Equivalence of Definitions of Generator of Module

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $R$ be a ring.

Let $M$ be an $R$-module.

Let $S \subseteq M$ be a subset of $M$.


The following definitions of the concept of Generator of Module are equivalent:

Definition 1

$S$ is a generator of $M$ if and only if $M$ is the submodule generated by $S$.

Definition 2

$S$ is a generator of $M$ if and only if $M$ has no proper submodule containing $S$.


Proof

Definition 1 implies Definition 2

By definition of generated submodule, it follows that:

$\ds M := \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} }$

Suppose that $M'$ is a proper submodule of $M$ such that $S \subseteq M'$.

It follows that there exists $x \in M \setminus M'$.

Hence, $x \notin M'$.

By definition of intersection, it follows that:

$\ds x \notin \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} } = M$

This is a contradiction.

Hence, the condition of Definition 2 is satisfied.

$\Box$


Definition 2 implies Definition 1

Let $M'$ be a submodule of $M$ that contains $S$.

By hypothesis, it follows that $M'$ is not a proper submodule of $M$.

It follows that $M' = M$.

Then:

$\ds \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} } = \bigcap \set { M } = M$


By definition of generated submodule, it follows that $M$ is generated by $S$.

$\blacksquare$