Actually a simple question: Does the following hold?
Let $E/F$ be a finite field extension and let $K$ be another field. Do we have $$[E:F] \ge [E \cap K : F \cap K]?$$
Or which assumptions do we need for this to be true (e.g. $E/F$ is a Galois extension or similar)?
I thought about arguing with vector space basis, but $F \cap K$ is a smaller underlying field, increasing the number of basis vectors, and $E \cap K$ is a smaller vector space, decreasing the number of basis vectors.
Otherwise I have thought about $[E:F] [F:F\cap K] = [E:E\cap K][E\cap K:F\cap K]$ and then proving $[E:E\cap K] \ge [F:F\cap K]$.
This need not hold in general.
Let $E=\mathbb{Q}(\sqrt[3]{2},\omega)$, where $\omega$ is a primitive cubic root of unity (a root of $x^2+x+1$). Let $F=\mathbb{Q}(\sqrt[3]{2})$, and let $K=\mathbb{Q}(\omega\sqrt[3]{2})$.
We have $[E:F]=2$, since $F\neq E$ and adding the root of quadratic yields $F$.
We also have $K\cap F=\mathbb{Q}$, and $K\cap E=K$. Thus, $$[K\cap E:K\cap F] = [\mathbb{Q}(\omega\sqrt[3]{2}):\mathbb{Q}]=3,$$ since $\omega\sqrt[3]{2}$ is a root of $x^3-2$, which is irreducible over $\mathbb{Q}$.
Thus, we have $$[E:F] = 2 \lt 3 = [E\cap K:F\cap K].$$
Note that in this example $E$ is Galois over $F$, since it is a quadratic extension of a field of characteristic $0$, so being Galois does not help.
Note that in general, we know that if $F\subseteq E$ and both $E$ and $L$ are subfields of some larger field, then $[EL:FL]\leq [E:F]$.
So when you have $E=FK$, we get $$[E:F]=[FK:F(F\cap K)]\leq [K:F\cap K] = [E\cap K:F\cap K],$$ which means that you will only get the inequality you want if this is an equality. That's how I constructed my counterexample, as $\mathbb{Q}(\sqrt[3]{2},\omega)$ is the compositum of $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\omega)$; but the compositum is of degree $2$ over each of them, whereas they are both of degree $3$ over their intersection.