Does the separable degree equal the L.C.M.? If $F\leq E\leq K$ is any sequence of field extensions, let $\newcommand{\mon}{\textrm{Mon}} \newcommand{\ri}{\rightarrow}\newcommand{\tr}{\textrm} \mon_K(E/F)$ denote the set of monomorphisms $:E\ri K$ that fix $F$.
Given any field, $F$, and any finite extension, $E/F$, one way of defining the separable degree $[E:F]_s$ is as the cardinality of $\mon_N(E/F)$ where $N$ is any extension of $E$ that is normal over $F$. It turns out that this equals $\max\{|\mon_K(E/F)|: K \;\tr{is any extension of E}\}$. My question: is it also true that $|\mon_K(E/F)|$ divides $[E:F]_s$ for all extensions, $K$ of $E$?
The motivation for this question is that an accompanying 'inverse' observation does in fact hold true - namely that $|\tr{Aut}(E/F)|=\min\{|\mon_K(E/F)|: K \;\tr{is any extension of E}\}$ divides $|\mon_K(E/F)|$ for all extensions, $K$ of $E$ (I tried modifying the proof I found of this fact to answer the main question, but it didn't work).
 A: Take an irreducible quintic polynomial $f$ over $\Bbb Q$ with the following property: there are two distinct roots $\alpha,\beta$ of $f$ such that $\Bbb Q(\alpha,\beta)$ contains no other roots of $f$ (this is the case if e.g. $f$ has full Galois group $S_5$). Consider the fields $F=\Bbb Q, E=\Bbb Q(\alpha), K=\Bbb Q(\alpha,\beta)$. By assumption, $\alpha$ and $\beta$ are the only roots of $f$ contained in $K$, so $|\operatorname{Mon}_K(E/F)|=2$ but $[E:F]_s=5$.
Edit: Let $L$ be a splitting field of $f$ over $\Bbb Q$ and assume that $[L:\Bbb Q]>40$. Let $\alpha,\beta$ be any distinct roots of $f$ in $L$. Note that $[K:\Bbb Q]=[K:\Bbb Q(\alpha)][\Bbb Q(\alpha):\Bbb Q]\leq 4\cdot 5$ (in fact, we have equality here which can be shown by a similar version of the following argument). Assume that $K=\Bbb Q(\alpha,\beta)$ contains another root $\gamma$ of $f$. Now the splitting field $L$ of $f$ over $\Bbb Q$ is the same as the splitting field of the 'remaining polynomial' $f/((X-\alpha)(X-\beta)(X-\gamma))\in K[X]$ over $K$. This however is a degree $2$ polynomial, so $[L:K]\leq2$, hence $[L:\Bbb Q]=[L:K][K:\Bbb Q]\leq 2\cdot 4\cdot 5=40$ which contradicts our assumption. Therefore $K$ can only contain two roots of $f$.
TLDR: The more roots contained in $K$, the smaller the splitting field.
