Finitely generated module over Noetherian ring I read that these two proposition are equivalent:


*

*$R$ is a noetherian ring;

*Every submodule of a finitely generated $R-$module is finitely generated.


While I was thinking about it I made the reasoning below, that leads me to an incorrect conclusion, so I must be wrong somewhere. I hope someone can explain me, thank you in advance.
Calling $M$ the $R-$module, we have $M\simeq R/\mathfrak{i_1}\oplus \dots \oplus R/\mathfrak{i_n}$, so it's clear that if $R$ is noetherian, so is $M$; however if $R/\mathfrak{i_1}\oplus \dots \oplus R/\mathfrak{i_n}$ is noetherian for some ideals $\mathfrak{i_1},\dots , \mathfrak{i_n}$, then $R$ is not necessarily noetherian. That means that the fact that $M$ is noetherian doesn't imply that $R$ is noetherian too.
For example, I let the ring of polynomials $\Bbb R[T_1,T_2,\dots ]$ act on the group $\Bbb R^2$, with $T_im=0\; \forall i,\forall \;m\in \Bbb R^2$. I obtain a module which is isomorphic to $\Bbb R[T_1,T_2,\dots ]/(T_1,T_2,\dots)\oplus \Bbb R[T_1,T_2,\dots ]/(T_1,T_2,\dots)$, which is in fact $\Bbb R^2$. This module is noetherian since every sumbodule is a subspace of dimension $1$ but clearly $\Bbb R[T_1,T_2,\dots ]$ is not.

$R$ is an $R-$ module with respect itself so if it has unity then it will be finitely generated then by hypothesis you have that each sub module of $R$ is finitely generated. 
You can observe that each ideal $I$ of $R$ is an $R-$ submodule of $R$ so it will be finitely generated, then there exists a finite number of elements $a_1,\dots, a_n$ such that 
$I=(a_1,\dots , a_n)$ 
So $R$ is a noetherian ring. 
Now we suppose that $R$ is a noetherian ring. If $f_1,\dots, f_n$ are generators of $M$ then you can define 
$\psi: R\times \dots \times R\to M$ such that for each $(r_1,\dots r_n)$ you have that 
$\psi(r_1,\dots , r_n):=r_1 f_1+\dots +\dots r_nf_n$
The map is surjective so by first homomorphism theorem you have that 
$M\cong (R\times \dots \times R)/ ker(\psi)$
It is easily prove that if $R$ is a noetherian ring then $R\times \dots \times R$ is such that each its submodule is finitely generated. 
Each submodule of $(R\times \dots \times R)/ K$ is finetly generated where $K$ is submodule of $R\times \dots \times R$ so you have finished. 
