Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 20G (Number Fields)

The element $\alpha$ is a root ofBy the rational root theorem, any rational root of this monic polynomial must be one ofDirect substitution shows that none is a root. A reducible cubic over a field has a linear factor, so $f$ is an irreducible polynomial over $\mathbb{Q}$. It is therefore the minimal polynomial of $\alpha$, and

PutUsing ${\alpha }^3=5\alpha -8$, multiplication by $\beta$ on the ordered basis $(1,\alpha ,\beta )$ is determined byIts matrix, with these coordinate vectors as columns, isThe characteristic polynomial isBy the Cayley-Hamilton theorem, this monic integer polynomial annihilates $\beta$, so $\beta$ is an algebraic integer.

LetThe element $\alpha$ is also integral because it satisfies the monic polynomial $f$. Hence $A\subseteq {\mathcal{O}}_K$. Sincethe lattice $\mathbb{Z}[\alpha ]$ has index two in $A$. The given power-basis discriminant isso the discriminant-index formula for an integral lattice gives

If $m=[{\mathcal{O}}_K:A]$, thenBecause $307$ is prime, the square $m^2$ can divide $307$ only when $m=1$. Thusso $(1,\alpha ,\beta )$ is an integral basis of the cubic field of discriminant minus 307.

Finally, the relation ${\beta }^2=-4-\alpha +3\beta$ givesConsequently ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$, which will allow direct use of Dedekind's theorem.

The minimal polynomial of $\beta$ isand part (b) showed that ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$. Modulo $3$,The quadratic has no root in ${\mathbb{F}}_3$, so it is irreducible. The Dedekind factorization theorem therefore givesThe two factors are distinct prime ideal, with norms $3$ and $3^2=9$, respectively.

Modulo $2$,and the quadratic is irreducible over ${\mathbb{F}}_2$. HenceThis is a product of two distinct proper prime ideals, so $(2)$ is not prime. These decompositions are collected in prime ideals above two and three in the cubic field of discriminant minus 307.