$[F(a,b):F(a)]=[F(b):F]\iff F(a)\cap F(b)=F$ Let $F$ be a field and $a,b$ be elements of some algebraic extension of $F$. Is it true that $[F(a,b):F(a)]=[F(b):F]\iff F(a)\cap F(b)=F$? 
I have a proof for the forward implication:
Let $c\in F(a)\cap F(b)$. Then, $[F(a,b):F(a)]=[F(a,c,b):F(a,c)]\leq [F(c,b):F(c)]=\dfrac{[F(b):F]}{[F(c):F]}$, and so $[F(c):F]=1$ and $c\in F$.
Does the reverse implication also hold?

The implication doesn't hold. You have that
$$[F(a,b):F]=\underbrace{[F(a,b),F(a)]}_{=[F(b):F]}[F(a):F]=[F(a):F][F(b):F].$$
Now $[F(a)\cap F(b):F]$ divide $[F(a):F]$ and $[F(b):F]$. Therefore, $[F(a)\cap F(b):F]=1$ if $\gcd([F(a):F],[F(b):F])=1$.
Then, if $F(a,b)=F(a)F(b)$, it will hold since $$[F(a)F(b):F]=[F(a):F][F(b):F]\iff \gcd([F(a),F],[F(b);F])=1.$$
If $F(a,b)\neq F(a)F(b)$, your equality will not hold. I let you find such an example (and there are many).
