# Uniform complete splitting in Kummer fields *Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).* This lesson proves Proposition 5.2 of [*Artin's conjecture and the index of a prime*](artins-conjecture-and-the-index-of-a-prime.md): the number of primes \(p\in(x,2x]\) that split completely in the Kummer field \(K_q=\mathbb Q(\mu_q,a^{1/q})\) is \(O\bigl(x/(q(q-1)\log x)+x^{1-\delta_0}\bigr)\), uniformly for primes \(q\le\exp((\log x)^{0.3})\). The field \(K_q\) is not cyclotomic, but after adjoining \(\mu_{12q}\) it becomes an abelian extension of the cyclotomic field \(\mathbb Q(\mu_{12q})\), to which the zero-free strip of [*The zero-free strip*](the-zero-free-strip.md) applies; the strip then descends to \(\zeta_{K_q}\). A smoothed explicit formula, whose error depends on the field only through its degree and discriminant, converts the zero-free strip into the count. The uniformity in \(q\) rests on a zero count for Dedekind zeta functions with absolute constants, used as a named result. **Prerequisites.** [*Artin's conjecture and the index of a prime*](artins-conjecture-and-the-index-of-a-prime.md), Section 3 and Proposition 5.2; [*The zero-free strip*](the-zero-free-strip.md), Corollary 4.2; [*Cyclotomic fields*](course:NT-ANT/cyclotomic-fields), Theorem 12.2; [*The different and the discriminant*](course:NT-ANT/the-different-and-the-discriminant), (1)–(4), Proposition 14.2 and Theorem 14.3; [*Discriminants and integral bases*](course:NT-ANT/discriminants-and-integral-bases), Proposition 2.2; [*Hilbert's theorem 90 and Kummer theory*](course:NT-CFT/hilberts-theorem-90-and-kummer-theory), Theorem 3.3; [*Artin L-functions, conductors and discriminants*](course:NT-CFT/artin-l-functions-conductors-and-discriminants), Proposition 21.1 and Theorem 21.2; [*Explicit formulas for Dirichlet L-functions and primes in progressions*](course:NT-EXPF/NT-EXPF-02), Section 6 and Theorem 6.1; [*The Gamma function and Stirling's formula*](course:NT-ZETA/NT-ZETA-03), Section 5. ## 1. Degree, discriminant and the zero-free strip Fix an integer \(a\) with \(|a|>1\), and recall \(E_q=\mathbb Q(\mu_q)\), \(K_q=\mathbb Q(\mu_q,a^{1/q})\) from Lesson 1. Put \(\delta_0=10^{-6}\). **Lemma 1.1** (the Kummer fields [OpenAI-029, Lemma 9.1]). For every sufficiently large prime \(q\), the degree \(n_q=[K_q:\mathbb Q]\) and the absolute discriminant \(D_q\) satisfy \[ \begin{gathered} n_q=q(q-1),\\ D_q\le q^{q(q-2)}\bigl(q^q|a|^{q-1}\bigr)^{q-1},\\ \log D_q+n_q\le C_a\,n_q\log(2q),\quad(1.1) \end{gathered} \] and \[ \begin{gathered} \zeta_{K_q}(s)\ne0\\ (\operatorname{Re}s>1-\delta_0).\quad(1.2) \end{gathered} \] *Proof.* *Degree.* Choose a prime \(\ell\mid a\) and take \(q\ne\ell\), \(q>v_\ell(a)\). By Theorem 12.2 of [*Cyclotomic fields*](course:NT-ANT/cyclotomic-fields), \(|d_{E_q}|=q^{q-2}\), so \(\ell\) is unramified in \(E_q\) and every prime of \(E_q\) above \(\ell\) gives \(a\) the valuation \(v_\ell(a)\), which is not divisible by \(q\). Hence \(a\) is not a \(q\)th power in \(E_q\). Since \(\mu_q\subset E_q\), Kummer theory (Theorem 3.3 of [*Hilbert's theorem 90 and Kummer theory*](course:NT-CFT/hilberts-theorem-90-and-kummer-theory)) makes \(K_q/E_q\) cyclic, with \(\sigma\mapsto\sigma(a^{1/q})/a^{1/q}\) embedding its Galois group in \(\mu_q\); the degree divides \(q\) and is not \(1\), so it is \(q\), and \(n_q=q(q-1)\). *Discriminant.* Let \(\alpha=a^{1/q}\), with minimal polynomial \(f(T)=T^q-a\) over \(E_q\), and let \(B\) be the ring of integers of \(K_q\), \(A\) that of \(E_q\). The trace-dual of \(A[\alpha]\) is \(f'(\alpha)^{-1}A[\alpha]\) (Euler's lemma, in the proof of Proposition 14.2 of [*The different and the discriminant*](course:NT-ANT/the-different-and-the-discriminant)), and it contains the trace-dual \(J_{B/A}\) of the larger ring \(B\), by (1) there. Hence \(f'(\alpha)J_{B/A}\subseteq A[\alpha]\subseteq B\), that is, \(f'(\alpha)\in\mathfrak D_{B/A}\) by (4) there. Taking norms with (15) of Theorem 14.3, the relative discriminant divides \(N_{K_q/E_q}(f'(\alpha))A\), and by Proposition 2.2 of [*Discriminants and integral bases*](course:NT-ANT/discriminants-and-integral-bases) this is the discriminant of \(f\), \(\pm q^qa^{q-1}\). The tower formula (16) of Theorem 14.3 then gives \[ \begin{aligned} D_q&=|d_{E_q}|^{q}\,\operatorname N_{E_q/\mathbb Q}(\mathfrak d_{K_q/E_q})\\ &\le q^{q(q-2)}\bigl(q^q|a|^{q-1}\bigr)^{q-1}, \end{aligned} \] and taking logarithms gives the last bound in (1.1). *The zero-free strip.* Put \(F_q=\mathbb Q(\mu_{12q})\), a cyclotomic field containing \(\mu_{12}\), and \(\widetilde K_q=K_qF_q\). The extension \(\widetilde K_q/F_q\) is cyclic, being generated by \(a^{1/q}\) over a field containing \(\mu_q\). The extension \(\widetilde K_q/K_q\) is abelian, since restriction embeds its Galois group in the abelian group \(\operatorname{Gal}(F_q/\mathbb Q)\). By (4) of Proposition 21.1 and Theorem 21.2 of [*Artin L-functions, conductors and discriminants*](course:NT-CFT/artin-l-functions-conductors-and-discriminants), for an abelian extension the Dedekind zeta function of the top field is the product of the primitive Hecke \(L\)-functions of the characters of its Galois group: \[ \begin{gathered} \zeta_{\widetilde K_q}(s)=\prod_{\eta}L_{F_q}(s,\eta),\\ \frac{\zeta_{\widetilde K_q}(s)}{\zeta_{K_q}(s)}=\prod_{\eta'\ne1}L_{K_q}(s,\eta'), \end{gathered} \] with \(\eta\) running over the characters of \(\operatorname{Gal}(\widetilde K_q/F_q)\) and \(\eta'\) over the nontrivial characters of \(\operatorname{Gal}(\widetilde K_q/K_q)\), viewed as finite-order Hecke characters. Every \(L_{F_q}(s,\eta)\) has no zero in \(\operatorname{Re}s>1-\delta_0\) by Corollary 4.2 of Lesson 14, so neither has \(\zeta_{\widetilde K_q}\). Every \(L_{K_q}(s,\eta')\) is entire, \(\eta'\) being nonprincipal. A zero of \(\zeta_{K_q}\) in the half-plane would therefore be a zero of \(\zeta_{\widetilde K_q}\). \(\square\) The width \(\delta_0\) does not depend on \(q\); everything else that enters below depends on \(K_q\) only through \(n_q\) and \(D_q\). ## 2. A uniform zero count For a number field \(K\) of degree \(n_K\) and absolute discriminant \(d_K\), let \(N_K(T)\) be the number of zeros \(\rho=\beta+i\gamma\) of \(\zeta_K\) with \(0<\beta<1\) and \(|\gamma|\le T\), counted with multiplicity. **Named result N4** (Hasanalizade, Shen and Wong [HSW, Corollary 1.2]). For every number field \(K\) and every \(T\ge1\), \[ \begin{aligned} &\Bigl|N_K(T)-\frac T\pi\log\Bigl(d_K\Bigl(\frac T{2\pi e}\Bigr)^{n_K}\Bigr)\Bigr|\\ &\quad \le0.228\,(\log d_K+n_K\log T)\\ &\qquad +23.108\,n_K+4.520 . \end{aligned} \] It is used without proof in the programme. **Corollary 2.1.** There is an absolute constant \(C\) such that for every number field \(K\) and every real \(t\), \[ \begin{aligned} &\#\{\rho:\ t\le\gamma\le t+1\}\\ &\le C\bigl(\log d_K+n_K\log(2+|t|)\bigr).\quad(2.1) \end{aligned} \] *Proof.* For \(|t|\ge2\) the zeros counted lie in \(|t|-1\le|\gamma|\le|t|+1\), so their number is at most \(N_K(|t|+1)-N_K(|t|-1-\epsilon)\) for small \(\epsilon>0\). By N4 the main terms differ by at most \(\frac2\pi\log(d_K((|t|+2)/(2\pi))^{n_K})+O(1)\), and the errors add at most twice the right side of N4 at height \(|t|+1\). For \(|t|<2\) use \(N_K(3)\). Both are at most \(C(\log d_K+n_K\log(2+|t|))\). \(\square\) ## 3. A smooth prime-ideal estimate **Lemma 3.1** (smooth prime ideals [OpenAI-029, Lemma 9.2]). Fix \(0<\delta<\frac12\) and a nonnegative \(\phi\in C_c^\infty([\frac12,3])\). Let \(K\) be a number field of degree \(n\) and absolute discriminant \(D\) such that \(\zeta_K\) has no zero in \(\operatorname{Re}s>1-\delta\). For every \(x\ge2\), \[ \begin{aligned} &\Theta_{K,\phi}(x)\\ &\quad=\sum_{\mathfrak p}\sum_{m\ge1}(\log\operatorname N\mathfrak p)\,\phi\Bigl(\frac{\operatorname N\mathfrak p^m}x\Bigr)\\ &\quad\le C_{\phi}\bigl(x+x^{1-\delta}(\log D+n)\bigr),\quad(3.1) \end{aligned} \] with \(C_\phi\) independent of \(K\) and \(x\); the sum runs over the nonzero prime ideals of \(K\). *Proof.* Apply Theorem 6.1 of [*Explicit formulas for Dirichlet L-functions and primes in progressions*](course:NT-EXPF/NT-EXPF-02) to the trivial Hecke character of \(K\), for which \(L=\zeta_K\), \(Q=D\), the completed function has its two poles at \(0\) and \(1\), and \(G(s)=\Gamma_{\mathbb R}(s)^{r_1}\Gamma_{\mathbb C}(s)^{r_2}\), with the test function \(f(y)=\phi(y/x)\). Its Mellin transform is \(H(s)=\int_0^\infty f(y)y^s\,dy/y=x^s\Phi(s)\), with \(\Phi(s)=\int_0^\infty\phi(t)t^s\,dt/t\) entire and, by repeated integration by parts, \(|\Phi(\sigma+it)|\le C_{\phi,A}(1+|t|)^{-A}\) on \(0\le\sigma\le1\). Since \(x\ge2\), \(f(1)=0\), and \(f^\sharp(y)=y^{-1}f(y^{-1})\) vanishes for \(y\ge1\); so the dual prime sum and the conductor term disappear, and (6.3) there reads \[ \begin{aligned} &\Theta_{K,\phi}(x)=H(1)+H(0)-\sum_\rho H(\rho)\\ &\quad+\frac1{2\pi}\int_{\mathbb R}2\operatorname{Re}\frac{G'}G\bigl(\tfrac12+it\bigr)\\ &\qquad\times H\bigl(\tfrac12+it\bigr)\,dt , \end{aligned} \] the sum running over the zeros of the completed function, which are the zeros of \(\zeta_K\) with \(0<\operatorname{Re}\rho<1\). Here \(H(1)=x\Phi(1)\) and \(H(0)=\Phi(0)\). Every zero has \(\operatorname{Re}\rho\le1-\delta\), so \(|H(\rho)|\le x^{1-\delta}C_{\phi,3}(1+|\gamma|)^{-3}\); grouping the zeros in unit intervals of height and using (2.1), \[ \begin{aligned} &\sum_\rho|H(\rho)|\le C\,x^{1-\delta}\sum_{k\in\mathbb Z}(1+|k|)^{-3}\\ &\qquad\times\bigl(\log D+n\log(2+|k|)\bigr)\\ &\quad\le C\,x^{1-\delta}(\log D+n). \end{aligned} \] Finally \(\frac{G'}G\) is a sum of \(r_1+r_2\le n\) logarithmic derivatives of \(\Gamma_{\mathbb R}\) and \(\Gamma_{\mathbb C}\), each \(O(\log(2+|t|))\) on \(\operatorname{Re}s=\frac12\) (Section 5 of [*The Gamma function and Stirling's formula*](course:NT-ZETA/NT-ZETA-03)); so the integral is at most \(C\,x^{1/2}n\)\(\int(\log(2+|t|))|\Phi(\frac12+it)|\,dt\)\(\le C_\phi x^{1/2}n\le C_\phi x^{1-\delta}n\). \(\square\) Only the first-power terms of prime ideals of degree one matter in the application, but the estimate keeps all terms: they are nonnegative. ## 4. Proof of the uniform splitting bound *Proof of Proposition 5.2 of Lesson 1.* Fix a nonnegative \(\phi\in C_c^\infty([\frac12,3])\) with \(\phi\ge1\) on \([1,2]\). Let \(q\) be large enough for Lemma 1.1, and apply Lemma 3.1 to \(K_q\) with \(\delta=\delta_0\), which is allowed by (1.2). A rational prime \(p\in(x,2x]\) with \(p\nmid aq\) that splits completely in \(K_q\) has \(n_q\) prime ideals above it, each of norm \(p\), and their first-power terms contribute at least \(n_q\log p\ge n_q\log x\) to the nonnegative sum \(\Theta_{K_q,\phi}(x)\). The primes dividing \(aq\) are \(O_a(1)+1\) in number. With \(L=\log x\), (3.1) and (1.1) give \[ \begin{aligned} &\#\{p\ \text{prime}:x