The coefficient family and the plan of the zero-free argument

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson defines the sums \(B_m(Z)\) on which the proof of the zero-free strip rests, states the two estimates that the following lessons prove, and proves the final step: a power saving for one inverse Mellin integral excludes zeros of \(L_F(s,\eta)\) from a half-plane. The sums \(B_m(Z)\) are smoothed series over ideals \(\mathfrak A=\mathfrak c\mathfrak n^3\) of cubic Gauss sum coefficients twisted by the sextic symbol \(\chi_{\mathfrak A}(m)\). Lessons 6–10 bound their mean square over a box of rows \(m\). Lessons 11–14 insert them into an additive average, the probe, and evaluate that average by Poisson summation; its main term contains \(1/L^S(s,\eta)\). Comparing the upper bound with the evaluation gives the power saving.

Prerequisites. S-integers and sixth-power classes, Lemma 2.1 and Lemma 4.1; Sextic reciprocity and Kummer characters, Section 1; Gauss sums and the correction characters, (3.1), Proposition 5.1 and Lemma 6.1; Hecke L-functions and the Dedekind zeta function; Dirichlet series and Euler products, Lemmas A.2 and A.3; Schwartz functions and Fourier inversion, Section F4.

1. Incomplete L-functions

Keep \(F\supseteq\mu_{12}\), \(S\), \(R\), \(\eta\) and the allowed generators of Lessons 2–4. For \(\operatorname{Re}s>1\) put \[ \begin{gathered} \zeta_F^S(s)=\sum_{\mathfrak a}q_{\mathfrak a}^{-s},\\ \begin{aligned} L^S(s,\eta)&=\prod_{\mathfrak p\notin S}\bigl(1-\eta(\mathfrak p)q_{\mathfrak p}^{-s}\bigr)^{-1}, \end{aligned}\quad(1.1) \end{gathered} \] the sums and products being over exterior ideals and primes. The same symbols denote their meromorphic continuations to \(\mathbb C\) (Hecke L-functions and the Dedekind zeta function); \(L^S(s,\eta)\) is entire unless \(\eta\) is principal, when it has a simple pole at \(s=1\). It differs from \(L_F(s,\eta)\) by the finitely many factors \(1-\eta(\mathfrak p)q_{\mathfrak p}^{-s}\) with \(\mathfrak p\in S\) unramified for \(\eta\), whose zeros lie on \(\operatorname{Re}s=0\).

2. A Gaussian weight

For \(t>0\) put \[ \begin{aligned} &P_G(t)\\ &\quad =\frac1{2\sqrt\pi}\exp\Bigl(-\frac{(\log t)^2}4\Bigr).\quad(2.1) \end{aligned} \]

Lemma 2.1. For every \(\xi\in\mathbb C\), \(\int_0^\infty P_G(t)t^\xi\,\frac{dt}t=e^{\xi^2}\). For every real \(\sigma\) and \(t>0\), \(P_G(t)\le e^{\sigma^2}t^{-\sigma}\).

Proof. With \(v=\log t\) the integral is \(\frac1{2\sqrt\pi}\int_{\mathbb R}e^{-v^2/4+\xi v}\,dv\)\(=\frac{e^{\xi^2}}{2\sqrt\pi}\int_{\mathbb R}e^{-(v-2\xi)^2/4}\,dv=e^{\xi^2}\), first for real \(\xi\) and then for all \(\xi\) because both sides are entire. The bound follows from \(v^2/4\ge\sigma v-\sigma^2\). \(\square\)

3. The coefficient sums

Put \(\vartheta=\overline\Lambda\,\overline G\,\eta\) as in Lesson 4. For \(m\in R\) and \(Z>0\) define \[ \begin{gathered} \begin{aligned} B_m(Z)&=\sum_{\substack{\mathfrak c\ \text{squarefree}\\ \mathfrak n}}\frac{\gamma_2(\mathfrak c)}{q_{\mathfrak c}^{1/2}q_{\mathfrak n}}\\ &\quad \,\vartheta(\mathfrak A)\,\chi_{\mathfrak A}(m)\,P_G\Bigl(\frac{q_{\mathfrak A}}Z\Bigr), \end{aligned}\\ \mathfrak A=\mathfrak c\mathfrak n^3 ,\quad(3.1) \end{gathered} \] where \(\mathfrak c\) runs over squarefree exterior ideals and \(\mathfrak n\) over all exterior ideals, possibly sharing primes with \(\mathfrak c\). The symbol \(\chi_{\mathfrak A}(m)\) carries the unit mask of Lesson 3, so it vanishes when \(m\) and \(\mathfrak A\) share a prime.

Lemma 3.1 ([OpenAI-029, (4.2)–(4.3)]). The series (3.1) converges absolutely, and for every \(\sigma>1/2\) \[ \begin{aligned} &\sum_{\mathfrak c,\mathfrak n}\Bigl|\frac{\gamma_2(\mathfrak c)}{q_{\mathfrak c}^{1/2}q_{\mathfrak n}}\,P_G\Bigl(\frac{q_{\mathfrak A}}Z\Bigr)\Bigr|\\ &\quad\le e^{\sigma^2}Z^\sigma\,\zeta_F^S(\sigma+\tfrac12)\\ &\qquad\times\zeta_F^S(3\sigma+1).\quad(3.2) \end{aligned} \] Moreover \(B_0(Z)=P_G(1/Z)\).

Proof. For squarefree \(\mathfrak c\), \(|\gamma_2(\mathfrak c)|=1\): by (5.2) of Lesson 4 it is a product of prime Gauss sums of absolute value \(1\) and of roots of unity. Lemma 2.1 gives \(P_G(q_{\mathfrak A}/Z)\le e^{\sigma^2}Z^\sigma q_{\mathfrak c}^{-\sigma}q_{\mathfrak n}^{-3\sigma}\), and the remaining sum is at most \(\zeta_F^S(\sigma+1/2)\zeta_F^S(3\sigma+1)\), finite for \(\sigma>1/2\). For \(m=0\), \(\chi_{\mathfrak A}(0)=0\) unless \(\mathfrak A=R\), which forces \(\mathfrak c=\mathfrak n=R\). \(\square\)

The value of \(\vartheta(\mathfrak A)\chi_{\mathfrak A}(m)\) does not depend on any choice of generators; the factor \(\gamma_2(\mathfrak c)/(q_{\mathfrak c}^{1/2}q_{\mathfrak n})\) will be identified in Lesson 6 as the product of the local spherical Whittaker values of a cubic theta function at the idèle with components \(A=cn^3\).

4. Scales, rows and their two parts

The comparison uses the exponents \[ \begin{gathered} a=\frac9{10},\qquad b=\frac{133}{1000},\\ M_0=a+b=\frac{1033}{1000},\\ h_0=1-a=\frac1{10},\\ X=Z^a,\qquad Y=Z^b .\quad(4.1) \end{gathered} \] For a compact set \(\mathscr K\subset F_S^\times\) put \[ \mathscr B_Z(\mathscr K)=R\cap\ell_{Z^{M_0}}\mathscr K , \] with \(\ell_T\) as in Lesson 2. On this set every local absolute value of \(\ell_{Z^{M_0}}^{-1}m\) is bounded above and away from zero; in particular \(q_{(m)}\asymp_{\mathscr K}Z^{M_0}\). For \(m\ne0\) factor its exterior ideal as \[ \begin{gathered} (m)_S=\mathfrak h(m)\,\mathfrak d(m),\\ \mathfrak h(m)=\prod_{v_{\mathfrak p}(m)\ge2}\mathfrak p^{v_{\mathfrak p}(m)},\\ \mathfrak d(m)=\prod_{v_{\mathfrak p}(m)=1}\mathfrak p .\quad(4.2) \end{gathered} \] The ideal \(\mathfrak d(m)\) is squarefree and prime to \(\mathfrak h(m)\), and every exponent of \(\mathfrak h(m)\) is at least \(2\): \(\mathfrak h(m)\) is powerful.

Proposition 4.1 (the reflected second moment [OpenAI-029, Proposition 4.1]). Fix a compact set \(\mathscr K\subset F_S^\times\). For every \(\epsilon>0\), every \(Z\ge2\) and all \(P,D\ge1\), \[ \begin{aligned} &\sum_{\substack{m\in\mathscr B_Z(\mathscr K)\\ P\le q_{\mathfrak h(m)}<2P\\ D\le q_{\mathfrak d(m)}<2D}}|B_m(Z)|^2\\ &\quad \ll_{\mathscr K,\epsilon}Z^\epsilon P^{1/2}\Bigl(D+\frac{D^2P}Z\Bigr),\quad(4.3) \end{aligned} \] and consequently \[ \begin{aligned} &\sum_{m\in\mathscr B_Z(\mathscr K)}|B_m(Z)|^2\\ &\quad \ll_{\mathscr K,\epsilon}Z^{2M_0-1+\epsilon}.\quad(4.4) \end{aligned} \] The estimate is uniform when \(\mathscr K\) is multiplied by an element of a fixed compact subset of \(F_S^1\).

Lessons 6–10 prove Proposition 4.1. The deduction of (4.4) from (4.3) is short: a nonempty dyad has \(P,D\ge1\) and \(PD\ll Z^{M_0}\), so \(P^{1/2}D\le PD\ll Z^{M_0}\le Z^{2M_0-1}\) (as \(M_0>1\)) and \(P^{1/2}D^2P/Z=(PD)^2/(ZP^{1/2})\)\(\ll Z^{2M_0-1}\); there are \(O((\log Z)^2)\) dyads.

5. The probe and its two evaluations

Lesson 11 defines, for weights \(W_0,W_1\) fixed there, an average \[ I_Z=I(Z^a,Z^b,Z) \] which pairs \(B_m(Z)\), for \(m\) in a compact annular box of norm \(\asymp XY=Z^{M_0}\), with a normalized sum of \(\chi_{\mathfrak s}(d)\,\mathrm e(-md/s)\) over exterior ideals \(\mathfrak s\) of norm \(\asymp Y\) and residues \(d\) modulo \(\mathfrak s\). The rest of the proof consists of three statements.

The exponent \(s-8/15\) in (5.1) is \(a/2+s-1+h_0/6\); at \(s=1\) it is \(7/15\). The reflection exponent is \[ \begin{aligned} \frac{2M_0-1-b}2&=\frac{933}{2000}\\ &=\frac7{15}-\frac1{6000}.\quad(5.2) \end{aligned} \] So (P1) and (P2) together give, for large \(Z\), \[ f_\eta(Z)\ll Z^{7/15-1/20000}, \tag{5.3} \] using (P1) with \(\epsilon=10^{-5}\) and \(1/6000-10^{-5}>1/20000\), and \(0.0004>1/20000\).

6. From a power saving to a zero-free half-plane

Proposition 6.1 (Mellin continuation [OpenAI-029, Section 8.3]). Let \(\delta_1>0\) and let \(\mathcal H\) be holomorphic, bounded and nowhere zero on an open set containing the closed half-plane \(\operatorname{Re}s\ge1-\delta_1\). For real \(c,s_0\) put \[ \begin{gathered} \begin{aligned} f(Z)&=\frac1{2\pi i}\int_{(2)}Z^{s-c}e^{(s-s_0)^2}\\ &\quad \,\frac{\mathcal H(s)}{L^S(s,\eta)}\,ds \end{aligned}\\ (Z>0). \end{gathered} \] If \(0<\delta\le\delta_1\) and \(f(Z)\ll Z^{1-c-\delta}\) for all large \(Z\), then \(L^S(s,\eta)\) and \(L_F(s,\eta)\) have no zero in \(\operatorname{Re}s>1-\delta\).

Proof. Put \(\mathscr A(s)=e^{(s-s_0)^2}\mathcal H(s)/L^S(s,\eta)\). For \(\operatorname{Re}s=\sigma>1\), the Euler product gives \(|1/L^S(s,\eta)|\le\prod_{\mathfrak p}(1+q_{\mathfrak p}^{-\sigma})\le\zeta_F(\sigma)\), and \(|e^{(s-s_0)^2}|=e^{(\sigma-s_0)^2-(\operatorname{Im}s)^2}\). So \(\mathscr A\) is holomorphic on \(\operatorname{Re}s>1\) and decays like a Gaussian on vertical lines there, uniformly on compact ranges of \(\sigma\).

Small \(Z\). By Cauchy's theorem (Lemma A.2 of Dirichlet series and Euler products), the line of integration may be moved from \(2\) to any \(\sigma_R>2\), the horizontal segments vanishing by the Gaussian decay. On the new line \(|Z^{s-c}|=Z^{\sigma_R-c}\). Hence \(f(Z)=O_{A}(Z^{A})\) for \(0<Z\le1\) and every \(A>0\).

The transform. By the two bounds, \(\mathcal T(s)=\int_0^\infty f(Z)Z^{c-s}\,\frac{dZ}Z\) converges absolutely and locally uniformly on \(\operatorname{Re}s>1-\delta\): at infinity the integrand is \(O(Z^{1-\delta-\operatorname{Re}s})\), and at zero it is \(O(Z^A)\) for every \(A\). So \(\mathcal T\) is holomorphic there.

Identification. With \(Z=e^t\) and \(s=2+i\tau\), the definition of \(f\) reads \[ e^{-(2-c)t}f(e^t)=\frac1{2\pi}\int_{\mathbb R}e^{i\tau t}\mathscr A(2+i\tau)\,d\tau . \] The function \(h(\tau)=\mathscr A(2+i\tau)\) is a Schwartz function: \(\mathscr A\) is holomorphic with Gaussian decay on \(\frac32\le\operatorname{Re}s\le\frac52\), so Cauchy's estimates on discs of radius \(\frac12\) (Lemma A.2 of Dirichlet series and Euler products) bound every derivative of \(h\) by a Gaussian. The left side is the inverse Fourier transform of \(h\), and the Fourier transform of the left side is \(\tau\mapsto\mathcal T(2+i\tau)\). Fourier inversion on Schwartz functions ((F13) in Section F4 of Schwartz functions and Fourier inversion) gives \(\mathcal T(2+i\tau)=\mathscr A(2+i\tau)\) for all real \(\tau\). Both functions are holomorphic on \(\operatorname{Re}s>1\), so they agree there, because the zeros of a nonzero holomorphic function on a connected open set are isolated (Lemma A.3 of Dirichlet series and Euler products). On \(\operatorname{Re}s>1-\delta\) the function \(\mathscr A\) is meromorphic, by the continuation of \(L^S(s,\eta)\), and agrees with the holomorphic \(\mathcal T\) on an open subset of this connected half-plane; so \(\mathscr A\) has no pole there.

Conclusion. If \(L^S(s_1,\eta)=0\) with \(\operatorname{Re}s_1>1-\delta\), then \(s_1\) is a pole of \(\mathscr A\), because \(e^{(s-s_0)^2}\mathcal H(s)\) does not vanish there; this is impossible. A pole of \(L^S\) at \(s=1\) only makes \(\mathscr A\) vanish. Finally \(L_F(s,\eta)\) differs from \(L^S(s,\eta)\) by finitely many factors \(1-\eta(\mathfrak p)q_{\mathfrak p}^{-s}\) with \(|\eta(\mathfrak p)|=1\), whose zeros have real part \(0\). \(\square\)

Corollary 6.2. Statements (P1), (P2) and (H) imply Theorem 5.3 of Artin's conjecture and the index of a prime: \(L_F(s,\eta)\) has no zero in \(\operatorname{Re}s>1-1/20000\), in particular none in \(\operatorname{Re}s>1-10^{-6}\).

Proof. Apply Proposition 6.1 with \(c=8/15\), \(s_0=5/6\), \(\mathcal H=\mathcal H_\eta\), \(\delta_1=0.002\) and \(\delta=1/20000\), using (5.3) and (H); note \(1-c-\delta=7/15-1/20000\). The field \(F\) and the character \(\eta\) were arbitrary, and the number \(1/20000\) does not depend on them. \(\square\)

All constants and the points from which (P1), (P2) and (5.3) hold may depend on \(F\), \(S\), \(\eta\) and the weights; only the numerical saving \(1/20000\) is uniform, and that is all Corollary 6.2 needs.

7. A map of Lessons 6–14

8. Exercises

8.1. Compute \(\int_0^\infty P_G(t)\,\frac{dt}t\) and \(\int_0^\infty P_G(t)\,dt\).

8.2. Show that every powerful ideal has a unique expression \(\mathfrak r^2\mathfrak s^3\) with \(\mathfrak s\) squarefree, and deduce that the number of powerful exterior ideals of norm less than \(2P\) is \(O(P^{1/2})\).

8.3. Check the arithmetic \(a/2+1-1+h_0/6=7/15\) and \((2M_0-1-b)/2=933/2000\)\(=7/15-1/6000\).

8.4. In Proposition 6.1, suppose only that \(f(Z)\ll Z^{1-c}\). What does the proof give?

9. Solutions

8.1. Lemma 2.1 with \(\xi=0\) gives \(1\); with \(\xi=1\) it gives \(e\).

8.2. For a prime exponent \(e\ge2\), write \(e=2r+3s\) with \(s\in\{0,1\}\) equal to the parity of \(e\); this is possible and unique since \(e-3s\) is then even and nonnegative. The number of \(\mathfrak r^2\mathfrak s^3\) of norm less than \(2P\) is at most \(\sum_{\mathfrak s}\#\{\mathfrak r:q_{\mathfrak r}<(2P)^{1/2}q_{\mathfrak s}^{-3/2}\}\)\(\ll P^{1/2}\sum_{\mathfrak s}q_{\mathfrak s}^{-3/2}\ll P^{1/2}\), using that there are \(O(t)\) ideals of norm at most \(t\).

8.3. \(9/20+1/60=27/60+1/60\)\(=28/60=7/15\). \(2M_0-1-b=2.066-1-0.133\)\(=0.933\), and \(7/15-933/2000\)\(=(2800-2799)/6000=1/6000\).

8.4. The transform converges on \(\operatorname{Re}s>1\) only, and no zero is excluded; the argument needs a genuine power saving \(\delta>0\).

Sources