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Write $q=\#k_K=p^f$. For every $a\in k_K$, the polynomial $F(T)=T^q-T$ has $a$ as a residue root and $\overline{F^{\prime }}=-1$. By the Hensel lemma there is a unique element $[a]\in {\mathcal{O}}_K$ such thatThis defines the Teichmuller lift, including $[0]=0$. The two displayed properties determine each value uniquely by Hensel's uniqueness statement.

They also imply the standard Teichmuller lift properties. The elements $0,1$ already satisfy their defining equations, so $[0]=0$ and $[1]=1$. The product $[a][b]$ reduces to $ab$ and satisfies $([a][b]{)}^q=[a][b]$, so uniqueness gives $[ab]=[a][b]$. Likewise $[a{]}^p$ reduces to $a^p$ and is fixed by its $q$th power, givingFor $a\ne 0$, division of $[a{]}^q=[a]$ by $[a]$ gives $[a{]}^{q-1}=1$. Thus the nonzero Teichmuller representatives are precisely the prime-to-$p$ roots of unity lifting the nonzero residue classes. This is a multiplicative section of reduction; additivity is not asserted in mixed characteristic.

For the convergence, use the normalized discrete valuation $v_K$ with $v_K(\pi )=1$, and put $e=v_K(p)\ge 1$. A useful binomial estimate isIndeed, writing $u=v+\delta$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$, so all terms of $(v+\delta {)}^p-v^p$ have valuation at least $\min \{e+r,pr\}$. The ultrametric inequality gives the bound for their sum.

Finite fields are perfect, so the successive $p$th roots $x_n$ exist uniquely and satisfy $x_n^{p^n}=x$. Let $a_n=[x_n]$. Since $y_n-a_n$ lies in the maximal ideal, iterating the estimate $n$ times givesMultiplicativity, or the Frobenius identity above, gives $a_n^{p^n}=[x_n^{p^n}]=[x]$. ThereforeThis proves the approximation of Teichmuller lifts by iterated pth powers for completely arbitrary choices of the lifts $y_n$, including $x=0$. The original PDF confirms $x\in k_K$, correcting the capital $K_K$ in the TeX transcription.