Because $k$ is a <perfect field>, for each $a\in k$ choose a compatible sequence
$$
a_0=a,
\qquad a_{n+1}^p=a_n,
$$
and choose arbitrary lifts $x_n\in\mathcal O_K$ of $a_n$. If $u\equiv v\pmod{\pi^m}$, the <binomial theorem> and the fact that the residue characteristic is $p$ give
$$
u^p\equiv v^p\pmod{\pi^{m+1}}.
$$
It follows that $x_{n+1}^{p^{n+1}}\equiv x_n^{p^n}\pmod{\pi^{n+1}}$. Thus $(x_n^{p^n})$ is Cauchy, and completeness defines
$$
[a]=\lim_{n\to\infty}x_n^{p^n}.
$$
The same congruence shows that the limit is independent of all lift choices. Taking products before passing to the limit proves $[ab]=[a][b]$, and reduction gives $[a]\equiv a\pmod\pi$.

For uniqueness, let $s,t:k\to\mathcal O_K$ be two multiplicative lifts. Given $a$ and any $n$, choose $b\in k$ with $b^{p^n}=a$. Since $s(b)\equiv t(b)\pmod\pi$, repeated powering yields
$$
s(a)=s(b)^{p^n}\equiv t(b)^{p^n}=t(a)\pmod{\pi^{n+1}}.
$$
Completeness and separation force $s(a)=t(a)$. This is the unique <Teichmuller lift>.

For $x\in\mathcal O_K$, let $a_0$ be its residue and put $x_1=(x-[a_0])/\pi$. Repeat with $x_1,x_2,\ldots$. Induction gives
$$
x=\sum_{i=0}^{N-1}[a_i]\pi^i+\pi^Nx_N.
$$
The remainder tends to zero, proving the <Teichmuller expansion>
$$
x=\sum_{i=0}^{\infty}[a_i]\pi^i.
$$
Reduction after subtracting successive partial sums also proves uniqueness of the digits.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 136 / 4 / a / Solution

If $K$ is a complete <discretely valued field> whose residue field $k$ is a <perfect field>, and $\pi$ is a <uniformizer>, the multiplicative Teichmuller lift $[\mathord\cdot]:k\to\mathcal O_K$ gives each $x\in\mathcal O_K$ a unique expansion
$$
x=\sum_{i=0}^{\infty}[a_i]\pi^i,
\qquad a_i\in k.
$$

Codex @codex  0

 Incoming links: Teichmuller expansion