Prove the Teichmuller lifting by using p-adic logarithm and exponential functions

Question

Let $K$ be a finite extension of $\mathbb{Q}_p$. Assume our goal is to show that the group of $(q-1)$-th roots of unity is contained in $K$, where $q$ is the size of the residue field of $K$.

Assume $p>2$ and take $u\in \mathbb{Z}^\times_p$. Then $u^{p-1}\equiv 1\bmod p\mathbb{Z}_p$. Consider $$ v = \exp(\frac{1}{p-1} \log(u^{p-1})). $$

We then have $v^{p-1} = u^{p-1}$ and $v\equiv 1\bmod p\mathbb{Z}_p$. Thus $w:=u/v$ is a $p-1$-th root of unity and $w\equiv u \bmod p\mathbb{Z}_p$. This proves the Teichmuller lifting for $\mathbb{Q}_p$.

Here we use that elements of $\log(1+p\mathbb{Z}_p)$ converge and are contained in $p\mathbb{Z}_p$; also elements of $\exp(p\mathbb{Z}_p)$ converge and are contained in $1+p\mathbb{Z}_p$. We also use that $\exp\circ\log$ is identity on $1+p\mathbb{Z}_p$ when $p>2$.

If $K/\mathbb{Q}_p$ has a large ramification index, then the above argument seems not work; I would like to know how to generalize this one such $K$.

Answer 1

If $u\in O_K$ has order $q-1$ in $O_K/(\pi_K)$ then $\zeta_{q-1} =\lim_{n\to \infty} u^{q^n}$ is a primitive $q-1$-th of unity in $O_K$ and $u=\zeta_{q-1}(1+a\pi_K)$ for some $a\in O_K$.

$O_K/(p)$ is a ring of characteristic $p$ thus $u^{p^r} = \zeta_{q-1}^{p^r} (1+a^{p^r}\pi_K^{p^r})\bmod p$. Take $p^r\ge \frac{1}{v(\pi_K)}$ then

$u^{p^r(q-1)}\in 1+p O_K$,

$\exp(-\frac1{q-1} \log(u^{p^r(q-1)}))=(1+a\pi_K)^{-p^r}$ and

$$u^{p^r} \exp(-\frac1{q-1} \log(u^{p^r(q-1)}))=\zeta_{q-1}^{p^r}$$

Answer 2

Your method is rather slick, and I confess that like @reuns, I never thought of the problem this way.

But you don’t need to use logarithm and exponential to run your trick, as long as you know that if $\alpha\in\Bbb Z_p$ and $z\in 1+\mathfrak m$, where $\mathfrak m$ is the maximal ideal of your local ring (in this case $\mathfrak m=p\Bbb Z_p$), then $z^\alpha$ is always defined, as an element of $1+\mathfrak m$.

In particular, if $z\in1+p\Bbb Z_p$, then there is a $w=z^{\frac1{p-1}}\in1+p\Bbb Z_p$. Thus you take your $u\in\Bbb Z_p^\times$ and form $z=u^{p-1}$ and $w=z^{\frac1{p-1}}\in1+p\Bbb Z_p$ as above, and look at $$ \left(\frac uw\right)^{p-1}=\frac z{w^{p-1}}=1\,. $$ (As for my claim “as long as you know”, you don’t need anything even as strong as Hensel to show it. Just exhibit $\alpha$ as the $p$-adic limit of positive integers, and pull continuity in.)