# Confusion over finite generation and noetherian modules

There are a few relevant posts

Error in proof that submodules of f.g. modules are f.g.

Specific proof that any finitely generated $$R$$-module over a Noetherian ring is Noetherian.

Finitely generated modules over a Noetherian ring are Noetherian

I'm trying to straighten out noetherianess and finite generation in my brain. So, say $$M$$ is a finitely generated left $$R$$-module. That is

$$M=Rm_1 + \dots + Rm_k$$

Then it feels to me that every submodule ought to be f.g. since for any chain

$$0=N_0 \subset N_1 \subset \dots \subset N_n \subset \dots$$

We have $$M = \bigcup N_j$$ and so each $$m_i$$ has to appear in $$N_i$$ for some $$i$$. We can use this to conclude finite generation. Then, since every submodule if f.g. we can conclude that $$M$$ is noetherian. This is incorrect but I'm having trouble understanding why despite going through other posts.

Then, knowing the above is incorrect, it would suffice to show that, in particular,

$$R + R$$

is noetherian (then it would follow for any finite sum by induction). So, given any chain of submodules (which will look like the sum of submodules of $$R$$) we will have something like

$$0 \subset N_1 + N'_1 \subset \dots$$

and so we can first stabilize the chain

$$N_1 \subset \dots$$

since it is a chain of submodules in $$R$$ and similarly for the $$N'$$ and so the whole chain above must stabilize. Then, given this, there is clearly a map from $$R \to M$$ which is an isomorphism which would give the result?

• There are rings which are not Noetherian. In that case $M=R$ is finitely generated but has non-finitely generated submodules. – Lord Shark the Unknown Oct 27 '18 at 13:46
• That makes sense. I know that a counter example exists I just don't see the flaw in the logic above? – Aaron Zolotor Oct 27 '18 at 13:47

$$M=Rm_1 + \dots + Rm_k$$
$$0=N_0 \subset N_1 \subset \dots \subset N_n \subset \dots$$
We have $$M = \bigcup N_j$$ and so each $$m_i$$ has to appear in $$N_i$$.