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
Proof
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:
| \(\ds \map\alpha {x'}\) | \(=\) | \(\ds a_1\map\alpha {n'_1} + \cdots + a_n \map\alpha {n' _n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map\alpha {a_1 n'_1 + \cdots + a_n n' _n}\) | Definition:Linear Transformation |
As $\map\ker\alpha = 0$:
- $x' = a_1 n'_1 + \cdots + a_n n' _n$
$\Box$
$M$ 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:
| \(\ds x\) | \(=\) | \(\ds \map\beta x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map\beta {b_1 m_1 + \cdots + b_k m_k}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds b_1 \map\beta {m_1} + \cdots + b_k \map\beta {m_k}\) | Definition:Linear Transformation |
$\Box$
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:
| \(\ds \map\beta {x - \paren { b_1 m_1 + \cdots + b_l m_l } }\) | \(=\) | \(\ds \map\beta x - \paren { b_1\map\beta {m_1} + \cdots + b_l \map\beta {m_l} }\) | Definition:Linear Transformation | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map\beta x - \map\beta x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
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:
| \(\ds x\) | \(=\) | \(\ds \map\alpha {x'} + b_1 m_1 + \cdots + b_l m_l\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map\alpha {a_1 m'_1 + \cdots + a_k m'_k} + b_1 m_1 + \cdots + b_l m_l\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a_1 \map\alpha {m'_1} + \cdots + a_k \map\alpha {m'_k} + b_1 m_1 + \cdots + b_l m_l\) | Definition:Linear Transformation |
$\blacksquare$