Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 9E (Groups, Rings and Modules)

An ideal of $R/I$ has the form $J/I$ for an ideal $J\supseteq I$ of $R$. If $R$ is Noetherian, writeThenso every ideal of $R/I$ is finitely generated. Hence

The Hilbert basis theorem states that $R[X]$ is Noetherian whenever $R$ is Noetherian. The integers form a principal ideal domain, so every ideal of $\mathbb{Z}$ has one generator and $\mathbb{Z}$ is Noetherian. It follows that $\mathbb{Z}[X]$ is Noetherian.

For a nonsquare integer $d$, evaluation at $\sqrt{d}$ givesIt is surjective and its kernel is $(X^2-d)$. Therefore is Noetherian by the quotient result. This is the noetherianity of a quadratic integer order.

The coefficient condition says precisely that every nonconstant monomial has positive powers of both variables. ThusIf $c+XYf$ and $d+XYg$ lie in $R$, so do their sum and their productso $R$ is a subring of $K[X,Y]$.

For $n\ge 1$, letThen $I_n\subseteq I_{n+1}$. The containment is strict because $X{Y}^{n+1}\notin I_n$: multiplying a generator $X{Y}^j$ by an element of $R$ produces either a scalar multiple of $X{Y}^j$, or terms divisible by $X^2$. It cannot produce the monomial $X{Y}^{n+1}$ when $j\le n$. Henceis a strictly increasing ideal chain. The constant-plus-ideal non-Noetherian subring therefore satisfies