Past exam of the number theory course of the University of Oxford / 2011 / A / 3

The Fundamental theorem of symmetric polynomials states that every symmetric polynomial in $n$ variables over a field is a polynomial in the elementary symmetric polynomials, uniquely so. For the three roots, Vieta's formulas give $e_1=0$, $e_2=p$, $e_3=-q$.

The squared product of root differences is symmetric. Expansion in elementary symmetric polynomials givesOne can verify the displayed identity simply by substituting the definitions of $e_1,e_2,e_3$ and expanding; it is the cubic polynomial discriminant.

If the cubic is irreducible, its roots are distinct in characteristic zero, so $\delta \ne 0$. Its Galois group is a transitive subgroup of $S_3$, hence either $A_3$ or $S_3$. A permutation of the roots multiplies $\delta$ by its sign. The fixed-field assertion of the Fundamental theorem of Galois theory therefore gives $\delta \in \mathbb{Q}$ exactly when all group elements are even. ThusFor example, $f=x^3-3x-1$ has no rational root: neither $1$ nor $-1$ is a root, and these are the only possibilities by the rational root theorem. Its discriminant is $108-27=81$, so $\delta =\pm 9\in \mathbb{Q}$ and its group is $A_3$.

A finite solvable group is a group admitting a finite chain of normal subgroups with abelian successive quotients; equivalently its derived series eventually becomes trivial. The chainhas quotients $C_2,C_3,V_4$, all abelian. Thus $S_4$ is solvable. Every subgroup of a solvable group is solvable, because intersecting the chain with the subgroup gives factors isomorphic to subgroups of the original abelian factors. A Galois group of a polynomial of degree at most $4$ embeds in $S_4$ by its action on the roots, so it is solvable even if the polynomial is reducible or has repeated roots.

For the substitution, the characteristic-zero solvability by radicals theorem says that a polynomial has solvable Galois group if and only if its splitting field is contained in a finite tower formed by adjoining successive roots of elements of the preceding field. First place all roots $\alpha$ of $f$ in such a tower. Adjoin a primitive $m$th root of unity and, for every nonzero $\alpha$, an $m$th root ${\beta }_{\alpha }$. These are radical adjunctions, and all roots of $x^m-\alpha$ are ${\beta }_{\alpha }{\zeta }_m^j$. Zero roots require no adjunction. The resulting radical tower therefore contains the splitting field of $f(x^m)$. The theorem implies