Free algebra (Boolean algebra)

Question

Could someone give me a simple explanation of Free Algebra on $\kappa$. How to construct free($\omega$).

here is it says

http://en.wikipedia.org/wiki/Free_Boolean_algebra

free($\omega$) is equal to the collection of all clopen of cantor space. How ?

Answer 1

This is my understanding of free boolean algebras, or equivalently free boolean rings:

The free commutative ring on a set $S$ is the polynomial ring $\mathbb{Z}[\{X_i\}_{i \in S}]$. In order to make it boolean, we have to ensure $2=0$ and $X_i^2=X_i$. That is, we take $B(S): =\mathbb{F}_2[\{X_i\}_{i \in S}]/(X_i^2-X_i)_{i \in S}$. This is easily seen to be a boolean ring. In fact, it is the free boolean ring on $S$, i.e. $S \mapsto B(S)$ is left adjoint to the forgetful functor from boolean rings to sets.

Observe that $B(S) = \bigotimes_{i \in S} \mathbb{F}_2[X_i]/(X_i^2-X_i)$ (tensor product of $\mathbb{F}_2$-algebras), and the Chinese Remainder Theorem implies $\mathbb{F}_2[X_i]/(X_i^2-X_i) \cong \mathbb{F}_2 \times \mathbb{F}_2$. In particular, we have $\dim_{\mathbb{F}_2} B(S) = 2^{|S|}$ when $S$ is finite; more precisely we have $B(S) \cong \mathbb{F}_2^{2^{|S|}}$ (using that tensor products commute with finite products).

For general $S$, we have that $B(S)$ is the colimit of the $B(T)$, where $T$ runs through the finite subsets of $S$. It follows that $\mathrm{Spec} B(S)$ (the corresponding Stone space) is the limit of the discrete spaces $\mathrm{Spec} B(T) \cong \{0,1\}^T$, i.e. that $\mathrm{Spec} B(S) = \{0,1\}^S$. For countable infinite $S$ this is the Cantor space, and corresponds to the colimit of the boolean rings $\mathbb{F}_2^{2^n}$ with transition maps $A \mapsto (A,A)$, which may be denoted as $\mathbb{F}_2^{2^{\infty}}$. This is analoguous to the Bratteli diagram of the AF-algebra corresponding to the Cantor space, which is basically the same with $\mathbb{F}_2$ replaced by $\mathbb{C}$.