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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.

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3 Answers 3

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"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.

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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.

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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 $K_{p}$.

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

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