Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 18H (Galois Theory)

If $\theta =0$, the splitting field is $K$ and its Galois group is trivial, hence cyclic. Otherwise choose $\alpha$ with ${\alpha }^n=\theta$. The roots are $\alpha {\zeta }_n^j$, so the splitting field is $K(\alpha )$ because ${\zeta }_n\in K$. The roots are distinct since the characteristic does not divide $n$, making this a Galois extension. Each automorphism satisfies $\sigma (\alpha )=c_{\sigma }\alpha$ with $c_{\sigma }\in ⟨{\zeta }_n⟩$. Because that root-of-unity group is fixed by all automorphisms, $c_{\sigma \tau }=c_{\sigma }{c}_{\tau }$. Thus $\sigma \mapsto c_{\sigma }$ is an injective homomorphism into a cyclic group. The splitting-field Galois group is cyclic, of order dividing $n$.

An automorphism of $L({\zeta }_n)$ fixing $K({\zeta }_n)$ fixes $K$. Because $L/K$ is a normal extension, it preserves $L$, so restriction gives a homomorphismIts kernel fixes both $L$ and ${\zeta }_n$, hence every element of their compositum. Therefore restriction is injective, proving the asserted subgroup identification. Normality is the key reason the restriction maps into automorphisms of $L$ rather than just embeddings into a larger field.

A primitive sixth root of unity satisfieswhich is irreducible over $\mathbb{Q}$ because its discriminant is $-3$. For the other polynomial choose $\alpha =i{3}^{1/6}$, so ${\alpha }^6=-3$ and ${\alpha }^3=-i\sqrt{3}$. Then ${\zeta }_6=(1-{\alpha }^3)/2\in \mathbb{Q}(\alpha )$, so this field already contains all six roots. The Eisenstein criterion at $3$ proves $x^6+3$ irreducible, giving a splitting field of degree six.

Put $r(\alpha )={\zeta }_6^2\alpha$ and $s(\alpha )={\zeta }_6\alpha$. The first automorphism fixes ${\zeta }_6$ and has order three. The second sends ${\alpha }^3$ to its negative, so sends ${\zeta }_6$ to ${\zeta }_6^{-1}$; hence $s^2=1$ and $srs=r^{-1}$. These six distinct transformations exhaust the Galois group. ThereforeThe presence of sixth roots of unity inside the splitting field is why this particular binomial has degree six rather than a generic twelve-degree splitting field.