Finite generation, ACC, and maximal-condition characterizations of Noetherian modules

Assume condition 1 and let $(N_n)$ be ascending. The union $N:={\bigcup }_n{N}_n$ contains zero. Any two of its elements lie in a common stage, so their combination $ru+v$ lies there too. Thus $N$ is a submodule by [L2]. By condition 1, $N$ has a finite generating set $S$. A finite subset of this ascending union lies in one stage: start with stage zero for the empty set, and when adjoining one element take the larger of the previous stage and a stage containing that element. Hence $S\subseteq N_k$ for some $k$, so $N=⟨S{⟩}_R\subseteq N_k\subseteq N$. For $n\ge k$, $N_k\subseteq N_n\subseteq N=N_k$. This proves condition 2 without countably many choices.

Assume condition 2 and DC. Let $\mathcal{F}$ be a nonempty set of submodules. If it has no maximal member, the relation $ARB$ defined by $A⊊B$ is entire on $\mathcal{F}$. Fix $A_0\in \mathcal{F}$. By [L3] there is a sequence $(A_n)$ in $\mathcal{F}$ with $A_n⊊A_{n+1}$ for every $n$, contrary to condition 2. Thus $\mathcal{F}$ has a maximal member, proving condition 3.

Assume condition 3 and fix a submodule $N\le M$. Its finitely generated submodules form a set $\mathcal{G}$ containing the zero submodule, generated by the empty set. Condition 3 gives a maximal $L\in \mathcal{G}$. Fix a finite generating set $S$ of $L$. If $L\mathrm{\ne }N$, an element $x\in N\setminus L$ makes $⟨S\cup \{x\}{⟩}_R$ a finitely generated submodule of $N$ strictly containing $L$, a contradiction. Hence $L=N$, proving condition 1 without DC.

Steps 1.1, 1.2 and 1.3 prove the stated implications. Only step 1.2 uses DC.