# Show $\text{Gal}(K_\infty/\mathbb Q)\cong \mathbb Z_p^{\times}$

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.

• @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 3 '15 at 21:49
• 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 3 '15 at 21:51
• @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 3 '15 at 21:59
• 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 4 '15 at 12:26
• @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 5 '15 at 20:15