Please help me to prove the following resut:
Let $R=\prod_{n\in\mathbb{N}}\mathbb{Z}$ and $M$ be $R$ regarded as $R$-module is usual way.
Then the submodule $N=\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ is not finitely generated
Hint: for every $n\in\mathbb{N}$, $e_n=(0,…,0,1,0,…)\in N$ (the $1$ is on the $n-$th position).
Thanks in advance
On
There is an interesting characterization of finitely generated modules. Call a family $\mathscr{F}$ of submodules of $M$ directed if, for every $A,B\in\mathscr{F}$, there exists $C\in\mathscr{F}$ such that $A+B\subseteq C$.
We can also consider $\sum\mathscr{F}$, the least submodule of $M$ containing all submodules belonging to $\mathscr{F}$.
Theorem. A module $M$ is finitely generated if and only if, for every directed family $\mathscr{F}$ of submodules of $M$, if $\sum\mathscr{F}=M$, then $M\in\mathscr{F}$.
Proof. ($\implies$) Exercise. ($\impliedby$) Consider the family $\mathscr{F}$ of all finitely generated submodules of $M$. Then clearly $\mathscr{F}$ is directed and $\sum\mathscr{F}=M$. Then $M\in\mathscr{F}$ is finitely generated. □
For your problem, consider the family $\mathscr{F}=\{A_n:n\ge1\}$, where $A_n$ is the submodule of $N$ spanned by $\{e_1,e_2,\dots,e_n\}$. Since $\mathscr{F}$ is clearly directed, being totally ordered by inclusion, and $\sum\mathscr{F}=N$, we conclude that $N$ is not finitely generated, because $N\notin\mathscr{F}$.
The condition in the theorem above can be easily dualized. A family $\mathscr{F}$ of submodule is called filtered if, for every $A,B\in\mathscr{F}$, there exists $C\in\mathscr{F}$ such that $C\subseteq A\cap B$.
Definition. A module $M$ is finitely cogenerated when, for every filtered family $\mathscr{F}$ of submodules of $M$, if $\bigcap\mathscr{F}=\{0\}$, then $\{0\}\in\mathscr{F}$.
For instance, a module $M$ is artinian if and only if, for every submodule $L$ of $M$, $M/L$ is finitely cogenerated. Compare with the theorem about noetherian modules: a module $M$ is noetherian if and only if every submodule $L$ of $M$ is finitely generated.
$N$ is the module generated by $e_i,i\in\mathbb{N}$, this implies that every $x\in N$ is of the form $x=l_{i_1}e_{i_1}+...+l_{i_m}e_{i_m}, {i_1}<i_2..<i_m$, we deduce if $m>i_m$ the $m$-coordinate of $x$ is zero. Suppose it is finitely generated by $u_1,...,u_n$ remark that for every $i=1,...,n$, there exists $n_i$ such that for every $m>n_i$ the $m$-coordinate of $u_i$ is zero. Let $M>sup\{n_i\}$, for every $x$ in the nodule generated by $u_1,...,u_n$ and $m>M$, the $m$-coordinate of $x$ is zero. Contradiction since $e_m\in N$ and the $m$-coordinate of $e_m$ is $1$.