Problems with understanding the proof of noetherian ring If $M$ is an $R$-module, the the following are equivalent:
1. M is finitely generated
2. M satisfies the ascending chain condition
3. Every non-empty set of submodules of M contains at least one maximal element
The proof: 3-> 1 Suppose 1 is false, so there is a submodule M' of M that is not finitely generated. We will show that the set of $S$ of all finitely generated submodules of $M'$ does not have a maximal element, which will be a contradiction. Suppose $S$ does not have a maximal element L. Since L is finely generated and $L \subsetneq M'$ and M' is not finitely generated, there is an $a\in M'$ such that $a \neq L$. The $L' = L + Ra$ is an element of S that strictly contains the presumed maximal element L, a contradiction.
My problem with understanding this proof is that why we can assume that submodules of M' also contain a maximal element when the statement is M contains a least element. I don't think submodules of M' is necessarily a submodule of M.
 A: 
Suppose 1 is false, so there is a submodule M' of M that is not finitely generated." 

That is not the negation of "$M$ is finitely generated" unless you are talking about $M=M'$. Using the notation $M'$ basically serves no purpose.
So, you are not really having any problems in your proof of 3 implying 1.


@MooS isn't a noetherian ring a ring that satisfies any of the statement above?

No, definitely not. It occurs to me you might be muddling a few problems together.
For any right $R$ module $M$, consider the following statements:


*

*The poset of submodules of $M$ satisfies the ascending chain condition.

*The poset of submodules of $M$ satisfies the maximal condition

*All submodules of $M$ are finitely generated

*$M$ is finitely generated.
The first three are always equivalent, and the first three always imply the fourth, but #4 is not equivalent to the others.
If, in addition, $R$ is right Noetherian, then the fourth condition becomes equivalent to the first three.
Now in your original post, you have been talking about conditions 1, 2, and 4 (on my list), but when you went to do your proof, you immediately started by negating 3 (on my list.)  So this may be the source of your confusion.
If you stop using $M'$ and replace it with $M$, you will have given a correct proof that a Noetherian module is finitely generated. On the other hand, if you want to prove 1$\implies$3 (on my list) then you can use exactly the same argument for an arbitrary submodule $M'$ of $M$, as long as you realize that conditions $1$ and $2$ pass to submodules.

I don't think submodules of M' is necessarily a submodule of M. 

A submodule of a submodule of $M$ is a submodule of $M$: being a submodule is a transitive relation. It's straightforward to check. Maybe you are thinking of ideals in ring theory, because being an ideal is not normally a transitive relation.
