$N_1$ and $N_2$ are noetherian $R$-modules. Does it imply that $N_1+N_2$ is noetherian?

Let $M$ be an $R$-module. Let $N_1\subseteq M$ and $N_2\subseteq M$ be submodules of $M$. I think that if $N_1$ and $N_2$ are noetherian $R$-modules, then so is $N_1+N_2$. Please tell me if there is any mistake in the idea.

Proof Idea:

1. Let $V\subseteq N_1+N_2$ be a submodule of $N_1$ and $N_2$. We claim that $V$ is of the form $$V=N_1'+N_2'$$ where $N_1'$ and $N_2'$ are submodules of $N_1$ and $N_2$. Define the modules $N_1'$ and $N_2'$ as $$N_1'=\{n_1\in N_1: n_1+n_2\in V \text{ for some }n_2\in N_2\}$$ $$N_2'=\{n_2\in N_2: n_1+n_2\in V \text{ for some }n_1\in N_1\}$$ Then $V=N_1'+N_2'$.

2. Since $N_1$ is noetherian, $N_1'$ is finitely generated. Similarly $N_2'$ is finitely generated. Therefore, $V=N_1'+N_2'$ is finitely generated.

So we have proved that any arbitrary submodule of $N_1+N_2$ is finitely generated. Hence $N_1+N_2$ is noetherian.

• – Crostul Apr 29 '17 at 7:00

1 Answer

It is not true that a general submodule of $N_1+N_2$ has the form $N_1'+N_2'$ where $N_j'\subseteq N_j$. For example consider the real vector space $\Bbb R^2$ and let $N_1$, $N_2$ and $V$ be spanned by $(1,0)$, $(0,1)$ and $(1,1)$.

For a sound proof, take an increasing sequence of submodules $(M_n)$. Consider the sequences of modules $(M_n\cap N_1)$ and $((M_n+N_1)/N_1)$. The first is contained within $N_1$, the second within $(N_1+N_2)/N_1\cong N_2/(N_1\cap N_2)$ which is Noetherian (why?). Both these sequences stabilise at some stage. Show that the sequence $(M_n)$ also stabilises from then on.