Primitive element in $\mathbb{F}_9$, constructed with $f(t)=t^2 + 1$

Asked Jan 25 '19 at 19:28

Active Jan 25 '19 at 21:32

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Given $f(t) = t^2 + 1$ over $\mathbb{F}_9$, I first showed that $f(t)$ is irreducible over this field. $f(t)$ is not a primitive polynomial because $t^8 \ne 1$. So now I'm trying to find a primitive element $\alpha$ of $\mathbb{F}_9$.

$\mathbb{F}_9 = \{0,1,-1,t,t+1,t-1,-t,-t+1,-t-1 \}$. I know that I have to find a non-zero element in this set with order $8$ ($=9-1$). It's clear that $1,-1,t,-t$ cannot be primitive elements of $\mathbb{F}_9$. Let's look at the other four elements:

$(t+1)^8 = ((t+1)^2)^3 = (2t)^3=-t^3 = t;$

$(t-1)^8 = \dots= -t;$

$(-t+1)^8 = \dots = -t;$

$(-t-1)^8 = \dots = t;$

Does this mean that there are no elements of order $8$ and therefor no primitive elements in $\mathbb{F}_9$?

edited Jan 25 '19 at 21:32

egreg

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asked Jan 25 '19 at 19:28

MyWorld

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A booboo with the exponents: $(t+1)^8=((t+1)^2)^4$ because $2\cdot4=8$. But $2\cdot3=6$ only. – Jyrki Lahtonen Jan 25 '19 at 19:39
Anyway, $\Bbb{F}_9^*$ is cyclic of order eight, and you listed all the elements of order four or lower. – Jyrki Lahtonen Jan 25 '19 at 19:40
Check again $\pm1\pm t$ are all of order eight. – Jyrki Lahtonen Jan 25 '19 at 19:54
See here. (looking for a better duplicate target...) – Jyrki Lahtonen Jan 25 '19 at 19:55

Primitive element in $\mathbb{F}_9$, constructed with $f(t)=t^2 + 1$

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