This is problem 2.1 of Falko Algebra I Fields and Galois Theory. I have finished the proof but it looks very unsatisfying.
$E/K$ is a field extension. $L_1,L_2$ are intermediate fields. $L_1L_2$ denotes the composite field formed by $L_1$ and $L_2$. Suppose $[L_1:K],[L_2:K]<\infty$. Prove if $[L_1L_2:K]=[L_1:K][L_2:K]$, then $L_1\cap L_2=K$.
I can show by presenting the basis of $L_1L_2/K$. If $K\subset L_1\cap L_2$ is proper, this means $L_1\cap L_2$ has non-trivial overlap(i.e. I can start removing redundant basis elements from either $L_1$ or $L_2$). This will show $[L_1L_2:K]<[L_1:K][L_2:K]$.
This proof lacks elegance though simple. Is there a proof based on purely showing algebraic manipulation to complete the proof(i.e. showing either $[L_1\cap L_2:K]\leq 1$, $[L_1L_2: L_1\cap L_2]=[L_1:K][L_2:K]$ or other equivalent relations)?
Why should I expect this is the case without looking at the basis?(i.e. Assume I do not know the proof but I want to know the reason why this algebraic relation leads to this special vector space construction.)
If $[L_1:K],[L_2:K]$ are relatively prime, this becomes trivial. In my view point $L_1L_2=L_1\otimes_K L_2$, relatively primeness implies for simple extensions $L_1,L_2$ over $K$, the $K-$algebra can be generated by the simple tensor of the two generators. How do I generalize this statement for arbitrary case by assuming finite extension say $L_1=K(a_1,\dots, a_m),L_2=K(b_1,\dots b_n)$ where I assume $m,n$ are minimal generator and $[L_1:K]=m_1,[L_2:K]=m_2$ which may not have any straightforward relationship with $m$ and $n$ respectively?(i.e. I want a straightforward proof showing that for $[L_1:K],[L_2:K]$ relatively prime, I can present an explicit construction of basis of $L_1\otimes_K L_2$ This statement does not say $[L_1:K],[L_2:K]$ must be relatively prime here.