Primitive elements of GF(8) I'm trying to find the primitive elements of $GF(8),$ the minimal polynomials of all elements of $GF(8)$ and their roots, and calculate the powers of $\alpha^i$ for $x^3 + x + 1.$ 
If I did my math correct, I found the minimal polynomials to be $x, x + 1, x^3 + x + 1,$ and $x^3 + x^2 + 1,$ and the primitive elements to be $\alpha, \dots, \alpha^6 $
Would the powers of $\alpha^i$ as a polynomial (of degree at most two) be: $\alpha, \alpha^2, \alpha+ 1, \alpha^2 + \alpha, \alpha^2 + \alpha + 1,$ and $\alpha^2 + 1$?
Am I on the right track? 
 A: Those are all correct. Here's everything presented in a table:
$$\begin{array}{lll}
\textbf{element} & \textbf{reduced} & \textbf{min poly} \\
0 & 0 & x \\
\alpha^0 & 1 & x+1 \\
\alpha^1 & \alpha & x^3+x+1 \\
\alpha^2 & \alpha^2 & x^3+x+1 \\
\alpha^3 & \alpha+1 & x^3+x^2+1 \\
\alpha^4 & \alpha^2+\alpha & x^3+x+1 \\
\alpha^5 & \alpha^2+\alpha+1 & x^3 + x^2 + 1 \\
\alpha^6 & \alpha^2+1 & x^3 + x^2 + 1 \\
\end{array}$$
A: What you have is correct. Some general facts that you can use to check that you have all of the polynomials are the following:
Let $p$ be prime and $q=p^k$.
$\textbf{Fact 1:}$ In $GF(q)$ the product of all monic irreducible polynomials of degree dividing $n$ is $x^{q^n}-x$.
$\textbf{Fact 2:}$ The minimal polynomial of any element of $GF(q)$ is a monic irreducible polynomial over $GF(p)$ whose degree divides $k$.
Set $p=2$, $k=3$. Then $q = 8$.
Using fact 2, we see that the minimal polynomial of any element of $GF(8)$ is a monic irreducible polynomial over $GF(2)$ whose degree is $1$ or $3$. 
Using fact 1, we see that the product of all monic irreducible polynomials of degree $1$ or $3$ over $GF(2)$ is $x^8-x$.
You can check that the product of the set of polynomials you listed is $x^8-x$ over $GF(2)$.
