S-integers and sixth-power classes

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

Lessons 2–14 prove Theorem 5.3 of Artin's conjecture and the index of a prime: a zero-free strip of fixed width for every finite-order Hecke \(L\)-function over a cyclotomic field \(F\) containing \(\mu_{12}\). The proof sums over ideals of \(F\) while using generators of those ideals in local fields, and it needs residue symbols that satisfy reciprocity with no correction factors. This lesson builds the ring in which that is possible. A finite set \(S\) of places is fixed so that the ring of \(S\)-integers \(R\) is a principal ideal domain whose additive group is a self-dual lattice. The group of sixth-power classes at \(S\) carries an alternating pairing from Hilbert symbols; the \(S\)-units form a self-orthogonal subgroup, and a complementary isotropic subgroup \(B\) selects one generator of every ideal up to sixth powers of units.

Prerequisites. The adèle ring of a number field, Theorems 2.2 and 2.3; Idèles and the idèle class group, Theorem 3.3; Additive characters, self-dual measures and Poisson summation on the adèles, Proposition 5.1 and Theorems 5.4 and 5.5; Dirichlet's unit theorem, Theorem 9.4; Places of number fields in extensions and the product formula, Theorem 5.2; The multiplicative group of a local field, Theorem 4.1; Hilbert symbols and local conics, Proposition 11.1, Theorem 11.2 and Proposition 11.3; Power residue symbols and reciprocity laws, Theorem 23.3; Hilbert's theorem 90 and Kummer theory, Theorem 3.3; Hilbert and ring class fields, and quadratic prime forms, Theorem 19.1.

1. The field and the set S

Throughout Lessons 2–14, \(F\) is a cyclotomic field containing \(\mu_{12}\). It is totally imaginary; write \([F:\mathbb Q]=2r\), so \(F\) has \(r\) complex places. At a finite place \(v\) normalize \(|\varpi_v|_v=(\operatorname N v)^{-1}\) for a uniformizer \(\varpi_v\); at a complex place use the square of the ordinary modulus. These are the absolute values of the product formula (Theorem 5.2 of Places of number fields in extensions and the product formula). Let \(\psi=\prod_v\psi_v\) be the standard additive character of \(\mathbb A_F/F\) of Additive characters, self-dual measures and Poisson summation on the adèles, and fix a finite-order Hecke character \(\eta\) of \(F\).

Fix a finite set \(S\) of places of \(F\) containing

Any further finite set may be added; the choice is made once, before any of the constructions below. The ring of \(S\)-integers \[ \begin{aligned} &R=\mathcal O_S=\{a\in F:v_{\mathfrak p}(a)\ge0\\ &\qquad\text{for every prime}\ \mathfrak p\notin S\} \end{aligned} \] is then a principal ideal domain: its class group is the quotient of the class group of \(F\) by the subgroup generated by the classes of the primes in \(S\), and that subgroup is the whole class group. All constants in Lessons 2–14 may depend on \(F\), \(S\) and \(\eta\).

Write \(S_f\) for the finite places in \(S\), \(s_f=|S_f|\), and \[ \begin{gathered} U_S=R^\times,\qquad F_S=\prod_{v\in S}F_v,\\ |y|_S=\prod_{v\in S}|y_v|_v,\\ F_S^1=\{y\in F_S^\times:|y|_S=1\}. \end{gathered} \] An exterior ideal is a nonzero ideal of \(R\); its norm is \(q_{\mathfrak a}=\operatorname N\mathfrak a=|R/\mathfrak a|\), extended multiplicatively to fractional ideals. For \(a\in F^\times\) put \(q_a=|a|_S\). On \(F_S\) put \[ \begin{gathered} \mathrm e(x)=\prod_{v\in S}\psi_v(x_v),\\ \widehat f(y)=\int_{F_S}f(x)\,\mathrm e(-xy)\,dx, \end{gathered} \] with \(dx\) the product of the self-dual local measures.

2. The S-integer lattice

Lemma 2.1 ([OpenAI-029, Lemma 3.1]). The diagonal image of \(R\) in \(F_S\) is a discrete cocompact subgroup, equal to its own annihilator for the pairing \(\mathrm e(xy)\), of covolume \(1\). For every Bruhat–Schwartz function \(f\) on \(F_S\), \[ \sum_{m\in R}f(m)=\sum_{\nu\in R}\widehat f(\nu). \tag{2.1} \] If \(\mathfrak a=aR\) is an exterior fractional ideal, then \(q_a=q_{\mathfrak a}\), \((aR)^\perp=a^{-1}R\), and \[ \begin{aligned} &\sum_{m\in aR}f(m)\\ &\quad =q_{\mathfrak a}^{-1}\sum_{\nu\in a^{-1}R}\widehat f(\nu).\quad(2.2) \end{aligned} \]

Proof. Put \(K^S=\prod_{v\notin S}\mathcal O_v\), with zero components at \(S\). Outside \(S\), \(\mathcal O_v\) is its own annihilator, so its self-dual measure is \(1\) and \(\widehat{\mathbf 1_{\mathcal O_v}}=\mathbf 1_{\mathcal O_v}\) (Proposition 5.2 of Additive characters, self-dual measures and Poisson summation on the adèles). An element of \(F\) whose components outside \(S\) are integral is exactly an element of \(R\), so \(F\cap(F_S\times K^S)=R\).

Discreteness and compactness. The diagonal \(F\) is discrete in \(\mathbb A_F\) (Theorem 2.2 of The adèle ring of a number field). For a compact neighbourhood \(C\) of \(0\) in \(F_S\), the set \(F\cap(C\times K^S)\) is finite, so shrinking \(C\) isolates \(0\) in \(R\). By strong approximation away from one complex place (Theorem 2.3 there), \(F+(F_S\times K^S)=\mathbb A_F\); hence \(F_S/R\cong\mathbb A_F/(F+K^S)\), a quotient of the compact group \(\mathbb A_F/F\).

Self-duality. The annihilator of \(F\) in \(\mathbb A_F\) is \(F\), and \(\mathbb A_F/F\) has volume \(1\) for the self-dual measure (Theorem 5.4 there). The annihilator of \(F+K^S\) is \(F\cap(K^S)^\perp=F\cap(F_S\times K^S)=R\), so the annihilator of \(R\) in \(F_S\) is \(R\). A lattice and its annihilator have reciprocal covolumes for self-dual measure; hence the covolume is \(1\).

Poisson summation. Apply adelic Poisson summation (Theorem 5.5 there) to \(f\otimes\mathbf 1_{K^S}\). Only elements of \(R\) occur on either side, and the factor \(\mathbf 1_{K^S}\) is its own transform; this is (2.1). Multiplication by \(a\) on \(F_S\) has modulus \(|a|_S\), and scaling (2.1) gives (2.2) once \(q_a=q_{\mathfrak a}\) is known. By the product formula, \(|a|_S=\prod_{v\notin S}|a|_v^{-1}=\operatorname N(aR)\). \(\square\)

3. Hilbert symbols of orders 2, 3 and 6

Let \(\operatorname{Art}_v\colon F_v^\times\to\operatorname{Gal}(F_v^{\mathrm{ab}}/F_v)\) be the local reciprocity map in which a uniformizer acts as arithmetic Frobenius on unramified extensions. For \(n\in\{2,3,6\}\) and \(a,b\in F_v^\times\) put \[ \begin{aligned} (a,b)_{n,v}&=\frac{\operatorname{Art}_v(b)(a^{1/n})}{a^{1/n}}\\ &\in\mu_n .\quad(3.1) \end{aligned} \] In the notation of Hilbert symbols and local conics, where the first argument enters reciprocity, this is the symbol with the arguments exchanged; since that symbol satisfies \((a,b)(b,a)=1\) (Proposition 11.1 there), (3.1) is its inverse. Proposition 11.1 also gives multiplicativity in each argument, dependence only on the classes modulo \(F_v^{\times n}\), and \((a,-a)_{n,v}=1\). The tame formula (Proposition 11.3 there) becomes, at a place \(v\nmid n\), for a unit \(u\) and a uniformizer \(\varpi\), \[ \begin{aligned} (u,\varpi)_{n,v}&\equiv u^{(\operatorname Nv-1)/n}\pmod v,\quad(3.2) \end{aligned} \] and \((u,u')_{n,v}=1\) for two units. The symbols of different orders are related by \((a,b)_{6,v}^3=(a,b)_{2,v}\) and \((a,b)_{6,v}^2=(a,b)_{3,v}\): with \(\beta^6=a\), the element \(\beta^3\) is a square root and \(\beta^2\) a cube root of \(a\).

Lemma 3.1. For every place \(v\), \(n\in\{2,3,6\}\) and \(x\in F_v^\times\), \((x,x)_{n,v}=1\). Consequently \((x,y)_{n,v}(y,x)_{n,v}=1\).

Proof. From \((x,-x)_{n,v}=1\), \((x,x)_{n,v}=(x,-1)_{n,v}^{-1}\). The field contains \(\zeta_{12}\), and \(-1=\zeta_{12}^6\) is a sixth power, so \((x,-1)_{n,v}=1\). The second statement follows by expanding \((xy,xy)_{n,v}=1\). \(\square\)

On the finite group \(\mathcal V=F_S^\times/F_S^{\times6}\) define \[ H_S(x,y)=\prod_{v\in S}(x_v,y_v)_{6,v}. \tag{3.3} \] At a complex place every element is a sixth power, so only the finite places of \(S\) contribute. Let \(E\) be the image of \(U_S\) in \(\mathcal V\). For \(T>0\) let \(\ell_T\in F_S^\times\) have components \(1\) at the finite places of \(S\) and \(T^{1/(2r)}\) at every complex place; then \(|\ell_T|_S=T\) and \(\ell_T\ell_{T'}=\ell_{TT'}\).

4. Allowed generators

Lemma 4.1 (generators with fixed power classes [OpenAI-029, Lemma 3.2]). The pairing \(H_S\) on \(\mathcal V\) is nondegenerate and alternating. The map \(U_S/U_S^6\to\mathcal V\) is injective, its image satisfies \(E^\perp=E\), and there is an isotropic subgroup \(B\) with \(\mathcal V=E\oplus B\).

Every exterior fractional ideal \(\mathfrak a\) has a generator \(a\) whose class \([a]\in\mathcal V\) lies in \(B\); two such generators differ by an element of \(U_S^6\). They can be chosen with \[ a/\ell_{q_{\mathfrak a}}\in\mathcal K_{\mathrm{bal}} \tag{4.1} \] for one fixed compact subset \(\mathcal K_{\mathrm{bal}}\subset F_S^1\), independent of \(\mathfrak a\).

Proof. It suffices to treat the two prime-power parts: for \(\ell\in\{2,3\}\) put \(\mathcal V_\ell=F_S^\times/F_S^{\times\ell}\), with the pairing \(\prod_{v\in S}(\cdot,\cdot)_{\ell,v}\). The Chinese remainder theorem gives \(\mathcal V\cong\mathcal V_2\oplus\mathcal V_3\), and under it \(H_S^3\) and \(H_S^2\) are the pairings of order \(2\) and \(3\).

Dimensions. At a finite place, Theorem 4.1 of The multiplicative group of a local field gives \((F_v^\times:F_v^{\times\ell})=\ell\,|\mu_\ell(F_v)|/|\ell|_v\)\(=\ell^{\,2+[F_v:\mathbb Q_\ell]\mathbf 1_{v\mid\ell}}\), since \(\mu_\ell\subset F\). All places above \(\ell\) lie in \(S\), and their local degrees add up to \(2r\); complex places contribute nothing. Hence \[ \dim_{\mathbb F_\ell}\mathcal V_\ell=2s_f+2r. \tag{4.2} \] The \(S\)-unit theorem (Theorem 9.4 of Dirichlet's unit theorem) gives \(U_S\cong\mu(F)\times\mathbb Z^{\,r+s_f-1}\); since \(\mu(F)\) is cyclic and contains \(\mu_\ell\), \[ \dim_{\mathbb F_\ell}U_S/U_S^\ell=r+s_f. \tag{4.3} \]

Injectivity. Let \(u\in U_S\) be an \(\ell\)th power at every place of \(S\). The extension \(F(u^{1/\ell})/F\) is abelian (Kummer theory, Theorem 3.3 of Hilbert's theorem 90 and Kummer theory). It is split at every place of \(S\), and unramified at every prime outside \(S\) because \(u\) is a unit there and \(\ell\) is invertible. Since \(F\) has no real places, it is unramified everywhere, so it lies in the Hilbert class field of \(F\) (Theorem 19.1 of Hilbert and ring class fields, and quadratic prime forms), and its Galois group is a quotient of the class group in which the classes of the primes of \(S\) map to their Frobenius elements, which are trivial. Those classes generate the class group, so \(F(u^{1/\ell})=F\) and \(u=t^\ell\) with \(t\in F\). The valuations of \(t\) outside \(S\) are those of \(u\) divided by \(\ell\), hence zero: \(t\in U_S\).

Isotropy of the units. For \(u,u'\in U_S\), every symbol \((u,u')_{\ell,v}\) at a prime \(v\notin S\) equals \(1\), because both arguments are units and \(v\nmid\ell\). Hilbert's product formula (Theorem 23.3 of Power residue symbols and reciprocity laws) then gives \(\prod_{v\in S}(u,u')_{\ell,v}=1\). By Lemma 3.1 the pairing is alternating, and by Theorem 11.2 of Hilbert symbols and local conics, applied at each place of \(S\), it is nondegenerate on \(\mathcal V_\ell\). The image of \(U_S\) is an isotropic subspace of dimension \(r+s_f\), half of (4.2). An isotropic subspace of half the dimension of a nondegenerate space satisfies \(W^\perp\supseteq W\) and \(\dim W^\perp=\dim\mathcal V_\ell-\dim W=\dim W\), so \(W^\perp=W\).

A complement. In a nondegenerate alternating space over \(\mathbb F_\ell\), written additively with pairing \(\langle\cdot,\cdot\rangle\), every subspace \(W\) with \(W^\perp=W\) has an isotropic complement. Take a basis \(w_1,\dots,w_k\) of \(W\); the space has dimension \(2k\). By nondegeneracy choose \(y_1,\dots,y_k\) with \(\langle w_i,y_j\rangle=\delta_{ij}\), and put \(y_j'=y_j+\sum_{i<j}\langle y_i,y_j\rangle w_i\). Exercise 5.4 shows that the \(y_j'\) are pairwise orthogonal and still satisfy \(\langle w_i,y_j'\rangle=\delta_{ij}\); so they are independent modulo \(W\) and span an isotropic complement. Combining the complements for \(\ell=2,3\) through the Chinese remainder theorem gives \(B\), and the statements for \(\ell=2,3\) give injectivity and \(E^\perp=E\) for \(\mathcal V\).

Generators. Let \(\mathfrak a=a_0R\). Decompose \([a_0]=eb\) with \(e\in E\), \(b\in B\), choose \(\varepsilon\in U_S\) representing \(e^{-1}\), and put \(a=\varepsilon a_0\); then \([a]=b\in B\). If \(a'\) is another generator with \([a']\in B\), then \(a'/a\in U_S\) and \([a'/a]\in E\cap B=\{1\}\); by injectivity \(a'/a\in U_S^6\).

Balance. By Lemma 2.1, \(|a|_S=q_{\mathfrak a}\), so \(a/\ell_{q_{\mathfrak a}}\in F_S^1\). The group \(F_S^1\times\prod_{v\notin S}\mathcal O_v^\times\) is open in the idèles of norm one and meets the diagonal \(F^\times\) in \(U_S\). Its image in the compact group \(C_F^1\) (Theorem 3.3 of Idèles and the idèle class group) is an open, hence closed, hence compact subgroup, and it projects onto \(F_S^1/U_S\). So \(F_S^1/U_S\) is compact, and so is \(F_S^1/U_S^6\), since \(U_S^6\) has finite index in \(U_S\). Choose a compact set \(\mathcal K_{\mathrm{bal}}\subset F_S^1\) mapping onto \(F_S^1/U_S^6\), and multiply \(a\) by the element of \(U_S^6\) that moves \(a/\ell_{q_{\mathfrak a}}\) into it. \(\square\)

Definition 4.2. A generator of an exterior fractional ideal whose class in \(\mathcal V\) lies in \(B\) is allowed. An allowed generator satisfying (4.1) is balanced.

Allowed generators are unique up to \(U_S^6\), and the product of allowed generators of two ideals is an allowed generator of their product. Balanced representatives are used whenever an estimate involves the \(S\)-components of a generator; in identities it is often convenient to use the product of the allowed generators of the prime factors, which differs from the balanced one by an element of \(U_S^6\).

5. Exercises

5.1. Show that \(R\) has only finitely many elements in any set \(\ell_TK\) with \(K\subset F_S\) compact, and that their number is \(O_K(1+T)\).

5.2. Let \(F=\mathbb Q(\mu_{12})\), so \(r=2\). Suppose \(S_f\) consists of the two primes above \(2\) and \(3\) (this field has class number one). Compute \(\dim_{\mathbb F_\ell}\mathcal V_\ell\) and \(\dim_{\mathbb F_\ell}U_S/U_S^\ell\) for \(\ell=2,3\).

5.3. Show that if \(a\) is an allowed generator of \(\mathfrak a\) and \(\varepsilon\in U_S\), then \(\varepsilon a\) is allowed if and only if \(\varepsilon\in U_S^6\).

5.4. Prove the step of the complement construction: if \(W=\operatorname{span}(w_1,\dots,w_k)\) satisfies \(W^\perp=W\) and \(y_1,\dots,y_k\) satisfy \(\langle w_i,y_j\rangle=\delta_{ij}\), then the vectors \(y_j'=y_j+\sum_{i<j}\langle y_i,y_j\rangle w_i\) are pairwise orthogonal and still satisfy \(\langle w_i,y_j'\rangle=\delta_{ij}\). Why does the sign of the correction matter for \(\ell=3\) but not for \(\ell=2\)?

6. Solutions

5.1. By Lemma 2.1, \(R\) has a compact fundamental domain \(D\subset F_S\) of volume \(1\). For distinct \(m\in R\cap\ell_TK\) the translates \(m+D\) are disjoint and lie in \(\ell_TK+D\). At a complex place the ordinary modulus of a point of \(\ell_TK+D\) is at most \(CT^{1/(2r)}+C'\), so its projection lies in a disk of area \(\ll(T^{1/(2r)}+1)^2\); at the finite places of \(S\) the projections lie in fixed compact sets. Hence \(\operatorname{vol}(\ell_TK+D)\)\(\ll_K(T^{1/(2r)}+1)^{2r}\ll1+T\), and the number of such \(m\) is at most this volume.

5.2. Here \(s_f=2\) and (4.2) gives \(\dim\mathcal V_\ell=2\cdot2+2\cdot2=8\), while (4.3) gives \(\dim U_S/U_S^\ell=2+2=4\), half of it, as the proof requires.

5.3. \([\varepsilon a]=[\varepsilon][a]\) with \([\varepsilon]\in E\) and \([a]\in B\). This lies in \(B\) exactly when \([\varepsilon]\in E\cap B=\{1\}\), that is, when \(\varepsilon\) is a sixth power at every place of \(S\); by the injectivity in Lemma 4.1 this means \(\varepsilon\in U_S^6\).

5.4. Since \(W\) is isotropic, \(\langle w_l,y_j'\rangle=\langle w_l,y_j\rangle=\delta_{lj}\). Let \(m<j\). In \(\langle y_m',y_j'\rangle\), the terms pairing two vectors of \(W\) vanish; the terms \(\langle w_i,y_j\rangle\) with \(i<m<j\) vanish; and \(\langle y_m,w_i\rangle=-\langle w_i,y_m\rangle=-\delta_{im}\). Hence \[ \begin{aligned} \langle y_m',y_j'\rangle&=\langle y_m,y_j\rangle\\ &\quad +\sum_{i<j}\langle y_i,y_j\rangle\langle y_m,w_i\rangle\\ &=\langle y_m,y_j\rangle-\langle y_m,y_j\rangle\\ &=0 . \end{aligned} \] With a minus sign in the correction the same computation gives \(2\langle y_m,y_j\rangle\), which is zero for \(\ell=2\) but not in general for \(\ell=3\).

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