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I am not sure this question is proper for this site, but there is no other places that I can get an answer. So if anyone can give an answer for this, it would be very helpful to me.

Let $F$ be a finite extension of $\mathbb{Q}_\ell$. Let $F_1$ is a finite Galois extension of $F$ and $F_\infty$ is a Galois extension of $F$ which is also an extension of $F_1$. Suppose we know $\mathrm{Gal}(F_\infty/F_1)\cong \mathbb{Z}_p$ or $\mathbb{Z}_p\times \mathbb{Z}_p$ for some $p\neq \ell$. With these assumption, why the extension $F_\infty/F_1$ should be unramified? The only information I know is the result comes from class field theory, but I cannot catch it.

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Hint: By class field theory, ramified abelian extensions of $F_1$ correspond to subgroups of the unit group of the ring of integers of $\mathcal{O}_{F_1}$, which is isomorphic to $\mu(F_1) \times \mathbf{Z}/\ell^a \times \mathbf{Z}_\ell^d$, a finite group times a pro-$\ell$ group.

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    $\begingroup$ And the extension has Galois group isomorphic to the corresponding subgroup of $\mathcal{O}_{F_1}^\times$. $\endgroup$
    – Aurel
    Commented Dec 11, 2013 at 16:03
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    $\begingroup$ @Timo Keller So if $F_\infty/F_1$ is a ramified extension, then its Galois group must be a subgroup of $\mu(F_1)\times \mathbf{Z}/\ell^a\times \mathbf{Z}^d_\ell$ which is impossible, right? $\endgroup$
    – Kevin.lijh
    Commented Dec 11, 2013 at 16:16
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    $\begingroup$ Quotient group, not subgroup. $\endgroup$
    – user19475
    Commented Dec 11, 2013 at 16:28
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    $\begingroup$ Yes, the Galois group is isomorphic to the quotient by the corresponding subgroup, my comment was incorrect. $\endgroup$
    – Aurel
    Commented Dec 11, 2013 at 17:15
  • $\begingroup$ @TimoKeller I am really sorry, but could I get some references about that correspondence? What I can find in my literature is just the correspondence between unramified extensions and norm groups. $\endgroup$
    – Kevin.lijh
    Commented Dec 11, 2013 at 17:49

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