Proof. Let $x=y+u{\pi }^k$ with $u\in O_K$. Then

$$\begin{array}{llll}\hfill x^p& =\sum _{i=0}^p\left(\genfrac{}{}{0ex}{}{p}{i}\right)y^{p-i}{(u{\pi }^k)}^i& \hfill & \\ \hfill & =y^p+\sum _{i=1}^p\left(\genfrac{}{}{0ex}{}{p}{i}\right)y^{p-1}{(u{\pi }^k)}^i& \hfill & \end{array}$$

Since $O_K∕\pi O_K$ has characteristic $p$, we have $p\in \pi O_K$. Thus

$$\left(\genfrac{}{}{0ex}{}{p}{i}\right){(u{\pi }^k)}^i{y}^{p-i}\in {\pi }^{k+1}{O}_K\forall i\ge 1,$$

hence $x^p\equiv y^{p}mod{\pi }^{k+1}$. □

Proof of Theorem 5.2. Let $a\in k$. For each $i\ge 0$ we choose a lift $y_i\in O_K$ of $a^{\frac{1}{p^i}}$, and we define

$$x_i:=y_i^{p_i}.$$

We claim that ${(x_i)}_{i=1}^{\infty }$ is a Cauchy sequence and its limit is independent of the choice of $y_i$.

By construction, $y_i\equiv y_{i+1}^{p}mod\pi$. By Lemma 5.4 and induction on $k$, we have $y_i^{p^k}\equiv y_{i+1}^{p^{k+1}}$ and hence $x_i\equiv x_{i+1}mod{\pi }^{i+1}$ (take $i=p$). Hence ${(x_i)}_{i=1}^{\infty }$ is Cauchy, so $x_i\to x\in O_K$.

Suppose ${(x_i^{\prime })}_{i=1}^{\infty }$ arises from another choice of $y_i^{\prime }$ lifting $a_i^{\frac{1}{p^i}}$. Then ${(x_i^{\prime })}_{i=1}^{\infty }$ is Cauchy, and $x_i^{\prime }\to x^{\prime }\in O_K$. Let

$$x_i^{″}=\{\begin{array}{cc}{x}_i\hfill & i\text{ even}\hfill \\ x_i^{\prime }\hfill & i\text{ odd}\hfill \end{array}.$$

Then $x_i^{″}$ arises from lifting

$$y_i^{″}=\{\begin{array}{cc}{y}_i\hfill & i\text{ even}\hfill \\ y_i\hfill & i\text{ odd}\hfill \end{array}.$$

Then $x_i^{″}$ is Cauchy and $x_i^{″}\to x$, $x_i^{″}\to x^{\prime }$. So $x=x^{\prime }$ and hence $x$ is independet of the choice of $y_i$. So we may define $[a]=x$.

Then $x_i=y_i^{p^i}\equiv {(a^{\frac{1}{p^i}})}^{p^i}\equiv amod\pi$. Hence $x\equiv amod\pi$. So (i) is satisfied.

We let $b\in k$ and we choose $u_i\in O_K$ a lift of $b^{\frac{1}{p^i}}$, and let $z_i:=u_i^{p^i}t$. Then $\underset{i\to \infty }{\lim }{z}_i=[b]$.

Now $u_i{y}_i$ is a lift of ${(ab)}^{\frac{1}{p^i}}$, hence

$$[ab]=\underset{i\to \infty }{\lim }{x}_i{z}_i=(\underset{i\to \infty }{\lim }{x}_i)(\underset{i\to \infty }{\lim }{z}_i)=[a][b].$$

So (ii) is satisfied.

If $\mathrm{characteristic}K=p$, $y_i+u_i$ is a lift of $a^{\frac{1}{p^i}}+b^{\frac{1}{p^i}}={(a+b)}^{\frac{1}{p^i}}$. Then

$$\begin{array}{llll}\hfill [a+b]& =\underset{i\to \infty }{\lim }{(y_i+u_i)}^{p^i}& \hfill & \\ \hfill & =\underset{i\to \infty }{\lim }{y}_i^{p^i}+u_i^{p^i}& \hfill & \\ \hfill & =\underset{i\to \infty }{\lim }{x}_i+z_i& \hfill & \\ \hfill & =[a]+[b]& \hfill & \end{array}$$

Easy to check that $[0]=0$, $[1]=1$, and hence $[\bullet ]$ is a ring homomorphism.

Uniqueness: let $\varphi :k\to O_K$ be another such map. Then for $a\in k$, $\varphi (a^{\frac{1}{p^i}})$ is a lift of $a^{\frac{1}{p^i}}$. It follows that

$$\begin{array}{llll}\hfill [a]& =\underset{i\to \infty }{\lim }\varphi {(a^{\frac{1}{p^i}})}^{p^i}& \hfill & \\ \hfill & =\underset{i\to \infty }{\lim }\varphi (a)& \hfill & \\ \hfill & =\varphi (a)\square & \hfill & \end{array}$$

Proof.

$$\begin{array}{llll}\hfill a\in k^{\times }& ⟹a\in {𝔽}_{p^n}^{\times }\text{ for some }n& \hfill & \\ \hfill & ⟹{[a]}^{p^n-1}=[a^{p^n-1}]=[1]=1\square & \hfill & \end{array}$$

Proof. Since $K=\mathrm{Frac}(O_K)$, it suffices to show $O_K\cong k[[t]$. Fix $\pi \in O_K$ a uniformiser, and let $[\bullet ]:k\to O_K$ be the Teichmüller map and define

$$\begin{array}{llll}\hfill \phi :k[[t]]& \to O_K& \hfill & \\ \hfill \phi \left(\sum _{i=0}^{\infty }{a}_i{t}^i\right)& =\sum _{i=0}^{\infty }[a_i]{\pi }^i& \hfill & \end{array}$$

Then $\phi$ is a ring homomorphism since $[\bullet ]$ is, and it is a bijection by Proposition 3.4(ii). □