It is known that every finite field of the same order $p^k$ are isomorphic. So, $F_p[x]/\langle q(x)\rangle$ leads to the same field for any choice of irriducible k-degree $q(x)$ over $F_p[x]$. But, if I wanted that the root of $q(x)$ used for the extension to be also a primitive element of $F_{p^k}$ (i.e., an element of order $p^{k}-1$) , does the choice of $q(x)$ makes the difference? I mean, can a root of a particular $q(x)$ not be a primitive element for $F_{p^k}$ but a root of $p(x)$ (another irreducible polynomial of degree $k$ over $F_p[x]$ used for the construction of $F_{p^k}$ ) be a primitive element for $F_{p^k}$ ?
I hope I have been clear enough. I have some doubts about it.