Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 20G / b

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 $cI\subset R$.