Let $K$ be an extension of $\mathbb{Q}_p$ which contains a primitive $n$-th root unity. Also, assume that $p$ does not divide $n$.
Let $L/K$ be a cyclic and totally ramified extension of degree $n$, and $L'/L$ be a cyclic and unramified extension of degree $n$.
Some result about cyclic extension states that there is an element $\beta \in L'$ such that $L' = L(\beta)$ with $\beta^n \in L$, and $n$ is the smallest power of $\beta$ which lies in $L$.
Question: Is there a way of finding an element $\tilde{\beta} \in L$ such that $L = K(\tilde{\beta})$ with $\tilde{\beta}^n \in K$, and $n$ being the smallest power with that property, which uses $\beta^n \in L$?
I am not so sure about this point but from my computations with Sage, it seems to be not enough to just take $\tilde{\beta} = \beta^n$ (because its minimal polynomial does not necessarily split and therefore, the extension of $K$ generated by $\beta^n$ is not totally ramified in this case). This is where I am stuck now.
Any help is appreciated!
Since $L/K$ is totally tamely ramified then $L= K(\pi_K^{1/n})$ and the residue field of $K$ thus of $L$ is $\Bbb{F}_q$ whence $L' = L(\zeta_{q^n-1})$.
Let $\sigma\in Gal(L'/L), \sigma(\zeta_{q^n-1})=\zeta_{q^n-1}^q$ (the canonical lift of the Frobenius of the residue fields, there is a canonical one because $L'/L$ is unramified, which is automatically cyclic)
$$a = \sum_{m=0}^{n-1} \zeta_n^{-m}\sigma^m(\zeta_{q^n-1})\in K(\zeta_{q^n-1})$$ $\sigma^m(a) = \zeta_n^m a$, the $L$-conjugates of $a$ are $\zeta_n^m a$ so that its minimal polynomial is $X^n-a^n \in K[x]$ and $K(\zeta_{q^n-1})=K(a), L'=L(a)$.