\(\displaystyle{1)\implies 2)}\) : Suppose that \(\displaystyle{_{R}\,M}\) is a noetherian module.
Let \(\displaystyle{N}\) be a submodule of \(\displaystyle{_{R}\,M}\) . If \(\displaystyle{N=\left\{0\right\}}\), then the
submodule \(\displaystyle{N}\) is finitely generated. Let now \(\displaystyle{N\neq \left\{0\right\}}\). There is \(\displaystyle{x\in M}\)
such that \(\displaystyle{x_1\in N}\) and \(\displaystyle{x_1\neq 0}\) . If \(\displaystyle{R\,x_1=N}\), it's ok. If \(\displaystyle{R\,x_1\neq N}\), then
there is \(\displaystyle{x_2\in M}\) such that \(\displaystyle{x_2\in N}\) and \(\displaystyle{x_2\notin R\,x_1\,\,\,,R\,x_1+R\,x_2\subseteq N}\) .
If \(\displaystyle{R\,x_1+R\,x_2=N}\), then \(\displaystyle{N=<\left\{x_1,x_2\right\}>}\) .
On the other hand, if \(\displaystyle{R\,x_1+R\,x_2\neq N}\), then \(\displaystyle{x_3\in N\,,x_3\notin R\,x_1+R\,x_2}\)
for some \(\displaystyle{x_3\in M}\) . Following the same progress, we have an increasing sequence \(\displaystyle{\left(I_{n}\right)_{n\in\mathbb{N}}}\) , where
\(\displaystyle{I_{n}=\sum_{i=1}^{n}R\,x_{i}}\), of submodules of \(\displaystyle{_{R}\,M}\) . Since \(\displaystyle{_{R}\,M}\) is a noetherian module,
there is \(\displaystyle{k\in\mathbb{N}}\) such that \(\displaystyle{N=\sum_{i=1}^{k}I_{k}\implies N=<\left\{x_1,...,x_k\right\}>}\) .
\(\displaystyle{2)\implies 3)}\) : Let \(\displaystyle{S}\) be a nonempty family of submodules of \(\displaystyle{_{R}\,M}\) . Consider the
partial ordered set \(\displaystyle{\left(S,\subseteq\right)}\) and the result follows from \(\displaystyle{\rm{Zorn's}}\) lemma.
\(\displaystyle{3)\implies 1)}\) : Suppose that \(\displaystyle{\left(I_{n}\right)_{n\in\mathbb{N}}}\) is an increasing sequence of
submodules of \(\displaystyle{_{R}\,M}\). By setting \(\displaystyle{S=\left\{I_{n}:n\in\mathbb{N}\right\}}\) , we get that \(\displaystyle{S}\) is a
nonempty family of submodules of \(\displaystyle{_{R}\,M}\) and according to \(\displaystyle{3)}\) there is maximal element \(\displaystyle{I_{k}}\) .
If \(\displaystyle{n\geq k}\), then \(\displaystyle{I_{k}\subseteq I_{n}\implies I_{k}=I_{n}}\), so the left \(\displaystyle{R}\) - module \(\displaystyle{_{R}\,M}\)
is a noetherian module.
Therefore, \(\displaystyle{1)\iff 2)\iff 3)\iff 1)}\) . In order to complete this proof,
we'll prove that \(\displaystyle{1)\implies 4)}\) is true, cause the converse is apparently true.
\(\displaystyle{1)\implies 4)}\) : Suppose that \(\displaystyle{_{R}\,M}\) is a noetherian module. If \(\displaystyle{N}\) is a submodule of \(\displaystyle{_{R}\,M}\)
and \(\displaystyle{\left(K_{n}\right)_{n\in\mathbb{N}}}\) is an increasing sequence of submodules of \(\displaystyle{N}\), then the sequence
\(\displaystyle{\left(K_{n}\right)_{n\in\mathbb{N}}}\) is also an increasing sequence of the noetherian module \(\displaystyle{_{R}\,M}\), so :
there is \(\displaystyle{i\in\mathbb{N}}\) such that \(\displaystyle{K_{j}=K_{i}\,,\forall\,j\geq i}\) and thus \(\displaystyle{_{R}\,N}\) is a noetherian module too.
Now, let \(\displaystyle{N}\) be a submodule of \(\displaystyle{_{R}\,M}\) and let \(\displaystyle{_{R}\,\left(M/N\right)}\) be the quotient module.
Consider an increasing sequence \(\displaystyle{\left(K_{n}\right)_{n\in\mathbb{N}}}\) of submodules of \(\displaystyle{_{R}\,\left(M/{N}\right)}\) .
Then, for each \(\displaystyle{n\in\mathbb{N}}\), there is a submodule \(\displaystyle{M_{n}}\) of \(\displaystyle{_{R}\,M}\) such that
\(\displaystyle{N\subseteq M_{n}\subseteq M}\) and \(\displaystyle{K_{n}=M_{n}/{N}}\) . Due to the fact that \(\displaystyle{K_{n}\subseteq K_{n+1}\,,\forall\,n\in\mathbb{N}}\),
we get \(\displaystyle{M_{n}\subseteq M_{n+1}\,,\forall\,n\in\mathbb{N}}\) and thus: