Equivalence of Definitions of Noetherian Module

From ProofWiki
Jump to navigation Jump to search



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:

$\sequence {a_1} \subsetneq \sequence {a_1, a_2} \subsetneq \sequence {a_1, a_2, a_3} \subsetneq \dotsb$

is strictly increasing.

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.

$\Box$


Also see