explicit uniformizer for the false Tate extension Let $p$ be an odd prime and let $n\geq 1$. Set $K=\mathbb{Q}_p(\zeta_{p^n})$,
$L=\mathbb{Q}_p(\sqrt[p^n]{p})$, and $M=KL$. I claim that $M$ is totally ramified of degree $\phi(p^n)p^n$ (the proof simultaneously shows that $K$ and $L$ are linearly disjoint). Assume not, then there exists an unramified extension $W\subseteq M$ with $[W:\mathbb{Q}_p]=f>1$ ($f$ being necessarily a power of $p$). By the structure theorem for unramified extensions of $\mathbb{Q}_p$, we know that $W=\mathbb{Q}_p(\zeta_{r})$
where $r=p^f-1$. The extension $K$ is totally ramified over $\mathbb{Q}_p$ and therefore linearly disjoint from $W$. We thus have
$$
\mathbb{Q}_p\subseteq K \subsetneqq KW \subseteq M=K(\sqrt[p^n]{p})
$$
By Galois theory, since $f>1$, it follows (this also holds true if $K$ and $L$ are not linearly disjoint) that $\sqrt[p]{p}\in KW$ and therefore
$\mathbb{Q}_p(\zeta_p,\sqrt[p]{p})\subseteq KW$. But $KW$ is abelian over $\mathbb{Q}_p$ and $\mathbb{Q}_p(\zeta_p,\sqrt[p]{p})$ is not. Contradiction.
Q1: How to construct (systematically in $p$ and $n$) an explicit uniformizer of $M$ (for $n>1$ of course)?
Q2 Is it possible to find some uniformizer $\pi'$ of $K$ (so $v_p(\pi')=\frac{1}{\phi(p^n)}$) such that $K(\sqrt[p^n]{\pi'})=M$ (for $n>1$ of course) ?
Remark: Note that Q2 is equivalent to find (by Lagrange's resolvent) $\beta\in M$ such that
$$
\sum_{j=0}^{p^n-1} \sigma^j(\beta)\zeta_{p^n}^j,
$$
is a uniformizer of $M$. Here $\sigma$ is a generator of $Gal(M/K)$. This seems to boil down to some difficult linear algebra.
 A: I think the answer to question Q2 is no. Indeed, Lemma 3 of Birch's paper in Cassels and Fröhlich tells us that two extensions $k(\sqrt[n]{a}),k(\sqrt[n]{b})$ of a field $k$ (of characteristic prime to $n$, and) containing a primitive $n$-th root of unity, coincide if and only if
$$
a=b^r\cdot c^n
$$
with $(r,n)=1$. In your setting, if we had $K(\sqrt[p^n]{p})=K(\sqrt[p^n]{\pi'})$ for a suitable uniformiser $\pi'$ of $K$, taking $\pi'$-adic valuations would give
$$
\phi(p^n)\equiv r\pmod{p^n}
$$
for a suitable $r$ prime to $p$, which is impossible.
A: I gave this question to a student as a summer project. Thanks to Mercio's answer to a similar question posted on StackExchange (link: https://math.stackexchange.com/questions/954731/ ) we have worked out how to find explicit uniformizers of $\mathbb{Q}_p(\zeta_{p^2},\sqrt[p]{p})$ for all odd p. We have written a short article on our calculations, which can be found here: http://arxiv.org/abs/1912.01656 Please let us know if you have any suggestions and/or comments for us.
