free algebraFreeAlgebra2007-03-21 13:41:482007-10-16 23:58:56Definitionfree generating set\usepackage{amssymb,amscd}
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\newcommand{\real}{\mathbb{R}}Let $\mathcal{K}$ be a class of algebraic systems (of the same type $\tau$). Consider an algebra $A\in \mathcal{K}$ \PMlinkname{generated by}{SubalgebraOfAnAlgebraicSystem} a set $X=\lbrace x_i\rbrace$ indexed by $i\in I$. $A$ is said to be a \emph{free algebra} over $\mathcal{K}$, with \emph{free generating set} $X$, if for any algebra $B\in \mathcal{K}$ with any subset $\lbrace y_i\mid i\in I\rbrace \subseteq B$, there is a homomorphism $\phi:A\to B$ such that $\phi(x_i)=y_i$.
If we define $f:I\to A$ to be $f(i)=x_i$ and $g:I\to B$ to be $g(i)=y_i$, then freeness of $A$ means the existence of $\phi:A\to B$ such that $\phi\circ f=g$.
Note that $\phi$ above is necessarily unique, since $\lbrace x_i\rbrace$ generates $A$. For any $n$-ary polynomial $p$ over $A$, any $z_1,\ldots,z_n \in \lbrace x_i\mid i\in I\rbrace$, $\phi(p(z_1,\ldots,z_n))=p(\phi(z_1),\ldots,\phi(z_n))$.
For example, any free group is a free algebra in the class of groups. In general, however, free algebras do not always exist in an arbitrary class of algebras.
\textbf{Remarks}.
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\item
$A$ is free over itself (meaning $\mathcal{K}$ consists of $A$ only) iff $A$ is free over some equational class.
\item
If $\mathcal{K}$ is an equational class, then free algebras exist in $\mathcal{K}$.
\item
Any term algebra of a given structure $\tau$ over some set $X$ of variables is a free algebra with free generating set $X$.
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