free algebra
Let $\mathcal{K}$ be a class of algebraic systems (of the same type $\tau $). Consider an algebra^{} $A\in \mathcal{K}$ generated by (http://planetmath.org/SubalgebraOfAnAlgebraicSystem) a set $X=\{{x}_{i}\}$ indexed by $i\in I$. $A$ is said to be a free algebra^{} over $\mathcal{K}$, with free generating set $X$, if for any algebra $B\in \mathcal{K}$ with any subset $\{{y}_{i}\mid i\in I\}\subseteq B$, there is a homomorphism^{} $\varphi :A\to B$ such that $\varphi ({x}_{i})={y}_{i}$.
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 $\varphi :A\to B$ such that $\varphi \circ f=g$.
Note that $\varphi $ above is necessarily unique, since $\{{x}_{i}\}$ generates $A$. For any $n$ary polynomial^{} $p$ over $A$, any ${z}_{1},\mathrm{\dots},{z}_{n}\in \{{x}_{i}\mid i\in I\}$, $\varphi (p({z}_{1},\mathrm{\dots},{z}_{n}))=p(\varphi ({z}_{1}),\mathrm{\dots},\varphi ({z}_{n}))$.
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.
Remarks.

•
$A$ is free over itself (meaning $\mathcal{K}$ consists of $A$ only) iff $A$ is free over some equational class.

•
If $\mathcal{K}$ is an equational class, then free algebras exist in $\mathcal{K}$.

•
Any term algebra of a given structure^{} $\tau $ over some set $X$ of variables is a free algebra with free generating set $X$.
Title  free algebra 

Canonical name  FreeAlgebra 
Date of creation  20130322 16:51:05 
Last modified on  20130322 16:51:05 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 08B20 
Synonym  free algebraic system 
Related topic  TermAlgebra 
Defines  free generating set 