Let $F$ be a finite field and $f(x)\in F[X]$ be an irreducible polynomial of degree $n$. Let $\alpha$ be a root of $f(x)$. So $E:=F[\alpha]$ is a finite field of order $|F|^{n}$.

We know that $(E^{\ast},.)$ is cyclic group of order $|F|^{n}-1$. My question is that what can we say about generators of $(E^{\ast},.)$ with respect to $\alpha$ and $f(x)$? Can we explicitly find a generator with respect to $\alpha$ and $f(x)$?

For example, if $|F|=2$ and $2^{n}-1$ is prime number, then $\alpha$ is a generator (actually each non-identity element in $E^{\ast}$ is a generator.)

If $|F|=2$ and $f(x)=x^{4}+x^{3}+x^{2}+x+1$, then order of $\alpha$ is $5$, which means that $\alpha$ is not a generator of $(E^{\ast},.)$.

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    $\begingroup$ What do you mean by "explicitly" finding $a$? No, I don't think there is some magic formula for it; one needs to do computations, generally speaking. There is a special class of irreducible polynomials, called primitive, for which any root has maximal period. Again, there is no generally known formula how to find such polynomials. $\endgroup$ – Dima Pasechnik Feb 24 '14 at 21:18
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    $\begingroup$ Can we find explicitly a generator of $F^*$? $\endgroup$ – abx Feb 24 '14 at 21:25
  • $\begingroup$ @abx: To repeat Dima's question: what exactly do you mean by explicitly find? $\endgroup$ – Derek Holt Feb 25 '14 at 12:21

See for example https://en.wikipedia.org/wiki/Conway_polynomial_(finite_fields). There is plenty of theory and practice in the area, going back to Richard Parker.


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