Lifting additive characters Let $K$ a finite extension of $\mathbb{Q}_p$ ($p$ prime different from 2) and let $G_K$ the absolute Galois group of $K$. 
Let $\bar{u} : G_K \longrightarrow \mathbb{F}_p$  a continuous additive character. Is it always possible to lift $\bar{u}$ to an additive character $u : G_K \longrightarrow \mathbb{Z}_p$ ?
I know the answer is yes when $K$ does not contains the $p^{th}$-root of unity. What happens in the other case ?
 A: By local class field theory, such a character corresponds to a character
$K^{\times} \to \mathbb F_p$.  Now $K^{\times} \cong \mathbb Z \times \mathcal O_K^{\times}.$  Certainly a character $\mathbb Z \to \mathbb F_p$ can be lifted
to a character $\mathbb Z \to \mathbb Z_p$.  So the question is whether 
$\mathcal O_K^{\times} \to \mathbb F_p$ can be lifted.  (This corresponds to
the restriction of your Galois character to inertia.)
Now $\mathbb O_K^{\times} \cong \mu \times \Gamma$, where $\Gamma$ is isomorphic to a product of copies of $\mathbb Z_p$, and $\mu$ is the subgroup of roots of unity in $K$.  Again, a character $\Gamma \to \mathbb F_p$ can always be lifted, so the question is whether a character $\mu \to \mathbb F_p$ can be lifted
to a character $\mu \to \mathbb Z_p$.
Since $\mu$ is finite, this is possible if and only if $\mu \to \mathbb F_p$ is trivial.  This will be automatic if and only if $\mu$ contains no elements of order $p$, i.e. if and only if $K$ contains no $p$-power roots of unity.
So, if $K$ contains $p$-power roots of unity, then you have to check whether or not your given character $K^{\times} \to \mathbb F_p$ is trivial on these roots of unity.  It lifts to a character $K^{\times} \to \mathbb Z_p$ if and only if it is trivial on them.
