Simple module

[Papapetros Vaggelis]

Let \(\displaystyle{_{R}M}\) be a left \(\displaystyle{R}\) - module.

Show that the following conditions are equivalent :

1. \(\displaystyle{M}\) is a simple \(\displaystyle{R}\) - module .

2. \(\displaystyle{\forall\,m\in M-\left\{0\right\}, M=\langle{m\rangle}}\) .

3. \(\displaystyle{M\simeq R/I}\), where \(\displaystyle{I}\) is a maximal ideal of \(\displaystyle{R}\) .

Note : The ring \(\displaystyle{R}\) is associative with unity \(\displaystyle{1_{R}}\) .

[Tsakanikas Nickos]

Hello, Vaggelis!

\( (i) \implies (ii) : \) Suppose that \( M \) is a simple \( R \)-module and consider an \( m \in M \smallsetminus \{ 0 \} \). Since \( M \) is simple, the only submodules of \( M \) are the trivial ones, and since \( m \) is non-zero, the submodule \( <m> \) of \( M \) generated by \( m \) has to be equal to \( M \).

\( (ii) \implies (iii) : \) Pick \( m \in M \smallsetminus \{ 0 \} \) and consider the map \( f \ \colon R \longrightarrow Rm \ , \ r \mapsto rm \). By hypothesis, \( Rm = \, <m> \, = M \), and it is immediate to observe that \( f \) is an \( R \)-module epimorphism. Hence, by the first isomorphism theorem for modules, we have that \[ R / Ker(f) \cong M \]We will show that \( I = Ker(f) \) is a maximal ideal of \( R \).

• \( I \neq M \), because \( f(1_{R})=m \neq 0 \)
• Let \( J \) be an ideal of \( R \) such that \( I \subseteq J \subseteq R \) and suppose that \( I \neq J \). Consider \( j \in J \smallsetminus I \) and observe that \( jm \neq 0 \), because \( j \notin I = Ker(f) \). By hypothesis, \( <jm> = M \). Therefore, there exists \( k \in R \) such that \( kjm = m \). Hence \[ kjm = m \implies \left( 1_{R} - kj \right) m = 0 \implies 1_{R} - kj \in I \subset J \implies 1_{R} - kj \in J \implies 1_{R} \in J \]because \( j \in J \) and \( J \) is an ideal of \( R \). Thus \( J = R \).

Conclusively, \( M \cong R/I \), where \( I = Ker(f) \) is a maximal ideal of \( R \).

\( (iii) \implies (i) : \) Suppose that \( M \cong R/I \), where \( I \) is a maximal ideal of \( R \). We know that there is a bijection between ideals of \( R / I \) and ideals of \( R \) containing \( I \). But since \( I \) is a maximal ideal, the only ideals of \( R \) containing \( I \) are \( I \) itself and \( R \). Now, recall that submodules of a ring coincide with the ideals of the ring. By the previous discussion, it follows that \( R/I \) is a simple \( R \)-module (as its only submodules are \( I / I \cong 0 \) and \( R / I \)) and therefore \( M \) is a simple \( R \)-module.