|
Let $\mathcal{K}$ be a class of algebraic systems (of the same type $\tau$ ). Consider an algebra $A\in \mathcal{K}$ generated by a set $X=\lbrace x_i\rbrace$ 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 $\lbrace y_i\mid i\in I\rbrace \subseteq B$ , 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.
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 $X$ .
|