Htykuut

Let $K = F_2(x)$ and $q_t(x) = x^2-x-t in K[x]$. Let $E$ be its splitting field. I need to prove that $Gal_{E/K} simeq mathbb{Z_2}$ and that $q_t$ is not solvable by radicals.

Any hint?

In general, $ X^p - X - t $ is never solvable by radicals over $ mathbb F_p(t) $ (for prime $ p $), and it always has Galois group $ C_p $.

To see that the Galois group is $ C_p $, let $ gamma $ be a root in a splitting field and note that the other roots are precisely $ gamma + k $ for $ k in mathbb F_p $, so the Galois group is generated by translations of the form $ gamma to gamma + k $ for $ k in mathbb F_p $. This means that there is no intermediate field between $ mathbb F_p(t) $ and the splitting field $ L $ of $ X^p - X - t $ (since by degree considerations it has no root in $ mathbb F_p(t) $) and we deduce that the polynomial is irreducible in $ mathbb F_p(t)[X] $.

To see that it is never solvable by radicals, it's easy to see this is implied by the following lemma through induction: (the algebraic closure is taken on purpose, and is applied only to $ mathbb F_p $)

Lemma. If $ K supset mathbb{bar{F}}_p(t) $ is a finite extension of $ mathbb {bar{F}}_p(t) $ over which $ X^p - X - t $ is irreducible, then for any prime $ q $, any $ y in K $ and any $ alpha notin K $, $ alpha^q in K $; $ X^p - X - t $ remains irreducible over $ K(alpha) $.

Proof. In the case $ q neq p $, note that by a general result in field theory, $ X^q - y $ is either irreducible in $ K[X] $ or it has a root in $ K $. Since $ K $ contains all roots of unity of order coprime to $ p $, this is equivalent to $ X^q - y $ completely splitting over $ K $, and thus either $ alpha in K $ or $ y $ has degree $ q $ over $ K $. We can see it's impossible for there to be a degree $ p $ subextension of $ K(y^{1/q})/K $ by a divisibility argument.

In the case $ q = p $, the extension $ K(alpha)/K $ is a purely inseparable extension, thus there is an equality of separability degrees $ [K(alpha): mathbb{bar{F}}_p(t)]_s = [K : mathbb{bar{F}}_p(t)]_s $. If $ K(alpha) $ contained a root of $ X^p - X - t $, then its separability degree over $ mathbb{bar{F}}_p(t) $ would be strictly greater than that of $ K $, so it follows that $ K(alpha) $ does not contain a root of $ X^p - X - t $, and by extension we conclude $ X^p - X - t $ is irreducible over $ K(alpha) $ by a similar Galois theory argument we used above.

Finally, notice that solvability in radicals over $ mathbb{bar{F}}_p(t) $ is equivalent to solvability in radicals over $ mathbb F_p(t) $, since the former is a radical extension of the latter (obtained by adjoining all roots of unity of order coprime to $ p $).