Let $K/F$ be a field extension, $L/F$ and $M/F$ finite subextensions of $K/F$ and $LM$ the composite of $L$ and $M$. I'm trying to prove that $[LM:F] = [L:F][M:F]$ implies the trivial intersection $L\cap M=F$, and that the converse holds when $[L:F]=2$ or $[M:F]=2$.
In order to prove the trivial intersection of the fields, I want to show that $[L\cap M:F]=1$. From the hypothesis, we can obtain $$[LM:L] = [M:F] = [M:L\cap M][L\cap M:F],$$ and $$[LM:M] = [L:F] = [L:L\cap M][L\cap M:F],$$ but I don't if that is usefull for what I'm trying to prove. And I don't know how to prove the converse part. I would appreciate any help you can give me.
I'll draw this diagram of fields in a slightly odd manner, since I'm limited by the MathJax.
$$\require{AMScd}\begin{CD}LM @<<< L \\ @AAA @AAA\\M @<<<L\cap M @<<< F\end{CD}$$
Now we have $$[LM:L]=[M:F]=[M:L\cap M][L\cap M:F],$$ as you've already noted. Note now that $[LM:L]\le [M:L\cap M]$, so that $$[M:L\cap M][L\cap M:F]\le [M:L\cap M],$$ or $$[L\cap M:F]\le 1,$$ which of course implies that $L\cap M=F$.