Proposition 3

Let M be a finitely generated A-module. Every proper submodule of M is contained in a maximal submodule.

Let N be a proper submodule of M. Denote by $\mathscr{S}$ the set of proper submodules of M containing N, ordered by inclusion; let us prove that $\mathscr{S}$ is inductive. Let $\mathscr{F}$ be a totally ordered subset of $\mathscr{S}$. If $\mathscr{F}$ is empty, then N is an upper bound for $\mathscr{F}$ in $\mathscr{S}$. In the opposite case, denote by Q the union of the elements of $\mathscr{F}$. Then Q is a submodule of M. Let F be a finite generating subset of M. If Q were equal to M, then F would be contained in a submodule $P\in \mathscr{F}$, which would imply P = M, contrary to the definition of $\mathscr{F}$. We therefore have $Q\in \mathscr{S}$, which proves that $\mathscr{S}$ is inductive. Proposition 3 then follows from Corollary 1 of Set Theory, III, §2, No. 4, p. 155.