Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 1E (Groups, Rings and Modules)

Use the norm $N(a+b\sqrt{-n})=a^2+n{b}^2$. If $n$ is composite, compare $n=(\sqrt{-n})(-\sqrt{-n})$ with an integer factorisation; if $n$ is prime, $n$ is odd and $n+1=(1+\sqrt{-n})(1-\sqrt{-n})$ has a nontrivial integer factorisation. Norm considerations show the relevant factors are irreducible and nonassociate, so unique factorisation fails.

At the maximal ideal $\mathfrak{m}=(2,X)$, the images of $4,2X,X^2$ form a three-dimensional basis of $I/\mathfrak{m}I$ over ${\mathbb{F}}_2$, so two generators are impossible. More generally $(2^{n-1},2^{n-2}X,\dots ,2X^{n-2},X^{n-1})$ needs exactly $n$ generators by the same argument.