finitely generated moduleFinitelyGeneratedRModule2003-10-15 01:26:082008-10-20 10:22:38Definitionfinitely generatedcyclic modulefinitely generated modulespancyclic modulezero vector $\vec{0}$singleton\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}\PMlinkescapeword{cyclic}
A module $X$ over a ring $R$ is said to be {\em 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 \emph{cyclic} if it can be \PMlinkescapetext{spanned by} a singleton.
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\textbf{Examples}. Let $R$ be a commutative ring with 1 and $x$ be an indeterminate.
\begin{enumerate}
\item $Rx=\lbrace rx \mid r\in R \rbrace$ is a cyclic $R$-module generated by $\lbrace x \rbrace$.
\item $R\oplus Rx$ is a finitely-generated $R$-module generated by $\lbrace 1, x \rbrace$. Any element in $R\oplus Rx$
can be expressed uniquely as $r+sx$.
\item $R[x]$ is not finitely generated as an $R$-module. For if there is a finite set $Y$ \PMlinkescapetext{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 \PMlinkescapetext{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.)
\end{enumerate}