Ramification of the field extension $\mathbb{Q}_p(\zeta_{p^2},p^{1/p})$

Question

I am interested in the extension of local fields $\mathbb{Q}_p(\zeta_{p^2},p^{1/p})/\mathbb{Q}_p$. Is it totally ramified?

Here are the partial results that I have:

1. It is a field extension of degree $p^2(p-1)$. To see that, note that it suffices to show $p^{1/p} \not \in \mathbb{Q}_p(\zeta_{p^2})$. Indeed, otherwise $\mathbb{Q}_p(p^{1/p}) \subset \mathbb{Q}_p(\zeta_{p^2})$, thus $\mathbb{Q}_p(p^{1/p})/\mathbb{Q}_p$ is an abelian extension, hence contains its conjugate $\zeta_p \cdot p^{1/p}$, hence also contains $\zeta_p$, which is of degree $p-1$ a contradiction.

2. It's Galois group is an explicite group of order $p^{2}(p-1)$, its element its uniquely determined by its action on $\zeta_{p^2}$ and $p^{1/p}$.

3. Its ramification index is at least $p(p-1)$, indeed, $\mathbb{Q}_p(\zeta_{p^2})/\mathbb{Q}_p)$ is a totally ramified extension of degree $p(p-1)$

But I can not figure out whether it has chance to have an unramified subextension.

Answer 1

So you’re taking a $p$-th root of $p$ in $K=\mathbb{Q}_p(\zeta_{p^2})$, a local field that contains all the $p$-th roots of unity. Let $L=K(p^{1/p})$, it has degree $p$ over $K$.

As $K$ is totally ramified over $\mathbb{Q}_p$, $L$ is totally ramified over $\mathbb{Q}_p$ iff $L/K$ is totally ramified. Suppose (for the sake of contradiction) that it is false.

Then $L/K$ is unramified of degree $p$, and thus contained in a cyclotomic extension. Therefore $L/\mathbb{Q}_p$ is contained in a cyclotomic extension, and again we find a contradiction as $\mathbb{Q}_p(p^{1/p})$ is not Galois.