Past exam of the number theory course of the University of Oxford / 2008 / A / 3 / Solution

A finite extension is Galois when it is both normal and separable; over $\mathbb{Q}$ separability is automatic, so it is equivalently a splitting field of a polynomial with distinct roots. Its Galois group consists of the field automorphisms fixing $\mathbb{Q}$. If $G=\mathrm{Gal}(L/\mathbb{Q})$, the Fundamental theorem of Galois theory gives mutually inverse inclusion-reversing bijectionsThey satisfy $[L:L^H]=|H|$ and $[L^H:\mathbb{Q}]=[G:H]$. The intermediate extension $K/\mathbb{Q}$ is Galois exactly when $H=\mathrm{Gal}(L/K)$ is normal, and then $\boxed{\mathrm{Gal}(K/\mathbb{Q})\cong G/H}$ via restriction.

Every permutation of the four roots permutes the three unordered partitions into two pairs and hence permutes the three ${\beta }_j$. Thus each coefficient of $g(X)={\prod }_{j=1}^3(X-{\beta }_j)$ is a symmetric polynomial in the four roots. The Fundamental theorem of symmetric polynomials expresses it in the elementary symmetric functions, which are rational coefficients of $f$. Hence $g\in \mathbb{Q}[X]$. Each product ${\alpha }_i{\alpha }_j$, $i<j$, occurs exactly twice in ${\beta }_1+{\beta }_2+{\beta }_3$, so that sum is $2q$. Therefore the quadratic coefficient is $\boxed{-2q}$.

For the given cubic write $g(X)=(X-2{)}^3+12$. The translated cubic $Y^3+12$ is Eisenstein at three, hence irreducible. It has one real root and a nonreal conjugate pair. Its Galois group is transitive on three roots, so its order is divisible by three; complex conjugation is a transposition. A subgroup of $S_3$ with both a three-cycle and a transposition is all of $S_3$. HenceLet $G=\mathrm{Gal}(L/\mathbb{Q})\le S_4$. Irreducibility of the quartic makes the root action transitive, so four divides $|G|$. The cubic splitting field $K$ lies inside $L$ by the definitions of the ${\beta }_j$, and restriction onto its Galois group gives a quotient of order six. Thus six also divides $|G|$. Since $|G|$ divides 24, it is 12 or 24. If it were 12, the supplied index-two result would give $G=A_4$. The restriction kernel would then be a normal subgroup of $A_4$ of index six, contrary to the other supplied result. Thus