Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 18F (Galois Theory)

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

PutThe four roots are $\alpha ,-\alpha ,\beta ,-\beta$. If $\delta \in \mathbb{Q}(\alpha )$, thenbecause $g$ is irreducible. Hence the Galois group has order four and acts transitively, hence regularly, on the roots.

Let $\tau$ be the automorphism taking $\alpha$ to $\beta$. Since $\tau (\delta )=\pm \delta$,If $\tau (\delta )=\delta$, then $\tau$ is an involution. Together with the distinct involution taking $\alpha$ to $-\alpha$, it givesIn the ordered root set $(\alpha ,-\alpha ,\beta ,-\beta )$ these generators act as double transpositions.

If $\tau (\delta )=-\delta$, thenso $\tau$ has order four andThis is the degree-four case in the Galois group of an irreducible even quartic.

If $\delta \notin \mathbb{Q}(\alpha )$, thenLetThis is the nonidentity automorphism of $K/\mathbb{Q}(\alpha )$. By extending an embedding that sends $\alpha$ to $\beta$, and composing with $s$ if necessary, chooseThenThe eight elementsare distinct, and exhaust the Galois group. Thus the dihedral Galois action on four radical roots gives

Now suppose $c$ is a square in $\mathbb{Q}$. Then $\delta \in \mathbb{Q}$, so every automorphism fixes $\delta$. We are in case (i) with $\tau (\delta )=\delta$, and therefore

Conversely, suppose the group is the Klein four-group. Label the rootsThe three nonidentity elements in its transitive action are the three double transpositions. Each fixes $\delta$: they send the ordered pair $(\alpha ,\beta )$ respectively towhose products all equal $\alpha \beta$. Hence every element of the Galois group fixes $\delta$. By the Artin fixed-field theorem,Therefore $c={\delta }^2$ is a square in $\mathbb{Q}$. This proves the Klein-four criterion for an irreducible even quartic: