wildly ramified extensions and $p$-power roots of unity Let $K$ be a finite extension of $\mathbf{Q}_p$. I have this vague intuition that $K$ having a lot of wild ramification is closely related to $K$ being close to containing high degree $p$-power roots of unity.
I was wondering if this intuition is true and if maybe the following result (which would formalise it) is true :
Let $e_K$ be the ramification degree of $K$ over $\mathbf{Q}_p$ then $p^n$ divides $e_K$ if and only if $K(\zeta_p)$ contains $p^n$ roots of unity where $\zeta_p$ is a $p$-root of unity.
If this is false is there a way to salvage the result/my intuition ?
 A: The correct statement (for $p$ odd) is as follows. If $K/\mathbf{Q}_p$ is abelian, then $p^n | e_K$ if and only if $L(\zeta_p)$ contains $\zeta_{p^{n+1}}$ for some unramified extension $L/K$. Similarly, for $p$ odd, $p^{n-1}(p-1) | e_K$ if and only if $K$ contains $\zeta_{p^n}$ for some unramified extension $L$. It is false without the requirement that one can pass to an unramified extension, as can be seen by "crossing" the cyclotomic extension with an unramified one, as in the answer of nguyen (although there are $p+1$ cyclic extensions of $\mathbf{Q}_p$ of degree $p$ rather than $3$). 
It also fails without the abelian hypothesis; there are extensions $K$ with arbitrarily large ramification such that $L \cap \mathbf{Q}_p(\zeta_{p^{\infty}}) = \mathbf{Q}_p$ for all unramified extensions $L/K$. For example, let $E/\mathbf{Q}_p$ be an unramified quadratic extension, and consider $\mathbf{Z}/p^n \mathbf{Z}$ extensions of $E$ on which $\mathrm{Gal}(E/\mathbf{Q}_p)$ acts by minus one.
A: First, I think that your guess should be "$K(\zeta_p)$ contains $\zeta_{p^{n+1}}$", not just $\zeta_{p^n}$. Second, I think there is a counterexample, with $K$ being a cyclic extension of $Q_p$ of degree $p$, totally ramified ($p$ odd). Actually, by local CFT, $Q_p$ admits exactly 3 cyclic extensions $K_i$ of degree $p$, because $Q^*_p/(Q^*_p)^p$ considered (in additive notation) as a vector space over the prime field $F_p$ has dimension 2. The first extension, say $K_0$, is the unramified one. The 2 others, say $K_1$ and $K_2$, are necessarily totally ramified, and their compositum contains $K_0$. Add $\zeta_p$ to get fields $L_i=K_i(\zeta_p)$. If your guess were founded, we would have $L_i=Q_p(\zeta_{p^2})$ for $i=1, 2$, hence also for $L_0$ : impossible, since $L_0$ is unramified over $Q_p(\zeta_p)$.
