A module $X$ over a ring $R$ is said to be finitely generated if there is a finite subset $Y$ of $X$ such that $Y$ spans $X$ . Let us recall that the span of a (not necessarily finite) set $X$ of vectors is the class of all (finite) linear combinations of elements of $S$ ; moreover, let us recall that the span of the empty set is defined to be the singleton consisting of only one vector, the zero vector $\vec{0}$ . A module $X$ is then called cyclic if it can be spanned by a singleton.

Examples. Let $R$ be a commutative ring with 1 and $x$ be an indeterminate.

1. $Rx=\lbrace rx \mid r\in R \rbrace$ is a cyclic $R$ -module generated by $\lbrace x \rbrace$ .
2. $R\oplus Rx$ is a finitely-generated $R$ -module generated by $\lbrace 1, x \rbrace$ . Any element in $R\oplus Rx$ can be expressed uniquely as $r+sx$ .
3. $R[x]$ is not finitely generated as an $R$ -module. For if there is a finite set $Y$ spanning $R[x]$ , taking $d$ to be the largest of all degrees of polynomials in $Y$ , then $x^{d+1}$ would not be in the spanning set of $Y$ , assumed to be $R[x]$ , which is a contradiction. (Note, however, that $R[x]$ is finitely-generated as an $R$ -algebra.)