Submodules of finite modules over a Noetherian ring are finite by induction

We show first that for every $n\in \mathbb{N}$ and the set $X_n:=\{1,\dots ,n\}$ every submodule $P\le R^{(X_n)}$ is finitely generated, by induction on $n$. For $n=0$ the module $R^{(\mathrm{\varnothing })}=0$ has $\{0\}=⟨\mathrm{\varnothing }{⟩}_R$ as its only submodule, so the claim holds. Let $n\ge 1$ and suppose the claim known for $n-1$. The coordinate map $\pi :R^{(X_n)}\to R$, $\pi ({\sum }_{i\in F}{r}_i{e}_i):=r_n$, is well defined by uniqueness of the expressions in [L3] and is a module homomorphism, because the expressions of $f+g$ and $rf$ have coefficientwise entries. Hence $\pi (P)$ is a submodule of the regular module ${}_{R}R$, that is an ideal of $R$ by [L6], and since $R$ is Noetherian [L1] and [L2] provide a finite generating list $\pi (P)=⟨a_1,\dots ,a_r{⟩}_R$; choose $p_1,\dots ,p_r\in P$ with $\pi (p_j)=a_j$ for each $j$, a selection from a finite list. Similarly $\ker \pi =\{f:f_n=0\}$ is the image of $R^{(X_{n-1})}$ under the coefficientwise inclusion, and $P\cap \ker \pi$ corresponds to a submodule $Q\le R^{(X_{n-1})}$, which is finitely generated by the induction hypothesis, say $Q=⟨q_1,\dots ,q_s{⟩}_R$; let $q_1^{\mathrm{\prime }},\dots ,q_s^{\mathrm{\prime }}$ be the corresponding elements of $P\cap \ker \pi$. Then $P=⟨p_1,\dots ,p_r,q_1^{\mathrm{\prime }},\dots ,q_s^{\mathrm{\prime }}{⟩}_R$: it contains the displayed elements, and for $p\in P$ the element $\pi (p)={\sum }_j{\lambda }_j{a}_j$ with ${\lambda }_j\in R$ satisfies $\pi (p-{\sum }_j{\lambda }_j{p}_j)=0$, so $p-{\sum }_j{\lambda }_j{p}_j={\sum }_k{\mu }_k{q}_k^{\mathrm{\prime }}$ for some ${\mu }_k\in R$ by [L3]. This completes the induction.

Let $m_1,\dots ,m_n\in M$ satisfy $M=⟨m_1,\dots ,m_n{⟩}_R$, as exists by finite generation of $M$ [L3]. By [L4] the set map $e_i\mapsto m_i$ on the standard basis of $R^{(X_n)}$ extends to a homomorphism $\phi :R^{(X_n)}\to M$. Its image is a submodule of $M$ by [L5] containing each $m_i$, hence containing $⟨m_1,\dots ,m_n{⟩}_R=M$ by [L3], so $\phi ({\phi }^{-1}(N))=N$ and $\phi$ is surjective. The preimage ${\phi }^{-1}(N)$ is a submodule of $R^{(X_n)}$: it is nonempty because $0$ maps to $0\in N$, and it is closed under addition and under scalars because $\phi$ is additive and $R$-linear and $N$ is a submodule. By step 1.1 it is finitely generated, say ${\phi }^{-1}(N)=⟨u_1,\dots ,u_t{⟩}_R$. Then $N=⟨\phi (u_1),\dots ,\phi (u_t){⟩}_R$: each $\phi (u_i)$ lies in $N$, and for $y\in N$ surjectivity gives $u\in R^{(X_n)}$ with $\phi (u)=y$, so $u\in {\phi }^{-1}(N)$; writing $u={\sum }_i{\lambda }_i{u}_i$ with ${\lambda }_i\in R$ by [L3] and applying $\phi$ expresses $y={\sum }_i{\lambda }_i\phi (u_i)$.

Combining steps 1.1 and 2.1, every submodule $N$ of a finitely generated module $M$ over a commutative Noetherian ring $R$ is generated by the finite list $\phi (u_1),\dots ,\phi (u_t)$; equivalently, the module $M$ is Noetherian in the sense of [L2]. The only selections used were from finite lists: the finite generating list of $M$, the finite lists of generators of $\pi (P)$, the generators of the finitely many submodules met in the induction, and the finitely many lifts $p_j$; the induction runs on $\mathbb{N}$. Hence no choice principle is used.