Galois group of $x^6+1$ over $\mathbb{F}_2$

Question

The question:

Find the Galois group of $x^6+1$ over $F=\mathbb{F}_2=\{0,1\}$.

EDIT: I have improved my attampt and would like for further help, thanks.

My attampt:

Let $E$ be the extension field of $x^6+1$ over $\mathbb{F}_2$. Notice that $x^6+1=(x+1)^2(x^2+x+1)^2$. $1$ is a double root of that polynomial so the Galois group of $x^6+1$ is identical to the galois group of $$f(x):=x^2+x+1$$ Neither $0$ or $1$ are roots of $f$. So assume $f$ has roots $t_1,t_2$, thatmay be equal to each other.

1. If $t_1\ne t_2$, let $\sigma\in\text{Aut}(E/F)$, $$ f(\sigma(t_i))=\sigma(f(t_i))=\sigma(0)=0 \\ \Rightarrow \sigma(t_i)\in\{t_1,t_2\} $$ Thus, $\text{Aut}(E/F)$ has $2$ elements: $id$ and $t_i\mapsto t_{1-i}$. In that case $\text{Aut}(E/F)\cong \mathbb{F}_2$.
2. If $t_1=t_2$ then $\text{Aut}(E/F)=\{id\}$.

Thus, I just need to know know whether $t_1$ equals $t_2$ or not to solve that problem.

Answer 1

Hint: Over ${\Bbb F}_2$: $x^6+1 = (x^3+1)^2$ by freshman's dream.

Answer 2

Note that $x^6+1=(x^3+1)^2=(x-1)^2(x^2+x+1)^2$. The polynomial $x^2+x+1$ is irreducible over $\mathbb{F_2}$ and hence $\mathbb{F_2}/(x^2+x+1)\cong\mathbb{F_4}$ is an extension field of $\mathbb{F_2}$. In that field $x^2+x+1$ has a root (the element $x+(x^2+x+1)$ is a root), and since $\mathbb{F_4}/\mathbb{F_2}$ is a normal extension all roots of $x^2+x+1$ are in $\mathbb{F_4}$. So $\mathbb{F_4}$ is a splitting field of $x^6+1$ (a minimal field which contains all roots of $x^6+1$) and the Galois group is $\mathbb{Z}/2\mathbb{Z}$.