How would I find the Field's elments generator?

Question

Suppose we would construct a Field $F=GF(2^4)$ by using $f(x)=x^4+x^3+x^2+x+1$. In this case the generator is $\alpha=x+1$. Why $\alpha$ here is equal to $x+1$ how would I find this?

Answer 1

you need to check that the order of $x+1$ is $15$. This will show that it is the generator.

The possible orders of any element in $\mathbb{F_{2^4}}$ is $1,3,5,15$. So now you should check which of these is the order of $x+1$. For example, \begin{align*} (x+1)^5 & = (x+1)(x+1)^4\\ & = (x+1)(x+1)^2(x+1)^2\\ & = (x+1)(x^2+1)^2 && ((a+b)^2 \equiv a^2+b^2 \text{ in } \mathbb{F_{2^4}})\\ & = (x+1)(x^4+1)\\ & = (x+1)(x^3+x^2+x) && (\because x^4+x^3+x^2+x+1 \equiv 0 \text{ in } \mathbb{F_{2^4}})\\ &= x^4+x\\ & = x^3+x^2+1\\ & \neq 1. \end{align*} So the order of $x+1$ is NOT $5$. Similarly rule out that the order cannot be $1$ or $3$.