is $x^2+x+1 \in \mathbb{F}_2$ solvable by radicals?

Question

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

The splitting field of $x^2+x+1$ over $\mathbb{F}_2$ is $\mathbb{F}_2^2\simeq \mathbb{F}_2[x]/(x^2+x+1)$. — Jack D'Aurizio, Apr 13 '17 at 14:38
In general, if $\Phi_n(x)$ is the $n$-th cyclotomic polynomial, its splitting field over $\mathbb{F}_p$ is $\mathbb{F}_p^d$ with $d$ being the least integer such that $n$ is a divisor of $p^d-1$. — Jack D'Aurizio, Apr 13 '17 at 14:39
So here $d=2$. But now how do I verify if it has a tower of radicals? @JackD'Aurizio — usere5225321, Apr 13 '17 at 14:45
Modulo $p$ "solvability" is more complicated than characteristic $0$. Instead of $p$-th roots we've got to allow ourselves to adjoin roots of $X^p-X+1$. — ancient mathematician, Apr 13 '17 at 14:46
Sorry I gave away my Galois Theory books. There are exercises in Lang's Algebra, though, that show that in characteristic $p$ a GG of order $p$ arises from a $X^p-X+a$. — ancient mathematician, Apr 13 '17 at 15:21
@ancientmathematician: That is, indeed, a standard result/exercise. For a local solution see here. — Jyrki Lahtonen, Apr 13 '17 at 15:24

Jyrki Lahtonen · Accepted Answer · 2017-04-13T15:25:52.637

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.

is $x^2+x+1 \in \mathbb{F}_2$ solvable by radicals?

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