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I'm starting to learn about $\mathbb{Z}_p$-extensions and looking at extensions $\mathbb{Q}(\zeta_{p^\infty})$ which have galois group $\mathbb{Z}_p^{\times}$ over the rationals. To get a $\mathbb{Z}_p$-extension, there's an isomorphism $\mathbb{Z}_p^{\times}\simeq(\mathbb{Z}/p)^{\times}\times\mathbb{Z}_p$ that allows us to drop to the galois group of the field fixed by $(\mathbb{Z}/p)^{\times}$ over the rationals. I'm struggling to understand this isomorphism. Could I get some help understanding this?

amWhy
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Krill
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    This is only true for odd primes $p$, for the record. – Thorgott Jun 09 '22 at 00:39
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    You need to prove that $a\mapsto (1+p)^a$ is (uniformly) continuous in the $p$-adic topology on $\Bbb{Z}$, so it stays well-defined for $a\in \Bbb{Z}p$, and that (for $p$ odd) $(1+p)^\Bbb{Z}$ is dense in $1+p\Bbb{Z}_p$, from which $1+p\Bbb{Z}_p \cong \Bbb{Z}_p$ the isomorphism being $a\mapsto (1+p)^a$. Finally by Hensel lemma $\Bbb{Z}_p^\times = \langle \zeta{p-1}\rangle \times (1+p\Bbb{Z}_p)$. – reuns Jun 09 '22 at 00:40
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    In other words, the exact sequence $1\to 1+p\mathbb Z_p\to\mathbb Z_p^\times\to(\mathbb Z/p)^\times\to 1$ splits. – Kenta S Jun 09 '22 at 00:44
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    Compare also https://math.stackexchange.com/q/2081512/96384, https://math.stackexchange.com/q/1296665/96384, https://math.stackexchange.com/q/2507887/96384, https://math.stackexchange.com/q/3704137/96384, and possibly more. – Torsten Schoeneberg Jun 09 '22 at 02:11
  • See also https://mathoverflow.net/questions/30572/what-is-the-automorphism-group-of-the-additive-group-of-the-p-adic-integers – Nicky Hekster Jun 09 '22 at 12:42

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