Reference/advice for infinite Kummer extension $\mathbb{Q}_p(\sqrt[p^{\infty}]{p}, \zeta_{p^{\infty}})$ I am looking for help with a good reference for infinite Kummer extensions, or a check if I am on the right lines. In particular I need to refer to the Galois group of the field extension  $\mathbb{Q}_p(\sqrt[p^{\infty}]{p}, \zeta_{p^{\infty}})$ over $\mathbb{Q}_p$.
I know that Serre's Local Fields gives the extension $\mathbb{Q}_p( \zeta_{p^{\infty}})$ over $\mathbb{Q}_p$ and from there we have that the Galois group is $\mathbb{Z}^*_p$, i.e. it is generated by  $\langle \sigma\rangle$ such that for any $n$ we have $\sigma(\zeta_{p^n}) =(\zeta_{p^n})^{a}$ for some $a \in (\mathbb{Z}/ n\mathbb{Z})^*$
As the field $\mathbb{Q}_p(\sqrt[p^{\infty}]{p}, \zeta_{p^{\infty}})$  is given by $\bigcup_{n \geqslant 1}\mathbb{Q}_p(\sqrt[p^{n}]{p}, \zeta_{p^{n}})$ it seems that the Galois group is generated by two elements $\langle \sigma, \tau \rangle$ where for any $n \geqslant 1$ we have $\sigma$ acts as above on $\zeta_{p^n}$, and $\tau(\sqrt[p^{n}]{p}) = \zeta_{p^n}\cdot \sqrt[p^{n}]{p}$. 
Here is the gap in my understanding, I seem to recall that $\sigma$ should act trivially on $\sqrt[p^{n}]{p}$, and similiarly $\tau$ should act trivially on $\zeta_{p^n}$, in which case for any $n\geqslant 1$, if $\sigma(\zeta_{p^n}) =(\zeta_{p^n})^{a}$ we can recover the relation $\sigma^{-1}\tau\sigma = \tau^{-a}$ 
If this is correct, how do we justify that  $\sigma$ acts trivially on $\sqrt[p^{n}]{p}$ and that $\tau$ acts trivially on $\zeta_{p^n}$? Am I missing something obvious? Or am I on the wrong track?
Thanks in advance for any help. If it is standard stuff please feel free to just give a reference rather than go to any trouble with a detailed reply
 A: There are two questions here, one explicit, one implicit. Let’s take a fixed $n$ here to avoid the typographical mess, much as possible.
We have two extensions of $\Bbb Q_p$, $A=\Bbb Q_p(\zeta)$, where $\zeta$ is a chosen primitive $p^n$-th root of $1$, an extension of degree $(p-1)p^{n-1}$, abelian and in fact cyclic except in the case $p=2$. And we have $R=\Bbb Q_p(\beta)$, where $\beta$ is a chosen $p^n$-th root of $p$. Not normal at all, but “radical” of degree $p^n$, so I’ve called it $R$. And of course $K=AR$ is normal over $\Bbb Q_p$.
The implicit question, which I thought, at dinner, you must have asked explicitly, is whether $[K:\Bbb Q_p]=(p-1)p^{2n-1}$, or in other words whether $A\cap R=\Bbb Q_p$. Do you know that? I spent quite a bit of time proving this to my own satisfaction by nonelementary means, but maybe you have an elementary proof.
I’ll grant a positive answer to the above implicit question; the explicit question now is how a generator $\sigma$ of $\text{Gal}^A_{\Bbb Q_p}$ (let’s ignore the case $p=2$) interacts with an element $\tau$ of the Galois group that leaves $\zeta$ fixed but sends $\beta$ to $\zeta\beta$.
At this point we have to be careful: $\sigma$ has been defined originally only on $A$, and it has many extensions to all of $K$; but it’s not hard to cook up a $\bar\sigma$ that agrees with $\sigma$ on $A$ and leaves $\beta$ fixed. Notice that there’s no question about $\tau$: it was actually defined as an element of $\text{Gal}^K_A$, which of course is naturally a subgroup of the whole Galois group. But if you take another root of $p$, ay $\beta'=\zeta^m\beta$, it’s just as good as $\beta$ is, but $\bar\sigma$ will not leave it fixed. If you were hoping that $\sigma$ left every conjugate of $\beta$ fixed, that is a forlorn hope, ’cause the Galois group is not the product of the separate groups, it’s their semidirect product.
Let me know what you think about the degree question.
