This entry is a collection of examples of modules over rings. Unless otherwise specified in the example, $M$ will be a module over a ring $R$ .

• Any abelian group is a module over the ring of integers, with action defined by $n\cdot g$ for $g\in G$ given by $n\cdot g=\sum_{i=1}^n g$ .
• If $R$ is a subring of a ring $S$ , then $S$ is an $R$ -module, with action given by multiplication in $S$ .
• If $R$ is any ring, then any (left) ideal $I$ of $R$ is a (left) $R$ -module, with action given by the multiplication in $R$ .
• Let $R=\mathbb{Z}$ and let $E = \{2k \mid k \in \mathbb{Z}\}$ . Then $E$ is a module over the ring $\mathbb{Z}$ of integers. Further, define the sets $B = E \times E$ and $C = E \times \{0\}$ and $D = \{0\} \times E$ . Then $B$ , $C$ , and $D$ are modules over $\mathbb{Z} \times \mathbb{Z}$ , with action given by $a \cdot x = (a \cdot x_1, a \cdot x_2)$ if $x = (x_1,x_2)$ even if the product is redefined as $a \cdot x_1 = 0$ and $a \cdot x_2 = 0$ , but now the identity element is $(1,1)$ . However by our new product definition $a \cdot x = (a \cdot x_1, a \cdot x_2) = (0,0)$ even if $a = (1,1)$ , the ring identity element originally In the more general definition of module which does not require an identity element $\bf{1}$ in the ring and does not require ${\bf 1} \cdot m = m$ for all $m \in M$ , we observe that ${\bf 1} \cdot m \neq m$ in this example just constructed. (one of the purposes of this comment is to show that all modules need not be unital ones).
• Yetter-Drinfel'd module.