E ∩ K = F implies [EK : F] = [E : F][K : F]?

Question

I have come across a problem that shows E ∩ K = F if [EK : F] = [E : F][K : F]. I wonder if the converse holds. I know this reduce to showing the the F-basis of E is a K-basis if E ∩ K = F (if this is true), but is stuck here. Thanks.

Answer 1

No. Consider the degree $6$ extension $L=\Bbb Q(\sqrt[3]2,e^{2\pi i/3})$ of $\Bbb Q$, the splitting field of $x^3-2$ over $\Bbb Q$. Then $L=EF$ where $E=\Bbb Q(\sqrt[3]2)$ and $F=\Bbb Q(e^{2\pi i/3}\sqrt[3]2)$. Both of these have degree $3$ and $E\cap F=\Bbb Q$.