Equivalence of Definitions of Noetherian Module
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Theorem
Let $A$ be a commutative ring with unity.
Let $M$ be an $A$-module.
The following definitions of the concept of Noetherian Module are equivalent:
Definition 1
$M$ is a Noetherian module if and only if every submodule of $M$ is finitely generated.
Definition 2
$M$ is a Noetherian module if and only if it satisfies the ascending chain condition on submodules.
Definition 3
$M$ is a Noetherian module if and only if it satisfies the maximal condition on submodules.
Proof
Definition 1 implies Definition 2
Assume $N_1 \subseteq N_2 \subseteq N_3 \subseteq \cdots$ is an increasing sequence of submodules of $M$.
By Definition 1 $N := \bigcup_{i \mathop = 1}^\infty N_i$ is a finitely generated submodule of $M$.
Suppose that $N$ is generated by $a_1, \dotsc, a_k \in N$.
For all $i \in \set {1, \dots, k}$, there is some $j_i \in \N$, such that $a_i \in N_{j_i}$.
For $j := \max \set {j_1, \dots, j_k}$, we have:
- $a_1, \dotsc, a_k \in N_j$
Hence $N_j = N$.
This shows, that the increasing sequence stabilizes, as desired.
$\Box$
Definition 2 implies Definition 1
Let $N$ be a submodule of $M$.
Aiming for a contradiction, suppose $N$ is not finitely generated.
Any finitely generated submodule of $N$ is not equal to $N$.
So we can inductively choose a sequence:
- $a_i \in N \setminus \sequence {a_1, \dots, a_{i - 1} }$.
The chain:
- $\sequence {a_1} \subsetneq \sequence {a_1, a_2} \subsetneq \sequence {a_1, a_2, a_3} \subsetneq \dotsb$
This contradicts Definition 2.
Thus $N$ is finitely generated.
$\Box$
Definition 2 iff Definition 3
This follows by Increasing Sequence in Ordered Set Terminates iff Maximal Element.
$\Box$