$|F|=q$ (power of a prime) , $|F(u)|=q^a , |F(v)|=q^b$ , and $F=F(u) \cap F(v)$ , then is it true that $|F(u) \cap F(v)|=q^{gcd(a,b)}$ ?

Question

Let $F$ be a finite field , $|F|=q$ (power of a prime) , $|F(u)|=q^a , |F(v)|=q^b$ , and

$F=F(u) \cap F(v)$ , then is it true that $|F(u) \cap F(v)|=q^{gcd(a,b)}$ ?

Writing $|F(u) \cap F(v)|=q^d$ , I can see that $d|a , d|b$ , so $d|gcd(a,b)$ , but I am unable to conclude $d=gcd(a,b)$ . Please help . Thanks in advance