number of primitive elements in $\mathbb{F}_9$

Question

How can we find number of primitive elements in the field $\mathbb{F}_9$.

$\mathbb{Z}_3[x]/(x^2+1) = \mathbb{F}_9$ and $α^4=1$. I could not improve the solution. I missunderstand the conception. can you help me.

thanks.

Answer 1

F9={0,1,2,α,α+1,α+2,2α,2α+1,2α+2}

Number of Primitive elements: φ(9-1)=φ(8)=2.(3-1)=4

So, I checked one by one. that has order 8.

Primitive elements are: {α+1,α+2,2α+1,2α+2}

{example: 1+α is primitive because the order of 1+α is 8. (1+α)^4=2. then 2^2=1.}

solved by Assoc.Prof @Jyrki Lahtonen and other mathematician 's helps and comments.