Let $E/F/\mathbb Q_p$ be a tower of finite extensions of $p$-adic fields, and fix an algebraic closure. Let $\pi\in F$ be a uniformiser, let $p\nmid s$, and let $\alpha$ be a root of $X^s-\pi$. Then $F(\alpha)/F$ is totally (tamely) ramified of degree $s$. Consider the compositum $E(\alpha)=EF(\alpha)$. Then $E(\alpha)/E$ has degree dividing $s$ (in particular, it is tame). Is it true that $E(\alpha)/E$ is also totally ramified?
A way to show total ramification of $E(\alpha)/E$ would be if $\alpha$ had Eisenstein minimal polynomial over $E$. Its minimal polynomial over $F$ is $X^s-\pi$, and we know that the minimal polynomial over $E$ divides this, but this is as far as I have gotten.
Is there perhaps some obvious counterexample?
No, try $F=\Bbb{Q}_3,E =\Bbb{Q}_3(3^{1/2}),\alpha = (-3)^{1/2}$ then $E(\alpha)/E= E(i)/E$ is unramified.