Proposition

Let $M$ be an $A$-module. If every non-empty set of finitely generated submodules of $M$ has a maximal element, then $M$ is Noetherian.

Solution

By Proposition 6.2[Atiyah], it suffices to show that every submodule of $M$ is finitely generated.

Let $N$ be a submodule of $M$. If $N = \{ 0 \}$, then we are done. Suppose otherwise. Let $x_1 \in N \setminus \{ 0 \}$. If $N = \{ x_1 \}$, then we are done.

Having picked $x_1, \cdots, x_n$ for some $n \in \mathbb{N}$, we pick $x_{n + 1}$ to be an element in $N \setminus \ev{x_1, \cdots, x_n}$ if possible.

Then $S = \{ \ev{x_1}, \ev{x_1, x_2}, \cdots \}$ is a (possibly infinite) nonempty set of finitely generated submodules of $M$. We are given that $S$ must have a maximal element $\ev{x_1, \cdots, x_k}$ for some $k$. The obvious inclusion relation $\ev{x_1} \subsetneq \ev{x_1, x_2}, \cdots$ implies that $S$ must be finite because otherwise $\ev{x_1, \cdots, x_k} \subsetneq \ev{x_1, \cdots, x_{k + 1}}$ which is a contradiction. Therefore, $N = \ev{x_1, \cdots, x_k}$, so $N$ is finitely generated.

By Proposition 6.2[Aityah], $M$ is Noetherian.