Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 20G (Number Fields)

For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}({\alpha }^{i-1}x)$ are integers, and duality gives $x={\sum }_i\mathrm{Tr}({\alpha }^{i-1}x)y_i$. Hence ${\mathcal{O}}_K\subset \sum \mathbb{Z}{y}_i$. Every ideal of $R$ is therefore a subgroup of a finitely generated free abelian group and is finitely generated over $\mathbb{Z}$, hence over $R$; so $R$ is Noetherian.

If $cI\subset R$, then $cI$ is an ideal of the Noetherian ring $R$, hence finitely generated; scaling back shows $I$ is finitely generated.

If $I=\sum R{x}_j$ with $x_j\in K$, clear the finitely many denominators relative to the $\mathbb{Q}$-basis $1,\alpha ,\dots ,{\alpha }^{n-1}$ to obtain $0\ne c\in \mathbb{Z}$ with $c{x}_j\in \mathbb{Z}[\alpha ]\subset R$. Thus $cI\subset R$.

Given a nonzero fractional $I$, scale it to an integral ideal $aI$. By (i), $(aI)J=(r)$ for some nonzero ideal $J$. Then $I^{-1}=a{r}^{-1}J$ is fractional and $I{I}^{-1}=R$.

If every fractional ideal is invertible and $I\subset R$ is nonzero, choose $I^{-1}$ and clear denominators: $0\ne c{I}^{-1}\subset R$. Then $I(c{I}^{-1})=(c)$ is principal.

Since $\beta$ is integral, $R[\beta ]$ is generated over $R$ by finitely many powers, hence is fractional. It is invertible by (c). Because it is a ring, $I^2=I$; multiplying by $I^{-1}$ gives $I=R$. Thus every $\beta \in {\mathcal{O}}_K$ lies in $R$, so $R={\mathcal{O}}_K$.

In $R=\mathbb{Z}[\sqrt{5}]$, take $I=J_1=(2,1+\sqrt{5})$ and $J_2=(2)$. Direct multiplication gives $I^2=(4,2+2\sqrt{5},6+2\sqrt{5})=2I=I{J}_2$, but $I\ne (2)$.