simple module

Definition

Let $M$ be an R-module. $M$ is called simple or irreducible if it has no non-zero, proper submodules.

Relation to cyclic modules

Every simple module is cyclic.

Proof

Let $M$ be simple. Let $X\subset M$ generate $M$. Then, for each $x\in X$, we have $⟨x⟩$ which is a submodule of $M$, therefore by assumption $⟨x⟩=0$ or $⟨x⟩=M$. If $⟨x⟩=0$ then it doesn’t contribute anything, and we don’t need it in the generating set, and we can consider $⟨X\backslash x⟩=M$. If $⟨x⟩=M$ then M is cyclic.

Converse is not true

Note, not every cyclic module is simple. Consider the $\mathbb{Z}$-module $\mathbb{Z}$. $\mathbb{Z}=⟨1⟩$, but $2\mathbb{Z}$ is a submodule, so $\mathbb{Z}$ is cyclic but not simple.

Examples

$\mathbb{Z}$-modules can just be thought of as abelian groups, so the the simple modules are the groups of prime order.

$\mathbb{C}$-modules are just $\mathbb{C}$ vector spaces, so the only simple modules are the 1 dimensional vector spaces, which are all isomorphic to $\mathbb{C}$.

The $\mathbb{C}[x]$ case is a lot more interesting.

$\mathbb{C}[x]$ is a PID. Therefore, from the classification of finitely generated modules of PID modules for any $\mathbb{C}[x]$-module $M$ (let $R=\mathbb{C}[x]$)

$$Mcong{R}^{\oplus b}\oplus R/(p_1^{r_1})\oplus \cdots \oplus R/(p_n^{r_n})$$

If $M$ is simple then it must be cyclic, so the only option are $Mcong\mathbb{C}[x]$ or $M=\mathbb{C}[x]/(p)$ for some prime $p$. We know that $\mathbb{C}[x]$ as a module has proper submodules (since these the ideals of $\mathbb{C}[x]$) so we are left with $\mathbb{C}[x]/(p)$. In order for this to be simple $p$ must be an irreducible polynomial, and since $\mathbb{C}$ is algebraically closed $p$ is of the form $(x-\lambda )$. Therefore, the simple $\mathbb{C}[x]$ modules are

$$\mathbb{C}[x]/(x-\lambda )\cong \mathbb{C}.$$