A certain inverse limit

Question

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

Answer 1

1. Yes. It is a consequence of Galois theory : Let $L|K$ be a finite Galois extension, of Galois group $G$. Then there is a decreasing bijection between distinguished subgroups of $G$, and subextensions $L|F|K$ such that $F|K$ is Galois, given by $H\vartriangleleft G \mapsto L^{H}$. Moreover, we have $\textrm{Gal}(L^H|K) \cong G/H$.
   Under the isomorphism $(\mathbb{Z}/p^n\mathbb{Z})^{\times} \cong (\mathbb{Z}/p\mathbb{Z})^{\times} \times (\mathbb{Z}/p^{n-1}\mathbb{Z})$, the subgroup $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is distinguished in $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$, as it's an abelian group. So the fixed subfield by this subgroup, which is $\mathbb{Q}_n$, is Galois, of group $\mathbb{Z}/p^n\mathbb{Z}$.
   ps : you actually wrote it. We only speak of Galois groups for Galois extensions, to avoid confusion.

2. You have the right explenation. The isomorphism $\textrm{Gal}(\mathbb{Q}_{\infty}|\mathbb{Q}) \cong \underset{\leftarrow}{\textrm{lim}} \ \textrm{Gal}(\mathbb{Q}_n|\mathbb{Q})$ apply an automorphism $\sigma$ to $(\sigma\vert_{\mathbb{Q}_n})$, and the maps $\textrm{Gal}(\mathbb{Q}_m|\mathbb{Q}) \rightarrow \textrm{Gal}(\mathbb{Q}_n|\mathbb{Q})$ are simply the restrictions, when $n\leq m$. So you want to prove that the diagrams $\require{AMScd}$ \begin{CD} \textrm{Gal}(\mathbb{Q}_m|\mathbb{Q}) @>{\textrm{res}}>> \textrm{Gal}(\mathbb{Q}_n|\mathbb{Q})\\ @V\sim VV @V\sim VV\\ \mathbb{Z}/p^m\mathbb{Z} @>{\textrm{res}}>> \mathbb{Z}/p^n\mathbb{Z} \end{CD} are commutatives. They are, here why : you hve isomorphisms $(\mathbb{Z}/p\mathbb{Z})^{\times}\times \mathbb{Z}/p^n\mathbb{Z} \cong (\mathbb{Z}/p^{n+1}\mathbb{Z})^{\times}\cong \textrm{Gal}(\mathbb{Q}(\zeta_{p^{n+1}})|\mathbb{Q})$, given by $(a \ \textrm{mod} \ p,b \ \textrm{mod} \ p^n) \mapsto (a^{p^n} + pb \ \textrm{mod} \ p^{n+1})$, and $a \mapsto (\zeta_{p^{n+1}}\mapsto \zeta_{p^{n+1}}^a)$. (in the first isomorphism, the $a^{p^n}$ is here only to assure that there is no dependance in the choice of $a$ modulo $p$) These isomorphisms make the diagrams

\begin{CD} \textrm{Gal}(\mathbb{Q}(\zeta_{p^{m+1}})|\mathbb{Q}) @>{\textrm{res}}>> \textrm{Gal}(\mathbb{Q}(\zeta_{p^{n+1}})|\mathbb{Q})\\ @V\sim VV @V\sim VV\\ (\mathbb{Z}/p^{m+1}\mathbb{Z})^{\times} @>{\textrm{res}}>> (\mathbb{Z}/p^{n+1})^{\times} \\ @V\textrm{quot}VV @V\textrm{quot}VV \\ \mathbb{Z}/p^m\mathbb{Z} @>{\textrm{res}}>> \mathbb{Z}/p^n\mathbb{Z} \end{CD} commutatives. Now, when you restrict the composition $\textrm{Gal}(\mathbb{Q}(\zeta_{p^{m+1}})|\mathbb{Q}) \overset{\sim}{\rightarrow} (\mathbb{Z}/p^{m+1}\mathbb{Z})^{\times} \overset{\textrm{quot}}{\rightarrow} \mathbb{Z}/p^m\mathbb{Z}$ to $\textrm{Gal}(\mathbb{Q}_m|\mathbb{Q})$, you precisely get the isomorphism $\textrm{Gal}(\mathbb{Q}_m|\mathbb{Q}) \cong \mathbb{Z}/p^m\mathbb{Z}$. Thus the first diagrams are indeed commutatives.