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.

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    $\begingroup$ Yes, this phenomenon is not uncommon. I think the smallest case is $p=3, k=2$. The polynomial $q(x)=x^2+1$ is irreducible modulo $3$ all right, but because $x^2+1\mid x^4-1$ the zeros of $q(x)$ have order four only. You need to use one of the irreducible factors of $x^4+1=(x^2-x-1)(x^2-x+1)$ to get a primitive polynomial. In characteristic two the smallest case is $k=4$, where $x^4+x^3+x^2+x+1$ is irreducible but not primitive. It isn't too difficult to give a formula for the number of primitive polynomials of a given degree. $\endgroup$ – Jyrki Lahtonen Jan 10 '18 at 23:00
  • $\begingroup$ Thank you for the answer. And, in the case that $\alpha$ was a root of $q(x)$ but not primitive in $F_{p^k}$, would I still have a normal basis of shape $\{\alpha,\alpha^{p},...,\alpha^{p^{k-1}}\}$ ? Or, to have a Normal basis, $\alpha$ has to be primitive ? $\endgroup$ – Bruce Wayne Jan 10 '18 at 23:23
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    $\begingroup$ No. The property of $\alpha$ generating a normal basis is independent from it being primitive. For example in the case $p=2, k=3$ the polynomials $x^3+x+1$ and $x^3+x^2+1$ are both primitive, but only the roots of the latter form a normal basis. Also, in the case $p=2, k=4$ the polynomial $x^4+x^3+x^2+x+1$ generates a normal basis, but it is not primitive. $\endgroup$ – Jyrki Lahtonen Jan 11 '18 at 7:09
  • $\begingroup$ Perfect! So, a last thing :), a root of an irreducible polynomial $q(x)$ (primitive or not) forms a polynomial basis over $F_{p^k}$ but it is not said that it forms a normal basis. Right? $\endgroup$ – Bruce Wayne Jan 11 '18 at 8:50
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    $\begingroup$ Correct. ${}{}$ $\endgroup$ – Jyrki Lahtonen Jan 11 '18 at 10:09

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