Short Exact Sequence Condition of Noetherian Modules

Theorem
Let $A$ be a commutative ring with unity.

Let:
 * $0 \longrightarrow M' \stackrel {\alpha} {\longrightarrow} M \stackrel {\beta} {\longrightarrow} M'' \longrightarrow 0$

be a short exact sequence of $A$-modules.

Then:
 * $M$ is Noetherian


 * $M'$ and $M''$ are Noetherian
 * $M'$ and $M''$ are Noetherian

Necessary condition
Assume $M$ is Noetherian.

$M'$ is Noetherian
Let $N'\subseteq M'$ be a submodule.

Then, $\alpha \sqbrk {N'}\subseteq M$ is a submodule by Image of Linear Transformation is Submodule.

As $M$ is Noetherian, $\alpha \sqbrk {N'}$ has a finite generator:
 * $\set {\map\alpha {n'_1}, \ldots, \map\alpha {n' _n}}$

We claim $\set {n' _1,\ldots n'_n}$ is a generator of $N'$.

Indeed, let $x'\in M'$.

Then, there are $a_1,\ldots, a_n\in A$ such that:

As $\map\ker\alpha = 0$:
 * $x' = a_1 n'_1 + \cdots + a_n n' _n$

$M''$ is Noetherian
Let $N\subseteq M$ be a submodule.

Then, $\beta ^{-1} \sqbrk {N''}\subseteq M$ is a submodule by Preimage of Submodule under Linear Transformation is Submodule.

As $M$ is Noetherian, $\beta ^{-1} \sqbrk {N''}$ has a finite generator:
 * $\set {m_1, \ldots, m_k}$

We claim $\set {\map\beta {m_1},\ldots, \map\beta {m_k}}$ is a generator of $N''$.

Indeed, let $x\in N$.

As $\image\beta = M''$, thee is a $x\in M$ such that:
 * $\map\beta x = x''$

Since $x\in\beta ^{-1} \sqbrk {N''}$, there are $b_1,\ldots, b_k\in A$ such that:
 * $x = b_1 m_1 + \cdots + b_k m_k$

Therefore:

Sufficient condition
Let $M'$ and $M''$ be Noetherian.

Let $N\subseteq M$ be a submodule.

Then, $\alpha ^{-1} \sqbrk N \subseteq M'$ is a submodule by Preimage of Submodule under Linear Transformation is Submodule.

As $M'$ is Noetherian, $\alpha ^{-1} \sqbrk N$ has a finite generator:
 * $\set {m'_1, \ldots, m'_k}$

Furthermore, $\beta \sqbrk N \subseteq M''$ is a submodule by Image of Linear Transformation is Submodule.

As $M''$ is Noetherian, $\beta \sqbrk N$ has a finite generator:
 * $\set {\map\beta {m_1}, \ldots, \map\beta {m_l}}$

We claim $\set {\map\alpha {m'_1},\ldots \map\beta {m'_k}, m_1, \ldots, m_l}$ is a generator of $N$.

Indeed, let $x\in N$.

As $\map\beta x \in \beta \sqbrk N$, there exist $b_1,\ldots, b_l\in A$ such that:
 * $\map\beta x = b_1\map\beta {m_1} + \cdots + b_l \map\beta {m_l}$

Therefore:

that implies:
 * $x - \paren { b_1 m_1 + \cdots + b_l m_l } \in \map\ker\beta = \image\alpha$

Thus, there is a $x'\in\alpha ^{-1} \sqbrk N$ such that:
 * $\map\alpha {x'} = x - \paren { b_1 m_1 + \cdots + b_l m_l }$

On the other hand, there are $a_1,\ldots,a_k\in A$ such that:
 * $x' = a_1 m'_1 + \cdots + a_k m'_k$

Therefore: