Comparison of degrees of field extensions

Question

Let $M$ be a field, and let $F$ and $K \subset L$ be subfields of $M$. Do we have the inequality $$[L \cap F \colon K \cap F] \leq [L \colon K] \, \text{?}$$ Using the multiplicativity of degrees, it is equivalent to $$[K \colon K \cap F] \leq [L \colon L \cap F] \, \text{.}$$ If the inequality holds, is the divisibility relation also true? If it does not, is the inequality true under some mild assumptions?

Answer 1

In general this inequality doesn't hold. Consider the case where $L=FK$ is the compositum of $F$ and $K$. Then we have $[L:L\cap F]=[FK:F]\leq [K:K\cap F]$, i.e. the reversed inequality! Now take $K,F$ to be fields that are not linearly disjoint over $F\cap K$ and such that all degrees are finite in order to get the strict inequality. For example: $K=\Bbb Q(\sqrt[3]{2}),F=\Bbb Q(\zeta\sqrt[3]{2})$, $L=FK$ where $\zeta\in\Bbb C$ is a primitive third root of unity. Then \begin{align} [L:L\cap F]&=2\\ [K:K\cap F]&=3 \end{align} These are probably not 'mild' assumption but: The inequality holds if for example $K,F$ are linearly disjoint over $K\cap F$ and $L,F$ are linearly disjoint over $L\cap F$. In that case we have: $$[K:K\cap F]=[KF:F]\leq[LF:F]=[L:L\cap F]$$ Since $[LF:F]=[LF:KF][KF:F]$, so we even have divisibility in this case.