Let $p$ be an odd prime and $n$ a positive integer. Let $\zeta_{p^{n+1}}$ be a primitive $p^{n+1}$-th root of unity. It can be shown that $Gal(\mathbb Q(\zeta_{p^{n+1}})/\mathbb Q)\cong (\mathbb Z/p\mathbb Z)^\times\times (\mathbb Z/p^n\mathbb Z)$.
Let $\mathbb Q_n= \mathbb Q(\zeta_{p^{n+1}})^{(\mathbb Z/p\mathbb Z)^\times}$ be the subfield of $\mathbb Q(\zeta)$ fixed by the subgroup $(\mathbb Z/p\mathbb Z)^\times$.
Then it's clear that $Gal(\mathbb Q_n/\mathbb Q)\cong \mathbb Z/p^n\mathbb Z$.
Now set $\mathbb Q_\infty=\bigcup_{n\ge 1}\mathbb Q_n$.
Questions- 1) Is $\mathbb Q_n/\mathbb Q$ a Galois extension for each $n$?
2) How can we justify the following:
$$Gal(\mathbb Q_\infty/\mathbb Q)\cong \varprojlim_n(\mathbb Z/p^n\mathbb Z)?$$
For 2), do I consider the inverse system $\big\{Gal(\mathbb Q_n/\mathbb Q);\phi_n^{n+1}\big\}$ where $\phi_n^{n+1}: Gal(\mathbb Q_{n+1}/\mathbb Q)\longrightarrow Gal(\mathbb Q_n/\mathbb Q)$ is the natural restriction homomorphism and the pair $\big(Gal(\mathbb Q_\infty/\mathbb Q),\psi_n\big)$ where the projections $\psi_n:Gal(\mathbb Q_\infty/\mathbb Q)\longrightarrow Gal(\mathbb Q_n/\mathbb Q)$ are the restriction mappings and then consider the homomorphism $$Gal(\mathbb Q_\infty/\mathbb Q)\longrightarrow \varprojlim_n Gal(\mathbb Q_n/\mathbb Q): \sigma\longmapsto (\sigma|\mathbb Q_n)_n$$
Is this an isomorphism via the universal property of the inverse limit?
Am I close?
Thanks
