Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 18F / Solution

The existence and uniqueness of splitting fields says that any two splitting fields of a polynomial $f\in F[X]$ are isomorphic by an isomorphism fixing $F$ pointwise. Inside a chosen algebraic closure, the subfield generated by all roots is unique.

For a polynomial $f\in \mathbb{Q}[X]$ with distinct roots, let $K$ be its splitting field. Its Galois group of a polynomial isthe group of field automorphisms of $K$ fixing $\mathbb{Q}$. It acts faithfully on the roots.

If $f$ is irreducible and $\alpha ,\beta$ are roots, the embedding $\mathbb{Q}(\alpha )\to K$ taking $\alpha$ to $\beta$ extends, because $K/\mathbb{Q}$ is normal and separable, to an element of the Galois group. Thus the action is transitive. Conversely, the roots belonging to one orbit have a product polynomial invariant under the whole Galois group, hence lying in $\mathbb{Q}[X]$. If there were more than one orbit, these orbit polynomials would give a nontrivial factorization of $f$. Therefore

Now let $g(X)=X^4+b{X}^2+c$ and let $\alpha$ be one root. The two roots of $Y^2+bY+c$ are ${\alpha }^2$ and $c/{\alpha }^2$. On choosing $\delta =\sqrt{c}$, the roots of $g$ areConsequently its splitting field is