The zero-free strip

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

This lesson contains the last part of the proof of the zero-free strip. The Poisson probe wrote the probe as a triple Mellin integral over the frequency classes \(u\), and The Euler product of the probe factored each Dirichlet series. Here the contours are moved. For \(u=1\) the numerator \(\zeta_F^S(6z)\zeta_F^S(w)\) has poles at \(z=\frac16\) and \(w=1\), and their double residue is the main term \(c_0f_\eta(Z)\) of statement (P2). For \(u\ne1\) the numerator is entire, and the contours can be moved left until the power of \(Z\) falls below the main term: directly for small frequencies, with the zero detection of Zero detection for the frequency characters for intermediate ones, and by a far rightward shift for large ones. Together with (P1) and (H), this proves Theorem 5.3 of Artin's conjecture and the index of a prime.

Prerequisites. The coefficient family and the plan of the zero-free argument, Sections 4–6; The Poisson probe, Lemma 1.1, Proposition 4.1 and Proposition 7.1; The Euler product of the probe, Proposition 5.1 and Corollary 5.2; Zero detection for the frequency characters, Lemmas 1.1 and 3.1 and Proposition 4.1; Sextic reciprocity and Kummer characters, Lemma 3.2; Hecke L-functions and the Dedekind zeta function, Theorem 10.1 and Corollary 10.4; Dirichlet series and Euler products, Lemmas A.2 and A.3.

1. The power of Z on a contour

Take \(X=Z^a\), \(Y=Z^b\) in (7.1) of Lesson 11, with \(a=\frac9{10}\), \(b=\frac{133}{1000}\), \(h_0=1-a=\frac1{10}\). Write \(\sigma_\xi,\sigma_w,\sigma_z\) for the real parts of \(\xi,w,z\). The modulus of \(X^{1/2-z}Z^{\xi+z-1}Y^{w-1}\) is \(Z^{E_*}\) with \[ \begin{aligned} &E_*(\sigma_\xi,\sigma_w,\sigma_z)=\frac a2+\sigma_\xi-1\\ &\qquad +h_0\sigma_z+b(\sigma_w-1),\quad(1.1) \end{aligned} \] and the factor \(q_u^{-z}\) has modulus \(q_u^{-\sigma_z}\). At \(w=1\), \(z=\frac16\) the power becomes \(Z^{C_*(\xi)}\) with \[ \begin{gathered} \begin{aligned} C_*(\xi)&=\frac a2+\xi-1+\frac{h_0}6\\ &=\xi-\frac8{15}, \end{aligned}\\ C_*(1)=\frac7{15}.\quad(1.2) \end{gathered} \] All comparisons below are with the reference power \(C_*(1)\).

On every fixed strip of real parts, (1.3) and (1.1) of Lesson 11 give, for every \(A>0\), \[ \begin{aligned} &\bigl|e^{(\xi+z-1)^2}M(z)\widehat W_1(w)\bigr|\\ &\quad\le C_A\,e^{-(\operatorname{Im}\xi+\operatorname{Im}z)^2}\\ &\qquad\times(1+|\operatorname{Im}z|)^{-A}\\ &\qquad\times(1+|\operatorname{Im}w|)^{-A}.\quad(1.3) \end{aligned} \] On the contours below every numerator \(L\)-function and zeta function grows at most polynomially in the heights, away from the poles that are extracted (Lemma 3.1 of Lesson 13 and the remark after it), \(\mathcal H_u\) is at most \(C_\epsilon q_u^\epsilon\) (Proposition 5.1 of Lesson 12), and on lines \(\operatorname{Re}\xi>1\) the reciprocal \(1/L^S\) is bounded by its absolutely convergent Euler product. Since \(\int_{\mathbb R}e^{-(T+v)^2}(1+|v|)^{-A}\,dv\)\(\le C_{A_1}(1+|T|)^{-A_1}\) once \(A\) is large, integrating in \(z\) and \(w\) first gives decay faster than any power in \(\operatorname{Im}\xi\) as well. These bounds justify every shift of an infinite contour below by Cauchy's theorem and the residue formula (Lemmas A.2 and A.3 of Dirichlet series and Euler products), the horizontal segments tending to zero.

2. The principal frequency

Let \(I_1(Z)\) be the term \(u=1\) of (7.1) of Lesson 11. By Proposition 5.1 of Lesson 12 its Dirichlet series is \(\zeta_F^S(6z)\zeta_F^S(w)\mathcal H_1(\xi,w,z)/L^S(\xi,\eta)\). Put \[ \begin{gathered} \delta_1=10^{-7},\\ \sigma_1=1+\delta_1,\\ \zeta_1=\frac16+\delta_1 . \end{gathered} \] Move the lines from \((\operatorname{Re}\xi,\operatorname{Re}w,\operatorname{Re}z)=(4,4,2)\) to \((\sigma_1,\sigma_1,\zeta_1)\): no pole is crossed, since \(\operatorname{Re}w>1\), \(\operatorname{Re}6z>1\) and \(1/L^S(\xi,\eta)\) stays in absolute convergence, and the second region of (5.1) of Lesson 12 contains the whole move. Then move \(w\) to \(\operatorname{Re}w=0.95\), crossing the simple pole of \(\zeta_F^S(w)\) at \(w=1\); call the remaining triple integral \(R_w(Z)\). In the residue at \(w=1\), move \(z\) to \(\operatorname{Re}z=0.16\), crossing the simple pole of \(\zeta_F^S(6z)\) at \(z=\frac16\); call the remaining integral \(R_z(Z)\). So \[ \begin{aligned} I_1(Z)&=c_0f_\eta(Z)+R_w(Z)\\ &\quad +R_z(Z),\quad(2.1) \end{aligned} \] where the double residue is computed in Proposition 2.2. The contours of the remainders are \[ \begin{gathered} \begin{aligned} R_w&:\ (\operatorname{Re}\xi,\operatorname{Re}w,\operatorname{Re}z)\\ &=(\sigma_1,0.95,\zeta_1), \end{aligned}\\ \begin{aligned} R_z&:\ (\operatorname{Re}\xi,w,\operatorname{Re}z)\\ &=(\sigma_1,1,0.16). \end{aligned} \end{gathered} \]

Lemma 2.1 (the principal remainders [OpenAI-029, (8.6)]). \(R_w(Z)\) and \(R_z(Z)\) are \(O\bigl(Z^{C_*(1)-0.0005}\bigr)\).

Proof. By (1.1) and (1.2), the powers of \(Z\) relative to \(C_*(1)\) are \[ \begin{aligned} &E_*(\sigma_1,0.95,\zeta_1)-C_*(1)\\ &\quad=(1+h_0)\delta_1-0.05\,b\\ &\quad=-0.00664989,\\ &E_*(\sigma_1,1,0.16)-C_*(1)\\ &\quad=\delta_1+h_0\bigl(0.16-\tfrac16\bigr)\\ &\quad=-0.000666566\ldots, \end{aligned} \] both below \(-0.0005\). On these contours \(|\zeta_F^S(6z)|\le\zeta_F(1+6\delta_1)\) or grows polynomially, \(\zeta_F^S(w)\) grows polynomially away from \(w=1\), \(1/L^S(\xi,\eta)\) is bounded, and \(\mathcal H_1\) is bounded; (1.3) makes the integrals converge. \(\square\)

Let \(R_S=\operatorname{Res}_{s=1}\zeta_F^S(s)\), which is positive: it is the residue of \(\zeta_F\) (Corollary 10.4 of Hecke L-functions and the Dedekind zeta function) times \(\prod_{\mathfrak p\in S}(1-q_{\mathfrak p}^{-1})\).

Proposition 2.2 (the double residue [OpenAI-029, (8.7)]). The double residue in (2.1) is \(c_0f_\eta(Z)\), with \[ \begin{gathered} c_0=\frac{R_S^2}6\,M\bigl(\tfrac16\bigr)\,\widehat W_1(1)\ne0,\\ \begin{aligned} f_\eta(Z)&=\frac1{2\pi i}\int_{(2)}Z^{C_*(\xi)}\\ &\quad e^{(\xi-5/6)^2}\frac{\mathcal H_\eta(\xi)}{L^S(\xi,\eta)}\,d\xi , \end{aligned}\quad(2.2) \end{gathered} \] the function \(f_\eta\) of (5.1) of Lesson 5.

Proof. The residue of \(\zeta_F^S(w)\) at \(w=1\) is \(R_S\), and that of \(\zeta_F^S(6z)\) at \(z=\frac16\) is \(R_S/6\). At \(w=1\), \(z=\frac16\), and \(q_1=1\), the remaining factors of (7.1) of Lesson 11 are \(X^{1/3}Z^{\xi-5/6}e^{(\xi-5/6)^2}M(\frac16)\widehat W_1(1)\)\(\mathcal H_1(\xi,1,\frac16)/L^S(\xi,\eta)\), and \(X^{1/3}Z^{\xi-5/6}=Z^{C_*(\xi)}\) because \(\frac a3-\frac56=-\frac8{15}\). The \(\xi\)-line, at \(\operatorname{Re}\xi=\sigma_1\), may be moved to \(\operatorname{Re}\xi=2\) in absolute convergence. Finally \(R_S>0\), \(\widehat W_1(1)=\int_0^\infty W_1(t)\,dt>0\), and \(M(\frac16)\ne0\) by Lemma 1.1 of Lesson 11. \(\square\)

3. The other frequencies

Split the terms \(u\ne1\) of (7.1) of Lesson 11 into dyads \(U\le q_u<2U\), \(U=2^j\), before any shift; each dyad has finitely many terms, and the whole sum converges absolutely on the initial lines. For a set \(\mathcal C\) of \(u\ne1\) in one dyad write \(\mathcal J_{\mathcal C}(Z)\) for its contribution. For \(u\ne1\) the Kummer character \(\kappa_u\) is not principal (Lemma 3.2 of Lesson 3), so \(L^S(w,\chi_\bullet(u))\) is entire, and by Lemma 3.1 of Lesson 13 and the remark after it, with the conductor bound of Lemma 1.1 there, \[ \begin{gathered} \begin{aligned} &|L^S(w,\chi_\bullet(u))|\\ &\quad \le C_\epsilon\,U^{1/2+\epsilon}(2+|\operatorname{Im}w|)^{A_F} \end{aligned}\\ (0.003\le\operatorname{Re}w\le4).\quad(3.1) \end{gathered} \] On \(\operatorname{Re}\xi=\sigma_1\), \(|L^S(\xi,\eta\overline{\chi_\bullet(u)})^{-1}|\)\(\le\prod_{\mathfrak p}(1+q_{\mathfrak p}^{-\sigma_1})\le\zeta_F(\sigma_1)\), uniformly in \(u\) and the height.

The basic contour. For each \(u\ne1\), move the lines, first \(\xi\), then \(z\), then \(w\), to \[ (\operatorname{Re}\xi,\operatorname{Re}w,\operatorname{Re}z)=(\sigma_1,\ 0.003,\ 0.4). \] No pole is crossed: \(\operatorname{Re}6z\ge2.4\), the numerator in \(w\) is entire, the denominator stays in absolute convergence, and the first region of (5.1) of Lesson 12 contains the move. If \(|\mathcal C|\le CU^\alpha\), then by (3.1), \(q_u^{-z}\) and Proposition 5.1 of Lesson 12, \[ \begin{aligned} |\mathcal J_{\mathcal C}(Z)|&\le C_\epsilon\,Z^{E_*(\sigma_1,0.003,0.4)}\\ &\quad \,U^{\alpha+1/2-0.4+\epsilon}.\quad(3.2) \end{aligned} \] The common relative exponent is \[ \begin{aligned} &E_*(\sigma_1,0.003,0.4)-C_*(1)\\ &\quad =10^{-7}+\frac7{300}-0.132601\\ &\quad =-0.109267566\ldots\quad(3.3) \end{aligned} \] For a general \(z\)-line the factor depending on \(U\) is \((Z^{h_0}/U)^{\operatorname{Re}z}\) times \(U^{\alpha+1/2+\epsilon}\): a shift of \(z\) to the right helps exactly when \(U>Z^{h_0}=Z^{0.1}\). This decides the three ranges.

Small frequencies, \(U\le Z^{0.09}\). Use (3.2) with \(\alpha=1\) (Lemma 1.1 of Lesson 13). The factor \(U^{1.1}\) adds at most \(0.09\cdot1.1=0.099\), so the relative exponent is at most \(-0.010267566\ldots\)

Intermediate frequencies, \(Z^{0.09}<U\le Z^{0.10001}\). For large \(Z\) these \(U\) exceed the threshold \(U_0\) of Proposition 4.1 of Lesson 13. For the exceptional frequencies \(u\in\mathcal E(U)\), use (3.2) with \(\alpha=\frac12\): the factor \(U^{0.6}\) adds at most \(0.10001\cdot0.6=0.060006\), leaving at most \(-0.049261566\ldots\) For each \(u\notin\mathcal E(U)\), cut the \(\xi\)-line at \(|\operatorname{Im}\xi|\le U^\rho\) and move this finite segment to \(\operatorname{Re}\xi=0.998\). On the rectangle so swept, (4.1) of Lesson 13 bounds the reciprocal by \(C_\epsilon U^\epsilon\). The new central segment contributes, summed over the at most \(CU\) good frequencies of the dyad, \[ \le C_\epsilon\,Z^{E_*(0.998,0.003,0.4)}\,U^{1+1/2-0.4+\epsilon}, \] and \(E_*\) has coefficient \(1\) in \(\sigma_\xi\), so the move gains \(Z^{0.998-\sigma_1}=Z^{-0.0020001}\). With the factor \(U^{1.1}\le Z^{0.110011}\), the relative exponent is at most \[ \begin{aligned} &-0.002+\frac7{300}-0.132601\\ &\qquad +0.10001\cdot1.1=-\frac{377}{300000}\\ &\quad =-0.001256666\ldots\quad(3.4) \end{aligned} \] The discarded tails of the line \(\operatorname{Re}\xi=\sigma_1\) and the horizontal segments at \(\operatorname{Im}\xi=\pm U^\rho\) are, by (1.3) integrated first in \(z,w\), at most \[ C_{A_1,\epsilon}\,Z^{E_*(\sigma_1,0.003,0.4)}\,U^{1+1/2-0.4+\epsilon-\rho A_1} \] for every \(A_1\): on the horizontal segments the power of \(Z\) is at most its value at \(\sigma_1\) and the reciprocal is bounded by (4.1) of Lesson 13, and on the tails it is bounded by absolute convergence. Since \(U>Z^{0.09}\), the choice \(A_1>1/\rho\) makes these terms smaller than the central ones.

Large frequencies, \(U>Z^{0.10001}\). Here \(Z^{h_0}/U\le Z^{-0.00001}\). Keep \(\operatorname{Re}\xi=\operatorname{Re}w=4\) and move \(z\) to the fixed line \(\operatorname{Re}z=R_0=350000\); the move stays in the Euler region, crosses no pole, and \(M\) is holomorphic there. Both \(L\)-functions are bounded by their Euler products, and with \(E_*(4,4,R_0)=3.849+0.1R_0\), \[ |\mathcal J_{\mathcal C}(Z)|\le C_\epsilon\,Z^{3.849+0.1R_0}\,U^{1-R_0+\epsilon}. \] The dyads sum geometrically; at the lowest one the power of \(Z\) is \[ \begin{aligned} &3.849+0.1R_0\\ &\qquad +0.10001(1-R_0+\epsilon)\\ &\quad =0.44901+0.10001\epsilon ,\quad(3.5) \end{aligned} \] which is below \(C_*(1)-0.01=0.45666\ldots\) The constants on the line \(\operatorname{Re}z=R_0\) are large but do not depend on \(Z\) or \(U\).

Proposition 3.1 (the nonprincipal frequencies [OpenAI-029, (8.12)]). For large \(Z\), \[ \begin{aligned} &\sum_{j\ge0}\mathcal J_{\{u\ne1:\,2^j\le q_u<2^{j+1}\}}(Z)\\ &\quad =O\bigl(Z^{C_*(1)-0.0004}\bigr).\quad(3.6) \end{aligned} \]

Proof. Choose the losses \(\epsilon\) so small that their total effect in the ranges \(U\le Z^{0.10001}\), including the \(O(\log Z)\) dyads, is below \(10^{-4}\) in the exponent of \(Z\). The smallest saving among the ranges is \(377/300000>0.0012\), in (3.4); the others are larger, and (3.5) gives a saving above \(0.01\). Each dyadic integral arose from its absolutely convergent initial form, and these bounds justify summing them. \(\square\)

4. Statement (P2) and the zero-free strip

Theorem 4.1 ((P2) [OpenAI-029, Section 8.3]). For large \(Z\), \[ \begin{aligned} I(Z^a,Z^b,Z)&=c_0f_\eta(Z)\\ &\quad+O\bigl(Z^{7/15-0.0004}\bigr). \end{aligned} \]

Proof. Proposition 7.1 of Lesson 11 writes the probe as the sum over \(u\) of the triple integrals; (2.1), Lemma 2.1, Proposition 2.2 and Proposition 3.1 bound everything except \(c_0f_\eta(Z)\) by \(O(Z^{C_*(1)-0.0004})\), and \(C_*(1)=\frac7{15}\). \(\square\)

Corollary 4.2 (the zero-free strip). For every cyclotomic field \(F\supseteq\mu_{12}\) and every finite-order Hecke character \(\eta\) of \(F\), \(L_F(s,\eta)\) has no zero in \(\operatorname{Re}s>1-\frac1{20000}\); in particular none in \(\operatorname{Re}s>1-10^{-6}\). This is Theorem 5.3 of Lesson 1.

Proof. Statement (P1) is Proposition 4.1 of Lesson 11, (P2) is Theorem 4.1, and (H) is Corollary 5.2 of Lesson 12. Corollary 6.2 of Lesson 5 deduces the corollary from these three statements. \(\square\)

The exponents used, before the arbitrarily small losses, are collected below; all are relative to \(C_*(1)=\frac7{15}\).

Contribution Relative exponent
Probe bound from reflection (P1) \(-1/6000=-0.000166\ldots\)
Principal remainder \(R_w\) \(-0.00664989\)
Principal remainder \(R_z\) \(-0.000666566\ldots\)
Frequencies \(U\le Z^{0.09}\) \(-0.010267566\ldots\)
Exceptional frequencies \(U\le Z^{0.10001}\) \(-0.049261566\ldots\)
Good central frequencies \(U\le Z^{0.10001}\) \(-0.001256666\ldots\)
Frequencies \(U>Z^{0.10001}\) \(-0.017656666\ldots\)

Comparing the first row with the others: the upper bound of (P1) and the evaluation of (P2) agree up to \(Z^{7/15-1/6000+\epsilon}\), which is what forces \(f_\eta(Z)\ll Z^{7/15-1/20000}\) and, by Proposition 6.1 of Lesson 5, the absence of zeros. The width \(\frac1{20000}\) depends on none of \(F\), \(S\), \(\eta\); the constants and the onset of "large \(Z\)" do.

5. Exercises

5.1. Derive (1.1) from the factors \(X^{1/2-z}Z^{\xi+z-1}Y^{w-1}\) with \(X=Z^a\), \(Y=Z^b\).

5.2. Check (3.3): \(h_0(0.4-\frac16)=\frac7{300}\) and \(b(0.003-1)=-0.132601\).

5.3. Verify (3.5), and find the smallest \(R_0\) for which the exponent (3.5) at \(\epsilon=0\) is below \(\frac7{15}-0.01\).

5.4. Explain why the principal frequency cannot be treated like the others: what would go wrong if one moved its \(w\)-line to \(0.003\)?

5.5. Show that \(\frac{933}{2000}=\frac7{15}-\frac1{6000}\), and that \(\frac1{6000}-10^{-5}>\frac1{20000}\).

6. Solutions

5.1. The modulus is \(Z^{a(1/2-\sigma_z)}Z^{\sigma_\xi+\sigma_z-1}Z^{b(\sigma_w-1)}\)\(=Z^{a/2+\sigma_\xi-1+(1-a)\sigma_z+b(\sigma_w-1)}\), and \(1-a=h_0\).

5.2. \(0.4-\frac16=\frac7{30}\), times \(\frac1{10}\) is \(\frac7{300}\); \(0.133\cdot0.997=0.132601\).

5.3. \(3.849+0.1R_0+0.10001\)\(-0.10001R_0=3.94901\)\(-0.00001R_0\); at \(R_0=350000\) this is \(0.44901\). The condition \(3.94901-0.00001R_0<0.456666\ldots\) holds for \(R_0>349234.33\ldots\), so the smallest integer is \(349235\); the chosen value leaves room.

5.4. At \(w=1\) the numerator \(\zeta_F^S(w)\) has a pole; moving the line past it produces the residue, which carries the main term. For \(u\ne1\) the numerator \(L^S(w,\chi_\bullet(u))\) is entire and no residue arises.

5.5. \(\frac7{15}=\frac{2800}{6000}\) and \(\frac{933}{2000}=\frac{2799}{6000}\). Then \(\frac1{6000}-\frac1{100000}=\frac{47}{300000}>\frac{15}{300000}=\frac1{20000}\).

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