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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?

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$\alpha(L)=\text{Im}\,\alpha$, and since the sequence is exact $\alpha$ must be mono (and injective) and thereby $\text{Im}\,\alpha$ is isomorphic to L and Noetherian.

It would be better to write:

$$\text{Im}\,\alpha\cap M_1 \subset \text{Im}\,\alpha \cap M_2 \subset...\subset \text{Im}\,\alpha\cap M_k \subset...$$ of submodules of $\text{Im}\,\alpha$, a sequence that is finite because $\text{Im}\,\alpha$ is Noetherian.

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  • $\begingroup$ 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... $\endgroup$ – user117449 Sep 28 '14 at 7:11

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