Cyclic extensions of local fields that are neither totally ramified nor unramified Suppose $K$ is a local field, say $\mathbb{Q}_p$, or something (note that cyclic extension of prime power of a local field answers this, somewhat, for specific local fields $K$). For a specific case, by general Galois theory (c.f. Galois group of a biquadratic quartic), it's known that if $x^4+ax^2+b$ is irreducible and $\alpha$ is a root, $b$ is not a square in $K$, and $b(a^2-4b)$ is a square in $K$, then $L=K(\alpha)$ is Galois over $K$ with Galois group $\mathbb{Z}/4\mathbb{Z}$. 
Under what conditions (for instance, is it possible to explicitly write down an extension?), is it true that the inertia group of $L/K$ is neither the entire Galois group nor trivial? Moreover, in general, for any local field, can we find a cyclic extension of the form $\mathbb{Z}/l^2\mathbb{Z}$ with $l$ prime that has inertia group neither the full group nor trivial (so equal to $\mathbb{Z}/l\mathbb{Z}$)?
 A: Welcome to the community. 
In my opinion the first question is quite vague. Are you asking for arbitrary Galois extensions, abelian...? 
With respect to the second question a fast answer is $K=\mathbb Q_3$, $\ell = 2$, and $L$ the decomposition field of $1 + 4 x - 4 x^2 - x^3 + x^4$. Here is a database of Galois groups of some local fields where one can check the claim.

Let me explain how I obtained that polynomial, based on Jyrki Lahtonen's answer in cyclic extension of prime power of a local field.
Let $p\neq \ell$ be primes, $p$ odd.
It is a consequence of Local Class Field theory that every 
$\mathbb Z/\ell^2$-extension of $\mathbb Q_p$ is a subfield of $K=\mathbb Q_p(\zeta_p, \zeta_{p^{\ell^2}-1})$. The Galois grup being 
$$
Gal(K/\mathbb Q_p)\simeq Gal(\mathbb Q_p(\zeta_{p})/\mathbb Q_p)\times Gal(\mathbb Q_p( \zeta_{p^{\ell^2}-1})/\mathbb Q_p)\simeq \mathbb Z/(p-1) \times \mathbb Z/\ell^2.
$$
The inertia subgroup $I$ of $G$ is $Gal(K/\mathbb Q_p(\zeta_{p^{\ell^2}-1}))$ 
corresponding
to $Gal(\mathbb Q_p(\zeta_p)/\mathbb Q_p)\times \{0\}$ under the (above) natural
isomorphism.
Notice that $G$ has a $\ell^2$-index subgroup $H$ not containing $I$ if and only if $\ell \mid p-1=[\mathbb Q_p(\zeta_p):\mathbb Q_p]$. In this case there are many such $H$ (as pointed out in the comments, see also Subgroups of a direct product). We will take such an $H$ satisfying $G/H$ being cyclic.
As a consequence the field $K^H$ fixed by $H$ answers your second question. 
For the case $p=3, \ell=2$ one has that $K=\mathbb Q_3(\zeta_3, \zeta_{80})=\mathbb Q_3(\zeta_{15})$. Notice that $K$ has 3 degree 2 subfields, $\mathbb Q_3(\zeta_5)$, $\mathbb Q_3(\sqrt{-3}, \sqrt{5})$ and $\mathbb Q_3(\zeta_{15} + \zeta_{15}^{-1})$. The first one is cyclic unramified over $\mathbb Q_3$ and the second one is a $C_2\times C_2$ extension of $\mathbb Q_3$. 
