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 $\mathrm{0}\mathrm{\to }{M}^{\mathrm{\prime }}\mathit{}\stackrel{𝜄}{\mathrm{\to }}\mathit{}M\mathit{}\stackrel{𝜋}{\mathrm{\to }}\mathit{}{M}^{\mathrm{\prime \prime }}\mathrm{\to }\mathrm{0}$ 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.∎