The connection between roots and primitive element

Question

given a field $F=GF(q)$ and an irreducible polynom of second degree $f(x)$ over $F$ I create the extention field $F'=GF(q^2)$. given $\beta\in F'$ a root of $f(x)$, is $\beta$ is primiive element of $F'$? is there a connection between them?

Answer 1

This is not true, consider the following counterexample. Let $F$ be the field obtained by adjoining a root of $X^2+1$ to $GF(3)$. Then this root is not a primitive element in $GF(9)$, because if $\alpha$ is such a root, then $\alpha^4=1$. However, there is a connection, suppose for example, that for some natural numbers $m$ and $q$, $\beta$ is a primitive element of $GF(q^m)$. Then we obtain $GF(q^m)$ by adjoining $\beta$ to $GF(q)$.