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For $a\in k$ and each $n$, choose a lift $x_n\in {\mathcal{O}}_K$ of the unique $p^n$th root $a^{p^{-n}}$, which exists because $k$ is a perfect field. DefineChanging $x_n$ by an element of the maximal ideal changes its $p^n$th power by an element whose valuation tends to infinity, so the limit exists and is independent of all choices. In characteristic $p$, the Frobenius endomorphism satisfies $(x+y{)}^{p^n}=x^{p^n}+y^{p^n}$, making $[\cdot ]$ a ring homomorphism lifting the identity on $k$. If $s:k\to {\mathcal{O}}_K$ is any other such section, thenfor every $n$, and the same limiting construction forces $s(a)=[a]$. This proves uniqueness of the Teichmuller lift.

Choose a uniformizer $t$. Repeatedly subtracting the lift of the residue and dividing by $t$ gives every $x\in {\mathcal{O}}_K$ a unique convergent Teichmuller expansionBecause the lift is a ring map, this identifies ${\mathcal{O}}_K$ with $k[[t]]$ and its fraction field with the Laurent series field $k((t))$. This is the equal-characteristic complete discretely valued field classification.

If $K$ is locally compact, its compact valuation ring has only finitely many disjoint residue-class balls. Thus $k$ is finite, as also follows from the local compactness criterion for a complete non-Archimedean field.

Suppose first that the valuation is discrete and the residue field $k$ is finite. For a uniformizer $\pi$, every quotient ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$ is finite, and completeness givesThis inverse limit is compact. Since ${\mathcal{O}}_K$ is a compact neighborhood of zero, $K$ is locally compact.

Conversely, local compactness gives a compact ball about zero, which can be rescaled to make ${\mathcal{O}}_K$ compact. Its distinct residue classes are disjoint open balls of radius below one, so compactness forces the residue field to be finite. Cover ${\mathcal{O}}_K$ by finitely many balls of some radius $r<1$. Applying the ultrametric inequality to centers lying in the maximal ideal produces $\rho <1$ such that every nonunit has absolute value at most $\rho$. Hence the value group has a largest value below one, and the valuation is discrete. This proves the local compactness criterion for a complete non-Archimedean field.

An algebraically closed valued field has an $n$th root of every element. If its valuation were discrete and $\pi$ were a uniformizer, then $v(\sqrt[n]{\pi })=1/n$ would contradict discreteness. Therefore an algebraically closed non-Archimedean field cannot be locally compact.