Example 4.4. Let $p=2$, $n=4$. Consider the polynomial $f=x^{2^{4}}-x=x^{16}-x$.
Then $\text{GF}(16)$ is the splitting field of $f$ over $\text{GF}(2)$. The irreducible factors of $f$ over $\text{GF}(2)$ have degrees 4, 2, and 1. A factorization of $f$ over $\text{GF}(2)$ is given by
$x^{16}-x = x(x-1)(x^{2}+x+1)(x^{4}+x+1)(x^{4}+x^{3}+1)(x^{4}+x^{3}+x^{2}+x+1)$.
Let $\alpha$, $\beta$, $\gamma$ be roots of $x^{4}+x+1$, $x^{4}+x^{3}+1$, $x^{4}+x^{3}+x^{2}+x+1,$ respectively. $\text{GF}(16)$ is constructed as $\text{GF}(2)(\alpha), \text{GF}(2)(\beta)$ or $\text{GF}(2)(\gamma)$. It is straightforward to check that $o(\alpha)=o(\beta)=15$ in the cyclic group of units of the splitting field of $f$ over $\text{GF}(2)$. So both $\alpha$ and $\beta$ are primitive elements of $\text{GF}(16)$.
On the other hand, from $\gamma^{{5}}-1=(1+\gamma+\gamma^{2}+\gamma^{3}+\gamma^{4})(\gamma-1)=0$, we can only conclude that the order of $\gamma$ divides 5, written as $o(\gamma)|5$. This implies $\gamma$ is not a primitive element. In other words, $\gamma$ generates a cyclic subgroup of order dividing 5 within $\text{GF}(16)^{\}$.
I came across this example while trying to study about finite fields in coding theory. I dont understand where every thing is coming from. I get that irreducible factors of f over Gf(2) must have degree 4,2 and 1 but I dont understand how they got the factorization given. Is this factorization unique?Also how did they conclude GF(16) is constructed as GF(2)(α),GF(2)(β) or GF(2)(γ). I also dont understand what they used to conclude o(α)=o(β)=15 and why o(γ) is different from the other two.
