Pure Mathematics Research-Number Theory（5）

Kronecker-Weber theorems

We now state the

global

Kronecker-Weber theorem。

$\text{Theorem}$

$\mathbb{Q}$

$\mathbb{Q}\left(\zeta_{m}\right).$

$\text{Theorem}$

$\mathbb{Q}_{p}$

$\mathbb{Q}_{p}\left(\zeta_{m}\right).$

$\text{Proposition}$

Let

$K / \mathbb{Q}$

be a finite abelian extension。 For each ramified prime

$\mathbb{Q},$

pick a prime

$\mathfrak{p} \mid p$

and let

$K_{\mathfrak{p}}$

be the completion of

$\mathfrak{p}$

the fact that

$K / \mathbb{Q}$

is Galois means that every

$\mathfrak{p} \mid p$

is ramified with the same ramification index； it makes no difference which

$\mathfrak{p}$

we pick

We have

$\operatorname{Gal}\left(K_{\mathfrak{p}} / \mathbb{Q}_{p}\right) \simeq D_{\mathfrak{p}} \subseteq \operatorname{Gal}(K / \mathbb{Q}),$

$K_{\mathfrak{p}}$

is an abelian extension of

$\mathbb{Q}_\mathfrak{p}$

and the local Kronecker-Weber theorem implies that

$K_{\mathfrak{p}} \subseteq \mathbb{Q}_{p}\left(\zeta_{m_{p}}\right)$

for some

$m_{p} \in \mathbb{Z}_{\geq 1}.$

Let

$n_{p}:=v_{p}\left(m_{p}\right),$

put

$m:={\prod}_{p} p^{n_{p}}$

this is a finite product

and let

$L=K\left(\zeta_{m}\right).$

We will show

$L=\mathbb{Q}\left(\zeta_{m}\right),$

which implies

$K \subseteq \mathbb{Q}\left(\zeta_{m}\right).$

The field

$L=K \cdot \mathbb{Q}\left(\zeta_{m}\right)$

is a compositum of Galois extensions of

$\mathbb{Q},$

and is therefore Galois over

$\mathbb{Q}$

with

$\operatorname{Gal}(L / \mathbb{Q})$

isomorphic to a subgroup of

$\operatorname{Gal}(K / \mathbb{Q}) \times \operatorname{Gal}\left(\mathbb{Q}\left(\zeta_{m}\right) / \mathbb{Q}\right),$

hence abelian

as recalled below， the Galois group of a compositum

$K_{1} \cdots K_{r}$

of Galois extensions

$K_{i} / F$

is isomorphic to a subgroup of the direct product of the

$\left.\operatorname{Gal}\left(K_{i} / F\right)\right).$

Let

$\mathfrak{q}$

be a prime of

lying above a ramified prime

$\mathfrak{p} \mid p;$

as above， the completion

$L_{\mathfrak{q}}$

$\mathfrak{q}$

is a finite abelian extension of

$\mathbb{Q}_{p},$

since

$L / \mathbb{Q}$

is finite abelian， and we have

$L_{\mathfrak{q}}=K_{\mathfrak{p}} \cdot \mathbb{Q}_{p}\left(\zeta_{m}\right).$

Let

$F_{\mathfrak{q}}$

be the maximal unramified extension of

$\mathbb{Q}_{p}$

$L_{\mathfrak{q}}.$

Then

$L_{\mathfrak{q}}/F_{\mathfrak{q}}$

is totally ramified

and

$\operatorname{Gal}\left(L_{\mathfrak{q}} / F_{\mathfrak{q}}\right)$

is isomorphic to the inertia group

$I_{p}:=I_{\mathfrak{q}} \subseteq \operatorname{Gal}(L / \mathbb{Q}),$

the

$I_{\mathfrak{q}}$

all coincide because

$L / \mathbb{Q}$

is abelian

It follows from that

$K_{\mathfrak{p}} \subseteq F_{\mathfrak{q}}\left(\zeta_{p^{n_{p}}}\right),$

since

$K_{\mathfrak{p}} \subseteq \mathbb{Q}_{p}\left(\zeta_{m_{p}}\right)$

and

$\mathbb{Q}_{p}\left(\zeta_{m_{p}} / p^{n_{p}}\right)$

is unramified， and that

$L_{\mathfrak{q}}=F_{\mathfrak{q}}\left(\zeta_{p^{n} p}\right),$

since

$\mathbb{Q}_{p}\left(\zeta_{m_{p}} / p^{n_{p}}\right)$

is unramified。 Moreover， we have

$F_{\mathfrak{q}} \cap \mathbb{Q}_{p}\left(\zeta_{p^{n_{p}}}\right)=\mathbb{Q}_{p},$

since

$\mathbb{Q}_{p}\left(\zeta_{p^{n_{p}}}\right) / \mathbb{Q}_{p}$

is totally ramified， and it follows that

$I_{p} \simeq \operatorname{Gal}\left(L_{\mathfrak{q}} / F_{\mathfrak{q}}\right) \simeq \operatorname{Gal}\left(\mathbb{Q}_{p}\left(\zeta_{p^{n_{p}}}\right) / \mathbb{Q}_{p}\right) \simeq\left(\mathbb{Z} / p^{n_{p}} \mathbb{Z}\right)^{\times}.$

Now let

be the group generated by the union of the groups

$I_{p} \subseteq \operatorname{Gal}(L / \mathbb{Q})$

for

$p \mid m.$

Since

$\operatorname{Gal}(L / \mathbb{Q})$

is abelian， we have

$\bigcup I_{p} \subseteq \prod I_{p},$

thus

$\# I \leq \prod_{p \mid m} \# I_{p}=\prod_{p \mid m} \#\left(\mathbb{Z} / p^{n_{p}} \mathbb{Z}\right)^{\times}=\prod_{p \mid m} \phi\left(p^{n_{p}}\right)=\phi(m)=\left[\mathbb{Q}\left(\zeta_{m}\right): \mathbb{Q}\right].$

Each inertia field

$L^{I_{p}}$

is unramified at

as is

$L^{I} \subseteq L^{I_{p}}.$

$L^{I} / \mathbb{Q}$

is unramified， and therefore

$L^{I}=\mathbb{Q},$

Thus

$[L: \mathbb{Q}]=\left[L: L^{I}\right]=\# I \leq\left[\mathbb{Q}\left(\zeta_{m}\right): \mathbb{Q}\right],$

and

$\mathbb{Q}\left(\zeta_{m}\right) \subseteq L, \text { so } L=\mathbb{Q}\left(\zeta_{m}\right)$

as claimed and

$K \subseteq L=\mathbb{Q}\left(\zeta_{m}\right).$

$\square$

To prove the local Kronecker-Weber theorem we first reduce to the case of cyclic extensions of prime-power degree。 Recall that if

$L_{1}$

and

$L_{2}$

are two Galois extensions of a field

then their compositum

$L:=L_{1} L_{2}$

is Galois over

with Galois group

$\operatorname{Gal}(L / K) \simeq\{\left(\sigma_{1}, \sigma_{2}\right):\left.\sigma_{1}\right|_{L_{1} \cap L_{2}}=\left.\sigma_{2}\right|_{L_{1} \cap L_{2}}\} \subseteq \operatorname{Gal}\left(L_{1} / K\right) \times \operatorname{Gal}\left(L_{2} / K\right).$

The inclusion on the

$\text{RHS}$

is an equality if and only if

$L_{1} \cap L_{2}=K.$

Conversely， if

$\operatorname{Gal}(L / K) \simeq H_{1} \times H_{2}$

then by defining

$L_{2}:=L^{H_{1}}$

and

$L_{1}:=L^{H_{2}}$

we have

$L=L_{1} L_{2}$

with

$L_{1} \cap L_{2}=K,$

and

$\operatorname{Gal}\left(L_{1} / K\right) \simeq H_{1}$

and

$\operatorname{Gal}\left(L_{2} / K\right) \simeq H_{2}.$

It follows from the structure theorem for finite abelian groups that we may decompose any finite abelian extension

into a compositum

$L=L_{1} \cdots L_{n}$

of linearly disjoint cyclic extensions

$L_{i} / K$

of prime-power degree。 If each

$L_{i}$

lies in a cyclotomic field

$\mathbb{Q}\left(\zeta_{m_{i}}\right),$

then so does

Indeed，

$L \subseteq \mathbb{Q}\left(\zeta_{m_{1}}\right) \cdots \mathbb{Q}\left(\zeta_{m_{n}}\right)=\mathbb{Q}\left(\zeta_{m}\right),$

where

$m:=m_{1} \cdots m_{n}.$

To prove the local Kronecker-Weber theorem it thus suffices to consider cyclic extensions

$K / \mathbb{Q}_{p}$

of prime power degree

$\ell^{r}.$

There two distinct cases：

$\ell \neq p$

and

$\ell=p .$

Kronecker-Weber theorem for

$\ell \neq p$

$\text{Proposition}$

$K / \mathbb{Q}_{p}$

$\ell^{r}$

$\ell \neq p.$

$\mathbb{Q}_{p}(\zeta_{m}).$

Let

be the maximal unramified extension of

$\mathbb{Q}_{p}$

then

$F=\mathbb{Q}_{p}\left(\zeta_{n}\right)$

for some

$n \in \mathbb{Z}_{\geq 1},$

The extension

is totally ramified， and it must be tamely ramified， since the ramification index is a power of

$\ell \neq p.$

we have

$K=F(\pi^{1 / e})$

for some uniformizer

$\pi,$

with

We may assume that

$\pi=-p u$

for some

$u \in \mathcal{O}_{F}^{\times},$

since

$F/ \mathbb{Q}_{p}$

is unramified： if

$\mathfrak{q} \mid p$

is the maximal ideal of

$\mathcal{O}_{F}$

then the valuation

$v_{\mathfrak{q}}$

extends

$v_{p}$

with index

$e_{\mathfrak{q}}=1,$

$v_{\mathfrak{q}}(-p u)=v_{p}(-p)=1.$

The field

$K=F(\pi^{1 / e})$

lies in the compositum of

$F((-p)^{1 / e})$

and

$F(u^{1 / e}),$

and we will show that both fields lie in a cyclotomic extension of

$\mathbb{Q}_{p}.$

The extension

$F(u^{1 / e}) / F$

is unramified， since

$v_{\mathfrak{q}}\left(\operatorname{disc}\left(x^{e}-u\right)\right)=0$

for

$p \nmid e,$

$F(u^{1 / e}) / \mathbb{Q}_{p}$

is unramified and

$F(u^{1 / e})=\mathbb{Q}_{p}(\zeta_{k})$

for some

$k \in \mathbb{Z}_{\geq 1}.$

The field

$K(u^{1 / e})=K \cdot \mathbb{Q}_{p}(\zeta_{k})$

is a compositum of abelian extensions， so

$K(u^{1 / e}) / \mathbb{Q}_{p}$

is abelian， and it contains the subexten-sion

$\mathbb{Q}_{p}((-p)^{1 / e}) / \mathbb{Q}_{p},$

which must be Galois

since it lies in an abelian extension

and totally ramified

since it is an Eisenstein extension

The field

$\mathbb{Q}_{p}((-p)^{1 / e})$

con-tains

$\zeta_{e}$

take ratios of roots of

$x^{e}+p$

and is totally ramified， but

$\mathbb{Q}_{p}(\zeta_{e}) / \mathbb{Q}_{p}$

is unramified

since

so we must have

$\mathbb{Q}_{p}(\zeta_{e})=\mathbb{Q}_{p}.$

Thus

$\mathbb{Q}_{p}((-p)^{1 / e}) \subseteq \mathbb{Q}_{p}((-p)^{1 /(p-1)})=\mathbb{Q}_{p}(\zeta_{p}).$

It follows that

$F((-p)^{1 / e})=F \cdot \mathbb{Q}_{p}((-p)^{1 / e}) \subseteq \mathbb{Q}_{p}(\zeta_{n}) \cdot \mathbb{Q}_{p}(\zeta_{p}) \subseteq \mathbb{Q}_{p}(\zeta_{n p}).$

We then have

$K \subseteq F(u^{1 / e}) \cdot F((-p)^{1 / e}) \subseteq \mathbb{Q}(\zeta_{k}) \cdot \mathbb{Q}(\zeta_{n p}) \subseteq \mathbb{Q}(\zeta_{k n p})$

and may take

$\square$

$\text{Lemma}$

$\mathbb{Q}_{p}((-p)^{1 /(p-1)})=\mathbb{Q}_{p}(\zeta_{p}).$

Let

$\alpha=(-p)^{1 /(p-1)}.$

Then

$\alpha$

is a root of the Eisenstein polynomial

$x^{p-1}+p,$

so the extension

$\mathbb{Q}_{p}(-p)^{1 /(p-1)}=\mathbb{Q}_{p}(\alpha)$

is totally ramified of degree

and

$\alpha$

is a uniformizer Let

$\pi=\zeta_{p}-1 .$

The minimal polynomial of

$\pi$

$f(x):=\frac{(x+1)^{p}-1}{x}=x^{p-1}+p x^{p-2}+\cdots+p,$

which is Eisenstein， so

$\mathbb{Q}_{p}(\pi)=\mathbb{Q}_{p}(\zeta_{p})$

is also totally ramified of degree

and

$\pi$

is a uniformizer。 We have

$u:=-\pi^{p-1} / p \equiv 1 \bmod \pi,$

is a unit in the ring of integers of

$\mathbb{Q}_{p}(\zeta_{p}).$

If we now put

$g(x)=x^{p-1}-u$

then

$g(1) \equiv 0 \bmod \pi$

and

$g^{\prime}(1)=p-1 \not \equiv 0 \bmod \pi,$

so we can lift

to a root

$\beta$

$\mathbb{Q}_{p}(\zeta_{p}).$

We then have

$p \beta^{p-1}=p u=-\pi^{p-1},$

$(\pi / \beta)^{p-1}+p=0,$

and therefore

$\pi / \beta \in \mathbb{Q}_{p}\left(\zeta_{p}\right)$

is a root of the minimal polynomial of

$\alpha.$

Since

$\mathbb{Q}_{p}(\zeta_{p}).$

is Galois， this implies that

$\alpha \in \mathbb{Q}_{p}(\zeta_{p}),$

and since

$\mathbb{Q}_{p}(\alpha )$

and

$\mathbb{Q}_{p}(\zeta_{p})$

both have degree

the two fields coincide。

$\square$

To complete the proof of the local Kronecker-Weber theorem， we need to address the case

$\ell=p.$

Kummer theory

Let

be a positive integer and let

be a field of characteristic prime to

that contains a primitive

th root of unity

$\zeta_{n}.$

While we are specifically interested in the case where

is a local or global field， in this section

can be any field that satisfies these conditions。

For any

$a \in K,$

the field

$L=K(\sqrt[n]{a})$

is the splitting field of

$f(x)=x^{n}-a$

over

the notation

$\sqrt[n]{a}$

denotes a particular

th root of

but it does not matter which root we pick because all the

th roots of

lie in

$f(\alpha)=f(\beta)=0$

then

$\alpha / \beta \in \zeta_{n}^{i} \in K$

for some

$0 \leq i<n$

and

$K(\alpha)=K(\beta)).$

The polynomial

is separable， since

is prime to the characteristic of

so L is a Galois extension of

and

$\operatorname{Gal}(L / K)$

is cyclic， since we have an injective homomorphism

$\begin{aligned} \operatorname{Gal}(L / K) & \hookrightarrow\left\langle\zeta_{n}\right\rangle \simeq \mathbb{Z} / n \mathbb{Z} \\ \sigma & \mapsto \frac{\sigma(\sqrt[n]{a})}{\sqrt[n]{a}}. \end{aligned}$

This homomorphism is an isomorphism if and only if

is irreducible。

Kummer’s key observation is that the converse holds。 In order to prove this we firstrecall a basic （but often omitted） lemma from Galois theory。

$\text{Lemma}$

$\operatorname{Aut}_{K}(L)$

$L \rightarrow L.$

Suppose not。 Let

$f:=c_{1} \sigma_{1}+\cdots+c_{r} \sigma_{r}=0$

with

$c_{i} \in L, \sigma_{i} \in \operatorname{Aut}_{K}(L),$

and

minimal； we must have

the

$c_{i}$

nonzero， and the

$\sigma_{i}$

distinct。 Choose

$\alpha \in L$

$\sigma_{1}(\alpha) \neq \sigma_{r}(\alpha)$

possible since

$\sigma_{1} \neq \sigma_{r}$

We have

$f(\beta)=0$

for all

$\beta\in L,$

and the same applies to

$f(\alpha \beta)-\sigma_{1}(\alpha) f(\beta),$

which yields a shorter relation

$c_{2}^{\prime} \sigma_{2}+\cdots+c_{r}^{\prime} \sigma_{r}=0,$

where

$c_{i}^{\prime}=c_{i} \sigma_{i}(\alpha)-c_{i} \sigma_{1}(\alpha)$

with

$c_{1}^{\prime}=0,$

which is nontrivial because

$c_{r}^{\prime} \neq 0,$

a contradiction。

$\square$

$\text{Corollary}$

$\langle\sigma\rangle$

$\zeta_{n}.$

$\sigma(\alpha)=\zeta_{n} \alpha$

$\alpha \in L.$

The automorphism

$\sigma$

is a linear transformation of

with characteristic polynomial

$\text{Lemma}$

this must be its minimal polynomial， since

$\{1, \sigma^{1}, \ldots, \sigma^{n-1}\}$

is linearly independent。 Therefore

$\zeta_{n}$

is eigenvalue of

$\sigma,$

and the lemma follows。

$\square$

$\text{Remark}$

$\text{Corollary}$

is a special case of Hilbert’s Theorem 90， which re-places

$\zeta_{n}$

with any element

of norm

$\mathrm{N}_{L / K}(u)=1;$

for example。

$\text{Lemma}$

$n \geq 1$

$\zeta_{n} \in K.$

$L=K(\sqrt[n]{a})$

$a \in K.$

Let

be a cyclic extension of degree

with

$\operatorname{Gal}(L / K)=\langle\sigma\rangle.$

$\text{Corollary}$

there exists an element

$\alpha \in L$

for which

$\sigma(\alpha)=\zeta_{n} \alpha.$

We have

$\sigma\left(\alpha^{n}\right)=\sigma(\alpha)^{n}=\left(\zeta_{n} \alpha\right)^{n}=\alpha^{n},$

thus

$a=\alpha^{n}$

is invariant under the action of

$\langle\sigma\rangle=\operatorname{Gal}(L / K)$

and therefore lies in

Moreover， the orbit

$\{\alpha, \zeta_{n} \alpha, \ldots, \zeta_{n}^{n-1} \alpha\}$

$\alpha$

under the action of

$\operatorname{Gal}(L / K)$

has order

$L=K(\alpha)=K(\sqrt[n]{a})$

as desired。

$\text{Definition}$

Let

be a field with algebraic closure

$\overline{K},$

let

$n \geq 1$

be prime to the characteristic of

and assume

$\zeta_{n} \in K.$

The

is the map

$\begin{aligned} \langle\cdot, \cdot\rangle: \operatorname{Gal}(\overline{K} / K) \times K^{\times} & \rightarrow\left\langle\zeta_{n}\right\rangle \\ (\sigma, a) & \mapsto \frac{\sigma(\sqrt[n]{a})}{\sqrt[n]{a}} \end{aligned}$

where

$\sqrt[n]{a}$

is any

th root of

$\in\overline{K}^{\times} .$

$\alpha$

and

$\beta$

are two nth roots of

then

$(\alpha / \beta)^{n}=1,$

$\alpha / \beta \in\left\langle\zeta_{n}\right\rangle \subseteq K$

is fixed by

$\sigma$

and

$\sigma(\beta) / \beta=\sigma(\beta) / \beta \cdot \sigma(\alpha / \beta) /(\alpha / \beta)=\sigma(\alpha) / \alpha,$

so the value of

$\langle\sigma, a\rangle$

does not depend on the choice of

$\sqrt[n]{a}.$

$a \in K^{\times n},$

then

$\langle\sigma, a\rangle= 1$

for all

$\sigma \in \operatorname{Gal}(\overline{K} / K),$

so the Kummer pairing depends only on the image of

$K^{\times} / K^{\times n};$

thus we may also view it as a pairing on

$\operatorname{Gal}(\overline{K} / K) \times K^{\times} / K^{\times n}.$

$\text{Theorem}$

$n \geq 1$

$\zeta_{n} \in K.$

$\begin{aligned} \Phi: K^{\times} / K^{\times n} & \rightarrow \operatorname{Hom}\left(\operatorname{Gal}(\overline{K} / K),\left\langle\zeta_{n}\right\rangle\right) \\ a & \mapsto(\sigma \mapsto\langle\sigma, a\rangle). \end{aligned}$

For each

$a \in K^{\times}-K^{\times n},$

if we pick an

th root

$\alpha \in \overline{K}$

then the extension

$K(\alpha) / K$

will be non-trivial and some

$\sigma \in \operatorname{Gal}(\overline{K} / K)$

must act nontrivially on

$\alpha.$

For this

$\sigma$

we have

$\langle\sigma, a\rangle \neq 1,$

$a \notin \operatorname{ker} \Phi ;$

thus

$\Phi$

is injective。

Now let

$f: \operatorname{Gal}(\overline{K} / K) \rightarrow\left\langle\zeta_{n}\right\rangle$

be a homomorphism， and put

$d:=\# \operatorname{im} f, H:=\operatorname{ker} f,$

and

$L:=\overline{K}^{H}.$

Then

$\operatorname{Gal}(L / K) \simeq \operatorname{Gal}(\overline{K} / K) / H \simeq \mathbb{Z} / d \mathbb{Z},$

is a cyclic extension of degree

and

$\text{Lemma}$

implies that

$L=K(\sqrt[d]{a})$

for some

$a \in K.$

If we put

and consider the homomorphisms

$\Phi(a^{m e})$

for

$m \in(\mathbb{Z} / d \mathbb{Z})^{\times},$

these homomorphisms are all distinct

because the

$a^{m e}$

are distinct modulo

$K^{\times n}$

and

$\Phi$

is injective

and they all have the same kernel and image as

their kernels have the same fixed field

because

contains all the

th roots of

There are

$\#(\mathbb{Z} / d \mathbb{Z})^{\times}=\# \operatorname{Aut}(\mathbb{Z} / d \mathbb{Z})$

distinct isomorphisms

$\operatorname{Gal}(\overline{K} / K) / H \simeq \mathbb{Z} / d \mathbb{Z},$

one of which corresponds to

and each corresponds to one of the

$\Phi(a^{m e}).$

It follows that

$f = \Phi(a^{m e})$

for some

$m \in(\mathbb{Z} / d \mathbb{Z})^{\times},$

thus

$\Phi$

is surjective。

$\square$

Given a finite subgroup

$K^{\times} / K^{\times n},$

we can choose

$a_{1}, \ldots, a_{r} \in K^{\times}$

so that the images

$\overline{a}_{i}$

of the

${a}_{i}$

$K^{\times} / K^{\times n}$

form a basis for the abelian group

this means

$A=\left\langle\overline{a}_{1}\right\rangle \times \cdots \times\left\langle\overline{a}_{r}\right\rangle \simeq \mathbb{Z} / n_{1} \mathbb{Z} \times \cdots \times \mathbb{Z} / n_{r} \mathbb{Z},$

where

$n_{i}|n$

is the order of

$\overline{a}_{i}$

For each

${a}_{i},$

the fixed field of the kernel of

$\Phi\left(\overline{a}_{i}\right)$

is a cyclic extension of

isomorphic to

$L_{i}:=K(\sqrt[n_i]{a_i}),$

as in the proof of

$\text{Theorem }$

The fields

$L_{i}$

are linearly disjoint over

because the

${a_i}$

correspond to independent generators of

and their compositum

$L = K(\sqrt[n_1]{a_1}, \cdots\sqrt[n_r]{a_r})$

has Galois group

$\operatorname{Gal}(L / K) \simeq A,$

an abelian group whose exponent divides

such fields

are called

Conversely， given an

-Kummer extension

we can iteratively apply

$\text{Lemma}$

to put

in the form

$L = K(\sqrt[n_1]{a_1}, \cdots,\sqrt[n_r]{a_r})$

with each

$a_{i} \in K^{\times}$

and

$n_{i}|n ,$

and the images of the

${a}_{i}$

$K^{\times} / K^{\times n}$

then generate a subgroup

corresponding to

as above。We thus have a 1-to-1 correspondence between finite subgroups of

$K^{\times} / K^{\times n}$

and （finite）

-Kummer extensions of

this correspondence also extends to infinite subgroups provided we put a suitable topology on the groups

So far we have been assuming that

contains all the

th roots of unity。 To help handle situations where this is not necessarily the case， we rely on the following lemma。

$\text{Lemma}$

$n \in \mathbb{Z}_{>1},$

$K=F(\zeta_{n}),$

$L=K(\sqrt[n]{a})$

$a \in K^{\times} .$

$\omega: \operatorname{Gal}(K / F) \rightarrow(\mathbb{Z} / n \mathbb{Z})^{\times}$

$\zeta_{n}^{\omega(\sigma)}=\sigma\left(\zeta_{n}\right).$

$\sigma(a) / a^{\omega(\sigma)} \in K^{\times n}$

$\sigma \in \operatorname{Gal}(K / F).$

Let

$G=\operatorname{Gal}(L / F),$

let

$H=\operatorname{Gal}(L / K) \subseteq G,$

and let

be the subgroup of

$K^{\times} / K^{\times n}$

generated by

The Kummer pairing induces a bilinear pairing

$H \times A \rightarrow\left\langle\zeta_{n}\right\rangle$

that is compatible with the Galois action of

$\operatorname{Gal}(K / F) \simeq G / H.$

In particular， we have

$\langle h, a^{\omega(\sigma)}\rangle=\langle h, a\rangle^{\omega(\sigma)}=\sigma(\langle h, a\rangle)=\langle h^{\sigma}, \sigma(a)\rangle=\langle h, \sigma(a)\rangle$

for all

$\sigma \in \operatorname{Gal}(K / F)$

and

$h \in H;$

the Galois action on

is by conjugation

lift

$\sigma$

and conjugate there

but it is trivial because

is abelian

$h^{\sigma}=h).$

The isomorphism

$\Phi$

induced by the Kummer pairing is injective，so

$a^{\omega(\sigma)} \equiv \sigma(a) \bmod K^{\times n}.$

$\square$

The local Kronecker-Weber theorem for

$\ell=p>2$

We are now ready to prove the local Kronecker-Weber theorem in the case

$\ell=p>2.$

$\text{Theorem}$

$K / \mathbb{Q}_{p}$

$p^{r}.$

$\mathbb{Q}_{p}(\zeta_{m}).$

There are two obvious candidates for

namely， the cyclotomic field

$\mathbb{Q}_{p}(\zeta_{p^{p^r}-1}),$

which is an unramified extension of degree

$p^{r},$

and the index

subfield of the cyclotomic field

$\mathbb{Q}_{p}(\zeta_{p^{r+1}}),$

which is a totally ramified extension of degree

$p^{r}$

the

$p^{r+1}$

-cyclotomic polynomial

$\Phi_{p^{r+1}}(x)$

has degree

$\phi(p^{r+1})=p^{r}(p-1)$

and remains irreducible over

$\mathbb{Q}_{p}).$

is contained in the compositum of these two fields then

$K \subseteq \mathbb{Q}_{p}(\zeta_{m}),$

where

$m:=(p^{p^{r}}-1)(p^{r+1})$

and the theorem holds。 Otherwise， the field

$K(\zeta_{m})$

is a Galois extension of

$\mathbb{Q}_{p}$

with

$\operatorname{Gal}\left(K(\zeta_{m} / \mathbb{Q}_{p}\right) \simeq \mathbb{Z} / p^{r} \mathbb{Z} \times \mathbb{Z} / p^{r} \mathbb{Z} \times \mathbb{Z} /(p-1) \mathbb{Z} \times \mathbb{Z} / p^{s} \mathbb{Z},$

for some

the first factor comes from the Galois group of

$\mathbb{Q}_{p}(\zeta_{p^{p^r}-1}),$

the second two factors come from the Galois group of

$\mathbb{Q}_{p}(\zeta_{p^{r+1}})$

note

$\mathbb{Q}_{p}(\zeta_{p^{r+1}}) \cap \mathbb{Q}_{p}(\zeta_{p^{p^r}-1})=\mathbb{Q}_{p}$

and the last factor comes from the fact that we are assuming

$K \nsubseteq \mathbb{Q}_{p}(\zeta_{m}),$

$\operatorname{Gal}(K(\zeta_{m}) / \mathbb{Q}_{p}(\zeta_{m}))$

is nontrivial and must have order

$p^{s}$

for some

$s \in[1, r],$

It follows that the abelian group

$\operatorname{Gal}(K(\zeta_{m}) / \mathbb{Q}_{p})$

has a quotient isomorphic to

$(\mathbb{Z} / p \mathbb{Z})^{3},$

and the subfield of

$K(\zeta_{m})$

corresponding to this quotient is an abelian extension of

$\mathbb{Q}_{p}$

with Galois group isomorphic

$(\mathbb{Z} / p \mathbb{Z})^{3}.$

$\text{Lemma}$

below， no such field exists。

$\square$

To prove that

$\mathbb{Q}_{p}$

admits no

$(\mathbb{Z} / p \mathbb{Z})^{3}$

-extensions our strategy is to use Kummer theory to show that the corresponding subgroup of

$\mathbb{Q}_{p}(\zeta_{p})^{\times} / \mathbb{Q}_{p}(\zeta_{p})^{\times p}$

given by

$\text{Theorem}$

must have

-rank

and therefore cannot exist。 For an alternative proof that uses higher ramification groups instead of Kummer theory。

$\text{Lemma}$

$\mathbb{Q}_{p}$

$(\mathbb{Z} / p \mathbb{Z})^{3}.$

Suppose for the sake of contradiction that

is an extension of

$\mathbb{Q}_{p}$

with Galois group

$\operatorname{Gal}(K / \mathbb{Q}_{p}) \simeq(\mathbb{Z} / p \mathbb{Z})^{3}.$

Then

$K/\mathbb{Q}_{p}$

is linearly disjoint from

$\mathbb{Q}_{p}(\zeta_{p}) / \mathbb{Q}_{p},$

since the order of

$G:=\operatorname{Gal}(\mathbb{Q}_{p}(\zeta_{p}) / \mathbb{Q}_{p}) \simeq(\mathbb{Z} / p \mathbb{Z})^{\times}$

is not divisible by

and

$\operatorname{Gal}(K(\zeta_{p}) / \mathbb{Q}_{p}(\zeta_{p})) \simeq(\mathbb{Z} / p \mathbb{Z})^{3}$

is a

$_{p}$

-Kummer extension。 There is thus a subgroup

$A \subseteq \mathbb{Q}_{p}(\zeta_{p})^{\times} / \mathbb{Q}_{p}(\zeta_{p})^{\times p}$

isomorphic to

$(\mathbb{Z} / p \mathbb{Z})^{3},$

for which

$K(\zeta_{p})=\mathbb{Q}_{p}(\zeta_{p}, A^{1 / p}),$

where

$A^{1 / p}:=\{\sqrt[p]{a}: a \in A\}$

here we identify elements of

by representatives in

$\mathbb{Q}_{p}(\zeta_{p})^{\times}$

that are determined only up to

th powers

For any

$a \in A,$

the extension

$\mathbb{Q}_{p}(\zeta_{p}, \sqrt[p]{a}) / \mathbb{Q}_{p}$

is abelian， so by

$\text{Lemma}$

we have

$\sigma(a) / a^{\omega(\sigma)} \in \mathbb{Q}_{p}(\zeta_{p})^{\times p}$

for all

$\sigma \in G,$

where

$\omega: G \stackrel{\sim}{\longrightarrow}(\mathbb{Z} / p \mathbb{Z})^{\times}$

is the isomorphism defined by

$\sigma(\zeta_{p})=\zeta_{p}^{\omega(\sigma)}.$

The field

$\mathbb{Q}_{p}(\zeta_{p})$

is a totally tamely ramified extension of

$\mathbb{Q}_{p}$

of degree

with residue field

$\mathbb{Z} / p \mathbb{Z};$

as shown in the proof of

$\text{Lemma}$

we may take

$\pi:=\zeta_{p}-1$

as a uniformizer。

For each

$a \in A$

we have

$v_{\pi}(a)=v_{\pi}(\sigma(a)) \equiv \omega(\sigma) v_{\pi}(a) \bmod p,$

thus

$(1-\omega(\sigma)) v_{\pi}(a) \equiv 0 \bmod p,$

for all

$\sigma \in G,$

hence for al

$\omega(\sigma) \in \omega(G)=(\mathbb{Z} / p \mathbb{Z})^{\times};$

for

this implies

$v_{\pi}(a) \equiv 0 \bmod p.$

Now

is determined only up to

th-powers， so after multiplying by

${\pi}{^{-}}^v{^\pi{^{(a)}}}$

we may assume

$v_{\pi}(a)=0,$

and after multiplying by a suitable power of

$\zeta_{p-1}^{p}=\zeta_{p-1},$

we may assume

$a \equiv 1\bmod \pi,$

since the image of

$\zeta_{p-1}$

generates the multiplicative group

$(\mathbb{Z} / p \mathbb{Z})^{\times}$

of the residue field。

We may thus assume that

$A \subseteq U_{1} / U_{1}^{p},$

where

$U_{1}:=\{u \equiv 1 \bmod \pi\}.$

Each

$u \in U_{1}$

can be written as a power series in

$\pi$

with integer coefficients in

and constant coefficient

We have

$\zeta_{p} \in U_{1},$

since

$\zeta_{p}= 1 + π,$

and

$\zeta_{p}^{b}=1+b \pi+O\left(\pi^{2}\right)$

for integers

$b \in [0, p - 1].$

For

$a \in A\subseteq U_{1},$

we can choose

so that for some integer

$c \in [0, p - 1]$

and

$e \in \mathbb{Z}_{\geq 2}$

we have

$a=\zeta_{p}^{b}(1+c \pi^{e}+O(\pi^{e+1})).$

For

$\sigma \in G$

we have

$\frac{\sigma(\pi)}{\pi}=\frac{\sigma(\zeta_{p}-1)}{\zeta_{p}-1}=\frac{\zeta_{p}^{\omega(\sigma)}-1}{\zeta_{p}-1}=\zeta_{p}^{\omega(\sigma)-1}+\cdots+\zeta_{p}+1 \equiv \omega(\sigma) \bmod \pi,$

since each term in the sum is congruent to

modulo

$\pi=(\zeta_{p}-1);$

here we are representing

$\omega(\sigma) \in(\mathbb{Z} / p \mathbb{Z})^{\times}$

as an integer in

Thus

$\sigma(\pi) \equiv \omega(\sigma) \pi \bmod \pi^{2}$

and

$\sigma(a)=\zeta_{p}^{b \omega(\sigma)}(1+c \omega(\sigma)^{e} \pi^{e}+O(\pi^{e+1})).$

We also have

$a^{\omega(\sigma)}=\zeta_{p}^{b \omega(\sigma)}(1+c \omega(\sigma) \pi^{e}+O(\pi^{e+1})).$

As we showed for

above， any

$u \in U_{1}$

can be written as

$u=\zeta_{p}^{b} u_{1}$

with

$u_{1} \equiv 1 \bmod \pi^{2}.$

Each interior term in the binomial expansion of

$u_{1}^{p}=\left(1+O(\pi^{2}\right))^{p}$

other than leading

is a multiple of

$p \pi^{2}$

with

$v_{\pi}(p \pi^{2})=p-1+2=p+1,$

and it follows that

$u^{p}=u_{1}^{p} \equiv 1 \bmod \pi^{p+1}.$

Thus every element of

$U_{1}^{p}$

is congruent to

modulo

$\pi^{p+1},$

and as you will show on the problem set， the converse holds， that is，

$U_{1}^{p}=\{u \equiv 1 \bmod \pi^{p+1}\}.$

We know from

that

$\sigma(a) / a^{\omega(\sigma)} \in U_{1}^{p},$

$\sigma(a)=a^{\omega(\sigma)}(1+O\left(\pi^{p+1}\right))$

and therefore

$\sigma(a) \equiv a^{\omega(\sigma)} \bmod \pi^{p+1}.$

For

this is possible only if

$\omega(\sigma)=\omega(\sigma)^{e}$

for every

$\sigma \in G,$

equivalently， for every

$\omega(\sigma) \in \sigma(G)=(\mathbb{Z} / p \mathbb{Z})^{\times},$

but then

$e\equiv 1\bmod(p - 1)$

and we must have

since

We have shown that every

$a \in A$

is represented by an element

$\zeta_{p}^{b}(1+c \pi^{p}+O(\pi^{p+1})) \in U_{1}$

with

$b, c \in\mathbb{Z} ,$

and therefore lies in the subgroup of

$U_{1} / U_{1}^{p}$

generated by

$\zeta_{p}$

and

$(1 + π^{p}),$

which is an abelian group of exponent

generated by

elements， hence isomorphic to a subgroup of

$(\mathbb{Z} / p \mathbb{Z})^{2}.$

But this contradicts

$A \simeq(\mathbb{Z} / p \mathbb{Z})^{3}.$

$\square$

$\text{Remark}$

In the proof of

$\text{Lemma}$

above， the elements of

$\mathbb{Q}_{p}(\zeta_{p})^{\times} / \mathbb{Q}_{p}(\zeta_{p})^{\times p}$

that lie in

are quite special。 For most

$a \in \mathbb{Q}_{p}(\zeta_{p})^{\times}$

the extension

$\mathbb{Q}_{p}(\zeta_{p}, \sqrt[p]{a}) / \mathbb{Q}_{p}$

will not be abelian， even though the extensions

$\mathbb{Q}_{p}(\sqrt[p]{a}) / \mathbb{Q}_{p}(\zeta_{p})$

and

$\mathbb{Q}_{p}(\zeta_{p}) / \mathbb{Q}_{p}$

both are， and we typically will not have

$v_{\pi}(a) \equiv 0 \bmod p$

consider

The key point is that we started with an abelian extension

$K / \mathbb{Q}_{p},$

$K(\zeta_{p})=K \cdot \mathbb{Q}_{p}(\zeta_{p})$

is an abelian extension containing

$A^{1 / p};$

this ensures that for

$a \in A$

the fields

$\mathbb{Q}_{p}(\zeta_{p}, \sqrt[p]{a})$

are abelian。

$\text{Remark}$

There is an alternative proof to

$\text{Lemma}$

that is much more explicit。One can show that for

the field

$\mathbb{Q}_{p}$

admits exactly

cyclic extensions of degree

the unramified extension

$\mathbb{Q}_{p}(\zeta_{p^{p}-1})$

and the extensions

$\mathbb{Q}_{p}[x] /(x^{p}+p x^{p-1}+p(1+a p)),$

for integers

$a \in[0, p-1];$

This implies that

$\mathbb{Q}_{p}$

cannot have a

$(\mathbb{Z} / p \mathbb{Z})^{3}$

extension， since this would imply the existence of

$p^{2}+p+1$

cyclic extensions of degree

one for each index

subgroup of

$(\mathbb{Z} / p \mathbb{Z})^{3}.$

For

there is an extension of

$\mathbb{Q}_{2}$

with Galois group isomorphic to

$(\mathbb{Z} / 2 \mathbb{Z})^{3},$

the cyclotomic field

$\mathbb{Q}_{2}(\zeta_{24})=\mathbb{Q}_{2}(\zeta_{3}) \cdot \mathbb{Q}_{2}(\zeta_{8}),$

so the proof we used for

will not work。 However we can apply a completely analogous argument。

$\text{Theorem}$

$K/\mathbb{Q}_{2}$

$2^{r}.$

$\mathbb{Q}_{2}(\zeta_{m}).$

The unramified cyclotomic field

$\mathbb{Q}_{2}(\zeta_{2^{2^{r}}-1})$

has Galois group

$\mathbb{Z} / 2^{r} \mathbb{Z},$

and the totally ramified cyclotomic field

$\mathbb{Q}_{2}(\zeta_{2^{r}+2})$

has Galois group

$\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2^{r} \mathbb{Z}$

up to isomorphism

Let

$m=(2^{2^{r}}-1)(2^{r+2}).$

is not contained in

$\mathbb{Q}_{2}(\zeta_{m})$

then

$\operatorname{Gal}(K(\zeta_{m}) / \mathbb{Q}_{2}) \simeq \begin{cases}\mathbb{Z} / 2 \mathbb{Z} \times(\mathbb{Z} / 2^{r} \mathbb{Z})^{2} \times \mathbb{Z} / 2^{s} \mathbb{Z} & \text { with } 1 \leq s \leq r \\ \text { or } \\ (\mathbb{Z} / 2^{r} \mathbb{Z})^{2} \times \mathbb{Z} / 2^{s} \mathbb{Z} \quad &\text { with } 2 \leq s \leq r\end{cases}$