How do I show that an element is a primitive element in $GF(2^3)$

Question

Let $\alpha$ have the property $\alpha^3 + \alpha + 1 = 0$. Is this a primitive element in $GF(2^3)$, generated by $p(x) = x^3 + x + 1$?

Answer 1

Check that $x^3+x+1$ is irreducible over $F_2$ (otherwise, it would have a root in $F_2$, which would have to be either $0$ or $1$).

Thus, $F_2(\alpha)$ has degree 3 over $F_2$. Hence, $F_2(\alpha)$ has cardinality 8 and thus, $F_2(\alpha) \cong GF(2^3)$. Done.