Kronecker–Weber and the maximal abelian extension of the rationals

Written by OpenAI GPT-6.1 Sol in Codex, Ultra effort, October 2026. Self-checked by the writing AI; no independent review is claimed. Public domain (CC0).

Lesson 20. Every finite abelian extension of the rationals is contained in a cyclotomic field. The ray fields of $\mathbf Q$ make this theorem a direct consequence of global class field theory. We identify those fields, compute the Artin map on arbitrary rational idèles, and derive the conductor–discriminant formula. These identifications also specify the Galois action on the roots of unity appearing in the Bost–Connes system.

We use Ray class fields, conductors and ideal reciprocity, the explicit local cyclotomic symbol in Corollary 9.5 of Explicit local reciprocity and existence, and the local conductor–discriminant theorem, Theorem 10.5 of Abelian ramification, conductors and Hasse–Arf. Theorem 12.1 and Theorem 12.3 of Cyclotomic fields prove cyclotomic irreducibility, degrees, inertia and arithmetic Frobenius. Quasi-characters and Hecke characters, especially Theorem 6.3, supplies the character terminology and its ideal/idèle convention. All these providers are written lessons.

Fix $\zeta_m=e^{2\pi i/m}$, put $U_m=(\mathbf Z/m\mathbf Z)^\times$, and write $\sigma_a(\zeta_m)=\zeta_m^a\quad(a\in U_m). \tag{1}$ The groups $U_1,U_2$ are trivial. Arithmetic reciprocity sends an unramified uniformizer to arithmetic Frobenius. Geometric reciprocity is its inverse. The cyclotomic character always records the exponent in (1); its definition does not change when reciprocity is inverted.

Prerequisite proof availability. The named results below identify specific programme lessons. The prerequisite record shows which results are proved in published lessons and which full proofs are still missing. A record or external reference is not a supplied proof; arguments using an unavailable prerequisite retain that dependency.

1. The rational ray class fields

For an integer $m\geq1$, the moduli $m\infty$ and $m$ differ by the real place condition. In the first, principal generators must be positive; in the second, either sign is allowed.

Proposition 20.1. Their ray groups and fields are $\begin{aligned} \operatorname{Cl}_{m\infty}(\mathbf Q)&\simeq U_m,& \mathbf Q_{m\infty}&=\mathbf Q(\zeta_m),\\ \operatorname{Cl}_{m}(\mathbf Q)&\simeq U_m/\langle-1\rangle,& \mathbf Q_m&=\mathbf Q(\zeta_m)^+. \end{aligned} \tag{2}$ Here the superscript $+$ means the subfield fixed by complex conjugation. The formulas include $m=1,2$.

Proof. Every fractional ideal of $\mathbf Q$ prime to $m$ has a unique positive rational generator $a$, whose numerator and denominator are coprime to $m$. Send that ideal to $a\bmod m$. Every unit residue occurs, by taking a positive integral representative. Its kernel consists exactly of the ideals with positive generator congruent to $1$ modulo $m$, giving the first group in (2). If the generator may have either sign, an ideal is ray principal exactly when its positive generator is congruent to $1$ or $-1$; this gives the second group.

Let $E=\mathbf Q(\zeta_m)$. The local symbol proves its conductor divides $m\infty$. Indeed, write $m=p^a b$, $p\nmid b$. At $p$, the field generated by the $b$-th roots is unramified, so local units act trivially on them. Units congruent to $1\pmod{p^a}$ act trivially on the $p^a$-th roots by Corollary 9.5. When $a=0$, all units suffice. At the real place, positive numbers act trivially. Thus the entire modulus subgroup is in the global reciprocity kernel, by local compatibility. Theorem 18.4 places $E$ in $\mathbf Q_{m\infty}$. Both degrees are $\varphi(m)$, the first by cyclotomic irreducibility and the second by the ray-group computation. They are equal.

Dropping the real place adds the class of a negative real component to the norm subgroup. Its symbol on $E$ is complex conjugation: local reciprocity at $\mathbf R$ identifies the sign quotient with $\operatorname{Gal}(\mathbf C/\mathbf R)$. Therefore the corresponding subfield is $E^+$. Its group is the quotient by $-1$ in (1). This proves the remaining field assertion, also when conjugation is already trivial. $\square$

The modulus in (2) need not be the field's least conductor. For example $\mathbf Q(\zeta_2)=\mathbf Q$, and for odd $m$ the fields of orders $m$ and $2m$ are equal. A conductor is a minimal condition, whereas a ray modulus is a specified condition.

2. The Kronecker–Weber theorem

Theorem 20.2 (Kronecker–Weber). Every finite abelian $L/\mathbf Q$ lies in $\mathbf Q(\zeta_m)$ for some positive integer $m$. Consequently $\mathbf Q^{\mathrm{ab}}=\mathbf Q^{\mathrm{cycl}} =\bigcup_{m\geq1}\mathbf Q(\zeta_m)=\mathbf Q(\mu_\infty). \tag{3}$

Proof. The extension has a finite global conductor, by Proposition 18.3. Enlarge it, if necessary, to a modulus $m\infty$. The ray-field containment theorem and (2) give $L\subseteq\mathbf Q_{m\infty}=\mathbf Q(\zeta_m)$. Every cyclotomic field is abelian by (1), so their union is contained in the maximal abelian extension. The reverse inclusion follows from the finite assertion, since every algebraic element in $\mathbf Q^{\mathrm{ab}}$ belongs to a finite abelian extension. $\square$

This proof identifies the containing field through its ray conditions. Section 7 gives a second proof using local Kronecker–Weber and finite inertia groups.

3. The explicit rational Artin map

The normalization of idèles in lesson 17 gives a topological isomorphism $C_{\mathbf Q}\simeq\mathbf R_{>0}\times\widehat{\mathbf Z}^{\times}. \tag{4}$ Explicitly, for an idèle $x=(x_\infty,(x_p))$, set $q=\prod_p p^{-v_p(x_p)}>0,\quad \epsilon=\operatorname{sgn}(x_\infty),\quad t=\epsilon qx_\infty>0,\quad u_p=\epsilon qx_p\in\mathbf Z_p^\times. \tag{5}$ Only finitely many factors define $q$. Multiplication by the principal idèle $\epsilon q$ yields the normalized representative $(t,u)$. A rational number that is a unit at every prime is $\pm1$, and positivity fixes its sign; hence the representative is unique. Continuity holds on the restricted-product neighborhoods where the finitely many valuations and the real sign are fixed.

Proposition 20.3. For $u\in\widehat{\mathbf Z}^{\times}$, compatible exponent notation gives $\operatorname{rec}_{\mathrm{arith}}(t,u)=\sigma_{u^{-1}}, \qquad \operatorname{rec}_{\mathrm{geom}}(t,u)=\sigma_u. \tag{6}$ For a prime $p$, the idèle $\varpi_p$ equal to $p$ at $p$ and $1$ elsewhere acts on $\mathbf Q(\zeta_m)$, $p\nmid m$, as $\sigma_p$ in the arithmetic convention. The idèle equal to $-1$ at the real place and $1$ at every finite place acts as complex conjugation.

Proof. A positive real component acts trivially by real local reciprocity. On $p^a$-th roots, the local component $u_p$ acts with exponent $u_p^{-1}$, by Corollary 9.5; every other finite unit component is unramified on those roots and acts trivially. Combining the prime-power roots gives the first formula on $\mathbf Q(\zeta_m)$, for every $m$. Local compatibility and (3) make it the full global formula. Inverting gives geometric reciprocity.

For $\varpi_p$, (5) gives $t=1/p$, $u_p=1$, and $u_\ell=1/p$ for $\ell\ne p$. Thus $u^{-1}\equiv p\pmod m$ when $p\nmid m$, proving the Frobenius assertion. No such residue $p\pmod m$ defines a unit when $p\mid m$; at the $p$-primary roots this particular idèle acts trivially, and at the prime-to-$p$ roots it acts with exponent $p$.

For the negative real idèle, (5) gives $t=1$, $u_p=-1$ at every finite prime. Its inverse is itself, so (6) is $\sigma_{-1}$, complex conjugation. $\square$

Corollary 20.4. The cyclotomic character is a canonical topological isomorphism $\chi_{\mathrm{cyc}}:\operatorname{Gal}(\mathbf Q^{\mathrm{ab}}/\mathbf Q) \xrightarrow{\sim}\widehat{\mathbf Z}^{\times}, \qquad \sigma_a\longmapsto a. \tag{7}$ Arithmetic reciprocity induces an isomorphism from $C_{\mathbf Q}/\mathbf R_{>0}$, and its composite with (7) is $u\mapsto u^{-1}$.

Proof. The finite isomorphisms (1) are compatible with reduction of exponents. Taking inverse limits, as proved in Proposition 1.3, gives (7) for the cyclotomic union, which is (3). Formula (6) is onto and has kernel exactly the positive real factor: a unit tuple acting trivially on every root has residue $1$ modulo every integer. Both (7) and its inverse are continuous by their finite quotient coordinates. $\square$

4. Dirichlet, Galois and Hecke characters

A Dirichlet character means a character of some $U_m$, with characters identified when they are inflations of the same primitive character. Its finite conductor $f(\chi)$ is the least modulus of that primitive character. When values on integers are used, the primitive character is extended by zero precisely at integers not coprime to $f(\chi)$.

Proposition 20.5. Dirichlet characters correspond bijectively to continuous complex characters of $\operatorname{Gal}(\mathbf Q^{\mathrm{ab}}/\mathbf Q)$, and to finite-order Hecke characters of $\mathbf Q$, by $\rho_\chi(\sigma_a)=\chi(a),\qquad \omega_\chi(t,u)=\chi(u^{-1}). \tag{8}$ The ideal character at an unramified prime is $\omega_\chi(\varpi_p)=\chi(p)$. Their finite conductor exponents agree. If $L/\mathbf Q$ is finite abelian and $X(L)$ is its character group inflated to (7), then $f_L=\operatorname{lcm}_{\chi\in X(L)}f(\chi) \tag{9}$ is its finite conductor and its least cyclotomic modulus. Its full global conductor includes $\infty$ exactly when $L$ is not totally real.

Proof. A continuous character from a profinite group to $\mathbf C^\times$ has open kernel. To check this, choose a small neighborhood of $1$ containing no nontrivial subgroup, for example $1/2<|z|<2$, $|\arg z|<\pi/2$. All powers force a subgroup element's modulus to be $1$, and a nonidentity circle element has a power outside that arc. A sufficiently small open subgroup of the profinite domain maps into this neighborhood and hence trivially. The resulting quotient is finite. Applying this to (7) shows that every character factors through some $U_m$.

A finite-order Hecke character kills $\mathbf R_{>0}$, since a connected group's continuous image in a finite group is trivial. Conversely a finite-image character on the unit factor in (4) defines such a Hecke character. Formula (6) gives exactly the inverse in (8), and gives $\chi(p)$ at $p\nmid f(\chi)$. Its real sign is $\chi(-1)$.

For the conductor, the Chinese remainder decomposition of $U_m$ gives a least exponent in each prime-power factor through which a character factors. These are the least exponents killing the local principal-unit groups; inversion changes neither the kernel nor the exponent. This proves agreement with local and Hecke conductors. The finite character lemma proved before section 4 of Hilbert's theorem 90 and Kummer theory says that the characters of $\operatorname{Gal}(L/\mathbf Q)$ separate its elements. Thus a local unit subgroup kills the field exactly when it kills every character. Its least exponent is their maximum, proving (9).

Equivalently $L\subseteq\mathbf Q(\zeta_m)$ precisely when every character factors modulo $m$, by the annihilator description of a fixed field. The least local exponents show that this is equivalent to $f_L\mid m$, proving minimality without assuming a chosen ambient modulus. Complex conjugation is $-1$ under (7). It is trivial on the field exactly when every $\chi\in X(L)$ is even, which is precisely that $L$ is totally real. The real conductor statement follows. $\square$

Multiplication in $X(L)$ means multiplication after inflation to a common modulus, followed by taking the primitive representative. Multiplying zero-extended arithmetic functions would impose an incorrect conductor on the result.

5. The global conductor–discriminant formula

Theorem 20.6. For a finite abelian number field $L/\mathbf Q$, $|d_L|=\prod_{\chi\in X(L)}f(\chi). \tag{10}$ The trivial character contributes $1$. These are finite conductors; an infinity factor is not an integer factor of the discriminant.

Proof. Fix a rational prime $p$, and let $D$ be its decomposition group in $G=\operatorname{Gal}(L/\mathbf Q)$. Its completion $E/\mathbf Q_p$ has group $D$, and there are $g=[G:D]$ primes above $p$. Restriction $\widehat G\to\widehat D$ is onto and each fiber has size $g$: a character of a subgroup of a finite abelian group extends by successively choosing a root for the value of a newly adjoined generator. Its kernel is $\widehat{G/D}$. The local conductor is computed by restriction to $D$. Theorem 10.5 therefore gives $\sum_{\chi\in X(L)}v_p(f(\chi)) =g\sum_{\eta\in\widehat D}a(\eta) =g\,v_p(\mathfrak d_{E/\mathbf Q_p}). \tag{11}$

This is also $v_p(d_L)$. To verify the local/global identification directly, tensor an integral basis and its trace matrix with $\mathbf Z_p$. The Chinese remainder theorem on the quotients by $p^n$, followed by completion, gives $\mathcal O_L\otimes\mathbf Z_p\simeq \prod_{\mathfrak P\mid p}\mathcal O_{L_{\mathfrak P}}.$ Trace on this product is the sum of the local traces. Choosing a basis in each factor makes the trace pairing block diagonal, so its determinant valuation is the sum of the local discriminant valuations. Integral changes of $\mathbf Z_p$-basis have unit determinant and do not alter that valuation. In the Galois case all $g$ local factors are isomorphic, giving the last term of (11). Equality of prime exponents proves (10); absolute value removes the archimedean signature sign. $\square$

For $L=\mathbf Q(\zeta_5)$, its four characters have conductors $1,5,5,5$, so $|d_L|=125$. For $L=\mathbf Q(\zeta_{12})$, the four characters are $1,\chi_{-3},\chi_{-4},\chi_{12}$, with conductors $1,3,4,12$; their product is $144$. Both agree with the discriminants calculated independently in the cyclotomic-fields lesson.

6. Cyclotomic values and the adelic component space

Put $e(r)=\exp(2\pi ir)$ when it denotes a complex number. In the Bost–Connes algebra the same notation labels a diagonal arithmetic generator; its value in a cooled extremal phase indexed by $v\in\widehat{\mathbf Z}^{\times}$ is $e(vr)$. The construction and classification of those states are proved in Phase transition in the Bost–Connes system, sections 3, 4, 6 and 18. We now specify the number theory in that formula.

Proposition 20.7. The values $e(a/N)=\zeta_N^a$, for rational residues $a/N$, generate $\mathbf Q^{\mathrm{ab}}$. For every compatible unit $b$, $\sigma_b\bigl(e(a/N)\bigr)=e(ba/N),\qquad \sigma_b\bigl(e(va/N)\bigr)=e(bva/N). \tag{12}$ The adelic component space used for the cyclotomic tower has a canonical topological identification $\mathbf Q^\times\backslash \bigl(\mathbf A_f^\times\times\{\pm1\}\bigr) \simeq \mathbf A_f^\times/\mathbf Q_{>0}^\times \simeq\widehat{\mathbf Z}^{\times}. \tag{13}$ Here a rational $q$ acts by $(a,\epsilon)\mapsto(qa,\operatorname{sgn}(q)\epsilon)$. It is the zero-dimensional space denoted $\operatorname{Sh}(\mathrm{GL}_1,\{\pm1\})$ in the Bost–Connes discussion. With the geometric reciprocity convention, a normalized unit idèle $b$ gives $\sigma_b$; its action on cyclotomic embedding labels in (13) is multiplication by $b$.

Proof. The values contain $\zeta_N=e(1/N)$ for every $N$, and every displayed value is a root of unity. Thus their generated field is the cyclotomic union, equal to (3). Equations (12) are exactly exponent multiplication in (1), including after replacing $a/N$ by a residue indexed by $v$.

For (13), every orbit has a representative with sign $+1$, obtained by multiplication by $-1$ when necessary. Two such representatives differ by a positive rational. For $a\in\mathbf A_f^\times$, multiplication by $q=\prod_p p^{-v_p(a_p)}$ produces a unit tuple. A positive rational unit everywhere is $1$, so that tuple is unique. The normalization is continuous on neighborhoods with fixed valuations, and its inverse is the inclusion of the compact unit tuples. This proves the topological identifications, not just a set bijection. The unit group is totally disconnected, so every connected component of this space is a point.

At level $N$, reduction of its unit coordinate is the label $v\in U_N$ of the embedding $\zeta_N\mapsto\zeta_N^v$. This identifies its inverse system with the complex points of the cyclotomic tower $\operatorname{Spec}\mathbf Q(\zeta_N)$. Composition with $\sigma_b$ sends that embedding to the one labeled $bv$, proving the Galois action. Deligne's chosen reciprocity sends a uniformizer to geometric Frobenius, so it is $\operatorname{rec}_{\mathrm{geom}}$. Formula (6) identifies its unit coordinate with precisely $\sigma_b$, as asserted. $\square$

For the arithmetic normalization the same $\sigma_b$ is the image of the unit idèle $b^{-1}$. Thus (12) uses the cyclotomic exponent $b$; substituting an arithmetic idèle coordinate for that exponent requires the inverse. The arithmetic symmetry formulas in The Bost–Connes Hecke algebra, section 6, use the exponent coordinate. Their operator algebra and state proofs remain in those lessons.

Hilbert's twelfth problem asks for explicit generators of abelian extensions through special values of functions. Roots of unity solve it over $\mathbf Q$ by (3). Imaginary quadratic fields motivate elliptic functions and complex multiplication; their general generation theory is beyond this lesson. The preceding lesson obtains its particular ring class fields by independently checked radical equations.

7. Exercises and complete solutions

Exercise 1 — The least cyclotomic modulus (easy)

Find the smallest $m$ containing $\mathbf Q(\sqrt{-7})$ and $\mathbf Q(\sqrt3)$ in $\mathbf Q(\zeta_m)$.

Solution. Their fundamental discriminants are $-7$ and $12$. The quadratic conductor calculation in lesson 18 gives finite conductors $7$ and $12$, respectively. Proposition 20.5 proves that each is its least cyclotomic modulus. Inclusion at $7$ is also exhibited by the quadratic Gauss sum $\sum_{a\bmod7}(a/7)\zeta_7^a$, whose square is $-7$, proved in Proposition 12.6 of Cyclotomic fields. At $12$, $\zeta_{12}+\zeta_{12}^{-1}=\sqrt3$. The real factor of the first field's conductor is $\infty$; it does not change the finite least modulus.

Similarly $\sqrt2=\zeta_8+\zeta_8^{-1}$, and $\sqrt5$ belongs to $\mathbf Q(\zeta_5)$ by its quadratic Gauss sum. Their least moduli are $8$ and $5$, as their conductors predict.

Exercise 2 — The negative real idèle (medium)

Compute the Artin image of the idèle equal to $-1$ at infinity and $1$ at every finite prime, in both conventions.

Solution. Its normalized representative under (5) is $(1,(-1)_p)$. Formula (6) sends it to $\sigma_{-1}$ in either convention, because this unit tuple is its own inverse. It sends every root of unity to its inverse, hence is complex conjugation. This idèle is not the diagonal principal element $-1$: the latter has $-1$ at the finite primes too and has trivial global symbol. Their product displays explicitly the cancellation of the real and finite signs on a principal idèle.

Exercise 3 — The real ray field (medium)

Prove that $\mathbf Q(\zeta_m)^+$ is the ray field for modulus $m$, including $m=1,2$.

Solution. In the ideal ray group, a positive generator represents the same ideal as its negative. Removing the positivity condition therefore replaces $U_m$ by $U_m/\langle-1\rangle$. The full ray field for $m\infty$ is $\mathbf Q(\zeta_m)$. Passing to modulus $m$ kills the image of the real sign, whose Artin symbol is $\sigma_{-1}$. The fixed field of that subgroup is exactly the real subfield. When $m>2$, this subgroup has order $2$, so the degree is $\varphi(m)/2$. When $m=1,2$, both groups and fields are trivial, and taking a half of $\varphi(m)$ would be incorrect.

Exercise 4 — A proof from local Kronecker–Weber (hard)

Prove the finite assertion in Theorem 20.2 using local Kronecker–Weber and the absence of a nontrivial abelian extension of $\mathbf Q$ unramified at every finite prime.

Solution. Let $S$ contain the finite primes ramified in $L/\mathbf Q$. At $p\in S$, local Kronecker–Weber puts each completion of $L$ in $\mathbf Q_p(\zeta_{p^{n_p}},\zeta_{b_p})$, with $p\nmid b_p$, for a sufficiently large $n_p$. The second factor is unramified. Therefore, after adjoining the $p^{n_p}$-th roots, the compositum of the completion with that cyclotomic local field is unramified over it: it is a subextension of the base change of the unramified factor.

Put $m=\prod_{p\in S}p^{n_p}$, $M=\mathbf Q(\zeta_m)$, and $T=LM$. At $p\in S$, the completion of $M$ is the product of the totally ramified $p$-primary cyclotomic extension and unramified prime-to-$p$ factors. The preceding observation makes $T/M$ unramified there. Outside $S$, both $L$ and $M$ are unramified, so their compositum is too. All these local assertions follow from separable residue-field extensions and their preservation under base change.

The group $G=\operatorname{Gal}(T/\mathbf Q)$ is abelian. Its inertia subgroup $I_p$ maps isomorphically onto the inertia of $M/\mathbf Q$: restriction is onto, and its kernel is relative inertia for $T/M$, already trivial. The cyclotomic inertia theorem gives $|I_p|=\varphi(p^{n_p})$. The subgroup $J$ generated by these finite inertia groups has fixed field unramified at every finite prime over $\mathbf Q$. Its fixed field is abelian, and is therefore $\mathbf Q$, since the narrow Hilbert class group of $\mathbf Q$ is trivial. Hence $J=G$. Because $G$ is abelian, multiplication of the finitely many inertia subgroups is onto, and $|G|\leq\prod_{p\in S}|I_p|=\prod_{p\in S}\varphi(p^{n_p}) =\varphi(m)=[M:\mathbf Q].$ But $M\subseteq T$ gives the reverse degree inequality. Thus $T=M$, so $L\subseteq M$, as required. If $S$ is empty, the narrow Hilbert argument already gives $L=\mathbf Q$. This proof uses no assertion about the class number of $M$; relative unramifiedness by itself would not imply $T=M$.

Editable edition

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References

The lesson proves both Kronecker–Weber routes, the rational Artin map, conductors and discriminants. Section 6 proves the cyclotomic values and the canonical topological component identification. The Bost–Connes state construction retains its separately identified programme prerequisite.

The arithmetic and geometric Frobenius conventions are compared in equations (6) and (13). The Connes–Marcolli reference is the author-posted draft, with its own section numbering and pagination.

The proof guide gives the lesson sequence and the exact prerequisite record. External references accompany the written arguments.