Uniform quartic zero-local Euler-product majorant

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Let d=(d_1,d_2,d_3,d_4) be positive integers, write D=d_1d_2d_3d_4, and set

$$\rho_p(0;d)=p^{-3}\#\{x\in\mathbf F_p^4:\textstyle\sum_i d_i x_i^4=0\}.$$

Assume

1. D is not a rational square;
2. for every ordered pair of distinct indices a,b, with complementary indices c,d, $d_a^3d_b d_c^2d_d^2\notin4\mathbf Q^{\times4}$.

Then, uniformly for B>=2,

$$\prod_{\substack{2<p\le B\\p\nmid D}}\rho_p(0;d) \ll (\log(2D))^{11}\ll_\epsilon D^\epsilon. \tag{1}$$

The implied constant in the first bound is absolute. One may replace its displayed exponent 11 by any larger fixed exponent without changing the application.

1. Exact local trace expansion

For odd p not dividing D, write

$$\rho_p(0;d)=1+\frac{p-1}{p^2}A_p(d). \tag{2}$$

If p=3 mod4, fourth powers and squares have identical counting measures on the finite field, and

$$A_p(d)=\left(\frac Dp\right). \tag{3}$$

Suppose p=1 mod4. Choose one prime ideal P=(pi) above p in Z[i], choose its primary generator pi=1 mod (1+i)^3, and let chi=(./P)_4. Put tau_j=G(chi^j). Additive orthogonality gives

$$A_p(d)=p^{-2} \sum_{\substack{j_1,\ldots,j_4\in\{1,2,3\}\\\sum j_i=0\pmod4}} \prod_i\tau_{j_i}\chi^{-j_i}(d_i). \tag{4}$$

There are 21 terms, so |A_p(d)|<=21. The exponent patterns are

$$(2,2,2,2),\quad (1,1,3,3),\quad(1,3,2,2),\quad(1,1,1,1),\quad(3,3,3,3),$$

with multiplicities 1,6,12,1,1. Using

$$\tau_2^2=p,\qquad \tau_1\tau_3=\chi(-1)p, \qquad \tau_1^4/p^2=J(\chi,\chi)^2/p=\pi^2/p,$$

we obtain the following character decomposition.

Let psi_R be the quartic Kummer Hecke character of Q(i) corresponding to the positive or negative rational number R, so $\psi_R(P)=(R/P)_4$. Let nu((pi))=pi^2/N(pi) for primary generators; this is a unitary Hecke character of infinity type 2 and modulus dividing (1+i)^3. Put $\eta=\nu\psi_D^{-1}$. Then

$$A_p(d)=\chi_D(p) +\sum_{S\in\mathcal P}\sum_{P\mid p}\psi_{R_S}(P) +\sum_{\{a,b\}}\sum_{P\mid p}\psi_{-R_{ab}}(P) +\sum_{P\mid p}\eta(P), \tag{5}$$

for split p, where:

• chi_D is the quadratic Dirichlet character of the squareclass of D;
• P (calligraphic) contains one representative of each of the three complementary pairs of 2-subsets of {1,2,3,4};
• $R_S=\prod_{i\in S}d_i^3\prod_{i\notin S}d_i$;
• the second sum runs over the six unordered pairs {a,b}, and $R_{ab}=d_a^3d_b d_c^2d_d^2$, with either orientation of {a,b}.

Conjugate orientations give conjugate Hecke characters, explaining the traces over the two prime ideals above p. At inert rational primes the Hecke prime ideals have norm p^2; they do not contribute a rational-prime term of order 1/p.

Here is a direct verification of the needed squared Jacobi identity. Reduction modulo P gives $J(\chi,\chi)\equiv\sum_x x^{(p-1)/4}(1-x)^{(p-1)/4}=0$, because every monomial has positive degree less than p-1. The usual Gauss-sum identity gives $|J|^2=p$, so J generates P. Pairing x and 1-x in its defining sum shows $J\equiv\chi(1/4)=\pm1\pmod2$. Thus J has odd real part and even imaginary part. Generators of P with this parity are unique up to sign, so their squares coincide. This proves J^2=pi^2 for a primary generator and also constructs nu directly as a Hecke character: choose the odd-real, even-imaginary generator of any odd ideal and square it, dividing by its norm. Multiplication preserves this parity and the sign disappears on squaring.

2. None of the eleven characters is principal

The Dirichlet character chi_D is nonprincipal by hypothesis 1. A rational positive number is a fourth power in Q(i) precisely when it is a rational fourth power. If R_S is a rational fourth power, reduction of its valuation vector modulo 2 shows that D is a square, excluded here.

A negative rational number -R, with R>0, is a fourth power in Q(i) precisely when R in 4 Q^{x4}. Indeed the fourth power of a+bi is real only when ab(a^2-b^2)=0; the negative case has a=+-b and value -4a^4. Thus hypothesis 2 excludes principality of the six mixed characters. Finally eta has nonzero infinity type 2, and so is never principal.

All finite conductors, and a common modulus for the imprimitive versions of these characters with every prime dividing 2D removed, have norm bounded by a fixed power of 2D. This follows either from quartic reciprocity or from the conductor-discriminant theorem applied to Q(i,sqrt[4]{R})/Q(i): a fourth-power-free representative of R can be chosen integral with absolute value bounded by a fixed power of D, the polynomial discriminant is -256 R^3, and the character conductor divides the relative discriminant. If x^4-R is reducible, take the monic minimal factor of the chosen root; its Gaussian-integral coefficients and the discriminant product formula show that its discriminant divides the full polynomial discriminant. The field discriminant divides that of this order. Twisting nu changes the conductor by only a fixed power of (1+i).

The character correspondence and the Jacobi infinity type are also stated in Ghidelli, Proposition 5.2, whose relevant passage has been checked. The polynomial discriminant argument gives the polynomial conductor bound needed here, without a sharp conductor formula.

A conductor bound in terms of the coefficient radical

Section 5 of the later independent record suggests the following sharper bound, checked here. Put $R=\operatorname{rad}(D)$ and $\pi=1+i$. Every finite Gaussian character in (5) has order dividing four, trivial infinity type, and conductor supported on the prime ideals over $2D$, by the checked Proposition 5.2. Since $\mathbb Z[i]$ is a PID, it is a residue-ring character. At an odd prime ideal, the principal units in each finite quotient form an odd-primary group; an order-dividing-four character kills them. At $\pi$, every unit $u\equiv1\pmod{\pi^9}$ is a fourth power locally: Hensel at $1$ for $T^4-u$ applies because $v_\pi(u-1)\ge9>2v_\pi(4)=8$. Reducing the local fourth root gives the same conclusion in every finite quotient. Thus the finite conductor divides

$$\pi^9\prod_{\substack{\mathfrak p\mid D\\\mathfrak p\nmid2}} \mathfrak p.$$

The Jacobi-square character has defining ideal dividing $(16)=\pi^8$, so twisting it by a finite character preserves this bound. A common imprimitive modulus, removing all prime ideals over $2D$, is therefore

$$\mathfrak m=\pi^9\prod_{\substack{p\mid R\\p\text{ odd}}}(p), \qquad N\mathfrak m\le512R^2. \tag{5a}$$

For the quadratic Dirichlet character, the analogous common presented modulus is $4R$. The estimates of Sections 3–4 apply to these enlarged moduli. Consequently (10), and hence (1), strengthen to

$$\sum_{\substack{2<p\le B\\p\nmid D}}\frac{A_p(d)}p \le11\log\log(3R)+O(1),\qquad \prod_{\substack{2<p\le B\\p\nmid D}}\rho_p(0;d) \ll(\log(3R))^{11}. \tag{5b}$$

No extra $\omega(R)$ loss is needed: the unwanted Euler factors are removed in the defining moduli before applying the lattice estimate. This refinement is not required by the original quartic proof; it is useful in the random omission alternative.

3. An elementary uniform bound for the relevant Hecke L-functions

Let psi be a nonprincipal unitary Hecke character of Q(i) with fixed infinity type m (here only 0 or 2) and integral modulus f, of norm Q. Since Z[i] is principal, it has the form

$$\psi((z))=\omega(z)(z/|z|)^m,$$

where omega is a Dirichlet character modulo f, extended by zero off the invertible residue classes, with the appropriate compatibility on units.

For each residue class modulo f, a lattice Riemann-sum estimate in the disk |z|<=sqrt X gives

$$\sum_{\substack{0<|z|\le\sqrt X\\z=a\pmod f}} (z/|z|)^m =\frac1Q\int_{|z|\le\sqrt X}(z/|z|)^m\,dz +O_m(\sqrt{X/Q}+1). \tag{6}$$

For completeness, the boundary cells contribute O(sqrt(X/Q)+1); in the interior, the gradient of the angular factor is O_m(1/|z|), whose disk integral is O_m(sqrt X), and the mesh diameter is O(sqrt Q). The cell containing zero costs O(1).

Summing (6) over the Q residue classes, and dividing by four for units, gives

$$A(X):=\sum_{N(a)\le X}\psi(a) \ll_m \sqrt{QX}+Q. \tag{7}$$

The main integral vanishes if m != 0. If m=0, it cancels because the finite character is nonprincipal. The same proof works for any enlarged, imprimitive modulus.

Split the Dirichlet series of L(s,psi) at N(a)=Q. The initial part is O(log(2Q)) because the number of Gaussian ideals of norm at most X is O(X). Partial summation with (7) bounds the remaining part by O_m(1), uniformly for real s>=1. Hence

$$|L(s,\psi)|\ll_m\log(2Q)\qquad(s\ge1). \tag{8}$$

For nonprincipal Dirichlet characters over Q, the corresponding estimate follows more simply by periodic cancellation, |sum_{n<=X}chi(n)|<=q, cutting the series at its modulus q.

4. Comparison of partial Euler products with L-values

For x>=3, set sigma=1+1/log x. For any of the above unitary characters,

$$\sum_{N(P)\le x}\frac{\Re\psi(P)}{N(P)} =\log|L(\sigma,\psi)|+O(1), \tag{9}$$

uniformly in the character and its modulus. Prime powers of exponent at least two cost O(1). For first powers, the absolute comparison below x is

$$\sum_{N(P)\le x}\frac{1-N(P)^{-1/\log x}}{N(P)} \le\frac1{\log x}\sum_{N(P)\le x}\frac{\log N(P)}{N(P)}\ll1.$$

The tail above x is also bounded:

$$\sum_{N(P)>x}N(P)^{-1-1/\log x}\ll1.$$

Both estimates use only the Chebyshev bound for prime ideals of Q(i), which follows immediately from the rational Chebyshev bound. The same argument applies to Dirichlet characters over Q.

Use imprimitive versions of all eleven characters which vanish at every prime dividing 2D. Their moduli are still polynomially bounded in D. Equations (5), (8), and (9), with the bounded contribution from inert prime ideals, give

$$\sum_{\substack{2<p\le B\\p\nmid D}}\frac{A_p(d)}p \le 11\log\log(2D)+O(1). \tag{10}$$

Here there is one Dirichlet character and ten Gaussian Hecke characters: three complementary 2-subset pairs, six mixed pairs, and the CM character.

Finally, from (2), log(1+u)<=u, and |A_p|<=21,

$$\log\prod_{\substack{2<p\le B\\p\nmid D}}\rho_p(0;d) \le\sum_{\substack{2<p\le B\\p\nmid D}}\frac{A_p(d)}p+O(1),$$

which proves (1). If a factor is zero, the assertion is automatic.

Status and scope

This proves the analytic majorant using the elementary local Gauss-sum identities and the standard quartic Kummer character correspondence. No prime number theorem, zero-free region, Siegel bound, or angular equidistribution is needed for this upper bound. The argument does not assert convergence of the partial Euler products; uniform boundedness is sufficient. Its application to the full powerful-number sumset is in the quartic theorem.

Tracking principal constituents

Appendix D of third independent record reproduces the uniform modulus, lattice cancellation and positive-factor argument above, and adds a useful bound without the regularity hypotheses. Put $R=\operatorname{rad}(D)$ and let $m$ count the principal constituents among the one rational quadratic character and nine finite Gaussian characters in (5), with multiplicity. The infinity-type-two character is never principal, so $0\le m\le10$. Then

$$\prod_{\substack{2<p\le Z\\p\nmid D}}\rho_p(0;d) \ll [\log(3R)]^{11-m}[\log(2Z)]^m. \tag{11}$$

Each nonprincipal constituent has the previously proved upper bound $\log\log(3R)+O(1)$ for its prime sum. For a principal constituent, bound its $L$-function by $\zeta$ or $\zeta_{\mathbb Q(i)}$; at $\sigma=1+1/\log Z$ its logarithm is at most $\log\log(2Z)+O(1)$. The same smoothing proves (11). For canonical tuples, including valuation-one targets at coefficient primes costs at most $4^{\omega(R)}$.

Here “no principal finite characters” must include the rational quadratic constituent, not only the nine Gaussian ones. For example, $(2^5,3^5,5^5,30^5)$ has square product and no principal Gaussian constituent, so $m=1$, not zero. In a complementary-pair character, at least one of the primes $2,3,5$ has argument valuation $2\pmod4$. For a mixed character, requiring valuations zero modulo four at both $3$ and $5$ would require the weights assigned to each of coordinates two and three to complement the weight at coordinate four to four; the pattern $(2,2,1,3)$ has insufficient repeated weights for this. Thus none of those Gaussian characters is principal.

Appendix G's additional lower-Hecke-$L$ route is not needed for (11) or for the universal selector: the baseline always has positive $L$-powers, and the selector needs only the rational quadratic two-sided estimate. No lower Gaussian $L$-bound is being assumed in either completion.