Is $x^2+x+1 \in \mathbb{F}_2$ solvable by radicals?
First of all, this polynomial is irreducible in $\mathbb{F}_2$. It is also separable since f'(x)=1.
But the zeros of f(x) are $(-1)^{2/3}, - (-1)^{1/3} \in \mathbb{C}$. So im unsure how i can define K, its field of decomposition.
Also, after defining K, is there a "tower" of radicals?
Thanks
Over a finite field $K$ every polynomial is solvable by radicals in the following sense. If $p(x)\in K[x]$, then the splitting field $L$ of $p$ over $K$ is another finite field. But any extension of finite fields, such as $L/K$, is cyclic because the Galois group is generated by the Frobenius automorphism $F(x)=x^q$, where $q=|K|$. Thus the Galois group is solvable.
We also get $L/K$ as a kind of root tower extension. This is trivially so, because every element of $L$ is a root of unity of order that is a factor of $|L|-1$.
The splitting field of your polynomial is the field of four elements. Its zeros are third roots of unity.
A catch is that it is often impossible to generate an extension of degree $n$ by adjoining an $n$th root of an element of the base field. You may need to use higher roots.