Totally ramified extension

A finite extension of local fields is totally ramified when its residue-field degree is one, equivalently when its ramification index equals its field degree.

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 $LM$.

Let $K={\mathbb{Q}}_3$, $L=K(i)$ with $i^2=-1$, and ${\alpha }^2=3(1+i)$. The extension $L/K$ is unramified quadratic, whereas $N=L(\alpha )$ is totally ramified quadratic over $L$, by the Eisenstein polynomial $T^2-3(1+i)$. Also $i={\alpha }^2/3-1$, so $N=K(\alpha )$. The Hensel lemma gives $r\in K$ with $r^2=-2$. Put $\beta =(1+i)/r$; then ${\beta }^2=-i$ and $\beta \overline{\beta }=-1$. Consequently $i\mapsto -i$, $\alpha \mapsto \beta \alpha$ defines an automorphism whose square fixes $i$ and negates $\alpha$. It generates the cyclic Galois group of order four. Its unique quadratic subfield is $L$, which is unramified. A decomposition as an unramified field and a totally ramified field would require a totally ramified quadratic subfield, and therefore cannot exist.

A finite extension $L/K$ of complete discretely valued fields is totally ramified exactly when $L=K(\alpha )$ for a root $\alpha$ of an Eisenstein polynomial over ${\mathcal{O}}_K$. In the reverse direction one may take $\alpha$ to be any uniformizer of $L$.