Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 18H (Galois Theory)

LetBoth have splitting field $\mathbb{Q}(\sqrt{2},\sqrt{3})$. On the roots of $f$, the Galois group has two orbits of size two and embeds asThe roots of $g$, namely $\pm \sqrt{2}\pm \sqrt{3}$, form one orbit and give the regular Klein-four subgroupThe subgroups are not conjugate because their orbit decompositions differ.

Let $F$ be the splitting field over ${\mathbb{F}}_p$ of $X^q-X$. Its derivative is $-1$, so it has $q$ distinct roots. The roots are closed under addition, subtraction, multiplication, and inversion, using $(x+y{)}^q=x^q+y^q$ and $(xy{)}^q=x^q{y}^q$. They therefore form a field with $q$ elements.

Every field with $q$ elements has multiplicative group of order $q-1$, so every element satisfies $x^q=x$; it is therefore a splitting field of the same polynomial and is unique up to isomorphism. If an irreducible factor over ${\mathbb{F}}_p$ has a root $\alpha \in {\mathbb{F}}_q$, its degree is $[{\mathbb{F}}_p(\alpha ):{\mathbb{F}}_p]$, which divides $d$ by the tower law. In particular it is at most $d$.

Dedekind's factorization/Frobenius theorem says that for a prime not dividing the discriminant, the degrees of the distinct irreducible factors modulo $p$ give the cycle lengths of an element of the Galois group.

The polynomial is irreducible over $\mathbb{Q}$ by the rational-root test followed by a comparison of possible monic quadratic factors. Its discriminant is $547^2$, a square, so its transitive Galois group lies in $A_4$. Modulo five,and the cubic has no root in ${\mathbb{F}}_5$, giving a three-cycle. A transitive subgroup of $A_4$ containing a three-cycle cannot be the Klein four group; hence