Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 1 / c / Solution

DefineThis makes $M[X]$ an $R[X]$-module.

Let $N\subseteq M[X]$ be an $R[X]$-submodule. For $n\ge 0$, let $L_n\subseteq M$ consist of zero and the leading coefficients of elements of $N$ of degree $n$. Each $L_n$ is an $R$-submodule, and multiplication by $X$ givesBecause $M$ is Noetherian, this chain stabilizes at some $n_0$, and each $L_n$ for $n\le n_0$ is finitely generated. Choose finitely many polynomials $f_{nj}\in N$ of degree $n$ whose leading coefficients generate $L_n$.

For $f\in N$ of degree $d$, if $d\le n_0$, subtract an $R$-linear combination of the $f_{dj}$ to lower its degree. If $d>n_0$, use $L_d=L_{n_0}$ and subtract a combination of $X^{d-n_0}{f}_{n_{0}j}$. Induction on degree expresses $f$ in terms of the finite collection $\{f_{nj}\}$. Therefore every submodule $N$ is finitely generated andThis is the module form of the Hilbert basis theorem.