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Let $\zeta_{p^n}$ be the primitive $p^n$-th root of unity where $p$ is a prime and $K_n=\mathbb Q(\zeta_{p^n})$ the $p^n$-th cyclotomic field. Let $K_\infty=\bigcup K_n$.

Could someone give a proof of the isomorphism $\text{Gal}(K_\infty/\mathbb Q)\cong \mathbb Z_p^{\times}$?

Many thanks in advance.

pete m
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  • By $\Bbb Z_p^\times$ do you mean the finite group of size $p-1$? Because the Galois group must be infinite, but that's just a finite group. – Gregory Grant May 03 '15 at 21:47
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    @GregoryGrant: No, $\mathbb{Z}_p$ means the $p$-adic integers (and in my opinion, should not be used to denote $\mathbb{Z}/p\mathbb{Z}$). Pete: are you familiar with inverse limits? – RghtHndSd May 03 '15 at 21:49
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    Oh ok, that makes more sense then. I agree there's a conflict of notation, but like it or not $\Bbb Z_p$ is used so often by so many people that it's probably a good idea to qualify it when using it for the $p$-adics. – Gregory Grant May 03 '15 at 21:51
  • Anyway, the $p$-adics are the inverse limit of the the groups $\Bbb Z/p^n\Bbb Z$ so this is not very surprising. – Gregory Grant May 03 '15 at 21:53
  • @RghtHndSd Yes I am. – pete m May 03 '15 at 21:54
  • Does this page help? http://en.wikipedia.org/wiki/Cyclotomic_character – Gregory Grant May 03 '15 at 21:55
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    @RghtHndSd My first thought was to consider how an arbitrary element $g$ of the galois group would act on $\zeta_{p^n}$. If $g(\zeta_{p^n})=\zeta_{p^n}^{a_g}$ then would $a_g$ be coprime to $p^n$? – pete m May 03 '15 at 21:59
  • @ Gregory Grant I think it might help- many thanks. – pete m May 03 '15 at 22:02
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    Pete, that's a good start. Surely $g(\zeta_{p^n})=\zeta_{p^n}^{a_n}$. And, yes, $\gcd(a_n,p)=1$, because otherwise a root of unity of order $p^n$ would be mapped to a lower order root of unity. One of the keys is to see the constraint between $a_n$ and $a_{n+1}$, Can you see what it is? – Jyrki Lahtonen May 04 '15 at 12:26
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    @Jyrki Would I be right in saying $\zeta_{p^{n+1}}^p=\zeta_{p^n}$ so $\zeta^{a_n}{p^n}=g(\zeta{p^n})=g(\zeta_{p^{n+1}}^p)=g(\zeta_{p^{n+1}})^p=(\zeta_{p^{n+1}}^{a_{n+1}})^p=\zeta_{p^n}^{a_{n+1}}$ and so $a_{n+1}\equiv a_n$ $\text{mod}(p^n)$. So would $(a_n)_n\in \mathbb Z_p$ and since each $a_n$ is coprime to $p$ then it is actually in $\mathbb Z_p^{\times}$? regards – pete m May 05 '15 at 20:15
  • Yes. I think that's the correct line of reasoning, Pete! – Jyrki Lahtonen May 05 '15 at 20:17
  • Many thanks for your hint Jyrki. – pete m May 05 '15 at 20:21
  • @Jyrki Lahtonen: why does $g$ map $\zeta_{p^n}$ to another primitive $p^n$-th root of unity? Is it because an element $g$ in the Galois group restricted to $K_n$ is a $\mathbb Q$-automorphism of $K_n$ so that the orders are preserved? Many thanks – user181834 May 06 '15 at 00:14
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    Correct, @user181834. $K_n$ is Galois over $\Bbb{Q}$, so any automorphism of $K_\infty$ must map the subfield $K_n$ to itself. Or, because as an automorphism $g$ is injective. If $g(\zeta_{p^n})$ were not primitive, we would have $g(\zeta_{p^n})^{p^{n-1}}=1$, So $$g(1)=1=g(\zeta_{p^n}^{p^{n-1}}),$$ contradicting injectivity of $g$. – Jyrki Lahtonen May 06 '15 at 05:20
  • @pete_m Related (not identical, of course): http://math.stackexchange.com/questions/104004 – Watson Jun 07 '16 at 10:14

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