Equivalence of Definitions of Generator of Module
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 $S$.
$\blacksquare$