What is the Stone space of the free sigma-algebra on countably many generators The Stone space of the free Boolean algebra on countably many generators is the Cantor space $2^\omega$. What is the Stone space of the free (Boolean) $\sigma$-algebra on countably many generators?
 A: Correct me if I'm wrong on this one. . . .
The free $\sigma$-algebra on countably many generators is realised as the power set of $\mathbb N$. The resulting Stone space is by definition the Stone-Cech compactification $\beta \mathbb N$. Namely the set of ultrafilters on $\mathbb N$ with the Stone topology. This answer sounds circular and unsatisfying until you realise how well studied that space is $-$ it's usually just called the Stone-Cech compactification. 
It would be more interesting if there is some noncircular description of the free $\sigma$-algebra on $\kappa$ many generators (algebra of all $\kappa$ or co-$\kappa$ subsets?) or it's Stone space.
Edit: Let $B$ be the free Boolean algebra on countably many generators. Then $B$ is isomorphic to the algebra of clopen subsets of the Cantor set $2^\omega$ which is the set of all numbers in $[0,1]$ whose ternary expansion has only $0$s and $1$s. (You should check this but) I believe $B$ is generated by the sets $0(n) = \{x_1 x_2 \ldots \in 2^\omega : x_n = 0\}$ and $1(n) = \{x_1 x_2 \ldots \in 2^\omega: x_n = 1\}$. 
There is an obvious map $F: B \to \mathcal P (\mathbb N)$ that takes each $0(n)$ to $\varnothing$ and $1(n)$ to $\{n\}$ and exchanges the roles of meets and joins. In other words $F(a \wedge b) = a \vee b$ and $F(a \vee b) = a \wedge b$. composing with the dual on the RHS we get a homomorphism $B \to \mathcal P(\mathbb N)$.
Exercise: Describe the associated map between Stone spaces
Exercise: Check the above agrees with the categorical theory of maps between free boolean algebras and $\sigma$ algebras. You might want to look up adjunctions from the category of Boolean algebras to the category of sigma-algebras and how this effects the properties of free objects.
One should not expect $B$ to be isomorphic to $ \mathbb N / \sim$ since the later is the field of clopens of the Stone-Cech remainder also called $\mathbb N ^*$ which is known to be non-metrisable. Since one Stone space is metrisable and the other not, the two fields are different.
