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

Suppose every submodule of $M$ is finitely generated. For an ascending chain $N_1\subseteq N_2\subseteq \cdots$, its union $N$ is a submodule. Finitely many generators of $N$ all lie in one $N_r$, so $N=N_r$ and the chain stabilizes. Thus the ascending chain condition holds.

If the ascending chain condition holds, any nonempty collection of submodules has a maximal member: otherwise, starting from one member and repeatedly choosing a strictly larger one constructs a nonstationary ascending chain.

Finally, assume the maximal condition. Among the finitely generated submodules of a given $N\subseteq M$, choose a maximal one $F$. If $F\ne N$, then $F+Rx$ is a larger finitely generated submodule for any $x\in N\setminus F$, a contradiction. Hence every $N$ is finitely generated. These are the three equivalent characterizations of a Noetherian module.

Let $n_1,\dots ,n_r$ generate $N$, and let the images of $m_1,\dots ,m_s$ generate $M/N$. For any $m\in M$, subtracting a suitable linear combination of the $m_j$ leaves an element of $N$, which is a combination of the $n_i$. Thus

If $M$ is Noetherian, every submodule of $N$ is a submodule of $M$, and submodules of $M/N$ correspond to submodules of $M$ containing $N$; hence both are Noetherian.

Conversely, suppose $N$ and $M/N$ are Noetherian. For any $P\subseteq M$, the intersection $P\cap N$ is finitely generated and the image $(P+N)/N$ is finitely generated. Lifting generators of the image and applying part i toshows that $P$ is finitely generated. Thus $M$ is Noetherian.

The exact sequenceand part ii show that a direct sum of two modules is Noetherian exactly when both summands are. Induction proves the assertion for every finite direct sum.

Each $R/I_j$ is Noetherian as an $R$-module because its $R$-submodules are precisely its ideals as a Noetherian ring. The diagonal maphas kernel ${\bigcap }_j{I}_j=0$. Hence $R$ is an $R$-submodule of a finite direct sum of Noetherian modules, so $R$ is a Noetherian $R$-module; equivalently, $R$ is a Noetherian ring.

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.