ell_b-2e_\lambda/3-S_0-B_0. The square of the extracted profile scalar contributes -(T_0-v-3\ell_b-e_\lambda)_+/2. Combining these exponents and the dimension-three separation norm proves Equation (5.34).□

Proof. Fix one character \nu. Lemma 5.2, with \Psi_{\rm base}=\nu, supplies a fixed finite family of literal characters for the reflection of \nu(A)\chi_A(m). Choose their common \mathfrak E,L before any good row prime varies. Partition the rows into the finitely many unit, S-valuation residue, and good-ray sectors from that lemma. This keeps every good local prime, including those whose row valuation is divisible by six.

For a row in the prescribed O-dyad, split its valuation-one primes into m_{\rm sup} supported on S and the squarefree good product R. Thus (m)=m_{\rm pow}m_{\rm sup}R on disjoint prime supports. Freeze the unit, the actual ideals m_{\rm pow},m_{\rm sup}, and then partition R into actual smooth annuli q_R\asymp Z^H, with H\ge0 and the unit in the bounded zero dyad. Keep this row cutoff in each reflected vector. Equation (5.35) gives

The fixed row restrictions, including coprimality with the frozen factors, are independent of the dual variables.

After the row-sector conversion, take \mathcal F in Definition 5.6 to be the good primes of m_{\rm pow}, with their valuations reduced modulo six. The primes of R have exponent one. Fix h_0, a class of R modulo M_{\rm ref}^2, and the local active and Ramanujan choices at the frozen primes. Every active frozen good prime divides m_{\rm pow} to exponent at least two. Consequently

This uses the actual powerful dyad’s upper comparison. The possible S-prime factors of m_{\rm pow} only increase its norm.

Apply Equation (5.1) at X=Z^N to the Mellin representation in Equation (5.31). Its dual sum is absolutely convergent. Insert fixed smooth dyadic partitions in the residual squarefree and cube norms, and fix the unit and ramified exponent of the dual index. These operations give the blocks of Definition 5.6 with uniformly bounded seminorms for W. No partition of the primal test V is needed: the compact profile here is the product of the dual and row cutoffs with the normalized inverse roots, and V^\sharp remains the kernel.

Fix a small \tau_{\rm ref}>0. The row lengths H,O, the active length A_0, and N lie in fixed bounded ranges. The same is true of S_0,N_0,B_0, since their prime products divide F_{\rm act}. The tail argument following Lemma 5.7 therefore removes all unretained whole dual blocks with any prescribed power saving, with a finite seminorm of V. Its polynomial cost includes the row and local counts below but not the discarded dual indices. Choose its kernel order first, and increase A,J in the statement so that \mathfrak M_{A,J}(V) dominates this tail seminorm and the block seminorm. This is possible because increasing A widens the supremum defining \mathfrak M_{A,J} and increasing J increases its weight.

Write e=e_\lambda. The source restriction k\ge-4 gives e\ge-4\log 3/\log Z. For a retained block,

Since v,\ell_b\ge0 and T_0 is bounded, this also bounds every retained dual length. There are only logarithmically many retained choices for the dual annuli and for k. Choose a nonnegative \epsilon_Z=O_{\rm fixed}(1/\log Z) large enough to cover the support offsets in Equations (5.37) and (5.38), and to ensure e\ge-\epsilon_Z.

The exponent in Equation (5.34) satisfies

To see this, use \max(H,v+\ell_b)-\ell_b=\max(H-\ell_b,v) and discard the nonpositive kernel term. The first branch is at most H+2\epsilon_Z/3, since \ell_b,S_0,B_0\ge0. In the second branch, retention gives v\le T_0-3\ell_b-e+\tau_{\rm ref}, so it is at most T_0-S_0-B_0+5\epsilon_Z/3+\tau_{\rm ref}. Moreover,

by Equations (5.37) and (5.38). Thus every retained block, with its frozen row parts fixed, has exponent at most

It remains to count the frozen parts and the finite expansions. Every powerful ideal is uniquely x^2y^3 with y squarefree: at a prime, the exponent of y is the parity of the powerful exponent, and an odd positive exponent is at least three. Ideal counting and \sum_yq_y^{-3/2}<\infty therefore give

The squarefree m_{\rm sup} is a divisor of the fixed radical of S and has only finitely many choices. The local branches at the good primes of m_{\rm pow} have divisor-bounded multiplicity, as do their Ramanujan assignments. Rowwise divisor Cauchy and then summation of the local choices cost an arbitrarily small power of Z. The choices of h_0, fixed ray sectors, and units are finite, and the actual row and retained dual annuli cost only powers of \log Z. This counts the powerful contribution O/2 exactly once.

Choose \tau_{\rm ref} and the local small-power losses within the prescribed \epsilon allowance. Then increase the fixed-data threshold for Z so that the O(\epsilon_Z) terms lie within that allowance. The block bound, the preceding counts, and the already chosen tail saving prove Equation (5.36). The twist assertion follows from the norm-twist bound for \mathfrak M_{A,J} in Proposition 5.1.□

Proof. The assertion is immediate when J=0. The characters e(mz_j) are well defined on \mathbb C/\mathcal O by self-duality. Choose representatives in \mathbb C for the other cases. We use the Fourier transforms

Choose a nonnegative \phi\in C_c^\infty(\mathbb C) of integral one, supported in a sufficiently small disk about zero. Since |e(xy)-1|\le (4\pi/\sqrt3)|x||y|, its support can be chosen so that

Thus F(x)=4|\check\phi(x)|^2 is a nonnegative Schwartz function and F(x)\ge1 on the unit disk. If \widetilde\phi(y)=\overline{\phi(-y)}, Fourier inversion and the product formula, with convolution taken with respect to d\mu, give

In particular, \widehat F is bounded and supported in a disk of some fixed radius L. This is the bandlimited majorant we need.

Put R=\sqrt Q. Positivity of F first gives

All sums on the right converge absolutely. For any z\in\mathbb C, the Fourier transform of x\mapsto F(x/R)e(xz) at y is R^2\widehat F(R(y-z)). The factor R^2 is the Jacobian of the real two-dimensional dilation; no lattice-volume factor occurs because d\mu has covolume one on \mathcal O. Poisson summation therefore expands the right side as

For each fixed j, a term in the inner sum can be nonzero only if the lift z_k+n lies within distance L/R of z_j. The set of all lifts \{z_k+n:1\le k\le J,\ n\in\mathcal O\} is \delta-separated in \mathbb C. Distinct classes have this property by hypothesis, and two distinct lifts of one class differ by a nonzero Eisenstein integer, whose absolute value is at least one and hence at least \delta. The disks of radius \delta/3 about lifts in a disk of radius L/R are disjoint and lie in the concentric disk of radius L/R+\delta/3. Comparing their Euclidean areas bounds the number of these lifts by

The same bound holds with j and k interchanged. Taking absolute values in the Poisson expansion, using the fixed bound for \widehat F, and then using 2|a_ja_k|\le |a_j|^2+|a_k|^2, we obtain

Since R^2=Q, this is the asserted estimate. Replacing Q by a fixed multiple proves the final statement.□

Proof. For a primary s outside S, put r_s=q_s/Y and

The inverse is evaluated only on these elements prime to S. Here |\tau|=1, |\xi(s)|=1, and |\chi_s(b_*)|=1, because b_* is supported on S. Thus |c_\sigma(s)|\le1. The definition of A and the Gauss expansion give the exact identity

The sum is over primary ideals outside S. For every such s, the character \chi_s(d) is zero exactly when (d,s)>1, and has absolute value one otherwise. This includes nonsquarefree s: a local exponent divisible by six is the indicator of the units at that prime, not the constant function one. For s=1 the single residue class is included and \chi_1=1.

We verify that the fractions d/s for the contributing pairs are distinct modulo \mathcal O. Suppose that (d,s)=(d',s')=1 and d/s-d'/s'\in\mathcal O. Then ds'-d's is divisible by ss'. Reducing this divisibility modulo s and using the inverse of d modulo s gives s\mid s'. The symmetric argument gives s'\mid s. The primary generator of an ideal outside S is unique, so s=s'; the original congruence then gives d=d'\pmod s. The same conclusion includes the unit modulus. Changing a residue representative only translates its fraction by an element of \mathcal O.

Let 0<r_-<r_+ be fixed with \operatorname{supp}W_1\subset[r_-,r_+]. For two distinct contributing fractions and every n\in\mathcal O, the Eisenstein integer ds'-d's-nss' is nonzero. Hence

The last inequality uses q_s,q_{s'}\le r_+Y. Thus these fractions are \delta_Y-separated in \mathbb C/\mathcal O, where \delta_Y=\min(1,r_+^{-1})/Y\le1.

Writing the expanded polynomial as \sum_{s,d}a_{s,d}e(-md/s), with only the unit residue classes retained, its coefficient square mass is exactly

There is no dependence on v in this expression. Since \varphi_{\mathcal O}(s)\le q_s\le r_+Y, r_s\ge r_- on the support, and there are O(Y) ideals with q_s\le r_+Y, this mass is O(Y^{-1}). Lemma 6.1, applied to the points -d/s with separation \delta_Y, now proves the first bound. Finally q_{b_*}\ge1, so Q=q_{b_*}Y^2 implies Q+Y^2\le2Q, proving the second.□

Proof. Set X=Y=Z^{1/2} and Q=q_{b_*}XY=q_{b_*}Z. The function \Omega in Equation (6.2) is bounded and supported in a fixed compact subinterval of (0,\infty). Thus all rows in that identity lie in 0<q_m\le C_\Omega Q for a fixed constant C_\Omega.

The specialization M=N=1 of Lemma 5.8, followed by the finite ray decomposition defining B_{m,\sigma}, gives for every \epsilon_1>0

Indeed, q_{b_*} is fixed, so the row ball is a fixed multiple of the M=1 ball. The completed scale is Z, which is N=1 in that lemma. The lemma applies directly to the Gaussian profile in the completed integrals defining B; the fixed finite Fourier sum over ray characters costs only a fixed factor by the triangle inequality in the row Hilbert space. In particular this application retains the original row zero masks.

Apply Cauchy–Schwarz to the row sum in Equation (6.2). The factor (q_m/Q)^{-iv} has absolute value one, |\xi(m)|\le1 with its zero extension, and \Omega is bounded. Lemma 6.2 is uniform in v, while B is independent of v. Since W_0 is smooth and annular, \int_{\mathbb R}|\widehat W_0(iv)|\,dv<\infty. The finite sum over \sigma\in T therefore gives

Taking \epsilon_1=2\epsilon proves the proposition.□

Proof. There are four families with t>0: (e_0,l)=(0,2r+2),(1,2r),(0,2r+1),(1,2r+1), where r\ge0. In each family, increasing r by one and the first permitted m by one multiplies the summand of Equation (7.10) by R. In fact the phase ratio is a_p^2: the remaining ratio is one because \gamma_3^2=\omega_p, \omega_p^2=1, and \rho^6=1. Increasing m further multiplies a summand by V.

For transparency, the following table lists every nonzero summand after division by R^r. In its first row j is arbitrary; all other conditions are as displayed. Every omitted case is zero by Equation (7.8).

Here is a direct check of its entries. In the first family t=6r+6. For k=0, the principal Gauss lift is Q^t(1-Q^{-1}) for m\ge r+1; its extra boundary value at (j,m)=(5,r) is -Q^{t-1}. For k=1, the product \omega_p^{-1}G_1G_{-1}=Q allows only (j,m)=(0,r+1). These give the first three rows. In the second family t=6r+1. For k=0 the nonprincipal equality j'=t-1 gives the fourth row. For k=1, the principal Gauss lift has its negative boundary at j'=t, giving the fifth row, and its value Q^{t-1}(Q-1) for j'>t, giving the sixth row. The latter condition is exactly m\ge r+1_{j\le1}.

For the two odd families t=6r+3 and t=6r+4. Their k=0 lines require j'=t-1 and their k=1 lines require j'=t. The phase reductions are

The first is Equation (4.7) at p; the second follows from the first and \gamma_2\gamma_4=1. Substitution gives the last four rows, including both j=3 terms. Thus the table covers all five ramified valuations as well as j=0.

Summing r\ge0 gives (1-R)^{-1}, and the two unbounded m ranges give (1-V)^{-1}. These sums are precisely the formula for P_p^* in Equation (7.11). If t=0, then e_0=l=0. The k=0 terms give (1-V)^{-1}, while the k>0 terms require j=m=0 and give W_{\rm loc}/(1-W_{\rm loc}). This proves the formula for P_p as well.

It remains to justify the analytic assertions, especially where w_r is negative. Put \mathcal E_p=P_p^*+D. The contribution from t=0 is (1-V)^{-1}+W/(1-W), so an exact simplification gives

This expression has no 1-W denominator. At p\mid u, where D=W=0, it reduces to (1-V)\mathcal E_p. In both stated regions |R|<1, |V|<1, and |D|<1, uniformly away from one, so these are holomorphic local expressions.

For p\nmid u and \vartheta=(-w_r)_+, the j=0 row gives

In the first region 4-6x_r-6z_r\le-11/10 and \vartheta\le1/100. The extra factor 1-W in Equation (7.17) therefore leaves this contribution at most Q^{-27/25}. The term DW is at most Q^{-1-\epsilon_0}; the other terms in that equation are smaller. For p\mid u, the strict second-family term has exponent 1-x_r-w_r\le-\epsilon_0. The closest other exponents in the table are

and all remaining ones are no larger than -3/100. Consequently, with \epsilon_H=\min(\epsilon_0,1/50),

In the second region the same comparison gives the stronger bounds

The closest good-prime term here is Q^{1-x_r-w_r-6z_r}, and the closest ramified term is Q^{1-x_r-w_r}.

The sum over primes not dividing u converges normally, and the finite product over primes of u is O_\epsilon(q_u^\epsilon) by the divisor-product bound. The estimates are uniform in all imaginary parts and unit phases. When u=1, the good-prime tail above P_0 is O(P_0^{-163/200}) by the ideal count, which implies the asserted O(P_0^{-4/5}) estimate for the product. Finally extracting (1-V)^{-1}(1-W)^{-1}(1-D) at every prime gives Equation (7.13). The normal convergence of the remaining product proves its claimed continuation.□

Proof. Each maximum exists: the collection is finite, the zeros of each nonprincipal primitive L-function are discrete in compact rectangles, and absolute Euler convergence excludes zeros with real part greater than one. The sequence M_j(u) is nondecreasing and lies in [51/100,1]. If every one of its I-1 increments exceeded e, then

which is impossible. Choose an index with M_{i+1}(u)-M_i(u)\le e, and round M_i(u) down on the stated grid. This gives the two inequalities. When a>51/100, the maximum M_i(u) is attained by an actual zero, giving \rho. This reasoning allows a zero on the line one.

For |t|\le(3i+2)T_1, the closed disk centered at 2+it with radius 2-a-2e has real part at least a+2e. Its radius is less than 3/2<T_1, so every point in it has imaginary part of absolute value less than 3(i+1)T_1. The bound M_{i+1}(u)<a+2e therefore excludes every zero of every primitive function in this disk. Lemma 4.9 controls the concentric disk of radius 2-a-6e. Apply it with center 2+i\Im s, and use absolute Euler convergence farther right. Since Q_\psi\ll_{\mathcal A}U and (3+|t|)^2\ll_e U^{1/50}, the arbitrarily small power in that lemma can be chosen so that its contribution is U^{\epsilon_1/2}. The estimates for polynomial-size deleted Euler products following that lemma supply the other U^{\epsilon_1/2}.

For the last assertion, the primitive functional equation recalled in the proof of Lemma 4.8 [11, Equation (1.1)] relates the value at 1-a-6e+it to the conjugate primitive function at a+6e-it. The conjugate presentation is in \mathcal X_u. The conductor and gamma quotient contribute at most

The reflected primitive value has an arbitrarily small U-power by the preceding disk bound. Finally, 1-a-6e\ge-6e, so the upper bound for the deleted product is U^{6e+\epsilon_1}, after reducing the preliminary losses. This proves the reflected estimate. The ranges of i and a give O_e(1) pairs.□

Proof. We spell out the height restriction because a full infinite contour shift would not be justified by the bin. For a fixed annular W, a bounded real number \sigma, and a pure twist y^{i\omega}, Mellin inversion on a line c+\sigma>1 gives

Here \mathcal MW is the Mellin transform defined in Lemma 4.5. The twist occurs in the L-argument, not in the transform whose tails are estimated. Shift only the portion with added frequency at most the assigned fraction of T_1 to \Re(s+\sigma)=a+6e. The horizontal joins and the shifted portion remain in the zero-free rectangle of Lemma 8.1. Its bound gives U^{(a-1/2+6e)r+\epsilon_1} for the central portion.

Leave the remaining tails on the absolute line, which we may take to be \Re(s+\sigma)=2. There the reciprocal Euler product is absolutely bounded. On the inverse joins the reciprocal remains inside the buffered rectangle. Since the real join interval and the length range are bounded, the remaining arithmetic and scale factors on a join are O(U^B(1+|\Im s|)^J) for fixed B,J chosen before the external tail order. The pointwise Mellin bound in Equation (4.11), applied on that compact real interval with derivative order N+\lceil J\rceil+1, bounds each horizontal join by O(U^B T_1^{-N}). The separate vertical tails have the same bound by Equation (4.10). Both estimates apply to the untwisted profile, uniformly in the stated family. Because T_1=Z^\tau, \tau>0, and all lengths lie in a fixed bounded interval, a sufficiently large fixed N makes these errors smaller than the claimed bound. Taking e and \epsilon_1 small in the prescribed \epsilon proves Equation (8.1).

For the plain polynomial the same calculation uses L_{\rm orig} instead of its reciprocal. Shifting to \Re(s+\sigma)=a+6e gives

Alternatively shift its central portion to \Re(s+\sigma)=1-a-6e. The function is entire because the row is nonprincipal. The reflected estimate of Lemma 8.1 gives

The joins here need only the global upper strip bound and the negative-strip bound for the deleted product, since no reciprocal is present. These give fixed B,J as above, so the pointwise Mellin estimate bounds the joins and the integrated Fourier estimate bounds the absolute-line tails. Taking the better central bound, choosing N to dominate also the possibly negative exponent when m>1, and using the displayed height hypothesis proves Equation (8.2). A primitive conductor smaller than U reduces the reflected conductor factor. The original deleted Euler factors have already been included in Lemma 8.1.

There are only finitely many separated variables in any one L-argument. Assign each a fixed fraction of the single T_1/2 allowance; do not assign that allowance anew at successive shifts. The profile windows and the finite height orders used in the row estimates are fixed before N. Lemma 4.5 differentiates only the untwisted separating profiles when increasing N, which proves the last assertion.□

Proof. The truncated-inverse and Gamma-integral construction is a form of the classical zero detector; compare [21, Appendix C] and [14, Section 13.1]. The buffered row-dependent rectangle and the simultaneous lower bounds needed here are established below. Choose the zero \rho supplied by the bin and set

Let V_{\le} be a fixed smooth function equal to one on [0,1] and zero on [2,\infty). Define

The row character is nonprincipal, so L_{\rm orig} is entire. Moving the line to \Re z=-1/4 crosses only the pole of \Gamma(z) at zero, and its residue vanishes because L_{\rm orig}(\rho,\psi)=0. The global strip bound in Equation (4.14), together with the deleted Euler factors, bounds the new line by

Indeed \Re(\rho-1/4)\ge26/100>0, so the deleted product has an arbitrarily small U-power; the deliberately weaker U^2 includes the primitive conductor bound. The finite polynomial is bounded by the displayed power of D_* using the ideal count. Finally, the fixed polynomial in the added height is integrable against \Gamma(-1/4+iv), since (1+|\gamma+v|)^2\le(1+|\gamma|)^2(1+|v|)^2. Uniformly for 1\le t\le3/2,

Thus J=o(1). This Gamma integration uses only global upper bounds and consumes no reciprocal height allowance.

On the original line, absolute convergence and the Mellin formula for the exponential give

For 1<q_n\le D_*, the coefficient of \psi(n)q_n^{-\rho} is \sum_{l\mid n}\mu(l)=0. For n=1 it is one. When q_n>D_*, V_{\le}(2q_n/D_*)=0. Consequently insertion of 1-V_{\le}(2q_lq_k/D_*) removes exactly the unit contribution e^{-1/Y_*}=1+o(1). The resulting tail has absolute value 1+o(1). Multiplying it by a fixed smooth terminal cutoff equal to one for q_lq_k\le U^{21} and zero for q_lq_k\ge2U^{21} changes it by o(1); this follows from Y_*=U^{20}, exponential decay, and the ideal count.

Partition q_l\asymp D, q_k\asymp N by a fixed smooth dyadic partition. There are O((\log U)^2) relevant pairs, and their support satisfies

With x=q_l/D, y=q_k/N, the profile for a pair is

where \Omega is a fixed annular cutoff equal to one on the product of the supports of W_1,W_2, and

Every fixed logarithmic derivative is bounded uniformly in D,N,U. On a cutoff transition its argument is bounded, and a logarithmic derivative of the exponential is a polynomial in its argument times that exponential. Logarithmic Fourier inversion therefore gives

for every fixed J,N_0. The coefficient L^1-norm on the full line is also uniformly bounded. Trivial bounds for the two finite polynomials have a fixed U-power because their lengths are bounded by U^{21+o(1)}. After \tau>0 is fixed, choose N_0 so that the discarded Fourier tail is o(1).

The absolute value of the remaining sum of integrals is bounded below by a positive constant. The number of dyadic pairs and the uniform L^1-norm show that, for one pair and one |\nu|\le cT_1, the product of the two unnormalized blocks has absolute value \gg(\log U)^{-2}. Both profiles have the same height \gamma-\nu:

The first extra cutoff remains part of the inverse annular profile and creates no second frequency. Removing the factors D^{1/2-\sigma} and N^{1/2-\sigma} by central normalization gives

for sufficiently large U. The support inequalities give the two length bounds in Equation (8.3).

Apply Lemma 8.2 to these profiles, choosing its loss smaller than the present \epsilon. Comparison of the product lower bound with the two upper bounds gives

Since \delta\ge1/50, this implies m\le1/2+O(\epsilon) with an absolute constant. The support inequality then gives r\ge t-1/2-O(\epsilon). Dividing the product lower bound in turn by each individual upper bound gives the two individual lower bounds in Equation (8.4). Only the dyadic support errors are O(1/\log U); all other losses are arbitrarily small fixed powers. Lemma 4.5 permits the stated rowwise choices with a fixed polynomial height factor.□

Proof. We follow the higher-order large-sieve recursion of Blomer, Goldmakher, and Louvel [4, Section 3], giving the local verification for the present primary normalization. In the finite Fourier calculation the individual residue characters act on elements; their unit-trivial quotient is the character on ideals to which we apply Poisson summation.

For good ideals a,b, represented by their primary generators, put

This uses the original zero-extended symbol, also when a is not squarefree. It is multiplicative in both a and b. In every power below, including a multiple of six, a nonunit still has value zero. For M>0 let

For a sequence supported on \mathcal A_N write \|\lambda\|_2^2=\sum_{b\in\mathcal A_N}|\lambda_b|^2, and define

An empty support gives norm zero. These norms include all finite ray classes. We will prove the symmetric bound for D_1; the matrix in the lemma is its transpose with the two lengths exchanged.

The squarefree norm D_j is our target. We also need E_j, since Poisson summation produces arbitrary evaluation ideals. Opening the square in its defining sum and pairing the column characters will replace a row length M by a dual length of order N^2/M. A power decomposition then returns the unrestricted row sum to squarefree norms. It selects either the first-power or the second-power factor, so the exponents j\in\{1,2,4\} are kept together: this set is closed under j\mapsto2j\pmod6. We will use this cycle to improve a provisional exponent in the bound for D_j.

The paired summation formula. Use the fixed finite group

The primary normalization identifies these with the ray classes modulo 36\mathcal O: each unit orbit has a unique representative equal to one modulo 3. In particular t is multiplicative. This subdivision is fixed, since the primes above 2 and 3 belong to S.

For a global unit u and a good prime p, reduction modulo p gives \chi_p(u)=u^{(q_p-1)/6}. Since q_p\equiv1\pmod6, multiplication over the prime factors of b gives

The exponent is read modulo six: expanding a product of integers 1+6h proves the equality of exponents. Thus equal t-classes have the same restriction to the six global units. They also have the same residue modulo 4, which controls the reciprocity factor \mathcal R of Lemma 4.4.

Let b,c be coprime squarefree good ideals with t(b)=t(c) and let j\in J. The raw character on elements

is trivial on global units by Equation (9.2). It therefore defines a finite-order ray character on ideals by

For fractional ideals prime to bc, the residue symbols are evaluated on local unit fractions; the same unit cancellation makes this definition independent of the generator. At each good prime the sextic residue character has exact order six, since the residue multiplicative group is cyclic. Hence its powers j and -j are nontrivial. CRT shows that \Theta_{b,c}^{(j)} has raw conductor bc, and the ray character has the same conductor: omitting a prime would contradict this nontriviality by varying only its residue. It is extended by zero on ideals meeting bc. For the pair b=c=1 define \Psi_{1,1}^{(j)} to be principal.

Sextic reciprocity, with its factor fixed by t(b)=t(c), gives for every good ideal a

For coprime inputs the two reciprocity factors cancel; on the other inputs both sides are zero. With the Gauss sums already defined in the arithmetic preliminaries, CRT using z=cv+bw gives

where \eta_{t(b),j}=\mathcal R(b,c)^j depends only on the common class. Every \gamma_{\pm j} here has modulus one by prime Gauss orthogonality and CRT. Formula (9.4) retains \gamma_{-j} itself, including its factor at -1 under conjugation.

Choose once a nonnegative W\in C_c^\infty((0,\infty)) with W\ge1 on [1,2]. For the Fourier transform and self-dual measure in the arithmetic preliminaries, put w(z)=W(|z|^2) and define \mathscr W by

The transform is radial, smooth at zero, and rapidly decreasing. Thus

Here \mathcal M denotes the Mellin transform. The second assertion follows by integration by parts in the Mellin integral; all logarithmic derivatives are bounded at zero and rapidly decreasing at infinity. Scaling the Fourier transform gives M\mathscr W(M|y|^2) for z\mapsto W(|z|^2/M).

For bc\ne1, finite Fourier inversion for the primitive raw character is

For a nonunit y the left side vanishes because one nontrivial local character is summed against the trivial additive character. For the stated e and d\mu, the trace pairing on the basis (1,\omega) has matrix \left(\begin{smallmatrix}0&1\\1&-1\end{smallmatrix}\right), of determinant -1, and the d\mu-covolume of \mathcal O is one. Thus the dual of bc\mathcal O is (bc)^{-1}\mathcal O, while the covolume of bc\mathcal O is q_{bc}. Coset Poisson summation, followed by division by the six generators of each ideal, therefore gives

Both sums here are over all nonzero integral ideals. The division by six uses unit-triviality of \Theta_{b,c}^{(j)} and radiality of the weight. The zero frequency vanishes since this character is nontrivial.

Poisson transformation of the squared norm. For \lambda supported on one t-class in \mathcal A_N, set

We claim, for every \varepsilon>0,

The maximum runs over dyadic shells and is zero if empty. If N\le1, only the unit good ideal can occur and the annular ideal count proves the M term. Also Q_j=0 when M is below a fixed positive constant, because W has a fixed upper support endpoint and nonzero ideals have norm at least one.

Suppose N>1 and normalize \|\lambda\|_2=1. Then every pair in Q_j is nontrivial and q_{bc}\asymp N^2. Apply Equation (9.6) and take the triangle inequality over the dual ideals. For a small \delta>0 put Z=(2+MN)^\delta. Cauchy gives (\sum_b|\lambda_b|)^2\ll N. By Equation (9.5), the part with q_{\mathfrak a}>ZN^2/M is, for any A>1, at most

The same bound holds if the cutoff is below one. Taking A large makes this an arbitrary negative power error.

In the remaining sum write \mathfrak a=a_0\mathfrak s, with a_0 good and \mathfrak s supported on S, and choose one generator s_0 for each fixed \mathfrak s. Equations (9.3) and (9.4) give the separated factors

All \chi_b(s_0) have modulus one. Split a_0 into shells a_0\in\mathcal I_H, where H\ll ZN^2/(Mq_{\mathfrak s}), and insert Mellin inversion for \mathscr W on \Re w=\delta. Apart from a bounded factor depending on M,q_{a_0},q_{\mathfrak s}, the two coefficient sequences are

They are fixed across the a_0 sum and have absolute values \ll N^{-1/2+\delta}|\lambda_b|. Detecting (b,c)=1 by Möbius inversion and applying Cauchy in a_0 gives

Here the divisor bound absorbs the sum over \mathfrak d. Conjugating the entire first polynomial uses the same E_j norm. The factors with H<1 have just the unit evaluation ideal and satisfy E_j(H,N)\ll N. There are only a fixed power of \log(2+ZN^2) choices of \mathfrak s and of the dyadic shell, because S is fixed and a nonzero Q_j has M bounded below. Equation (9.5) integrates the t variable. Choosing \delta sufficiently small in terms of \varepsilon proves Equation (9.7), including the allowed global loss on its M term.

We now apply this paired estimate to E_j. Partition its inner sum by t, insert W, and open the square. Write the two squarefree indices as db,dc, where d is their gcd, (b,c)=1, and (d,bc)=1. Their residual classes agree because t is multiplicative. Index multiplicativity and Equation (9.3) give on every good a

Let \mathfrak s_S=\prod_{\mathfrak p\in S}\mathfrak p. Möbius inversion for (a,\mathfrak s_Sd)=1 writes a=\mathfrak r\mathfrak a with \mathfrak r\mid\mathfrak s_Sd. Choose one generator r_0 of each fixed \mathfrak r. The value of \Psi at \mathfrak r separates as \chi_b(r_0)^j\overline{\chi_c(r_0)^j}, so the common residual sequence is

It is supported on \mathcal A_{N/q_d} in one class. Apply Equation (9.7) with lengths M/q_{\mathfrak r} and N/q_d. The summed coefficient mass is at most

Writing \Delta_{\mathrm r}=q_{\mathfrak r} and \Delta_{\mathrm d}=q_d, we have 1\le\Delta_{\mathrm d}\le2N and 1\le\Delta_{\mathrm r}\le q_{\mathfrak s_S}\Delta_{\mathrm d}. We obtain the recursive interface

The maxima may be restricted to nonempty ranges. This is the combination of the gcd and Poisson steps in [4, Lemmas 3.2–3.3]; every sequence used here is common within its norm sum, and S remains fixed.

The initial and power bounds. Row inclusion gives D_j\le E_j. Hilbert space duality and sextic reciprocity, splitting one matrix variable by its finite t-class, give

The phases are unit-valued and the noncoprime entries are zero in both orientations. We also have the initial estimate

Indeed, for the squarefree row a, finite Fourier inversion gives on every b, including nonunits,

Cauchy costs at most \#(\mathcal O/(a))^\times/q_a\le1. The reduced fractions x/a from a\in\mathcal A_M are distinct modulo \mathcal O and separated by at least 1/(2M): a nonzero numerator of x/a-y/c-h has absolute value at least one, while |ac|\le2M; equality modulo \mathcal O forces the same primary denominator and residue. The dual of Lemma 6.1, with the coefficient generators in the ball q_b\le2N, proves Equation (9.10).

We need the following near-monotonicity, also used in the large-sieve recursions of [13, Lemma 4.4 and Remark 5] and [4, Lemma 3.1]:

Here is a direct verification. The assertion is immediate for a zero norm. Otherwise choose a maximizing unit coefficient vector for D_j(M_1,N), and split the row shell at 3M_1/2; one part carries at least half its mass. Put L=M_2/M_1. For the first part use good prime ideals of norms in (L,4L/3], and for the second use norms in (2L/3,L]. Then ap\in\mathcal A_{M_2} when p\nmid a. The fixed-field prime ideal theorem supplies P\gg_S L/\log L such primes. An ideal of norm at most 2x contains at most d_x=\log(2x)/\log(2L/3) of them. For S_a(v)=\sum_bv_bF_b(a)^j and p\nmid a, multiplicativity gives

The first coefficient is zero at p\mid b and cancels the factor at p otherwise. Squaring and summing over the chosen rows and primes gives

For the stated threshold with C sufficiently large, d_{M_1}+4d_N\le P/2, proving Equation (9.11).

We finish with the power decomposition and recursion of [4, Lemmas 3.4–3.5]. Suppose, simultaneously for j\in J, that for every positive loss

for M,N\ge1; the unit shells satisfy the same bound directly. The initial estimate gives this with \alpha=2. Write a good evaluation ideal uniquely as

where a_1,\ldots,a_5 are pairwise coprime and squarefree, and a_6 is arbitrary. Insert a_i\in\mathcal I_{X_i} and put X=\prod_iX_i; here X_i\ge1/2 and \prod_iX_i^i\asymp M. Choose \ell\in\{1,2\} with X_\ell=\max(X_1,X_2). After the other factors are fixed, the coefficient

has absolute value at most |\lambda_b|, and the remaining kernel is exactly that of D_{j\ell\bmod6}(X_\ell,N). Enlarging its fixed positive row restriction gives this norm bound. This includes the zero of the sixth power factor. The set J is closed under j\mapsto j\ell\bmod6. Counting the fixed ideals and forgetting their coprimalities in positive sums gives

To verify the factor, the number of fixed choices is O(X/X_\ell). For \ell=1 it is at most a constant times both (M/X_1)^{1/2}\ll(M/X_1)^{2/3} and (MX_2/X_1)^{1/3}\le M^{1/3}; for \ell=2 it is at most a constant times (MX_1^2/X_2^2)^{1/3}\le M^{1/3}\ll(M/X_2)^{2/3}. The number of boxes is a fixed power of \log(2+M).

Substituting Equation (9.12) into Equation (9.13) yields

The three terms before simplification are bounded by M^\alpha, M^{1/3}N, and (MN)^{2/3}; the last is bounded by one of the first two according as N\le M or N\ge M. Use this estimate in Equation (9.8). All auxiliary nonempty scales are bounded by fixed powers of 2+MN: here N/\Delta_{\mathrm d}\ge1/2 and H(N/\Delta_{\mathrm d})\ll_{S,\varepsilon}(2+MN)^{3+\varepsilon}. The unit column shell satisfies the same estimate directly. Thus preliminary losses can be chosen smaller to give any specified final loss. The two scale factors are

Consequently, for M,N\ge1,

Put h=2-3/(3\alpha-1)=(2\alpha-5/3)/(\alpha-1/3) and \beta=2-2/(3\alpha-1). If M\ge N^h, the middle term in Equation (9.14) is at most (MN)^{2/3}. If M<N^h, apply Equation (9.11) with a fixed multiple of N^h\log(2MN) in place of M. Equation (9.14) there gives (MN)^\varepsilon N^\beta, since 2(h+1)/3=\beta and h<\beta. Thus in both cases

Symmetry gives Equation (9.12) with \beta in place of \alpha, simultaneously for every j\in J. Starting from 2, the map \alpha\mapsto2-2/(3\alpha-1) decreases to 4/3. For a requested loss take finitely many iterations until the remaining exponent gap is smaller than a fixed fraction of that loss, and absorb it into (MN)^\varepsilon. Combining the bound and its symmetric form, using M^{4/3}\le(MN)^{2/3} when M\le N and the analogous inequality when N\le M, proves

Dyadic subdivision of both norm balls, with Cauchy across the column shells, converts Equation (9.15) to the same bound for the ball matrix with entries F_b(a)=\chi_a(b). The logarithmic factors are absorbed by a smaller preliminary loss, and the unit shells are bounded directly. The matrix in Equation (9.1) for \varsigma=1 is the transpose of this matrix at lengths (D,K). A complex matrix and its transpose have the same operator norm, by the adjoint identity followed by whole-vector conjugation. The bound is symmetric in the two lengths, so it gives precisely K+D+(KD)^{2/3}. Conjugating the entire inner sum gives \varsigma=-1, with unchanged zeros. Finally, a fixed row restriction removes nonnegative terms, and a fixed coefficient restriction is imposed by zero extension. This proves all assertions of the lemma.□

Proof. Fix a small preliminary loss \epsilon_0>0. Apply Proposition 8.3 with t=1 to every row in \mathcal B. Subdivide by its presentation labels in \psi_{u,\nu,\varsigma}(n)=\nu(n)\chi_n(u)^{\varsigma} and its dyadic pair of polynomial lengths. There are a fixed finite number of presentation labels and O_{\mathcal A}((\log U)^2) pairs. Within one such subdivision, the inverse length D=U^r and the pair (\nu,\varsigma) are fixed. In this subdivision write

for the inverse polynomial of that presentation. The witness gives

The notation W_u retains the permitted rowwise norm-profile parameters. The polynomial here is the original one for the displayed presentation, not the polynomial of its primitive inducing character. We first bound it when those parameters are fixed, and then handle their rowwise choice.

For each sixth-power-free row, factor its ideal uniquely as

Thus \mathfrak k is squarefree, (\mathfrak k,\mathfrak w)=1, and \mathfrak w is powerful, meaning that each of its positive prime valuations is at least two. There are only finitely many possibilities for \mathfrak s, because its valuations lie in \{0,\ldots,5\} on the fixed set S. Choose one generator s_{\mathfrak s} for each of these ideals. Writing w,k for the primary generators of \mathfrak w,\mathfrak k, there is a unique unit \upsilon such that

We fix \mathfrak s and \upsilon whenever a row sum is taken.

Insert dyadic ranges q_{\mathfrak w}\asymp V, including the unit ideal in the bounded range V=1. Only 1\le V\ll_{\mathcal A}U can occur, and there are O_{\mathcal A}(\log U) such ranges. The original annulus q_u\asymp U gives the actual residual annulus

We keep this restriction when forming the sum. It will be enlarged to the ball q_k\ll_{\mathcal A}U/V only when applying a positive row bound. Let \mathcal U_V denote the rows of the current witness subdivision in this powerful-part dyad, with all original physical restrictions retained.

The number of powerful ideals of norm at most V is O(V^{1/2}). Indeed, every powerful ideal has a unique expression \mathfrak v^2\mathfrak y^3 with \mathfrak y squarefree; the two factors need not be coprime. The ideal count from the arithmetic preliminaries gives

The last series converges by the same ideal count. For each fixed \mathfrak w in its dyad, the ball containing Equation (9.18) has O_{\mathcal A}(U/V) ideals. A nonempty annulus has U/V bounded below by a fixed positive constant, which also covers the unit residual ideal. Since the factorization of (u) and its unit are unique, the number of rows in this powerful-part dyad satisfies

Discarding any of the restrictions defining \mathcal U_V in this upper count can only add rows.

For fixed \mathfrak w,\mathfrak s,\upsilon, put v=\upsilon s_{\mathfrak s}w. Multiplicativity in the numerator gives the exact identity

It includes the zero extensions: if n meets either factor, both sides are zero, and otherwise it is ordinary multiplicativity. In particular, the inverse polynomial for a fixed profile W is the row sum in Lemma 9.1 with coefficients

The factor \mu(n) keeps the columns squarefree. The factor \chi_n(v)^{\varsigma} retains every original zero at (n,v)>1 and is fixed across the residual row k. The only remaining zero depending on both n and k is the zero of the displayed sieve kernel itself. The restriction (k,w)=1, and any restriction inherited from the physical row set for fixed external parameters, is a fixed row restriction in this application. We do not enlarge S to contain the primes of w. Consequently the constant in the sieve is independent of w, even though its fixed coefficient mask may have large norm.

The common annular support and ideal counting give, uniformly for the fixed profile parameters,

The same statement holds for each of the finitely many profile derivatives used below, with the corresponding fixed seminorm. Apply Lemma 9.1 after enlarging the positive residual row sum from its actual annulus to q_k\ll_{\mathcal A}U/V. A fixed enlargement makes the row bound at least one on every nonempty quotient range and includes the annular column support. Summing the result over the O(V^{1/2}) choices of \mathfrak w and the finitely many choices of \mathfrak s,\upsilon yields

All coefficient sequences in this derivation are fixed within the row sum to which the sieve is applied. They may differ for different frozen values of \mathfrak w, which are summed only after those positive bounds.

We now restore the rowwise choice of W_u in Equation (9.17). Here the detector was used with t=1, so D_*=U. After the presentation and dyadic pair are fixed, its inverse profiles have the form

where W_{\rm dyad} is the fixed smooth dyadic cutoff and \nu_F is the Fourier frequency from the detector. The factor V_{\le}(Dx/U) is common to the row sum; the varying parameters are \sigma\in[51/100,1] and |\omega|\le(3i+1)T_1. Differentiating in either parameter inserts a power of \log x, which is bounded on the fixed annulus. Cover these two parameter ranges by unit boxes. There are at most a fixed power of 1+T_1 such boxes. The parameter Sobolev inequality in Lemma 4.5 bounds the supremum on a box by a finite sum of integrals of squared profile derivatives. Sum over the rows before these integrals. At each fixed parameter value, differentiation changes only the fixed profile coefficient, inserting the allowed logarithmic weights or profile derivatives. Equation (9.21) therefore applies to every such integral. This proves

The derivative order and hence A_{\mathcal A} are fixed before choosing \tau. The height intervals and the total allowance for additional frequencies are exactly those in the shared detector estimates; the Sobolev step does not assign a new frequency allowance or change a contour. It does not permit a coefficient to be selected separately for each k.

It remains to combine this moment with the independent count Equation (9.19). Write

These are the parameters of the actual dyads. A nonempty dyad has 0\le o\le1+O_{\mathcal A}(1/\log U); the unit dyad has o=0. Since the witness lengths stay bounded, the arbitrarily small power of UD in Equation (9.22) can be written as an arbitrarily small power of U. Dividing that equation by the spike in Equation (9.17), and also using Equation (9.19), gives for this subdivision

after decreasing the sieve loss denoted by \epsilon_1 if necessary.

We compute the largest exponent in this expression first on 0\le o\le1 and 1/2\le r\le1. Put x=1-o and A=(1+x)/2. Then

The minimum of A with a maximum of three real numbers is the maximum of its minima with those numbers. The first such minimum is A-\delta r\le1-\delta/2. For fixed r, the second is the minimum of an increasing and a decreasing affine function of x. They cross at x=(1-\delta)r\in[0,1], so its maximum over x is

Both functions in the third minimum increase with x. Its maximum over x is therefore

The affine expression in r takes the values 1-\delta/2 and 4/3-\delta at the two endpoints. It follows, and equality is attained at one of those endpoints, that

This computation is uniform for 0\le\delta\le1.

The functions just used are maxima and minima of affine functions whose slopes are uniformly bounded on a fixed neighborhood of these compact ranges. Replacing the actual o by its nearest point in [0,1], and the actual witness r by its nearest point in [1/2,1], changes the exponent by at most O_{\mathcal A}(\epsilon_0+1/\log U), by Equation (9.17) and the actual dyad bounds. Thus this replacement is made only in the final optimization, not in the row sum or its sieve application. Choose \epsilon_0 and the preliminary sieve loss small enough in terms of the prescribed \epsilon. The O((\log U)^2) witness pairs and O(\log U) powerful-part dyads cost only another arbitrarily small power. Summing Equation (9.23) proves Equation (9.16).

Finally, the bin a=51/100 need not contain an actual zero, so no use of Proposition 8.3 is made there. Counting all elements in its physical annulus gives O_{\mathcal A}(U) rows. Since 1/50<1/3, the displayed formula for R gives R(1/50)=1, as claimed.□

Proof. Multiply the kernel bound by Equation (10.6). Integrating (1+|\boldsymbol t|)^{J-L} outside the box proves Equation (10.7) when L>N+J+m. On t_j=T, integration in the other m-1 coordinates gives

Taking L>N+J+m-1 proves Equation (10.8). Restricting to \mathcal D decreases these positive integrals, and a bounded real interval contributes only its length.□

Proof.

Section 8 gives a\le\beta_*, including the floor bin. Isolate the finite sum over \mathcal B on the absolute lines in Equation (10.5). Move z from 2 to z_0, keep w=3, and move s from 3 to \beta_*+20e. These moves stay in \mathcal D_1(\epsilon_0) for a fixed positive \epsilon_0, the two numerator arguments remain to the right of their poles, and the reciprocal stays in \Re s>\beta_*. Its global bound follows from Lemmas 4.9 and 4.10, using the conductor and deletion radical bounds in Section 8. The correction obeys Equation (10.4). Lemma 10.2 with T\to\infty therefore justifies these moves.

Now move w from 3 to 1-a-6e while s stays on that global line. Throughout the move,

The numerator character of every retained row is nonprincipal, so its L-function is entire. Also \Re w\ge-6e>-1/100 and \Re z=z_0. Thus the correction remains holomorphic in \mathcal D_1(\epsilon_0), for example with \epsilon_0=10e, and no pole is crossed. On a w-join the reciprocal is still global. The global upper strip bound for the numerator, the deleted-factor bound and Equation (10.4) give a majorant of the form Equation (10.6) on the axes integrated there. The trace estimate makes the joins tend to zero.

Before moving s farther, restrict the three original Mellin heights to the stated box. On the discarded part keep s on \Re s=\beta_*+20e. There the global reciprocal bound, the global numerator bound, the deleted factors and the all-height correction bound give Z^{B_\eta} times a fixed polynomial in the heights: the row range is bounded by q_u\ll Z^{d_{\max}}, and the number of rows is O(U). The integrated tail in Lemma 10.2 gives O_{\eta,N}(Z^{B_\eta}T_1^{-N}).

Move only the retained s segment to a+16e. On this rectangle,

and the other inequalities defining \mathcal D_1(10e) still hold. The real part of the reciprocal argument is at least a+16e>a+6e. The stated height box places both scalar arguments strictly inside |\Im v|\le(3i+2)T_1 for sufficiently large Z. Lemma 8.1 therefore excludes its zeros throughout the retained rectangle. Since 6z_0>1, the scalar zeta factor has no pole on this move. Thus no pole is crossed.

For an s-join, its imaginary coordinate is fixed at a constant multiple of T_1. The scalar denominator in Equation (10.5) depends only on s, so its buffered estimate remains valid when the \Im z and \Im w integrations are extended to their whole axes. On those extended axes use the global numerator estimate and Equation (10.4). The trace estimate then gives O_{\eta,N}(Z^{B_\eta}T_1^{-N}). No w- or z-join in this argument extends an s-axis that has been moved into a merely buffered region.□

Proof. Apply Lemma 10.3 to the original triple integral, using for its three box constants the corresponding c_k in Equation (10.14). These constants are positive and at most 1/2. The full correction and the fixed bin therefore give the retained integral and the asserted errors before any pointwise pieces are introduced. Any auxiliary integrations used for the central bound are subject to the additional hypotheses in the statement.

Now use the central decomposition and form the pointwise sets \mathcal B_{\lambda,\nu}. At each retained point apply the triangle inequality to the original row sum, and then apply Equation (10.13) to the sets present at that point. There are at most a fixed multiple of (1+\log Z)^D such sets pointwise. Bound each of their exponents by the supremum over \mathcal P_Z and integrate only the original holomorphic row sum. Thus neither measurability of an individual pointwise subdivision nor one common label set across contour points is needed. On the retained contours, Lemma 8.1 bounds the reciprocal by O_{\eta,e,\varepsilon_d}(U^{\varepsilon_d}). The tests have a uniformly bounded joint L^1 norm, since the transformation (\Im s,\Im z,\Im w)\mapsto(\Im(s+z),\Im z,\Im w) is invertible. The scalar factor \zeta_F^S(6z_0) is absolutely bounded. It remains to compute the real power of Z.

The outside factor, including q_u^{-z_0} and the real displacements in Equation (10.11), contributes

Equation (10.13) adds d(R+\delta/2+\varepsilon_c)+\ell(z_0-1/2)+g+\varepsilon_c. The reciprocal adds d\varepsilon_d. Subtracting C(\sigma_0)=\sigma_0+l_x/2-1+h/6 and using h=1-l_x+\ell gives Equation (10.15) and the remaining terms in Equation (10.16). The pointwise logarithmic multiplicity is bounded by Z^{\varepsilon_p} for sufficiently large Z. This proves the claim.□

Proof. For u=1, the scalar factor in the high identity is

Isolate this term on the absolute contours. Move to \Re s=\beta_*+e, \Re w=1+e and \Re z=1/6+e. The reciprocal stays in \Re s>\beta_*, and both zeta arguments stay to the right of one. The paths lie in \mathcal D_2, so the correction is holomorphic. The all-height majorant and Lemma 10.2 justify the horizontal limits.

Move w to 19/20, crossing its simple pole at w=1. In that residue move z to 33/200=1/6-1/600, crossing its simple pole at z=1/6. The unresidued w integral keeps \Re z=1/6+e. All these paths lie in \mathcal D_2, and M(z) is holomorphic for \Re z>0. Thus the two stated scalar poles are the only poles crossed. The reciprocal remains on its global line throughout, even when \eta is principal. Every extended vertical w- or z-axis used to bound a tail lies strictly on one side of its scalar pole; w=1 and z=1/6 occur only as residues, and the horizontal joins have large nonzero height. On those axes the zeta functions have fixed polynomial bounds. Consequently the integrated and trace estimates of Lemma 10.2 apply without an unremoved pole in their majorants. After the w residue, its Mellin factor has become the constant \widehat W_1(1). The remaining heights use the two-dimensional kernel in Equation (10.9), with (y_1,y_2)=(\Im(s+z),\Im z); its trace estimate justifies the subsequent z-join.

On the principal contours, Equation (10.17) and the outside powers give the raw exponent

For the unresidued w integral this is C(\beta_*)+(1+h)e-l_y/20; for the leftover z integral in the w residue it is C(\beta_*)+e-h/600. The reciprocal on \Re s=\beta_*+e has an arbitrarily small power of its fixed target conductor and a fixed polynomial in height by Lemmas 4.9 and 4.10. The tests integrate those height powers. Requesting the small powers in the correction and in A_\eta^{-1} to be sufficiently small gives the first two terms of Equation (10.19).

The product of the scalar residues is c_S: the residue of \zeta_F^S(6z) at 1/6 is one sixth of the residue of \zeta_F^S(v) at one. Positivity of c_S was proved when the normalization was defined.

At the double residue the outside power is

Equation (10.18) therefore makes the normalized double residue

When present, the error factor is estimated on this line; the global reciprocal bound and Gaussian give O_{\eta,e,\epsilon}(Z^{C(\beta_*)+e-\mu+\epsilon}). No continuation of that error factor is required.

Move only the main integral right to \Re s=2. The reciprocal is holomorphic for \Re s>\beta_* by the definition of \beta_* and absolute Euler convergence beyond one. At a principal pole its reciprocal has a zero, so that case adds no residue. The function H_\eta is holomorphic there by Lemma 7.1, and is bounded there by Equation (10.2). The global reciprocal estimate is polynomial in height uniformly on the fixed real strip, while the Gaussian is O(e^{-(\Im s)^2}). Rectangular contours therefore have vanishing horizontal sides, and the shifted integral is exactly f_\eta(Z). This also covers \beta_*=1. The three remainders give the asserted bound.□

Proof. Every row under consideration has nonprincipal numerator by the classification preceding Lemma 10.5. The denominator may be principal for a bounded unit row; its reciprocal is nevertheless holomorphic and bounded on \Re s=\beta_*+e by the principal specialization of Lemma 4.9. For all rows, the conductor and deletion radical bounds in Section 8, together with Lemma 4.10, give an arbitrarily small power of q_u and a fixed height polynomial for that reciprocal.

For a small dyad move its finite sum to (\Re s,\Re w,\Re z)=(\beta_*+e,1/2,z_0). The reciprocal stays global. The numerator is entire, and the scalar zeta argument stays to the right of one. These paths may be taken in \mathcal D_1(3/8): at the final point \Re(s+w)=\beta_*+e+1/2>1+3/8, and the other inequalities are immediate. The all-height correction bound and Lemma 10.2 justify the moves. On the final lines, Equation (4.14) and the deletion bound give U^{3/5+\epsilon} times a fixed height polynomial for the nonprincipal numerator. There are O(U) element rows in the dyad, including unit factors, and q_u^{-z_0}\ll U^{-z_0}. The correction is bounded by Equation (10.20). The tests integrate all fixed height powers. The resulting exponent outside the row power, relative to C(\beta_*), is

and the row power is 1+3/5-z_0=63/50. Requesting the component small powers to sum to the displayed \epsilon proves Equation (10.21). The row exponent is positive, so dyadic summation up to Z^{d_{\min}} gives Equation (10.22), after decreasing the preliminary losses.

For each fixed Z and large dyad, move its finite row sum to the absolute lines (2,2,v). The path from (3,3,2) remains in \Re s,\Re w\ge2 and \Re z\ge2, so the scalar factors are holomorphic there. For this fixed dyad choose D large enough to contain its rows in q_u\le Z^D; Equation (10.4) and the external trace bound make its horizontal limits vanish. Constants used only to justify this equality may depend on the fixed dyad. The estimate on the final lines is uniform in the dyad by Equation (10.23). The scalar L-factors are absolutely bounded there, as is \zeta_F^S(6z), and Equation (10.23) applies. The outside power, including the correction but not the row count, is

The O(U) rows and q_u^{-v} give U^{1-v}; the tests integrate the fixed height polynomial. This proves Equation (10.24). Because 1+\epsilon-v<0, summing its geometric dyadic tail gives Equation (10.25). For fixed \zeta>0, the coefficient of v in that exponent is -\zeta, proving the last assertion. The original row series is absolutely convergent on its starting lines, and the displayed bounds give an absolutely summable final tail, so the individual dyadic contour identities may be summed. The number v and the finite test orders it requires are fixed before Z tends to infinity.□

Proof. For the fixed target choose

Then (1+T_1)^{A_\eta}\ll_\eta Z^{m/4}, so the first term in Equation (11.1) is O_\eta(Z^{C(\beta_*)-3m/4}). Next choose a fixed integer N such that

This is possible because \tau_\eta>0 and B_\eta is finite. The second term is then bounded by the required power. Increase the lower threshold for Z after these choices. Both m and \omega were fixed before the target, whereas \tau_\eta,N and the threshold may depend on it. Since J_\eta and f_\eta do not contain T_1, this proves the asserted family-wide exponent without changing either function.□

Proof. Use l_x=l_y=h=1/2, \ell=0 and C=C_{\mathrm I}. The exact identity and local correction were verified at the start of Section 10; Equation (10.12) estimates its intermediate rows, and the principal and outer specializations there give the margins listed above. It remains to choose their losses.

Order of choices. We give a common loss budget. Put

The error-free central, small, principal, and large savings just proved are all at least m_0. Fix the displayed geometry, row ranges, and v=4000 before the target. Request loss \epsilon_* in the row count, numerator, local correction, buffered reciprocal, central multiplicity, and each principal or outer estimate. Reserve a further loss \epsilon_* for the O(\log Z) physical dyads. The witness-pair and powerful-part logarithms are already included in the requested row-count loss; the O_e(1) bins and fixed presentation choices cost constants.

To obtain the requested row-count loss, first choose its preliminary witness, dyadic, and sieve losses sufficiently small in terms of \epsilon_*. These choices are independent of the target. Indeed the witness length error has an absolute coefficient, and the affine slopes in the optimization of Proposition 9.2 are uniformly bounded. The target-dependent constants in the actual dyadic endpoints multiply only 1/\log U, which changes a fixed constant or lower threshold. Apply Lemma 8.2 on the enclosing pre-saturation ranges 0\le r,m\le22: the detector’s terminal product cutoff is O(U^{21}) before it proves the shorter witness lengths. Its number e_0 depends only on the requested dyadic loss and these bounded ranges. We may thus fix, still before the target,

With \epsilon_r=\epsilon_n=\epsilon_H=\epsilon_*, this gives \varepsilon_c\le15\epsilon_*. Since d\le501/1000<1, the explicit central losses in Equation (10.12), together with the reserved physical-dyad loss, are at most

The two principal losses are at most (3/2+1)\epsilon_* and 2\epsilon_*; the small-row loss is at most 2\epsilon_*. Each is less than m_0/4. The displayed large-row estimate already uses its requested loss. Therefore all these terms, after division by the fixed c_S, retain at least 3m_0/4=1/1600 before the height factor. We use only m=1/2400, leaving a further power 1/4800 available below.

It remains to check when the height hypotheses hold. Let \epsilon_{\rm det}>0 be the fixed dyadic loss chosen above, and, after fixing a target, let A_{{\rm det},\eta} dominate the finitely many orders in its uses of Lemma 8.2. Set

For 0<\tau\le\tau_{0,\eta} and sufficiently large Z,

This verifies the detector height condition uniformly over the physical range. The fixed factor from 1+Z^\tau\le2Z^\tau is absorbed by the lower threshold. We also impose \tau\le d_{\min}/100 as required by the bins. The internal tail orders used to establish the witnesses are chosen after this \tau and are separate from the final external order N below. Their increase does not change the retained profile or height orders. Any fixed constants introduced by those internal choices are absorbed into the unused power 1/4800 by increasing the lower threshold, which may depend on \eta and \tau.

For every final integer N\ge0, Lemma 10.3 bounds the discarded high portions and joins by O_{\eta,N}(Z^{B_\eta}T_1^{-N}), with B_\eta fixed before N. The bounded real and row ranges and the finite set of bins permit one such B_\eta for all the central dyads. Summing the O(\log Z) dyads can be absorbed by increasing B_\eta by one, again before N. Combining the preceding estimates gives

for some finite A_\eta,B_\eta, whenever 0<\tau\le\min\{d_{\min}/100,\tau_{0,\eta}\} and Z is sufficiently large. Both functions are independent of T_1. Proposition 6.3, applied with loss \omega and divided by the same fixed c_S, supplies

All hypotheses of Lemma 11.1 are now verified. It gives the saving m/2=1/4800 asserted in the proposition.□

Proof. Passing from a primitive Hecke character to one that it induces changes only finitely many factors 1-\eta(\mathfrak p)N\mathfrak p^{-s}. They are nonzero for \Re s>0, so the Hecke assertion extends to all finite-order characters. The same observation for factors 1-\chi(p)p^{-s} reduces the Dirichlet assertion to a primitive Dirichlet character \chi of conductor q.

Let \chi_{-3} be the quadratic character of conductor three, and let \eta be the finite-order Hecke character given by \mathfrak a\mapsto\chi(N\mathfrak a) on ideals coprime to 3q. It is a ray character: if \alpha\equiv1\pmod{q\mathcal O}, then N\alpha\equiv1\pmod q, so the norm character is trivial on the corresponding principal ray subgroup. Let S_{\mathbb Q} be the rational primes dividing 3q, and let \mathcal S_{\mathbb Q} be the primes of F above them. Superscripts by these sets denote deletion of those Euler factors.

For p\notin S_{\mathbb Q}, the local factors agree as follows. If p splits in F, then \chi_{-3}(p)=1, the two prime ideals have norm p, and the Hecke factor is (1-\chi(p)p^{-s})^{-2}. This is the product of the Dirichlet factors for \chi and \chi\chi_{-3}. If p is inert, then \chi_{-3}(p)=-1, the unique prime ideal has norm p^2, and its factor is

These exhaust the primes outside S_{\mathbb Q}. Absolute Euler convergence for \Re s>1, followed by uniqueness of meromorphic continuation, gives

The product character on the right may be imprimitive; deletion of S_{\mathbb Q} makes the identity independent of that choice. Every deleted factor is nonzero for \Re s>0.

Both Dirichlet factors are holomorphic in 0<\Re s<1. A zero of L(s,\chi) there would therefore give a zero of the Hecke factor, with no cancellation by a pole, and the assumed Hecke half-plane excludes it when \Re s>\sigma_0. Absolute Euler convergence handles \Re s>1. On s=1+it with t\ne0, neither Dirichlet factor has a pole, so the same product argument applies.

At s=1, a pole-zero cancellation could occur only if one primitive inducing character among \chi and \chi\chi_{-3} were principal. The other would then be \chi_{-3}. Its Dirichlet series converges at one by bounded partial sums of the nonprincipal periodic character, and grouping consecutive terms gives

The first integral follows by monotone convergence of the nonnegative paired integrands. Thus there is no zero in this last case either; a principal pole is allowed. This proves the strict half-plane assertion with no claim on its boundary.□

Proof of Theorem 3.1. Suppose that \beta_*>11/12. Lemma 7.1 and Equation (10.2) give the required holomorphic, nonzero H_\eta on \Re s>11/12 for every primitive target. The low and high bounds established in Proposition 11.2 and its proof verify Proposition 2.1 with

These two positive losses are independent of the target, while the allowed constants, excluded sets, and thresholds may depend on it. The continuation criterion contradicts the supposition. Therefore \beta_*\le11/12. By its definition and absolute Euler convergence in \Re s>1, every primitive finite-order Hecke L-function over F is zero-free in the strict half-plane \Re s>11/12, with the principal pole allowed.

Proposition 11.3, applied at the same boundary, extends this assertion to all finite-order Hecke characters and all Dirichlet L-functions, including \zeta(s). It preserves the strict half-plane and the allowed principal pole. This proves the theorem.□

Proof. For the fixed abelian extension of F corresponding to T, the prime ideal theorem in a fixed Frobenius class gives

This is the fixed-extension consequence of Chebotarev in [27, Theorem 1.1]; the extension and its conductor are fixed as x\to\infty. If \operatorname{supp}W\subset[a,b]\subset(0,\infty), the error o(x/\log x) is uniform for aP\le x\le bP as P\to\infty. Stieltjes integration by parts therefore changes the left side of Equation (13.3) by o(P^{1/6}/\log P) when d\pi_{1_T}(x) is replaced by dx/(|T|\log x). The resulting integral is

The last ratio tends uniformly to one. This proves the formula and its positivity. A fixed finite set is eventually outside the annular window. No error exponent uniform in the ray conductor is used.□

Proof. Write a residue modulo p^a as x_0+py, with x_0\bmod p and y\bmod p^{a-1}. The sum over y is zero unless p^{a-1}\mid k, and equals P^{a-1} otherwise. In the latter case write k=p^{a-1}k_0. The remaining normalized sum is

If 6\nmid a, its inner sum vanishes when p\mid k_0 and otherwise has absolute value \sqrt P. If 6\mid a, the inner sum is the sum of the additive character over the units, namely P1_{p\mid k_0}-1. These are exactly the two cases in the statement.□

Proof. Expanding both Gauss sums in the displayed Fourier transform leaves

Additive orthogonality makes the inner sum q_uq_v when the congruence holds and zero otherwise. If j=0 and a summand is nonzero, x and y are units modulo u and v. Reducing the congruence modulo u gives u\mid v, and reducing modulo v gives v\mid u. Since the generators are primary, u=v. The congruence then says x=y\bmod u and gives \varphi(u). This argument uses the zero masks also when a local character power is principal.

The divisibility by C is immediate. After division by C, the congruence for j=Ck is

Reduction modulo n_1 and n_2 gives, with zero values retained,

The product of the two unit factors is \mathcal R(n_1,n_2). It remains to identify the multiplicity of lifts of the congruence modulo C. This can be checked at each prime. If c=v_p(C) and neither residual modulus contains p, there is no additional lift. If, say, d=v_p(n_1)>0 and p\nmid n_2, then y\bmod p^c is free and the congruence uniquely determines x\bmod p^{c+d} from it. Reduction modulo p^c is exactly the common congruence in Equation (13.7). The case p\mid n_2 is symmetric, and this argument includes c=0. The Chinese remainder theorem thus gives a bijection with the common solutions used in L_C, even when C meets one residual modulus.

For the local calculation put k_p=\mathcal O/(p) and A=\chi_p^c, a character of k_p^\times extended by zero. Since at least one of n_1,n_2 is a unit at p, each solution modulo p has P^{c-1} lifts modulo p^c. If both are units and p\mid k, the unit variables are proportional and the field sum is A(n_1/n_2)(P-1). If p\nmid n_1n_2k, put y=(k/n_1)t and x=(k/n_2)(1+t). The field sum becomes

It is -A(n_1/n_2) for nonprincipal A and (P-2)A(n_1/n_2) for principal A. If p\mid n_1 and p\nmid n_2, the field equation forces x=k/n_2. For p\mid k its character is zero. For p\nmid k the remaining sum over y is zero unless A is principal, in which case it is P-1. The other case is symmetric. This proves all local formulas on the genuine residual locus. Extending them by the stipulated zero when both residuals meet p gives functions of the residual columns modulo p, each of absolute value at most P^c. Their product proves the final assertions for the artificial extension as well; no congruence-sum identity is asserted at a newly added pair.□

Proof. At primes of a and b, solving the congruence in Equation (13.6) gives respectively \chi_a(j)\overline{\chi_a(Eb)} and \overline{\chi_b(-j)}\chi_b(Da). These statements remain valid when a character of j is zero; only the displayed unit factors are inverted. At primes of DE change variables x'=bx\bmod D, y'=ay\bmod E. The congruence becomes Ex'-Dy'\equiv j\pmod{DE} and its character factor changes by \overline{\chi_D(b)}\chi_E(a). The remaining unit factor is

All denominators here are symbols of units by the hypotheses. The Chinese remainder theorem and the definition of F(D,E;j) complete the proof.□

Proof. For every good prime and every A\in\mathcal O, the zero convention gives the pointwise identity 1_{q\mid A}=1-\chi_q(A)^0. Expand its product over \mathcal Q, and apply Proposition 5.1 to each of the resulting finitely many completed sums. For a fixed h_0 and active radical, the Proposition uses the same c_F,c,d,\vartheta,\kappa_F, other local phases, and kernel whether an absent prime is omitted from the local set or included with exponent zero in its inactive branch. Thus the two inactive contributions at q combine with coefficient 1-(1-q_q^{-1})=q_q^{-1}. The active contribution comes only from the subtracted zero-mask transform and has local factor q_q^{-1/2}\chi_q(\lambda^4\mu)^{-2} and scalar -\omega_{q,0}=\tau_{q,2}^{+}\chi_q(\epsilon_q)^{-2}. Collecting these choices at all marked primes proves Equations (14.3) and (14.4). The inactive prime is absent from r, hence from that branch’s conductor and kernel scale. This is finite inclusion-exclusion between compatible branches, not a new theta identity.

If a marked prime is a base local prime, the mark forces it to divide nb^3, while the zero-extended base factor at that prime then vanishes, including for exponent zero. This proves the last assertion without any division of a zero symbol. It also covers the case where the primal squarefree and cube ideals share the marked prime.□

Proof. Write \mathcal H for the left side of Equation (14.7). First remove the complete moving row-mark mask:

For each row the number of such divisors is at most d_{\mathcal O}(k). Rowwise divisor Cauchy, followed by the positive sum over rows, therefore costs a permitted factor (KNBL)^\epsilon. In a fixed D-summand write P=DP' and k=Dk'. Squarefreeness gives the fixed restrictions (D,P')=(D,k')=1. Define

All these factors retain their zeros and |\beta_D|\le1. Factoring q_P^{-1/2}=q_D^{-1/2}q_{P'}^{-1/2} gives

Here q_{k'}\ll K/q_D, q_{P'}\asymp L/q_D, and all inherited fixed restrictions are retained. The outer D’s are the squarefree good divisors allowed by the split, so q_D\ll\min(K,L). In particular, no (P',k') mask is reinstated inside an individual D-summand. Its cancellation belongs to the complete Möbius sum already bounded by rowwise Cauchy.

For fixed D, the product of \beta_D(n,b) and the inner P'-sum is an arbitrary complex coefficient independent of k'. Apply Equation (5.27) with row length K/q_D. Use its notation

so B\asymp GT^2, and write c=rc' for its complete c-mask divisor. The ideals c,m,h are pairwise coprime and squarefree; no coprimality between g and t is imposed. The cited reduction already includes the natural k'-t restriction and the complete Möbius expansion of the k'-c mask. It does not reinstate a mask between the remaining row and c'.

For one C,G,T block its positive output, inserted in Equation (14.8), is at most

where the positive c',m,h-sum keeps the original support and

The factorization used here is, including all zeros,

The character already supplies the missing zeros at r,c', and m. Thus the r-factor is tuple-only, including its zero when (P',r)\ne1. The fixed D-phases remain inside \beta_D until this positive bound. The quadratic reduction has already separated the finitely many squarefree S-parts of m,h before sieving their good product; m,h in Equation (14.9) are the original ideals.

We may now discard |\beta_D|^2\le1. For fixed h,t,D,r, the P'-coefficients are independent of j=c'm, and ideal counting gives

The product j=c'm=n/r is squarefree primary and has norm O(N/q_r). It may contain a permitted prime of S other than \lambda, since only the quadratic column was stripped of its fixed S-part. Grouping c',m by j costs at most a divisor factor. Enlarge only this positive squarefree j-range. Cubic reciprocity, conjugation, and Equation (14.6), followed by the O(G/C) choices of h, give, on q_D\asymp D_0,\ q_r\asymp r_0,

The fixed mask at ht is a coefficient restriction in this cubic sieve, not a row-dependent mask.

The reduction supplies T\sum_t, with O(T) possible t, and the factor C/q_r. Together with the h-count, these contribute

There are O(r_0) possible r in its dyad, while the sum of q_D^{-1} over its dyad is O(1), up to a permitted small power. Both sieve factors increase when D_0,r_0 are replaced by 1, and NG/C^2\ll NB. Summing the logarithmically many dyads proves Equation (14.7). Subunit quotient ranges are empty, and bounded ranges are included by changing the fixed annular constants. This argument has introduced neither a (g,t)=1 condition nor a new pair-dependent mask.□

Proof. First remove the inactive slots by triangle inequality in the row Hilbert space. At such a slot the absolute coefficient mass is

uniformly under any fixed restriction, including for a bounded or zero-length list. It therefore suffices to prove a uniform bound with those primes fixed. Each such branch has its own conductor, cusp sector, and kernel scale, with the inactive prime absent from the conductor. Its earlier collision exclusion is now a fixed row restriction.

Fix f,\rho,k_{\rm pow},k_{\rm sup}, the non-slot local choices, h_0, and separate classes of the residual row product and active slot product modulo the full M_{\rm ref}^2. Also fix the unit and ramified exponent of the dual index. No non-slot j=4 label remains averaged. We use only the algebraic frozen-base extraction in Equation (5.32), inside the original sum over active tuples. Corollary 14.1 and the subsequent phase separation make that extraction simultaneous across the tuples: the base coefficient is common to the residual rows and tuples after their separate sectors are fixed, while the remaining scalar separates into a row factor and a tuple factor. We do not apply the numerical unmarked estimate separately to each tuple.

The structural extraction gives the common central coefficient

times normalized inverse-root factors and bounded coefficients. Its N_0-assignment extracts a forced j=4 prime only from the squarefree dual ideal, even if the cube ideal shares it; the entire shared cube part remains in its residual norm and coefficient. Its B_0-assignment extracts one occurrence from the cube ideal and leaves the factor q_p^{-1/2}. These are statements about the full source support, including the permitted primes of S. Every active marked column in Equation (14.3) contributes q_p^{-1/2}. Since q_P/Z^{z_a} lies in a fixed compact interval, this proves the per-tuple coefficient exponent in Equation (14.12).

We now record the precise marked instance of the common profile. Let F_{\rm act},N_F,B_F be the actual frozen products with log-norms A_0,N_0,B_0. Put R=k_{\rm res}, P=\prod_{i\ {\rm active}}p_i. The branch denominator and a supported dual index are

With the fixed block centers, define

Multiplicativity gives the exact identity

It remains true if N_F shares primes with b. All frozen factors in Y are their actual products, and every moving norm remains a coordinate. The actual row cutoff and the individual dual and slot cutoffs therefore place the kernel argument in

for one C_{\rm ker}\ge1 depending only on the fixed support intervals and the common fixed modulus. In particular this comparison has not used any shorter row added by positivity.

Put D_{\rm ker}=T_d-v-3\ell_b-e_\lambda. On this full product of actual annuli, Lemma 5.3 permits

to be factored from the entire homogeneous linear row vector, at a factor at most C_{\rm ker}^{1/4} in its fixed seminorm constants. The normalized profile has bounded fixed Euler seminorms; no inhomogeneous tuple seminorm is claimed to become small. The profile includes the normalized inverse roots from the structural extraction and every other common smooth norm window.

The tail is removed on the genuine, unseparated sums. If \log Z\ge2\log C_{\rm ker}/\tau_{\rm ref}, every whole dyad with center excess greater than \tau_{\rm ref} has actual kernel argument greater than Z^{\tau_{\rm ref}/2}. Apply Lemma 5.4 with lattice \lambda^{-4}\mathcal O, a=X/q_c^2, and U=Z^{\tau_{\rm ref}/2}. On the full source support, the coefficient bound used in Part I gives

Each base or marked local factor is at most q_p^{1/2} after its indicator is discarded. The bounded row, tuple, and local log-length ranges, the local branch counts, and 1+a^{-1} have a fixed polynomial cost Z^{B_{\rm tail}}, with B_{\rm tail} chosen independently of the kernel order. For any desired saving D>0, the choice

in that shell lemma makes the absolute total of these whole dyads O(Z^{-D}) times a finite test seminorm. Norm twists have the fixed polynomial height cost of that seminorm. The shell sum counts all discarded dual indices, so no count at a retained dual length is used for this tail. It is removed before Fourier absolutization.

For a retained dyad, apply the common-profile lemma and Equation (5.24) to the one joint profile in Equation (14.15). Its density is common to the entire current row Hilbert space, including the active-tuple sum: the actual row, dual, and individual slot norms are its coordinates, not parameters of separately chosen measures. The full fixed support boxes govern every invoked weighted Fourier norm and height cost. At a separated mode the norm powers have absolute value one and preserve the row, tuple, and dual coefficient independences. Only inside this separated nonnegative row norm may the row set be enlarged from the original annulus to q_R\ll Z^H, using the same separated formula on the added rows. The kernel is never evaluated there.

After extracting the common coefficient but before using m_{\rm ker}^2, the separated squared norm is bounded by

where

The ideals n,b retain their stated dyadic norms and fixed coefficient restrictions. After extracting common bounds, |a(\boldsymbol p)|,|\beta(n,b)|\le1.

We verify the two independence hypotheses before invoking the norm bound. Every residual prime has exponent j=1, every active slot is a whole-index j=0 mark, and all other active primes are frozen. The separate full M_{\rm ref}^2 classes fix the cusp coefficient and additive factor. Equation (14.5) cancels every row-slot phase, while the preceding marked-reflection discussion assigns all row-row, slot-slot, and frozen interactions to a row scalar or a tuple scalar. The angular conductor scalar separates in the same fashion. The remaining source and frozen local columns depend only on the full extracted dual index; their zeros at a frozen factor give fixed row, tuple, or dual restrictions. Finally the exact zero-preserving column identities leave only 1_{(P,k_{\rm res})=1} and 1_{(P,b)=1} in addition to the character zeros. Hence a is independent of k_{\rm res},n,b, and \beta is independent of k_{\rm res},\boldsymbol p, exactly as required in Equation (14.16). The function a need not retain product form at this norm step.

Apply Lemma 14.2 with

Aggregating tuples with the same P costs only a divisor factor. Since

the hybrid bound and the outside coefficient give exponent

Only now does the squared scalar m_{\rm ker}^2 supply the last term in Equation (14.14). This proves the fixed-part assertion.

Every powerful ideal is uniquely x^2y^3 with y squarefree. Ideal counting and \sum_yq_y^{-3/2}<\infty give O(Z^{O/2}) such ideals in the O-dyad. The supported squarefree part divides the radical of S f\mathfrak r_\rho, so it has O(Z^\epsilon) choices in the stated ranges; the local choices have the same divisor bound. Rowwise divisor Cauchy on the local expansion followed by the positive sum over these fixed parts therefore adds O/2+\epsilon once. The active tuples were already inside the hybrid norm and are not counted again. This proves the aggregate assertion.□

Proof. Put A=cn^3 and \varrho_i=q_{p_i}/P_i for i\notin J. Since q_A=q_cq_n^3 even when c,n share primes, and q_{p_{J^c}}=Z^{\ell'}\prod_{i\notin J}\varrho_i, Mellin inversion of the displayed integral defining B^J gives the physical smooth factor

Here V=W_{\mathrm G} is the Gaussian of Lemma 4.6. Empty products are one.

Use the fixed translate partition \chi from that lemma on r. On the annulus r=e^kx, retain the single joint profile

The logarithms of x and of all the \varrho_i lie in fixed compact sets on this support. Thus the logarithm of the Gaussian argument is k+O_{\rm fixed}(1). The derivative estimate in Lemma 4.6, with this bounded shift and the fixed slot windows, gives for every fixed \kappa_{\mathrm G}>0, A_{\rm cnt},B_{\rm cnt},C_{\rm ann}\ge0, D_{\rm tail}>0, and fixed seminorm order j,

The constants may depend on the fixed slot system, but not on its moving prime labels. On an annulus \log r=k+O(1), an absolute bound for the row \ell^2 norm is at most Z^{B_{\rm cnt}}e^{A_{\rm cnt}|k|}p_j(w_{k,J}) for some fixed exponents, by the elementary row, slot, and ideal counts. Choose C_{\rm ann} larger than the fixed logarithmic radius of \chi and discard only whole annuli with |k|>\kappa_{\mathrm G}\log Z+C_{\rm ann}. Summing their norm bounds and then squaring gives less than any prescribed power of Z^{-1}. Every annulus meeting |\log_Z r|\le\kappa_{\mathrm G} is retained.

For each remaining annulus, keep fixed individual annular cutoffs equal to one on the support of w_{k,J} and apply Lemma 4.5 to this whole profile. Its logarithmic Fourier inversion uses one coefficient density common to every row and every surviving slot tuple. The normalized slot ratios remain variables of the joint profile until separation; they do not index separately chosen densities. For fixed Fourier variables, each resulting slot coefficient is a product of the original arithmetic coefficient \overline{\eta(p_i)} and an individual annular cutoff and norm power. It remains bounded and independent of the row and of every other slot. The completed factor is an annular test at length

These retained lengths lie in a fixed bounded range. The weighted Fourier norms have a summable total by the displayed Gaussian estimate; the small annular weight stays in this homogeneous single-profile Fourier norm, not in an inhomogeneous tuple seminorm. Minkowski’s inequality passes the separated row norm through this common density. The existing q_c^{-1/2}q_n^{-1} normalization stays unchanged when the test scale is recentered, so no additional power of e^k is introduced. Choose \kappa_{\mathrm G} within the final \epsilon allowance. For each fixed \theta, Lemma 5.2 supplies the row-sector reduction of \nu_\sigma\theta\chi_\bullet(m) with every original zero mask. It covers all nonzero element rows, including those meeting S; no factor \xi(m) is used in this row norm. The row ball depends only on the fixed rescaled tuple p_J, through Q=q_{b_*}X'Y', and is independent of the surviving marked slots and dual variables. The physical slot sets remain disjoint and their separated coefficients are product-form. Finite triangle over \theta and Lemma 14.3 therefore give the exponent in Equation (14.14) at this actual N_*, taking the retention parameter \tau_{\rm ref} and the independently prescribed output and separation losses all at most a small \epsilon_0>0. We use the exponent E_{\rm ref} after the fixed row parts are summed; their count is already included in E_{\rm ref}+\epsilon_0, with principal exponent O/2, and is not counted again.

Here f=1 and the puncture is \rho=1, so the only moving non-slot primes are those of the row. After the finite unit and reflection-sector splits in Lemma 14.3, freeze its powerful part of norm length O and its valuation-one part supported on S, and split its residual squarefree good product into actual dyads of norm \asymp Z^H. Let \epsilon_Z=O_{\rm fixed}(1/\log Z) be nonnegative and large enough to contain all fixed annular and fixed-conductor logarithmic offsets in this calculation. Then

The actual row and dual annular cutoffs are kept in one joint profile while its common Fourier density is separated. The small kernel amplitude is extracted from that homogeneous density before the row norm is squared, and only then may the positive sieve sum be enlarged, as in Equation (5.24). In particular the kernel saving in Equation (14.14) uses this H; it is not reevaluated on any shorter rows added by the positive enlargement.

In the notation of that energy estimate, every moving non-slot prime counted in A_0 divides the powerful row part to exponent at least two. The comparison of its exact norm with the powerful dyad, and any fixed-conductor contribution, therefore gives

The available dual length at the actual completed scale is

Since M'+\ell'-1=-3d, direct subtraction gives

As in the proof of the reflected energy, dual ranges below a fixed negative length are negligible after a positive Mellin shift, and the bounded boundary range costs an arbitrarily small power. The source dual ideals n,b in these dyads retain the support of Equation (5.2): they may share primes and may contain the permitted primes of S; no additional S-mask or coprimality condition between n and b is imposed. On each remaining dual dyad, put y=v+3\ell_b+e_\lambda. Retention gives y\le T_d+\epsilon_0, while the nonzero unit ranges give v,\ell_b,e_\lambda\ge-\epsilon_Z after increasing its fixed constant. It follows that

Write s_{\rm hyb}=\min\{v,z_a,(v+z_a)/3\} for the quantity called u in Equation (14.13). If s_{\rm hyb}=z_a, then -S_0-B_0+z_a-s_{\rm hyb}\le0, the kernel term is nonpositive, and

Here the standalone O is the powerful-row logarithmic length. It follows that

Otherwise s_{\rm hyb}\ge v/2-O(\epsilon_Z), and the bounded unit ranges imply

The last inequality holds separately for y\le T_d and y\ge T_d. Here the squared kernel saving is obtained by first extracting \min\{1,Z^{(y-T_d)/4}\} from the common Fourier density and only then squaring the row norm. Relative to M', the saving before the remaining +z_a is exactly

The second numerator is nonnegative. Since z_a\le\ell', the energy in this branch is at most

After using O\ge-\epsilon_Z and absorbing \theta_N, the two upper bounds are

They are nonincreasing in \Delta_H. Since \Delta_H\ge-\epsilon_Z, their formal \Delta_H=0 values bound every actual dyad up to O(\epsilon_Z), whether or not an endpoint dyad occurs. Their maximum is therefore at most M'+((d-1/6)/4)_++O(\epsilon_0+\kappa_{\mathrm G}+\epsilon_Z). At d=1/6 one has \ell'=z_a=0 and the clipped unit range falls in the first branch; bounded negative unit dyads change only \epsilon_Z. Choose \epsilon_0 and \kappa_{\mathrm G} within the prescribed \epsilon allowance, and then increase the fixed-data lower threshold so that \epsilon_Z also lies within it. Summing the logarithmically many row and dual dyads proves the lemma.□

Proof. Majorize the row ball by a fixed nonnegative radial Schwartz function. Put r_s=q_s/Y' and, on the primary elements prime to S, c_\sigma(s)=1_{s\in\sigma}\chi_s(b_*)/(\tau\xi(s)), extended by zero elsewhere before the inverse character is evaluated. The exact annular rewriting of the polynomial is

Thus the expanded square has the common normalization Y'^{-3} and the joint annular profile contains the factors W_1(r_{s_1})r_{s_1}^{-1+i\nu}W_1(r_{s_2})r_{s_2}^{-1-i\nu}. Write s_1=Cn_1, s_2=Cn_2, with (n_1,n_2)=1. Poisson in the row variable has factor Q/(q_{s_1}q_{s_2}), whereas the complete transform of the two unnormalized Gauss sums is q_{s_1}q_{s_2}F(s_1,s_2;j). It therefore gives the exact prefactor Q/Y'^3 and the correlation F(Cn_1,Cn_2;Ck) of Equation (13.7). Here F again denotes the full correlation of Equation (13.6), not the local coefficient of Equation (7.1). The first Fourier kernel has argument comparable to q_Cq_k/P_a.

On coprime residuals the factor L_C in that correlation has |L_C|\le q_C. For the subsequent signed extension use its specified bounded extension, periodic modulo \operatorname{rad}C in both columns, including when C shares primes with a column. Its value is defined to be zero when a prime of C divides both columns; this is a definition of the extension, not a claim about the genuine correlation there. Before any factorwise estimate, insert the complete Möbius identity

and write n_i=dm_i. No coprimality condition on m_1,m_2 is then imposed. Nonzero terms have (d,Ck)=1. Throughout this step use the fixed ray extension of \mathcal R(n_1,n_2); a quotient of zero residue symbols would not be defined.

We specify a common fixed modulus for that coefficient. For k\ne0 write k=\zeta k_Sk_{\mathrm{good}} using the unit convention and the fixed S-prime generators of Lemma 4.1, with k_S=\prod_{\mathfrak p\in S}\pi_{\mathfrak p}^{e_{\mathfrak p}} and k_{\mathrm{good}}=\prod_{p\notin S}p^{e_p}. For a primary residual ideal m outside S, reciprocity gives

Every displayed good local power retains its zero on a shared prime, including when 6\mid e_p. Lemma 4.1 places the finitely many \kappa_{\zeta,\boldsymbol e} in a fixed ray group supported on S. Also \mathcal R(k_{\mathrm{good}},m) belongs to a fixed finite family in m.

Choose \mathfrak l_{\rm fixed} once to contain the primary-class modulus and the S zero masks, the moduli of every fixed ray coefficient in A_m, a period for \mathcal R in each variable, and the conductors of all the preceding \kappa_{\zeta,\boldsymbol e} characters. The extended coefficient is zero off the primary class or at a fixed nonunit before any inverse character is evaluated. This modulus is independent of C,d,k, and may have higher powers at S than \operatorname{rad}(k). After reciprocity the joint arithmetic coefficient in m_1,m_2 is periodic modulo

If k has a valuation e_p not divisible by six at a prime outside C and \mathfrak l_{\rm fixed}, call k nonexceptional. At that prime the first residual column has the nonprincipal zero-extended factor \chi_p(m_1)^{e_p} up to a unit scalar. The lift, the fixed phases, and the other prime factors are independent of m_1 modulo p. Its complete mean, and hence the complete joint mean by the Chinese remainder theorem, is zero. The substitution by d does not change this conclusion because (d,k)=1. Principal zero masks at valuations divisible by six remain present in the exceptional case.

The frequency k=0 contributes only the diagonal: the correlation forces n_1=n_2=1 and equals \varphi(C). There are O(Y') possible C of norm comparable to Y', each with \varphi(C)\le q_C\ll Y'. Its contribution is therefore O(Q/Y').

For k\ne0, split the Fourier argument into dyadic shells q_Cq_k/P_a\asymp R, with R=1,2,4,\ldots and the first shell including all arguments at most two. Choose dyad centers C_0,D_0\in\{1,2,4,\ldots\} for q_C\asymp C_0 and q_d\asymp D_0, with the first dyads containing the bounded unit ranges. Use the nominal positive residual scale

The actual residual norms lie in fixed multiples of N; the nominal value N is allowed to be below one. Fix arbitrary A>2 and B>2. For each fixed C,d,k, the one joint annular profile W_R for the two columns has the finitely many homogeneous single-profile seminorms used below bounded by O_{B,J_{\mathrm{Gr}}}((1+|\nu|)^{J_{\mathrm{Gr}}} R^{-B}), with one fixed J_{\mathrm{Gr}} chosen after these derivative orders and A,B. This follows by differentiating the first Fourier kernel, whose Schwartz decay absorbs every resulting power of R; derivatives of the norm powers cost a fixed power of 1+|\nu|. This factor R^{-B} remains in the single-profile Fourier estimate and occurs once in the linear lattice sum below; no small inhomogeneous tuple seminorm is used.

Let a_{C,d,k} be this periodic arithmetic coefficient. For a nonexceptional frequency its normalized complete Fourier transform, with normalization \mathcal Q^{-2} on the two residue variables, has absolute value at most q_C and vanishes at (0,0). Since every ideal of \mathcal O is principal, the period lattice has linear scale \mathcal Q^{1/2}. Four-dimensional lattice Poisson therefore gives, for this A,

For N\ge\mathcal Q, the zero frequency vanishes and the other dual vectors have linear scale (N/\mathcal Q)^{1/2}. Fourier decay of order 2A>4 bounds their sum by (\mathcal Q/N)^A. For N<\mathcal Q, use the point count in the two annuli. A nonempty subunit annulus has N bounded below by a fixed positive constant, so it still contains O(N) lattice points; below that constant it is empty. This proves Equation (15.5) without any primitivity or tensor-product assumption on a_{C,d,k}.

Put K_R=RP_a/C_0. The fixed shell comparison constants give q_k\le C_{\rm sh}K_R for a fixed C_{\rm sh}>0. Hence the nonzero frequency range is empty when K_R<C_{\rm sh}^{-1}. On a nonempty range, elementary lattice counting gives

There are therefore O(C_0D_0K_R) choices of C,d,k on these dyads. Multiplying Equation (15.5) by this count and by the common prefactor gives at most

For \Lambda=Y'/(RP_a) the required positive dyadic sum is

For \Lambda\ge1, split the two geometric tails at 2^n\asymp\Lambda; for \Lambda<1, the left side is bounded and the displayed right side is larger than a positive constant. Thus the nonexceptional frequencies contribute

after the R sum, on choosing a fixed B>2.

It remains to count the exceptional nonzero frequencies. They have (k)=a_1^6e, where e is sixth-power-free and supported on the primes of C and the fixed support. There are at most 6^{\omega(C)+O_S(1)}\ll_\epsilon q_C^\epsilon possible e. For each of them the shell bound q_k\le C_{\rm sh}K_R permits only q_{a_1}\le(C_{\rm sh}K_R/q_e)^{1/6}. If this range is nonempty, its upper bound is at least one, and the ideal count gives O_{\rm fixed}(K_R^{1/6}) choices; if q_e>C_{\rm sh}K_R it is empty. Unit factors cost only a fixed factor. Hence there are O_\epsilon(K_R^{1/6}q_C^\epsilon) such elements k, including the bounded subunit range of K_R.

Use K_R=RP_a/C_0, |L_C|\le q_C, and the trivial two-column point count. Before multiplication by Q/Y'^3, the contribution on these dyads is

The sums over C_0,D_0 converge after reducing the arbitrary small power, and the R sum converges for the same fixed B>2. This gives the term (Q/Y')P_a^{1/6}Z^\epsilon. Adding the diagonal and nonexceptional terms proves the proposition.□

Proof. For a fixed rescaled subset and tuple, put Q=q_{b_*}X'Y'. Its separation is Equation (6.2) with A_{m,\sigma,v}(Y') and B^J_{m,\sigma}(Z). Put r_J=q_{p_J}/Z^d; the fixed annular supports give 0<c_J\le r_J\le C_J<\infty uniformly in the tuple. Exactly,

In particular P_a\ge1 for Z\ge q_{b_*}^8. Enlarge the allowed fixed-data lower threshold so that Q,Y'\ge1 uniformly over the tuple ratios as well; this is possible because M'\ge1/2 and l_y-d\ge5/16. The remaining length inequalities are

The last one gives

and hence P_a^2/Y'\ll P_a^{1/6}. Proposition 15.2 therefore gives

Cauchy–Schwarz in the row sum, Lemma 15.1, and the integrability of (1+|v|)^{J_{\mathrm{Gr}}/2}|\widehat W_0(iv)| now bound this tuple’s unscaled separation by Z^\epsilon times

Since r_J is bounded, this is O((X')^{1/2}P_a^{1/12}Z^\epsilon).

For d=\sum_{i\in J}\ell_i there are at most Z^{d+\epsilon} rescaled tuples, the coefficient is O(Z^{-3d/2}), and (X')^{1/2}\asymp X^{1/2}Z^{-d/2}. The total additional exponent is consequently

Choose the small powers in the component estimates so that their sum lies within the prescribed \epsilon. Summing the finitely many subsets proves Equation (15.6). For the already defined C=C_{\mathrm{II}} of Equation (12.3), the normalization used for the principal signal satisfies

Thus, under the contradiction assumed in Part II, this low bound is smaller than Z^{C_{\mathrm{II}}(\beta_*)} by the exact power \Delta=\beta_*-7/8, apart from the arbitrarily prescribed \epsilon.□

Proof. In this region \vartheta=(-w_r)_+\le6e. The estimates in the proof of Lemma 7.1 sharpen to H_p-1=O(Q^{-1-10e}) for p\nmid u and H_p-1=O(Q^{-10e}) for p\mid u. Increase the fixed P_0 so that |H_p-1|<1/2 for all these primes. The holomorphic factor G_p in Equation (16.2), divided by H_p, now defines \mathcal B_p throughout this region. It is H_p, not the possibly singular raw factor P_p, that is bounded away from zero when w_r<0.

Consequently the tuple sum in Equation (16.3) factors as \mathcal H_{\eta,u}\prod_i\mathcal B_i on this dynamic region. Substituting \mathcal B_i=\mathcal Q_i+\mathcal D_i proves Equation (16.10). This factorization is not used to continue the correction outside the present region.

For p\nmid u, Equation (16.8) and the estimate for \mathcal E_p give

The four exponents are bounded respectively by -a-4e, -51/25+12e, 49/25-5a-68e, and a-51/25+12e. Division by the bounded H_p preserves this estimate. The number of primes in a slot is O(P_i) by the ideal count; hence its total contribution from p\nmid u, including Q^{z-1}, is O(P_i^{z_r-51/100+\epsilon}).

For p\mid u there are only divisor-many possible labels. A monomial in \mathcal E_p is multiplied by Q^{z-1+x} in Q^{z-1}\mathcal B_p, so after separating Q^z its exponent is x_r-1 plus the exponent of that monomial. For the non-tail boundary terms of the six-valuation table, these exponents at e=0 are

At positive e their changes are respectively +6e,-32e,(-26e,-48e),-42e,-80e. They are all strictly below -1/2 in the stated range. The common R term is smaller still: its exponent after separating Q^z is at most 49/25-1-5a-80e<-1/2. Additional valuation pairs and m values have ratios |R|\le Q^{-11/10} and |V|=Q^{-51/25}, so their geometric sums preserve these bounds. The strict second-family term with an additional V is also smaller than Q^{-1/2} after this normalization.

The sole remaining term is the strict (e_0,l,k,m)=(1,0,1,0) term, which occurs for j\ge2 and is -\eta(p)(Q-1)Q^{-x-w}. Its exponent after separating Q^z is -w_r. The explicit rescaling term -Q^{-w} has exponent after separating Q^z -1-w_r<-1/2, so it needs no further estimate.

Let \mathfrak f_u be the conductor of \psi_u^*. Write u=\epsilon_u\prod_{p\mid u}p^{j_p}, with \epsilon_u a unit. For primary n coprime to uS, reciprocity gives

The unit exponent is read modulo six. The prime formula extends to composite n: when A\equiv B\equiv1\pmod6,

Thus the unit factor depends only on q_n modulo 36, and the \mathcal R product depends only on n modulo 4. Together with the multiplicativity of \mathcal R and the local symbols, the displayed norm congruence makes the right side multiplicative on integral ideals prime to 6u; extend it to their fractional ideal group. A generator congruent to one modulo (36)\operatorname{rad}(u) is already primary and makes every displayed factor one. This character therefore has a ray presentation with that fixed 2,3 modulus times \operatorname{rad}(u). It agrees with the Kummer character \chi_\bullet(u) of Lemma 4.1 on ideals avoiding S. These characters induce the same primitive character: in the principal ideal ring \mathcal O, the Chinese remainder theorem supplies a representative avoiding the additional finite set S in every ray class of a common modulus. At a good prime p\mid u, hold the fixed class and all other good residues at one and vary the residue modulo p by the Chinese remainder theorem. The remaining local character \chi_p^{j_p} has exact order 6/\gcd(6,j_p)>1. Its conductor exponent at p is therefore exactly one: the displayed ray modulus has only the first power of p, and the character cannot descend to a modulus omitting p. This is a local assertion; the ideal character may still need the fixed normalization modulus at 2,3.

It follows that, for any set J_0 of distinct selected ramified labels,

The functional equation for the primitive numerator [11, Equation (1.1)], the assumed reflected bound, and Stirling give a cost q_{\mathfrak f_u}^{A_*}U^\epsilon(3+T_1)^C, where

At a selected p\mid u the primitive character already has value zero, so restoring the original local factor there multiplies by exactly one. The remaining deleted Euler factors have radical O_S(U) and cost at most U^{6e+\epsilon}, because w_r\ge-6e.

Expand a product of error slots into their coprime-prime terms, their already bounded ramified terms, and their strict ramified terms. For each fixed tuple in this triangle expansion, let J_0 be precisely its strict ramified labels. They are distinct because the slot supports are disjoint. Apply the single inequality Equation (16.11) to this entire J_0 before summing the labels. The numerator together with these local factors, including q_p^{z-1} at each selected prime, is then bounded by

because j_p\ge2 and -w_r-A_*=-1/2. Thus different selected labels use different factors of the conductor deficit, and the bounds hold simultaneously. Summing the divisor-many ramified labels costs U^\epsilon, while the coprime labels have the stronger exponent z_r-51/100. This proves Equation (16.9).

This argument uses triangle only on error labels. It never changes the physical row or the coefficients and masks in any main slot. The tuple-independent factor \mathcal H_{\eta,u} separately costs U^\epsilon by Lemma 7.1. Relative to the central scale P_i^{z_r-1/2} of a main slot, an error slot thus has no positive amplitude exponent in the later row count.□