Let $p$ be an odd prime number and consider the extensions $\mathbb{Q}_p\left((-p)^{\frac{1}{n}}\right)$, $\mathbb{Q}_p\left((-p)^{\frac{2}{n}}\right)$ of the $p$-adic field $\mathbb{Q}_p$.
- Are these fields two equal ?
- Are they totally ramified extension of $\mathbb{Q}_p$ ?
The field $\mathbb{Q}_p\left((-p)^{\frac{1}{n}}\right)$ is obtained by adding the roots of the polynomial $x^n+p$, which is Eisenstein and so it is totally ramified of degree $n$.
The field $\mathbb{Q}_p\left((-p)^{\frac{2}{n}}\right)$ is obtained by adding the roots of the polynomial $x^n+p^2$.
If $n$ is even, then we can write $x^n+p^2=(x^{n/2}+ip)(x^{n/2}-ip)$, where $i=\sqrt{-1} \in \mathbb{Q}_p$.
In other words, $\mathbb{Q}_p\left((-p)^{\frac{2}{n}}\right)=\mathbb{Q}_p\left((-p)^{\frac{1}{n/2}}\right)$.
So the field $\mathbb{Q}_p\left((-p)^{\frac{2}{n}}\right)$ is totally ramified of degree $n/2$.
So $\mathbb{Q}_p\left((-p)^{\frac{1}{n}}\right)$ and $\mathbb{Q}_p\left((-p)^{\frac{2}{n}}\right)$ both are totally ramified but $$\mathbb{Q}_p\left((-p)^{\frac{1}{n}}\right) \neq \mathbb{Q}_p\left((-p)^{\frac{2}{n}}\right).$$
Am I missing something for the case $n=$even ?
Thanks