= Cyclic quartic local extension without a totally ramified complement
{title2=$N=\mathbb Q_3(\alpha),\quad \alpha^4-6\alpha^2+18=0$}
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.
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