2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.