abstract algebra - $E/K$ is field extension. $L_1,L_2$ are intermediate fields. $[L_1L_2:K]=[L_1:K][L_2:K]$ implies $K=L_1cap L_2$.

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 $[L_1L_2:K]<[L_1:K][L_2:K]$.

Questions:

1. 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)?




2. 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.)




3. 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.

Answer

This indeed is a bit laboured. Suppose $L_1\cap L_2\ne K$.

Then $|L_1\cap L_2:K|=d>1$. Let $n_i=|L_i:K|$. Then $|L_i:L_1\cap L_2| =n_i/d$. Also $|L_1L_2:L_1|\le|L_2:L_1\cap L_2|=n_2/d$, so

$$|L_1L_2:L_1\cap L_2|=|L_1L_2:L_1||L_1:L_1\cap L_2|\le n_1n_2/d^2$$
and then
$$|L_1L_2:K|=|L_1L_2:L_1\cap L_2||L_1\cap L_2:K|\le n_1n_2/d

Here I used the principle $|LM:L|\le|M:K|$ where $L$ and $M$ are
extensions of $K$ and $LM$ is a compositum of $L$ and $M$ over $K$.
This boils down to noting that if $M=K(\alpha)$ then $LM=L(\alpha)$.