If $L_1$, $L_2$ are intermediate fields of $E/K$, then $[L_1 L_2 : K] = [L_1 : K] [L_2 : K] \implies L_1 \cap L_2 = K$.

Question

Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < \infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]\cdot[L_2:K] \implies L_1 \cap L_2 = K$.

Answer 1

One can proceed as follows:

1. If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words, $$ L_1L_2 = \text{span}_{L_2}(S) $$ This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)

2. Once you have that, it shows that $$ [L_1L_2:K] \leq |S| = [L_1:K] \qquad (\ast) $$ (This is one way to prove that $[L_1L_2:K] \leq [L_1:K][L_2:K]$) In your case, since equality holds in $(\ast)$, it also shows that $S$ is $L_2$-linearly independent.

3. Now since $K\subset L_1\cap L_2$, choose a basis $B$ for $L_1\cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $B\subset L_2$, so it must follows that $$ |B|=1 $$ and so $L_1\cap L_2 = K$.