# $0 \to L \to M \to N \to 0$ is a short exact sequence. $M$ is Noetherian $\iff$ $L$ and $N$ are Noetherian

Let $$0 \rightarrow L \xrightarrow{\alpha} M \xrightarrow{\beta} N \rightarrow 0$$ be a short exact sequence of $$A$$-modules ($$A$$ is a commutative ring).

Prove $$M \text{ is Noetherian} \iff L \text{ and } N \text{ are Noetherian.}$$

I don't understand the first part of $$\impliedby$$. It says:

Let $$M_1 \subset M_2 \subset \dots \subset M_k \subset \cdots$$ be an increasing chain of submodules of $$M$$. Then identifying $$\alpha(L)$$ with $$L$$ and taking intersection gives a chain $$L\cap M_1 \subset L \cap M_2 \subset \dots \subset L\cap M_k \subset \cdots$$ of submodules of $$L$$.

I don't get it, what does "identifying $$\alpha(L)$$ with $$L$$" even mean? I don't understand what $$L$$ is and therefore the chain of submodules of $$L$$ doesn't make sense.

Can someone bring clarity to this, what am I missing?

$\alpha(L)=\text{Im}\,\alpha$, and since the sequence is exact $\alpha$ must be mono (and injective) and thereby $\text{Im}\,\alpha$ is isomorphic to L and Noetherian.
$$\text{Im}\,\alpha\cap M_1 \subset \text{Im}\,\alpha \cap M_2 \subset...\subset \text{Im}\,\alpha\cap M_k \subset...$$ of submodules of $\text{Im}\,\alpha$, a sequence that is finite because $\text{Im}\,\alpha$ is Noetherian.
• Thank you so much, I understood it correct now. I dont get it how one can use $L$ for $Im(\alpha)$ when $L$ is the domain of the function, whatever... – user117449 Sep 28 '14 at 7:11