Verifying if a given polynomial is primitive polynomial Given a polynomial:  $f(x) = x^2 + 2x + 2$ over $GF(3)$. I want to know if i can use it to construct $GF(3^2)$.
My approach:

*

*This equation satisfies first condition: A primitive polynomial is irreducible.

*The second condition i'm trying to confirm is that i can use it generate $GF(3^2)$.

$\alpha^2 + 2\alpha + 2 \rightarrow \alpha^2 = \alpha + 1$.

*

*$\alpha^1 = \alpha$


*$\alpha^2 = \alpha+1$


*$\alpha^3 = \alpha*\alpha^2 \rightarrow1 + 2\alpha $


*$\alpha^4 = 2 $


*$\alpha^5 = 2\alpha$


*$\alpha^6 = \alpha+2$


*$\alpha^7 = 1$
Now Here $\alpha^8$ should be equal to $1$ instead of $\alpha^7 .$


*$\alpha^8 = \alpha$
What i am doing wrong , any help would be great.

It strikes me that our OP Khan Saab's calculations of the powers of $\alpha$ are OK through $\alpha^5$, but there is an error in $\alpha^6$.  With
$\alpha^2 = \alpha + 1, \tag 1$
we have, correctly, 
$\alpha^5 = 2 \alpha; \tag 2$
then
$\alpha^6 = 2 \alpha^2 = 2 \alpha + 2, \tag 3$
not $\alpha^6 = \alpha + 2$!
Continuing:
$\alpha^7 = 2\alpha^2 + 2\alpha = 2(\alpha + 1) + 2\alpha = \alpha + 2; \tag 4$
$\alpha^8 = \alpha^2 + 2\alpha = \alpha + 1  + 2\alpha = 1! \tag 5$
easy, you know the way it's supposed to be!
