Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 18H (Galois Theory)

Let $L$ be the splitting field of the separable polynomial. Its Galois group is $G={\mathrm{Aut}}_K(L)$, acting faithfully on its $n$ distinct roots, hence as a subgroup of $S_n$. An automorphism preserves each root's minimal polynomial. Conversely an isomorphism $K(\alpha )\to K(\beta )$ for roots of the same irreducible factor extends to an embedding of $L$ in an algebraic closure, and normality of the splitting field makes the image $L$ itself. Thus the root orbits are precisely the irreducible factors' root sets. The action is transitive exactly when the polynomial is irreducible.

If $n=p$ is prime, a transitive action has $p\mid |G|$ by orbit-stabilizer, so Cauchy theorem for groups supplies an element of order $p$. Its permutation must be a $p$-cycle. Conversely a $p$-cycle already acts transitively on all $p$ roots. This proves the stated equivalence for prime degree.

The factorization theorem modulo a good prime says that if an integer polynomial retains its degree and has square-free reduction modulo $p$, and that reduction's irreducible factor degrees are $d_1,\dots ,d_r$, its Galois group over $\mathbb{Q}$ contains a permutation with disjoint cycle lengths $d_1,\dots ,d_r$. Equivalently the finite-field Galois group, generated by Frobenius on the roots, occurs as an unramified decomposition subgroup. The degree-retention qualification matters for nonmonic polynomials.

For the given quintic, Eisenstein criterion at three proves irreducibility. Modulo two it factors asNeither factor has a root in ${\mathbb{F}}_2$, so both are irreducible, and they are distinct; the reduction is separable. Thus $G$ contains an element of cycle type $(2)(3)$, whose cube is a transposition. It is also transitive of degree five.

To prove that this forces the full symmetric group here, form the graph on the five roots with an edge for each conjugate in $G$ of that transposition. The graph is $G$-invariant, and transitivity makes its connected components all the same size. That size divides five; there is an edge, so the size cannot be one. Hence the graph is connected. Transpositions along the edges of a connected graph generate all transpositions (move along a path and conjugate successive swaps), and hence generate $S_5$. They all belong to $G$, giving