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.
Best Answer
"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.