Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 17I (Galois Theory)

A splitting field of a nonzero polynomial $f$ over $K$ is an extension in which $f$ splits into linear factors and which is generated over $K$ by all its roots. For a constant nonzero polynomial the splitting field is $K$; the zero polynomial does not have a root-generated splitting field in this standard definition.

Given two splitting fields $L,L^{\prime }$, enumerate the roots generating $L$ and extend a $K$-embedding one root at a time. If an embedding is defined on $K({\alpha }_1,\dots ,{\alpha }_j)$, the minimal polynomial of the next root divides $f$ over that field. Its image under the embedding therefore divides $f$ over the corresponding subfield of $L^{\prime }$, and has a root in $L^{\prime }$. The quotient construction $K(\alpha )\cong K[t]/(m_{\alpha })$ extends the embedding to that root. The polynomial need not be a separable polynomial.

At the end this gives a $K$-embedding $L\to L^{\prime }$. It sends the distinct roots of $f$ injectively to its roots, hence permutes the finite set of distinct roots and contains every root in its image. Since these generate $L^{\prime }$, the embedding is onto. Thus the splitting field is unique up to a (K)-isomorphism.

The quartic is irreducible by Eisenstein criterion at two, so its Galois group is transitive on its four roots. Its discriminant is $1616=16\cdot 101$, so five and seven are unramified primes for the polynomial. The squarefree factorizations areThe cubic modulo five has no root in ${\mathbb{F}}_5$, hence is irreducible. The quadratic modulo seven has discriminant five, a nonsquare there. The Frobenius cycle type theorem therefore supplies a three-cycle and a transposition in the group.

Transitivity implies that its order is divisible by four, and the three-cycle implies divisibility by three. Its order divides $24$, so it is $12$ or $24$. The only subgroup of index two in $S_4$ is $A_4$: an index-two subgroup is the kernel of a group homomorphism of a nontrivial homomorphism to $C_2$, and all transpositions are conjugate and generate $S_4$, forcing this homomorphism to be the sign. The exhibited transposition excludes $A_4$. Hence

Modulo three, the quintic has no linear factor: its value is $-1$ at every element of ${\mathbb{F}}_3$. The monic irreducible quadratics over this field are $t^2+1$, $t^2+t+2$ and $t^2+2t+2$. Division leaves respectively $-1$, $t-1$ and $t-1$, all nonzero. A reducible quintic has a factor of degree at most two, so it is irreducible modulo three and therefore over $\mathbb{Q}$. Its Galois group is transitive on five roots.

Modulo two its squarefree factorization isBoth factors are irreducible, so the Frobenius cycle type theorem gives an element that is a disjoint transposition times a three-cycle. Cubing it gives a transposition.

The prime-degree transposition criterion can be proved here: a transitive subgroup of prime degree containing a transposition is the whole symmetric group. Construct a graph on the five roots whose edges are the pairs swapped by all conjugates of this transposition. The group acts transitively on vertices and permutes connected components, so all component sizes are equal and divide five. An edge rules out size one; the graph is connected. Edge transpositions of any connected graph generate the full symmetric group, for instance by transporting swaps along a spanning tree. All belong to the Galois group, provingThe prime-degree condition is essential to the short transposition argument.