I am suppose to find the Galois group of the polynomial $f(x)=x^{12}-1$ of $\mathbb{F}_2$
First attempt: I noticed that a root of this polynomial is actually a $12$th root of $1$ and then the idea was to find the smallest field that extends $F_2$ and contains a $12$th root of $1$. Which means basically to find $n \in \mathbb{N}$ such that $2^n \equiv 1 \ mod \ 12$. However, I could not find such $n$. I would love to know why it didn't work here.
Second attempt: Since $f(x)=x^{12}-1=(x-1)^4(x^2+x+1)^4$ the splitting field extension for $f$ would be the same as $x^2+x+1$ which we know is $\mathbb{F}_4$ and then the Galois group is $C_2$.
I am pretty confident with my second solution but I would like someone just look at it and confirm. Apart of that, if someone can explain why the first attempt didn't work I would be also glad to hear!
Thank you everybody!
Your solution is correct and also indicates why your first approach didn't work. Since $\Bbb{F}_2$ has characteristic 2 there are no $2^{n}$th roots of $1$ for $n>0$ in any extension. The $(x-1)^4$ factor of $f(x)$ reflects that.