Noetherian and Artinian properties are inherited in short exact sequences
Theorem 1.
Let $M\mathrm{,}{M}^{\mathrm{\prime}}\mathrm{,}{M}^{\mathrm{\prime \prime}}$ be $A$modules and $\mathrm{0}\mathrm{\to}{M}^{\mathrm{\prime}}\mathit{}\stackrel{\mathit{\iota}}{\mathrm{\to}}\mathit{}M\mathit{}\stackrel{\mathit{\pi}}{\mathrm{\to}}\mathit{}{M}^{\mathrm{\prime \prime}}\mathrm{\to}\mathrm{0}$ a short exact sequence^{}. Then

1.
$M$ is Noetherian^{} if and only if ${M}^{\prime}$ and ${M}^{\prime \prime}$ are Noetherian;

2.
$M$ is Artinian if and only if ${M}^{\prime}$ and ${M}^{\prime \prime}$ are Artinian.
For $\Leftarrow $, we will need a lemma that essentially says that a submodule^{} of $M$ is uniquely determined by its image in ${M}^{\prime \prime}$ and its intersection^{} with ${M}^{\prime}$:
Lemma 1.
In the situation of the theorem, if ${N}_{\mathrm{1}}\mathrm{,}{N}_{\mathrm{2}}\mathrm{\subset}M$ are submodules with ${N}_{\mathrm{1}}\mathrm{\subset}{N}_{\mathrm{2}}$, $\pi \mathit{}\mathrm{(}{N}_{\mathrm{1}}\mathrm{)}\mathrm{=}\pi \mathit{}\mathrm{(}{N}_{\mathrm{2}}\mathrm{)}$, and ${N}_{\mathrm{1}}\mathrm{\cap}\iota \mathit{}\mathrm{(}{M}^{\mathrm{\prime}}\mathrm{)}\mathrm{=}{N}_{\mathrm{2}}\mathrm{\cap}\iota \mathit{}\mathrm{(}{M}^{\mathrm{\prime}}\mathrm{)}$, then ${N}_{\mathrm{1}}\mathrm{=}{N}_{\mathrm{2}}$.
Proof.
The proof is essentially a diagram chase. Choose $x\in {N}_{2}$. Then $\pi (x)=\pi ({x}^{\prime})$ for some ${x}^{\prime}\in {N}_{1}$, and thus $\pi (x{x}^{\prime})=0$, so that $x{x}^{\prime}\in \mathrm{im}\iota $, and $x{x}^{\prime}\in {N}_{2}$ since ${N}_{1}\subset {N}_{2}$. Hence $x{x}^{\prime}\in {N}_{2}\cap \iota ({M}^{\prime})={N}_{1}\cap \iota ({M}^{\prime})\subset {N}_{1}$. Since ${x}^{\prime}\in {N}_{1}$, it follows that $x\in {N}_{1}$ so that ${N}_{1}={N}_{2}$. ∎
Proof.
($\Rightarrow $): If $M$ is Noetherian (Artinian), then any ascending (descending) chain of submodules of ${M}^{\prime}$ (or of ${M}^{\prime \prime}$) gives rise to a similar sequence in $M$, which must therefore terminate. So the original chain terminates as well.
($\Leftarrow $): Assume first that ${M}^{\prime},{M}^{\prime \prime}$ are Noetherian, and choose any ascending chain ${M}_{1}\subset {M}_{2}\subset \mathrm{\dots}$ of submodules of $M$. Then the ascending chain $\pi ({M}_{1})\subset \pi ({M}_{2})\subset \mathrm{\dots}$ and the ascending chain ${M}_{1}\cap \iota ({M}^{\prime})\subset {M}_{2}\cap \iota ({M}^{\prime})\subset \mathrm{\dots}$ both stabilize since ${M}^{\prime}$ and ${M}^{\prime \prime}$ are Noetherian. We can choose $n$ large enough so that both chains stabilize at $n$. Then for $N\ge n$, we have (by the lemma) that ${M}_{N}={M}_{n}$ since $\pi ({M}_{N})=\pi ({M}_{n})$ and ${M}_{N}\cap \iota ({M}^{\prime})={M}_{n}\cap \iota ({M}^{\prime})$. Thus $M$ is Noetherian. For the case where $M$ is Artinian, an identical proof applies, replacing ascending chains by descending chains.
∎
References
 1 M.F. Atiyah, I.G. MacDonald, Introduction to Commutative Algebra, AddisonWesley 1969.
Title  Noetherian and Artinian properties are inherited in short exact sequences 

Canonical name  NoetherianAndArtinianPropertiesAreInheritedInShortExactSequences 
Date of creation  20130322 19:11:52 
Last modified on  20130322 19:11:52 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  5 
Author  rm50 (10146) 
Entry type  Theorem 
Classification  msc 16D10 