examples of modules

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

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Any abelian group is a module over the ring of integers, with action defined by $n\cdot g$ for $g\in G$ given by $n\cdot g=\sum_{i=1}^{n}g$ .
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If $R$ is a subring of a ring $S$ , then $S$ is an $R$ -module, with action given by multiplication in $S$ .
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If $R$ is any ring, then any (left) ideal $I$ of $R$ is a (left) $R$ -module, with action given by the multiplication in $R$ .
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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).
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Yetter-Drinfel’d module. (http://planetmath.org/QuantumDouble)