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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 ?

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up vote 4 down vote accepted

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.

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