Take a finite separable extension $L/K$ of non-archimedean local fields. For example, $K = \mathbb{Q}_p$ and $L$ some finite extension. We know that we can decompose $L/K$ into a tower $L/K_0/K$ where $K_0/K$ is unramified and $L/K_0$ is totally ramified.
Can we also do it the other way around? I.e., find $L/M/K$ with $M/K$ totally ramified and $L/M$ unramified?
And if not, can you perhaps give an example that illustrates this?
In a cyclic Galois extension $L/K$ of degree $q^2$, where $q$ is prime, there is only one intermediate field $K_0$ strictly between $L$ and $K$. If we have $e(L/K) = f(L/K) = q$, then there is an unramified extension of $K$ in $L$ with degree $q$ over $K$, so that extension must be $K_0$ and $L/K_0$ is totally ramified. If $M$ is an intermediate field such that $M/K$ is totally ramified then $M \not= L$ (since $e(L/K)= q < [L:K]$) and $M \not= K_0$ (since $e(K_0/K)= 1 < [K_0:K]$), so $M = K$, but $L/K$ is not unramified (since $e(L/K) = q > 1$). So no $M$ of the desired type exists.
I'll present some examples with $K = \mathbf Q_p$ and $q = 2$: cyclic degree $4$ Galois extensions of $\mathbf Q_p$ with $e = f = 2$. They have just one intermediate quadratic extension.
Example. Let $L = \mathbf Q_5(\sqrt[4]{50})$ and $K= \mathbf Q_5$. Setting $\alpha = \sqrt[4]{50} = \sqrt{5}\sqrt[4]{2}$, we have $\alpha^2 = \sqrt{50} = 5\sqrt{2}$, so $\alpha^2/5 = \sqrt{2}$ and $F := \mathbf Q_5(\alpha^2)= \mathbf Q_5(\sqrt{2})$ is unramified quadratic over $\mathbf Q_5$.
In $F[x]$, $x^4 - 50 = x^4 - (5\sqrt{2})^2 = (x^2-5\sqrt{2})(x^2+5\sqrt{2})$ and both quadratic factors are Eisenstein (since $5$ is prime in $F$). Thus $[L:F] = 2$, so $[L:\mathbf Q_5]=4$ and $L/\mathbf Q_5$ is a Kummer extension, so $F$ is the only intermediate quadratic extension.
Example. Let $p \equiv 1 \bmod 4$ and $r$ in $\mathbf Z_p$ be a quadratic nonresidue mod $p$. Set $L = \mathbf Q_p(\sqrt[4]{p^2r})$. By the same reasoning as in the previous example, $L/\mathbf Q_p$ is a Kummer extension of degree $4$ with $e = f = 2$. (The previous example is the special case $p = 5$ and $r = 2$.)