Let $(K,|\cdot|)$ be a field with a complete nonarchimedean valuation, $L/K$ be an extension of degree $n$. We know that $|\cdot|$ extends uniquely to a complete valuation on $L$. Let $\mathcal{O}_K$, $\mathcal{O}_L$ be their valuation ring respectively, and $\mathfrak{m}_K$, $\mathfrak{m}_L$ be the unique maximal ideal respectively. We have the ramification index $$ e := (|L^\times|: |K^\times|)$$ (the subgroup index of $|K^\times|$ in $|L^\times|$) and the inertia degree $$ f := [\mathcal{O}_L/\mathfrak{m}_L: \mathcal{O}_K/\mathfrak{m}_K]$$ (the degree of the residue field extension).
Neukirch, Algebraic number theory, p.150-151, proposition 6.8 proves that $ef \le n$, and $ef = n$ if $|\cdot|$ is discrete (see the remark after the proposition). This is remarkable enough because we know that when $|\cdot|$ is non-discrete, then $\mathcal{O}_K$ and $\mathcal{O}_L$ are not Dedekind domains. But I was still wondering if $ef<n$ can occur when $|\cdot|$ is nondiscrete (we restrict ourselves to the case $|\cdot|$ being complete all the time)?
Let $T$ be the maximal tamely ramified extension of $\mathbb{Q}_p$ with $p>2$, and $K$ be the completion of $T_p=T(\mu_{p^{\infty}})$.
It’s clear that $|K^{\times}|=p^{\mathbb{Q}}$: in particular, every ramification index above $K$ is one.
On the other hand, the residue field of $K$ is algebraically closed (since $K$ is the completion of a algebraic extension of $T$, whose residue field is algebraically closed), so there’s no inertia degree either for any algebraic extension of $K$.
Now let $\alpha \in K$ be algebraic over $\mathbb{Q}_p$. Then there’s some $\beta \in T_p$ which is closer to $\alpha$ than any of its $\mathbb{Q}_p$-conjugates. By Krasner’s lemma, $\mathbb{Q}_p(\alpha) \subset \mathbb{Q}_p(\beta) \subset T_p$.
In other words, $T_p$ is the algebraic closure of $\mathbb{Q}_p$ in $K$.
Thus, if we can find $\alpha \notin T_p$ algebraic over $\mathbb{Q}_p$, then $K(\alpha)/K$ fits the bill.
Take $\alpha=p^{1/p} \notin T$. The minimal polynomial of $\alpha$ over $T$ is $X^p-p$, which splits simply in $T(\alpha)$, and $T(\alpha)/T$ has Galois group $\mathbb{Z}/p\mathbb{Z}$ (because of degree).
Suppose $\alpha \in T_p$, then (as $T_p/T$ has Galois group $1+p\mathbb{Z}_p \cong \mathbb{Z}_p$) $T(\alpha)=T(\mu_{p^2})$, hence $\alpha$ is contained in some $L=\mathbb{Q}_p(\mu_{p^2(p^n-1)},p^{1/(p^n-1)})$. Write $M=p^n-1$.
Now, let $C=\mathbb{Q}_p(\mu_p)$, then $C(\alpha) \subset L$, so we have a natural map $Gal(L/C(\mu_{p^2M})) \rightarrow Gal(C(\alpha)/C)$ – but the former group has cardinality dividing $M$ (prime to $p$) while the second group is cyclic of order $p$, so this map is trivial and $C(\alpha) \subset C(\mu_{p^2M})$. In particular, $C(\alpha)$ is abelian over $\mathbb{Q}_p$, hence $\mathbb{Q}_p(\alpha)/\mathbb{Q}_p$ is Galois – we get a contradiction.