# Local Kronecker–Weber theorem implies the global one?

I was reading local Kronecker–Weber theorem implies global one in a course manual, but there's some parts I don't understand:

Assume local Kronecker–Weber theorem, that is , every finite abelian extension of $$\mathbb Q_p$$ lies in a cyclotomic field $$\mathbb Q_p(\zeta_m).$$

Now let $$K/\mathbb Q$$ be a finite abelian extension, for each ramified prime $$p$$ of $$\mathbb Q$$, pick a prime $$\mathfrak p|p$$ in $$K$$ and let $$K_{\mathfrak p}$$ be its completion. The extension $$K_{\mathfrak p}/\mathbb Q_p$$ is finite abelian (its galois group is isomorphic to a subgroup of $$\mathrm {Gal}(K/\mathbb Q)$$ by previous theorem), then by assumption $$K_{\mathfrak p}\subset \mathbb Q_p(\zeta_{m_p})$$ for some integer $$m_p\geq 1$$. Now let $$e_p=v_p(m_p)$$ and let $$m = \prod_p p^{e_p}$$.(this is finite since it ranges over ramified primes)

Let $$L=K(\zeta_m)$$, then $$L$$ is Galois and abelian.(Since its Galois group is isomorphic to a subgroup of $$\mathrm{Gal(K/\mathbb Q)}\times\mathrm{Gal}(\mathbb Q(\zeta_m)/\mathbb Q)$$). Let $$\mathfrak q$$ be a prime of $$L$$ lying above one of our choosen $$\mathfrak p|p$$, then the completion $$L_\mathfrak q$$ is a finite abelian extension of $$\mathbb Q_p$$.

Let $$F$$ be the maximal unramified extension of $$\mathbb Q_p$$ in $$L_\mathfrak q$$(it's the union of all $$E \subset L_\mathfrak q$$ with $$E$$ finite unramified over $$\mathbb Q_p$$), Then $$L_\mathfrak q/F$$ is totally ramified, so its Galois group is isomorphic to the inertia group $$I_{\mathfrak q}$$. The field $$F$$ contains roots of unity $$\zeta_n$$ for all $$n|m$$ not divisible by $$p$$ (because extensions $$\mathbb Q_p(\zeta_n)$$ are all unramified), so $$L_\mathfrak q = F(\zeta_m) = F(\zeta_{p^{e_p}})$$.

My question is, why in the last paragraph $$L_\mathfrak q = F(\zeta_m)$$?

We have $$L_\mathfrak q = K_\mathfrak p(\zeta_m)\subset \mathbb Q_p(\zeta_{m_p}, \zeta_m)\subset F(\zeta_{m_p}, \zeta_m)= F(\zeta_{m_p}, \zeta_{p^{e_p}})= F(\zeta_{m_p})$$ clearly, but this is not the result I want so I don't know how to proceed.

"Your" notations are a bit confusing, but let us keep them, just adding $m'=m/p^{e_p}$, so that $p \nmid m'$. Then known classical results on cyclotomic extensions assert that $\mathbf Q_p (\zeta_m)/\mathbf Q_p$ is the compositum of the two linearly disjoint extensions $\mathbf Q_p (\zeta_{m'})$ and $\mathbf Q_p (\zeta_{p^{e_p}})$. In the same way, $L_\mathfrak q=K_\mathfrak p(\zeta_m)$ is the compositum of $\mathbf Q_p (\zeta_{p^{e_p}})$ and $\mathbf Q_p (\zeta_{m''})$ (containing $\mathbf Q_p (\zeta_{m'})$), with $p\nmid m''$. Concerning ramification, the branch $\mathbf Q_p (\zeta_{p^{e_p}})/\mathbf Q_p$ is totally ramified, whereas the branches $\mathbf Q_p (\zeta_{m''})/\mathbf Q_p$ and $L_\mathfrak q/\mathbf Q_p (\zeta_{p^{e_p}})$ are unramified, so that $Gal(L_\mathfrak q /\mathbf Q_p (\zeta_{m''}))\cong Gal(\mathbf Q_p (\zeta_m)/\mathbf Q_p (\zeta_{m'}))\cong Gal(\mathbf Q_p (\zeta_{p^{e_p}})/\mathbf Q_p)$ and the corresponding extensions are totally ramified.

By the above, the inertia subfield $F$ (bad notation, there should be an index indicating that this is a local field) coincides with $\mathbf Q_p (\zeta_{m''})$, and $L_\mathfrak q=F(\zeta_m)=F(\zeta_{p^{e_p}})$ . NB: things get clearer with a Galois diagram.

It's known that $$\mathbb Q_p(\xi_{m'})$$ is unramified over $$\mathbb Q_p$$ for $$m'$$ not divisible by $$p$$, so write $$m_p =p^{e_p}m'$$, we have $$\mathbb Q_p (\xi_{m'})\subset F$$, hence $$F( \xi_{p^{e_p}})=F(\xi_{p^{e_p}},\xi_{m'})\supset \mathbb Q_p(\xi_{m_p})\supset K_p$$.

Now $$L_q\supset F(\xi_{p^{e_p}})=F(\xi_{p^{e_p}}, \xi_m)\supset K_p(\xi_m) = L_q$$, so it's done.

This is a general result from the general theory of local number fields. Let $K$ be a number field and let $K(\alpha)/K$ be a finite extension of degree $n$. Let $p$ be a prime in $K$ and let $\mathfrak{p}$ be a prime above $p$ in $K(\alpha)$. Write $L=K(\alpha)$. Then $$L_{\mathfrak{p}}=K_{p}(\alpha),$$ where $K_{p}$ is the completion of $K$ at the prime $p$ and $L_{\mathfrak{p}}$ is the completion of $L$ at the prime $\mathfrak{p}$. To show this note first that $K_{p}(\alpha)\subseteq L_{\mathfrak{p}}$ because $\alpha\in L\subseteq L_{\mathfrak{p}}$ and $K_{p}\subseteq L_{\mathfrak{p}}$. Conversely, any element of $a\in L_{\mathfrak{p}}$ is a Cauchy-sequence of elements of $K(\alpha)$, say $(a_{i})_{i\in\mathbb{N}}$ with $a_{i}\in K(\alpha)$. Then we can write $$a_{i}=\sum\limits_{j=0}^{n-1}b_{ij}\alpha^{j},$$ where $b_{ij}\in K$. Then $(b_{ij})_{i\in\mathbb{N}}$ is a Cauchy-sequence w.r.t. to $v_{\mathfrak{p}}$ for each $j$ and therefore also a Cauchy-sequence w.r.t. to $v_{p}$ because $C\cdot v_{\mathfrak{p}}$ is an extension of $v_{p}$ for some positive constant $C$. Hence each sequence $(b_{ij})_{i\in\mathbb{N}}$ converges in $K_{p}$ and we conclude that the limit of $(a_{i})_{i\in\mathbb{N}}$ converges in $K_{p}(\alpha)$, i.e. $a\in K_{p}(\alpha)$, so $L_{\mathfrak{p}}\subseteq K_{p}(\alpha)$, whence $L_{\mathfrak{p}}=K_{p}(\alpha)$.

The proof also works in the case where $L=K(\alpha_{1},\ldots,\alpha_{n})$ to show that $L_{\mathfrak{p}}=K_{p}(\alpha_{1},\ldots,\alpha_{n})$ but do note that the $\alpha_{i}$ might not be linearly independent over $K_{p}$.

• My question is why $F(\zeta_m)=L_{\mathfrak q}$?
– CYC
Sep 12, 2018 at 6:13
• @CYC Ah, sorry, I guess I was too sleepy when reading your question. Sep 13, 2018 at 8:05