[Local Class Field Theory 4] The Maximal Abelian Extensions of Local Fields

We will construct the maximal abelian extension of a local field $K$ in this section. It gives us the solution of the Hilbert's twelfth problem for local fields, note that by using this result, we can solve the Hilbert's twelfth problem for $\mathbb{Q}$, which is a corollary of the global Kronecker-Weber's theorem. We will see that the global Kronecker-Weber's theorem can be deduced from the local Kronecker-Weber's theorem.

As in the previous section, we will fix a local field $K$, and its Lubin-Tate's tower

$$K\subset K_1\subset ...\subset K_n\subset K_\infty$$

The goal if this section is to prove that $K^{\text{ab}}=K_\infty K^{\text{un}}$, where $K^{\text{ab}}$ is the maximal abelian extension of $K$, $K^{\text{un}}$ is the maximal unramified extension of $K$. We first need to get familiar with (un)ramified extensions for compositum of fields.

Lemma 4.1. Let $L$ be a finite unraimfied extension of $K_\infty$, then $L=K_\infty L'$ for some $L'$: finite unramified extension of $K_n$, for some $n$.

Proof. We recall that a finite extension $M/F$ is an unramified extension of fields if $e(M|F)=1$, and $M$ is separable over $F$. Hence, $L/K_\infty$ is unramified implies that $L/K_\infty$ is separable. Because $L/K_\infty$ is finite, this yields $L=K_\infty(\alpha)$, by the primitive element theorem. Let $f(X)=a_0+a_1X+...+X^n$ is a minimal polynomial of $\alpha$. This yields $\overline{f(X)}$ is irreducible and separable in $k_{K_\infty}$-the residue field of $K_\infty$. Also, because $K_\infty=\cup_{n\ge 1}K_n$, there exists some $n$ such that all $a_i\in K_n$. Let $L'=K_n(\alpha)$, one has $\overline{f(X)}$ is also irreducible in $k_{K_n}$, and it is separable there, since $k_{K_n}$ is finite, and hence, perfect. From this, one can see $L'/K_n$ is unramified. And $L=K_\infty L'$. (Q.E.D)

Lemma 4.2. Let $L'$ be a finite unramified extension of $K_n$, then $L'=K_nL''$, for some unramified extension $L''$ over $K$.

Proof. Because $K_n$ is also a local field, and because $K_n/K$ is totally ramified, we have $k_K=k_{K_n}$. Also $L'/K_n$ is an unramified extension implies that $L'=K_n(\zeta)$, where $\zeta$ is a $q^m-1$-th root of unity in $K_n$. This yiels $\zeta$ is also $q^m-1$-th root of unity in $K$. Hence $K(\alpha)/K$ is unramified. Let $L'':=K(\alpha)$, we have $L'=K_nL''$ (Q.E.D)

By the two lemmas above, we obtain

Corollary 4.3. Let $L$ be a finite unramified extension of $K_\infty$, then $L\subset K_\infty K^{\text{un}}$.

Proof. By Lemma 4.1, we have $L\subset K_\infty L'$, for some $L'/K_n$ is unramified extension, for some $n$. By Lemma 4.2, $L=K_nL''$, for some $L''/K$ is unramified extension. This yields $L\subset K_\infty K^{\text{un}}$. (Q.E.D)

We will need more about Galois groups (extensions) of compositum of fields.

Lemma 4.4. Let $F$ be a field, $L/F, K/F$ is finite Galois extensions, with $L,K\subset \overline{F}$, then so is $L\cap K/F, LK/F$.

Proof. Obvious.

Lemma 4.5. Let $F, L,K$ be as above, then 

$$Gal(KL/F)=\{(\sigma,\psi)\in Gal(L/F)\times Gal(K/F)|\sigma_{L\cap K}=\psi_{L\cap K}\}$$

Proof. It can be seen that the map $Gal(KL/F)\to Gal(K/F)\times Gal(L/F)$ defined by

$$\sigma\mapsto (\sigma_{|K}, \sigma_{|L})\subset \{(\sigma,\psi)\in Gal(L/F)\times Gal(K/F)|\sigma_{L\cap K}=\psi_{L\cap K}\}$$

is an injective map. We will prove that this map is in fact an isomorphism, by deducing they have the same number of elements.

Let $L\cap K=M$, $k:=\#Gal(KL:K), l:=\#Gal(KL/L), n:=\#Gal(M/F)$ and $A:=Gal(KL/K), B:=Gal(KL/L)$, then it can be seen that $A\cap B=id_M, (KL)^{AB}=M$. And hence, $Gal(KL/M)\cong A\times B$. This yields $\#Gal(KL/M)=kl$. And from this, one obtains $[L:M]=k, [K:M]=l, \#Gal(KL/F)=nkl$.

Due to Lemma 4.4, and the primitive element, we can write $M=F(\alpha), K=F(\alpha, \beta), L=F(\alpha,\gamma)$. Let $\alpha_1:=\alpha,...,\alpha_n$ be all Galois conjugates of $\alpha$, similarly for $\beta_1,...,\beta_l$ and $\gamma_1,...,\gamma_k$. From this, the group in the LHS of the statement is

$$\{(\sigma,\psi)|\sigma(\alpha)=\alpha_i, \sigma(\beta)=\beta_j, \psi(\alpha)=\alpha_i,\psi(\gamma)=\gamma_k\}$$

So, in particular, we have this group has $nkl$ elements. This yields the injective map in the beginning of the proof is an isomorphism. (Q.E.D)

We are now ready for the second corollary.

Corollary 4.6. Let $L/K_\infty$ be a finite abelian extension of exponent $m$, then $L\subset L_tK_m$, for $K_m/K_\infty$: an unramified extension of degree $m$, and $L_t/K_\infty$ is a totally ramified abelian extension.

Proof.  Consider $G:=Gal(LK_m/K_\infty)$, by the previous lemma, this group is also finite, abelian. Let $\sigma$ be an element in this group, then $((\sigma|_L)^m=(\sigma)|_{K_m})^m=id$, so we get $G$ is an abelian group of exponent $m$.

Let $\tau\in G$ such that $\tau|_{K_m}$ is the Frobenius element, then $\tau^m=1$. And hence, we can express $G=\langle \tau\rangle \times H$. Let $L_t:=(LK_m)^{\langle \tau \rangle}$, then $Gal(LK_m/L_t)=\langle \tau \rangle$. Also, $Gal(LK_m/K_m)=H$, since $Gal(K_m/K_\infty)=\langle \tau \rangle$.

Hence, by the previous lemma, $L_t\cap K_m=K_\infty$. Let $L'$ be an intermediate field between $L_t$ and $K_\infty$, such that $L_t/L'$ is totally ramified and $L'/K_\infty$ is unramified. From the hypothesis, one gets $Gal(L'/K)$ is a cyclic group of exponent $m$, i.e. $L'$ is also a subfield of $K_m$. This yields $L'=K_\infty$. And hence, $L_t/K_\infty$ is an abelian totally ramified extension, and $L\subset L_tK_m$. (Q.E.D)

The following result is more difficult to prove, because it is quite long, and require results of ramification groups (See for example Milne's note on CFT (Lemma 4.9)).

Fact 4.7. Let $L$ be an abelian totally ramified extension of  $K$, if $K_\infty \subset L$, then $K_\infty = L$.

This result yields $K_\infty$ is a maximal abelian totally ramified extension of $K$. We are now ready for the main result

Theorem 4.8. $K^{\text{ab}}=K_\infty K^{\text{un}}$. And $Gal(K^{\text{ab}}/K)\cong \mathscr{O}_K^\times \times \hat{\mathbb{Z}}$.

Proof. Let $K'$ be any a finite abelian extension of $K$. Then $L:=K'K_{\infty}$ is a finite abelian extension of $K_{\infty}$. By Corollary 4.6, $L\subset L_tK_m$, for some $L_t/K_\infty$ is a finite abelian totally ramified extension, and $K_m/K_{\infty}$ is a finite unramified extension. By Fact 4.7, $L_t=K_{\infty}$, and by Corollary 4.3, $K_m\subset K_{\infty} K^{\text{ab}}$. This yields $K'\subset K_{\infty} K^{\text{un}}$. Hence, $K^{\text{ab}}\subseteq K_{\infty} K^{\text{un}}$. The other inclusion is clear. Hence, $K^{\text{ab}}=K_{\infty} K^{\text{un}}$.

For the second statement, we can see that $K^{\text{un}}\cap K_{\infty}=K$, hence, the statement about Galois group follows from (infinite version) of Lemma 4.5.

(Q.E.D)