Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 16F (Galois Theory)

If $f$ is an irreducible polynomial, any two roots $\alpha ,\beta$ have the same minimal polynomial. Thus the map $\mathbb{Q}(\alpha )\to \mathbb{Q}(\beta )$ carrying $\alpha$ to $\beta$ is a field embedding. The extension theorem for field embeddings extends it to an embedding of the splitting field $L$ into an algebraic closure. Since $L/\mathbb{Q}$ is normal, its image is $L$, so it is an element of the Galois group.

Conversely, factor $f$ into irreducible factors over $\mathbb{Q}$. An automorphism preserves the vanishing of each factor, so it cannot send a root of one factor to a root of another. Since there are no repeated roots, different factors have disjoint root sets. Transitivity on all roots therefore forces a single irreducible factor. This proves the equivalence.

Label the distinct roots ${\alpha }_1,\dots ,{\alpha }_n$. Define $\tau (\sigma )$ by $\sigma ({\alpha }_i)={\alpha }_{\tau (\sigma )(i)}$. Composition respects this rule, so it is a group homomorphism. If its image permutation is the identity, $\sigma$ fixes all the roots; they generate $L$, so $\sigma$ is the identity. Thus $\tau$ is injective.

For $f=t(t^2-2)$, label the roots $0,\sqrt{2},-\sqrt{2}$. The splitting field is $\mathbb{Q}(\sqrt{2})$ and its nontrivial automorphism swaps precisely roots $2,3$. Hence $\boxed{\mathrm{im}\tau =⟨(23)⟩}$.

For $f=t^3-2$, Eisenstein criterion gives irreducibility. The real root $\alpha =\sqrt[3]{2}$ generates a degree-three real field. Adjoining a primitive cube root of unity ${\zeta }_3$ gives a further extension of degree two, since it is nonreal and satisfies $t^2+t+1=0$. Thus its splitting field has degree $6$. In characteristic zero this is a Galois extension, so its Galois group has order $6$. The injection into $S_3$ is therefore an isomorphism.

Let $\alpha =\sqrt[5]{3}$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb{Q}(\alpha ,\zeta )$. The field degrees $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$ and $[\mathbb{Q}(\zeta ):\mathbb{Q}]=4$ both divide $[L:\mathbb{Q}]$, while adjoining the two generators gives $[L:\mathbb{Q}]\le 20$. Consequently $[L:\mathbb{Q}]=20$.

The cyclotomic field $\mathbb{Q}(\zeta )/\mathbb{Q}$ is Galois with cyclic group of order $4$. The Fundamental theorem of Galois theory gives a normal subgroupThus $N\cong C_5$ and the chain $\boxed{G▹N▹\{1\}}$ has abelian successive quotients, proving that $G$ is a solvable group. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.