Generator of $\mathbb{F}_{q}^*$ where $q$ is not prime.

Asked Jan 01 '19 at 11:54

Active Jan 01 '19 at 12:25

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I'm trying to find the generator of the multiplicative group $\mathbb{F}_{8}^*$, where $\mathbb{F}_8$ is a field. If the order of the field is prime, then it is easy since in that case the field would be isomorphic to the integers modulo, say p. However, for cases where the order of the field is not prime, I'm stuck. Can anyone help me out?

Thanks in advance.

edited Jan 01 '19 at 12:25

toric_actions

asked Jan 01 '19 at 11:54

3

By $\Bbb F_q^x$ do you mean the multiplicative group of nonzero elements of $\Bbb F_q$? This is usually denoted as $\Bbb F_q^\times$ or $\Bbb F_q^*$. When $q=8$, it has seven elements, and seven is a prime number, so any element other than the identity is a generator. – Angina Seng Jan 01 '19 at 11:56
@Xenidia Can you please explain your multiplicative group $\mathbb{F}^x_q$? Is it the same as $\mathbb{F}^*_q$ – toric_actions Jan 01 '19 at 12:04
See also the answers to this question on $\Bbb{F}_q^{\times}$. In general, see also this duplicate. – Dietrich Burde Jan 01 '19 at 13:56
So $t + (t^3+t+1)$ is a generator of $\mathbb{F}_2[t]/(t^3+t+1)^\times \cong \mathbb{F}_8^\times$ – reuns Jan 01 '19 at 17:02

Generator of $\mathbb{F}_{q}^*$ where $q$ is not prime.

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