= Solution
Work with the usual commutative unital <Noetherian ring> convention. First prove that every <submodule> of $R^d$ is a <finitely generated module>. For $d=1$, a <submodule> is an <ideal>, hence finitely generated by Noetherianity. For the induction step, project a submodule $N\subseteq R^d$ onto its last coordinate. Its image is a finitely generated <ideal>; choose lifts in $N$ of its generators. The projection kernel lies in $R^{d-1}$ and is finitely generated by induction. The kernel generators together with the chosen lifts generate $N$, since subtracting a combination of the lifts leaves an element of the kernel.
Now choose a surjection $R^d\to M$ from a finite generating set of the <module> $M$. For any <submodule> $U\subseteq M$, its inverse image in $R^d$ is finitely generated by the result just proved. The images of those generators generate $U$.
\b[Every submodule of $M$ is finitely generated.] In particular $M$ is a <Noetherian module>: the union of an ascending chain is a submodule, and its finite generators all belong to a single member of the chain, forcing eventual stabilization.
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