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 $(E^{\ast},.)$.

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 $(E^{\ast},.)$.

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 $(E^{\ast},.)$.

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 $(E^{\ast},.)$.