Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 18J (Galois Theory)

Up to conjugacy, the transitive subgroups of the symmetric group on four points areRepresentatives are $⟨(1234)⟩$, the normal Klein four group of double transpositions, $⟨(1234),(13)⟩$, $A_4$, and $S_4$.

For the biquadratic extension $L=K(\sqrt{a},\sqrt{b})$, the assumptions say that the square classes of $a$ and $b$ are independent. In particular $\sqrt{b}\notin K(\sqrt{a})$: writing $\sqrt{b}=u+v\sqrt{a}$ and comparing the coefficient of $\sqrt{a}$ after squaring would force either $b$ or $ab$ to be a square in $K$. Hence

The field is the splitting field of $(X^2-a)(X^2-b)$ and the characteristic is not two, so it is Galois. Its four automorphisms independently choose the signs of $\sqrt{a}$ and $\sqrt{b}$, givingBy the Galois correspondence, the intermediate-field lattice isEach of the three middle fields has degree two over $K$, and distinct middle fields intersect in $K$.

Squaring and eliminating the inner radical givesThus $\alpha ,\beta$ are roots ofand $\gamma ,\delta$ are roots ofBoth polynomials are irreducible by the Eisenstein criterion at $3$, so these are the four minimal polynomials over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{3})$, the relevant radicands areThe norms of $a$ and $b$ to $\mathbb{Q}$ equal $6$, so neither can be a square in $K$. Also $(u+v\sqrt{3}{)}^2=6$ with $u,v\in \mathbb{Q}$ would force $uv=0$, followed by either $u^2=6$ or $v^2=2$, both impossible. Thus $6$ is not a square either.

Similarly, over $K^{\prime }=\mathbb{Q}(\sqrt{6})$ the radicands $3+\sqrt{6}$ and $3-\sqrt{6}$ have nonsquare norm $3$, while $(u+v\sqrt{6}{)}^2=3$ would force either $u^2=3$ or $v^2=1/2$. Hence $3$ is not a square in $K^{\prime }$. Both extensions therefore satisfy all the hypotheses of part (b).

Let $E=\mathbb{Q}(\alpha ,\beta )$. Sincepart (b) gives $[E:\mathbb{Q}]=8$. The roots of $X^4-6X^2+6$ are $\pm \alpha ,\pm \beta$, all in $E$, so $E$ is its splitting field and is Galois. Its action on the four roots is transitive; the order-eight entry in the list from part (a) is $D_8$. Therefore

For an explicit dihedral Galois action on four radical roots, defineThen $r^4=s^2=1$ and $srs=r^{-1}$. The subgroup lattice is determined byHere $⟨r^2⟩$ lies in all three order-four subgroups; $⟨s⟩,⟨r^{2}s⟩$ lie in $⟨r^2,s⟩$; and $⟨rs⟩,⟨r^{3}s⟩$ lie in $⟨r^2,rs⟩$.

Reversing inclusions under the Galois correspondence gives the complete field latticeThe incidence is likewise reversed: $\mathbb{Q}(\sqrt{2})$ lies in $\mathbb{Q}(\sqrt{2},\sqrt{3})$; $\mathbb{Q}(\sqrt{3})$ lies in that field, $\mathbb{Q}(\alpha )$, and $\mathbb{Q}(\beta )$; and $\mathbb{Q}(\sqrt{6})$ lies in that field, $\mathbb{Q}(\alpha +\beta )$, and $\mathbb{Q}(\alpha -\beta )$. The hint verifies the last two quartic fields through $\alpha +\beta =\sqrt{2}\gamma$ and $\alpha -\beta =\sqrt{2}\delta$.