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?
Best Answer
$$M=Rm_1 + \dots + Rm_k$$
Then it feels to me that every submodule ought to be f.g. since for any chain
$$0=N_0 \subset N_1 \subset \dots \subset N_n \subset \dots $$
Wrong
We have $M = \bigcup N_j$ and so each $m_i$ has to appear in $N_i$.
We do not know that the union of any chain will be the entire module we are looking at. So we aren't able to apply that argument.