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Sextic reciprocity and Kummer characters

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

The coefficient sums of Lesson 5 contain sextic residue symbols \(\chi_{\mathfrak A}(m)\), whose modulus \(\mathfrak A\) varies with the summation and whose argument \(m\) varies with the row. Two facts make such sums tractable. First, for allowed generators in the sense of S-integers and sixth-power classes the sextic reciprocity law holds exactly, with no supplementary factor; this lets the symbol be read in either direction. Second, for a fixed argument \(u\) the symbol \(\mathfrak a\mapsto\chi_{\mathfrak a}(u)\) is a Hecke character of finite order, a Kummer character, whose conductor is bounded by the norm of the radical of \(u\). This lesson proves both, and parametrizes the frequencies \(h\in R\setminus\{0\}\) that will appear after Poisson summation by a Kummer character and a sixth power.

Prerequisites. S-integers and sixth-power classes, Lemma 2.1, (3.1)–(3.3), Lemma 3.1 and Lemma 4.1; Power residue symbols and reciprocity laws, Propositions 23.1 and 23.2 and Theorem 23.3; Hilbert's theorem 90 and Kummer theory, Theorem 3.3; The global reciprocity law, Theorem 16.4; Quasi-characters and Hecke characters, Propositions 6.1 and 6.4; Hensel's lemma, squares and roots of unity in p-adic fields, Corollary 1.2; The multiplicative group of a local field, Corollary 4.2; Counting ideals of bounded norm, Theorem 15.3.

1. Sextic residue symbols

Keep the field \(F\supseteq\mu_{12}\), the set \(S\), the ring \(R=\mathcal O_S\) and the subgroup \(B\) of Lesson 2. Let \(\mathfrak p\) be an exterior prime, that is, a prime ideal of \(R\), with norm \(q_{\mathfrak p}\). Since \(\mathfrak p\nmid6\), reduction modulo \(\mathfrak p\) is injective on \(\mu_{12}\), so \(12\mid q_{\mathfrak p}-1\). The sextic residue character is \[ \begin{gathered} \begin{aligned} \chi_{\mathfrak p}(x)&\equiv x^{(q_{\mathfrak p}-1)/6}\\ &\quad \pmod{\mathfrak p}\quad(x\in R\setminus\mathfrak p), \end{aligned}\\ \chi_{\mathfrak p}(x)=0\quad(x\in\mathfrak p),\quad(1.1) \end{gathered} \] with values in \(\mu_6\subset F\); it is the power residue symbol \((x/\mathfrak p)_6\) of Proposition 23.1 of Power residue symbols and reciprocity laws. For every integer \(j\) the notation \(\chi_{\mathfrak p}(x)^j\) means the usual power on units and \(0\) on nonunits, also when \(6\mid j\); this unit mask is used throughout the course. For an exterior ideal \(\mathfrak a=\prod_{\mathfrak p}\mathfrak p^{k_{\mathfrak p}}\) put \[ \begin{gathered} \chi_{\mathfrak a}(x)=\prod_{\mathfrak p\mid\mathfrak a}\chi_{\mathfrak p}(x)^{k_{\mathfrak p}},\\ \chi_R(x)=1,\quad(1.2) \end{gathered} \] with the same convention for powers of \(\chi_{\mathfrak a}\). Since \(12\mid q_{\mathfrak p}-1\), the exponent \((q_{\mathfrak p}-1)/6\) is even and \[ \chi_{\mathfrak p}(-1)=1 . \tag{1.3} \]

2. Reciprocity without supplementary factors

Proposition 2.1 ([OpenAI-029, (3.3)–(3.5)]). Let \(\mathfrak p\ne\mathfrak p'\) be exterior primes with allowed generators \(p,p'\). Then \[ \chi_{\mathfrak p}(p')=\chi_{\mathfrak p'}(p). \tag{2.1} \] Let \(\mathfrak a\) be an exterior ideal with allowed generator \(a\), let \(m\in R\setminus\{0\}\) with \(mR=\mathfrak b\), and write \(m=\epsilon b\) with \(b\) an allowed generator of \(\mathfrak b\) and \(\epsilon\in U_S\). Then \[ \chi_{\mathfrak a}(m)=\chi_{\mathfrak a}(\epsilon)\,\chi_{\mathfrak b}(a), \tag{2.2} \] both sides being zero when \(\mathfrak a\) and \(\mathfrak b\) have a common prime, and for \(\epsilon\in U_S\) \[ \chi_{\mathfrak a}(\epsilon)=H_S(\epsilon,a)^{-1}. \tag{2.3} \] In particular \(\chi_{\mathfrak a}(\epsilon)\) depends on \(\mathfrak a\) only through the class \([a]\in B\).

Proof. Apply Hilbert's product formula (Theorem 23.3 of Power residue symbols and reciprocity laws) to the symbols \((p,p')_{6,v}\) of (3.1) of Lesson 2. At an exterior prime other than \(\mathfrak p,\mathfrak p'\) both arguments are units, and the symbol is \(1\). At \(\mathfrak p\), the tame formula (3.2) of Lesson 2 with \(p'\) a unit and \(p\) a uniformizer gives \((p',p)_{6,\mathfrak p}=\chi_{\mathfrak p}(p')\), hence \((p,p')_{6,\mathfrak p}=\chi_{\mathfrak p}(p')^{-1}\) by Lemma 3.1 of Lesson 2. At \(\mathfrak p'\) the same formula gives \((p,p')_{6,\mathfrak p'}=\chi_{\mathfrak p'}(p)\). The product of the symbols at \(S\) is \(H_S(p,p')=1\), because \([p],[p']\in B\) and \(B\) is isotropic. The product formula therefore reads \(\chi_{\mathfrak p}(p')^{-1}\chi_{\mathfrak p'}(p)=1\), which is (2.1).

For (2.2), allowed generators are unique up to \(U_S^6\), and the product of allowed generators of the prime factors of \(\mathfrak b\) is an allowed generator of \(\mathfrak b\); since \(\chi_{\mathfrak p}\) is trivial on sixth powers of units, we may take \(b\) to be that product, and likewise \(a\). Multiplicativity of \(\chi_{\mathfrak a}\) in its argument and of \(\chi_{\mathfrak a}\) in \(\mathfrak a\) reduce \(\chi_{\mathfrak a}(b)=\chi_{\mathfrak b}(a)\) to (2.1) for pairs of distinct primes; if a prime divides both ideals, both sides vanish by the unit mask. Finally \(\chi_{\mathfrak a}(m)=\chi_{\mathfrak a}(\epsilon)\chi_{\mathfrak a}(b)\).

For (2.3), apply the product formula to \((\epsilon,a)_{6,v}\). At an exterior prime \(\mathfrak p\) with \(v_{\mathfrak p}(a)=k\), write \(a=p^ku\) with a local unit \(u\); then \((\epsilon,a)_{6,\mathfrak p}=(\epsilon,p)_{6,\mathfrak p}^k(\epsilon,u)_{6,\mathfrak p}=\chi_{\mathfrak p}(\epsilon)^k\) by (3.2) of Lesson 2. The product over exterior primes is \(\chi_{\mathfrak a}(\epsilon)\), the product over \(S\) is \(H_S(\epsilon,a)\), and the total is \(1\). \(\square\)

Without the normalization of generators, (2.1) acquires the factor \(\chi_{\mathfrak p}(\epsilon)\) of (2.3) for each unit adjustment; this is the supplementary factor that the choice of \(B\) removes (Exercise 5.2).

3. Kummer characters

Lemma 3.1 (Kummer and class characters [OpenAI-029, Lemma 3.3]).

  1. For \(u\in R\setminus\{0\}\) there is a finite-order Hecke character \(\kappa_u\) of \(F\) such that \(\kappa_u(\mathfrak p)=\chi_{\mathfrak p}(u)\) for every exterior prime \(\mathfrak p\nmid u\). Its local component at a place \(v\) is \(y\mapsto(u,y)_{6,v}\).
  2. Let \(\mathfrak f_u\) be the conductor of \(\kappa_u\). Then \[ \begin{aligned} \operatorname N\mathfrak f_u&\le C_{F,S}\operatorname N\operatorname{rad}(uR)\\ &\le C_{F,S}\,q_u ,\quad(3.1) \end{aligned} \] and the same holds, with another constant, for the conductor of \(\eta\kappa_u\) and of \(\eta\overline{\kappa_u}\).
  3. \(\kappa_u=\kappa_v\) if and only if \(u/v\in F^{\times6}\).
  4. Every character \(\theta_B\) of the finite group \(B\) extends to a finite-order Hecke character \(\theta\), unramified outside \(S\), with \(\theta(\mathfrak a)=\theta_B([a])\) for every exterior ideal \(\mathfrak a\) with allowed generator \(a\).

The Kummer character \(\chi_\bullet(u)\) is the ideal function \(\mathfrak a\mapsto\chi_{\mathfrak a}(u)\); it equals \(\kappa_u\) on exterior ideals prime to \(u\) and vanishes on those meeting \(u\).

Proof. 1. Let \(L=F(u^{1/6})\). Since \(\mu_6\subset F\), Kummer theory (Theorem 3.3 of Hilbert's theorem 90 and Kummer theory) makes \(L/F\) cyclic and gives the character \(\kappa_u(\sigma)=\sigma(u^{1/6})/u^{1/6}\) of its Galois group; its kernel, as a function of \(u\), is \(F^{\times6}\). Composing with the global reciprocity map (Theorem 16.4 of The global reciprocity law), whose local components are the maps \(\operatorname{Art}_v\), gives a character of the idèle class group of finite order, whose component at \(v\) is \(y\mapsto\operatorname{Art}_v(y)(u^{1/6})/u^{1/6}=(u,y)_{6,v}\). At an exterior prime \(\mathfrak p\nmid u\), the extension is unramified and the image of a uniformizer \(\varpi\) is the Frobenius; by (3.2) of Lesson 2, \((u,\varpi)_{6,\mathfrak p}\equiv u^{(q_{\mathfrak p}-1)/6}\), so \(\kappa_u(\mathfrak p)=\chi_{\mathfrak p}(u)\) (equivalently, Proposition 23.2 of Power residue symbols and reciprocity laws).

2. At an exterior prime \(\mathfrak p\), every principal unit is a sixth power: \(X^6-y\) with \(y\equiv1\pmod{\mathfrak p}\) has the simple residue root \(1\), because \(6\) is a unit, and Corollary 1.2 of Hensel's lemma, squares and roots of unity in p-adic fields lifts it. So the local component \(y\mapsto(u,y)_{6,\mathfrak p}\) is trivial on \(1+\mathfrak p\), and the conductor exponent is at most \(1\) (Proposition 6.1 of Quasi-characters and Hecke characters). If \(u\) is a unit at \(\mathfrak p\), the tame formula shows that the symbol of two units is \(1\), so the component is unramified. At a place \(v\in S\), the component depends only on the class of \(u\) in the finite group \(F_v^\times/F_v^{\times6}\); the finitely many possible local characters have a largest conductor. Hence \(\operatorname N\mathfrak f_u\) is at most a constant times the product of \(q_{\mathfrak p}\) over exterior primes dividing \(u\), which is \(\operatorname N\operatorname{rad}(uR)\le\operatorname N(uR)=q_u\). Multiplying by \(\eta\), whose conductor is supported in \(S\), changes only the components at \(S\), again among finitely many possibilities.

3. The reciprocity map of Theorem 16.4 maps the idèle class group onto \(\operatorname{Gal}(L/F)\) for every finite abelian \(L/F\), so two Hecke characters obtained by composing Galois characters with it are equal exactly when the Galois characters are equal. Apply this to the compositum of \(F(u^{1/6})\) and \(F(v^{1/6})\): \(\kappa_u=\kappa_v\) if and only if \(\sigma(u^{1/6})/u^{1/6}=\sigma(v^{1/6})/v^{1/6}\) for every \(\sigma\), that is, \((u/v)^{1/6}\) is fixed by every \(\sigma\), that is, \(u/v\in F^{\times6}\).

4. Extend \(\theta_B\) to \(\mathcal V=E\oplus B\) by \(\theta_S(eb)=\theta_B(b)\), and view it as a character of \(F_S^\times\) trivial on \(F_S^{\times6}\). It is continuous: \(F_v^{\times6}\) is open at each finite place (Corollary 4.2 of The multiplicative group of a local field) and equal to \(F_v^\times\) at each complex place. It is trivial on \(U_S\), whose image lies in \(E\). For an exterior ideal \(\mathfrak a=aR\) with \(a\) allowed, put \(\theta(\mathfrak a)=\theta_S(a)\); this is independent of the allowed generator and multiplicative. Define an idèle character by \(\theta_S^{-1}\) on \(F_S^\times\) and, at each exterior prime, the unramified character sending a uniformizer to \(\theta(\mathfrak p)\). For \(x\in F^\times\), write \(x=\epsilon a\) with \(a\) an allowed generator of \(xR\) and \(\epsilon\in U_S\); the value of the idèle character at \(x\) is \(\theta(xR)\theta_S(x)^{-1}=\theta_S(a)\theta_S(\epsilon a)^{-1}=1\). So it is a Hecke character of finite order, unramified outside \(S\), with the stated values. \(\square\)

The frequency decomposition

Choose a complete set \(\mathcal U\) of representatives, modulo \(U_S^6\), of the nonzero \(S\)-integers whose valuations at all exterior primes lie in \(\{0,1,\dots,5\}\), with \(1\in\mathcal U\). For \(u\in\mathcal U\) put \(q_u=|u|_S\).

Lemma 3.2 (frequencies [OpenAI-029, (3.7)]). With allowed generators \(r\), the map \[ \begin{gathered} (u,\mathfrak r)\longmapsto[ur^6]\\ \begin{aligned} &\text{from}\ \mathcal U\times\{\text{exterior ideals}\}\\ &\quad \ \text{to}\ (R\setminus\{0\})/U_S^6 \end{aligned}\quad(3.2) \end{gathered} \] is a bijection. Among \(u\in\mathcal U\), only \(u=1\) has \(\kappa_u\) principal, and distinct \(u\) give distinct characters \(\kappa_u\), also after multiplication by \(\eta\) or by \(\overline\eta\).

Proof. Given \(h\ne0\), write \(v_{\mathfrak p}(h)=6m_{\mathfrak p}+j_{\mathfrak p}\) with \(0\le j_{\mathfrak p}\le5\) at each exterior prime, and put \(\mathfrak r=\prod\mathfrak p^{m_{\mathfrak p}}\) with allowed generator \(r\). Then \(u_0=h/r^6\) is an \(S\)-integer with exterior valuations \(j_{\mathfrak p}\); replacing it by its representative in \(\mathcal U\) changes \(u_0r^6\) only by \(U_S^6\). Conversely the exterior valuations of a class \([ur^6]\) determine \(\mathfrak r\) and the valuations of \(u\), and then \(u\) modulo \(U_S^6\). Changing \(r\) within its allowed class does not change the image. So (3.2) is bijective.

If \(\kappa_u\) is principal, Lemma 3.1(3) gives \(u=t^6\) with \(t\in F\). The exterior valuations of \(u\) are divisible by \(6\) and lie in \([0,5]\), so they vanish, and so do those of \(t\): \(t\in U_S\), \(u\in U_S^6\), and \(u=1\). If \(\kappa_u=\kappa_{u'}\), then \(u/u'\) is a sixth power; the differences of exterior valuations are divisible by \(6\) and lie in \([-5,5]\), hence vanish; so \(u/u'\) is the sixth power of an element with vanishing exterior valuations, an \(S\)-unit, and \(u/u'\in U_S^6\), so \(u=u'\). A common factor \(\eta\) cancels from an equality of twisted characters. \(\square\)

4. A count of frequencies

Lemma 4.1 ([OpenAI-029, (8.1)]). For \(U\ge1\), \[ \begin{gathered} \#\{u\in\mathcal U:U\le q_u<2U\}\ll U,\\ \operatorname N\mathfrak f_{\eta\overline{\kappa_u}}\ll U\quad(U\le q_u<2U). \end{gathered} \]

Proof. The ideal \(uR\) determines \(u\) up to the finite group \(U_S/U_S^6\), and the number of exterior ideals of norm less than \(2U\) is \(O(U)\): exterior ideals correspond to integral ideals of \(\mathcal O_F\) of the same norm, and (16) of Theorem 15.3 of Counting ideals of bounded norm bounds their number. The conductor bound is (3.1). \(\square\)

5. Exercises

5.1. Show that for \(\zeta\in\mu_6\) and an exterior prime \(\mathfrak p\), \(\chi_{\mathfrak p}(\zeta)=\zeta^{(q_{\mathfrak p}-1)/6}\), and deduce \(\chi_{\mathfrak p}(\zeta_{12}^2)=1\) whenever \(72\mid q_{\mathfrak p}-1\).

5.2. Let \(p\) be an allowed generator of \(\mathfrak p\) and \(\epsilon\in U_S\). Show that \(\chi_{\mathfrak p'}(\epsilon p)=\chi_{\mathfrak p}(p')\,H_S(\epsilon,p')^{-1}\) for every exterior prime \(\mathfrak p'\ne\mathfrak p\) with allowed generator \(p'\).

5.3. Show that \(\kappa_{uv}=\kappa_u\kappa_v\), and that \(\kappa_{\epsilon^6}\) is principal for \(\epsilon\in U_S\).

5.4. Show that if \(u\in R\setminus\{0\}\) is squarefree as an exterior ideal (no exterior prime divides it twice), then the conductor of \(\kappa_u\) is divisible by every exterior prime dividing \(u\).

6. Solutions

5.1. \(\zeta\) is a unit at \(\mathfrak p\) and \(\chi_{\mathfrak p}(\zeta)\equiv\zeta^{(q_{\mathfrak p}-1)/6}\pmod{\mathfrak p}\); both sides are sixth roots of unity, and reduction is injective on \(\mu_6\). If \(72\mid q_{\mathfrak p}-1\), then \(12\mid(q_{\mathfrak p}-1)/6\), and \(\zeta_{12}^2\) has order \(6\), so the power is \(1\).

5.2. \(\chi_{\mathfrak p'}(\epsilon p)=\chi_{\mathfrak p'}(\epsilon)\chi_{\mathfrak p'}(p)\). By (2.1), \(\chi_{\mathfrak p'}(p)=\chi_{\mathfrak p}(p')\), and by (2.3) with \(\mathfrak a=\mathfrak p'\), \(\chi_{\mathfrak p'}(\epsilon)=H_S(\epsilon,p')^{-1}\).

5.3. \((uv,y)_{6,v}=(u,y)_{6,v}(v,y)_{6,v}\) at every place, by multiplicativity of the symbol in its first argument; and \(\epsilon^6\) is a sixth power, so \(\kappa_{\epsilon^6}\) is trivial by Lemma 3.1(3).

5.4. If \(v_{\mathfrak p}(u)=1\), then \(u\) is a uniformizer at \(\mathfrak p\) and \(y\mapsto(u,y)_{6,\mathfrak p}\) is nontrivial on units: by the tame formula and Lemma 3.1 of Lesson 2, \((u,y)_{6,\mathfrak p}=(y,u)_{6,\mathfrak p}^{-1}\equiv y^{-(q_{\mathfrak p}-1)/6}\) for a unit \(y\), which is not identically \(1\) because the residue field group is cyclic of order divisible by \(6\). So \(\mathfrak p\) divides the conductor.

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