Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.