Finite modules over Noetherian rings are Noetherian

First suppose $M$ is cyclic, say $M=Rx$. Let $N\le M$ be a submodule and put $I:=\{r\in R:rx\in N\}\mathrm{.}$ Then $I$ is an ideal of $R$, so [L1] gives generators $r_1,\dots ,r_s$ for $I$. Every element of $N$ has the form $rx$ with $r\in I$, hence lies in the submodule generated by $r_{1}x,\dots ,r_{s}x$. Thus every submodule of a cyclic module is finitely generated, so every cyclic module over $R$ is Noetherian.

If $M=0$, its only submodule is $0$, generated by the empty set, so it is Noetherian. Otherwise let $M=⟨x_1,\dots ,x_n{⟩}_R$ with $n\ge 1$, and argue by induction on $n$. The base case $n=1$ is step 1.1. For $n>1$, set $M^{\mathrm{\prime }}:=⟨x_1,\dots ,x_{n-1}{⟩}_R$. Then $M\mathrm{/}{M}^{\mathrm{\prime }}$ is cyclic, generated by $x_n+M^{\mathrm{\prime }}$, so step 1.1 makes $M\mathrm{/}{M}^{\mathrm{\prime }}$ Noetherian. By the induction hypothesis, $M^{\mathrm{\prime }}$ is Noetherian.

For the induction step $n>1$, let $N\le M$. Since $M^{\mathrm{\prime }}$ is Noetherian, the submodule $N\cap M^{\mathrm{\prime }}$ is finitely generated. Since $M\mathrm{/}{M}^{\mathrm{\prime }}$ is Noetherian, the image $(N+M^{\mathrm{\prime }})\mathrm{/}{M}^{\mathrm{\prime }}\le M\mathrm{/}{M}^{\mathrm{\prime }}$ is finitely generated; choose lifts $y_1,\dots ,y_t\in N$ of its generators. Every element of $N$ differs from an $R$-linear combination of the $y_i$ by an element of $N\cap M^{\mathrm{\prime }}$, so $N$ is generated by those $y_i$ together with generators of $N\cap M^{\mathrm{\prime }}$. Thus every submodule of $M$ is finitely generated.

The zero case is in step 2.1, the one-generator case in step 1.1, and step 3.1 completes the induction for all remaining finite generating lists. Therefore $M$ is Noetherian.