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For a finite extension $L/K$ of local fields, let $e=e(L/K)$ be its ramification index and $f=f(L/K)=[k_L:k_K]$ its residue-field degree. An unramified extension has $e=1$. A totally ramified extension has $e=[L:K]$, equivalently $f=1$. A tamely ramified extension has separable residue extension and ramification index coprime to the residue characteristic.

Let $\bar{\theta }$ generate the finite extension $k_L/k_K$, and let $\bar{g}$ be its minimal polynomial. Lift $\bar{g}$ to a monic $g\in {\mathcal{O}}_K[X]$ and choose any lift $t\in {\mathcal{O}}_L$ of $\bar{\theta }$. Since finite fields are perfect fields, ${\bar{g}}^{\prime }(\bar{\theta })\ne 0$. The simple-root form of Hensel lemma, applied inside $L$, gives $\theta \in {\mathcal{O}}_L$ withSet $K_0=K(\theta )$. Its residue field contains $k_K(\bar{\theta })=k_L$, soThe equation $g(\theta )=0$ gives the reverse inequality. Thus $[K_0:K]=f$, its residue-field degree is $f$, and $e(K_0/K)=1$; hence $K_0/K$ is unramified. Since $k_{K_0}=k_L$, the extension $L/K_0$ has residue-field degree one and is totally ramified. This constructs the maximal unramified subextension of a local field extension.

Suppose first that $L/K$ is a totally ramified extension of degree $n$, and let $\alpha$ be a uniformizer of $L$. If the valuation on $L$ is normalized by $v_L(\alpha )=1$, then $v_L(K^{\times })=n\mathbb{Z}$. The value group of $K(\alpha )$ already contains both $n\mathbb{Z}$ and $1$, so its ramification index over $K$ is at least $n$. Hence $[K(\alpha ):K]\ge n$, and therefore $L=K(\alpha )$.

Letbe the minimal polynomial of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverwhich means $v_K(a_0)=1$. Thus $f$ is an Eisenstein polynomial.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the Eisenstein criterion makes that polynomial irreducible and its Newton polygon gives $v_L(\alpha )=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha )/K)\ge n$; equality with the field degree forces $e=n$ and residue-field degree one. Thus $K(\alpha )/K$ is totally ramified.

The degree of the residue field extension induced by a finite extension of complete discretely valued fields. Along with the ramification index, it satisfies $[L:K]=ef$. An unramified extension additionally has separable residue extension; a totally ramified extension has residue degree one.

Here $a\ge 0$ and $p\nmid m$. The prime-to-$p$ roots of unity generate an unramified extension of degree equal to the multiplicative order of $p$ modulo $m$; for $m=1$ this degree is one. Indeed, their distinct residues lift uniquely by Hensel lemma, and the Frobenius automorphism acts by raising them to $p$. The $p$-power roots of unity generate a totally ramified extension, because the translated cyclotomic polynomial ${\Phi }_{p^a}(1+T)$ is Eisenstein. The two fields have trivial intersection and their compositum has the same residue degree as the prime-to-$p$ field, by unramified base change preserves total ramification. Thus total ramification is equivalent to $p\equiv 1(\mathrm{mod}m)$.

If $L/K$ is an unramified extension and $M/K$ is a totally ramified extension of non-Archimedean local fields inside a fixed algebraic closure, then $LM/L$ is totally ramified of degree $[M:K]$. A uniformizer of $M$ has an Eisenstein polynomial over $K$. Since a uniformizer of $K$ remains a uniformizer in $L$, that polynomial remains Eisenstein over $L$ and generates $LM$.