TITLE: Spectrum of $\mathbb{Z}^\mathbb{N}$
QUESTION [16 upvotes]: Is anything known about the spectrum of $\mathbb{Z}^{\mathbb{N}}$? Notice that the fiber of $\mathrm{Spec}(\mathbb{Z}^{\mathbb{N}}) \to \mathrm{Spec}(\mathbb{Z})$ at a non-zero prime ideal $(p)$ is the spectrum of $\mathbb{Z}^{\mathbb{N}}/(p) \cong \mathbb{F}_p^{\mathbb{N}}$, which corresponds$^1$ to the set of ultrafilters on $\mathbb{N}$. Thus we only have to look at the generic fiber, which is the spectrum of the $\mathbb{Q}$-algebra $\mathbb{Z}^{\mathbb{N}} \otimes_{\mathbb{Z}} \mathbb{Q}$. This is isomorphic to a subalgebra of $\mathbb{Q}^{\mathbb{N}}$ consisting of those sequences of rational numbers whose denominators are bounded (with respect to suitable (not any) representations as fractions). In particular, every ultrafilter on $\mathbb{N}$ induces a prime ideal of $\mathbb{Z}^{\mathbb{N}} \otimes_{\mathbb{Z}} \mathbb{Q}$, but not everyone arises like this.
$^1$If $(F_i)_{i \in I}$ is family of fields, then there is a bijection between the (ultra)filters on $I$ and the (prime) ideals of $\prod_{i \in I} F_i$. It maps a filter $\mathcal{U}$  to the ideal $\{x \in \prod_{i \in I} F_i : \{i \in I : x_i = 0\} \in \mathcal{U}\}$.

REPLY [2 votes]: Here is a comment about the first bit of Eric Wofsey's answer, namely the natural map $X \to \beta \mathbb{N}$. $\beta \mathbb{N}$ is, in a precise sense, the "space of connected components" of $X$, as follows. There is a functor $\text{CRing} \to \text{Set}$ sending a commutative ring $R$ to the set of idempotents in $R$. This functor naturally lifts to Boolean rings: the set of idempotents acquires a Boolean ring structure via the usual multiplication with modified addition
$$a +' b = a + b - 2ab.$$
(This reduces to the usual addition if $R$ has characteristic $2$ but not in general.) I will call this ring $B(R)$. The spectrum of $B(R)$ in the sense of Stone duality is a Stone space called the Pierce spectrum of $R$, and so taking Pierce spectra gives a functor from affine schemes to Stone spaces.
Given any prime ideal $P$ of $R$, the homomorphism $R \to R/P$ induces a homomorphism $B(R) \to B(R/P)$, but since $R/P$ is an integral domain, $B(R/P) \cong \mathbb{F}_2$, and hence we get a homomorphism $B(R) \to \mathbb{F}_2$, or equivalently a point in the Pierce spectrum. This gives a natural continuous map
$$\text{Spec } R \to \text{Spec } B(R).$$
Now, why is it natural to think of $\text{Spec } B(R)$ as the space of connected components? The functor $B(R)$ is representable by the free commutative ring on an idempotent, namely
$$\mathbb{Z}[x]/(x^2 - x) \cong \mathbb{Z} \times \mathbb{Z}.$$
The corresponding affine scheme is "two points" $\text{Spec } \mathbb{Z} \coprod \text{Spec } \mathbb{Z}$. This is naturally a Boolean ring object in the category of affine schemes (true more generally for the coproduct $1 \coprod 1$ of two copies of the terminal object in any distributive category), and so maps into it naturally form a Boolean ring, which is just $B(R)$. The analogous construction for topological spaces produces for each space $X$ a natural Stone space $S(X)$ and a natural continuous map $X \to S(X)$ whose fibers (I think) are connected. Maybe it is the left adjoint to the inclusion of Stone spaces into spaces? 
From here it's not hard to see that the Pierce spectrum of any infinite product $\prod_{i \in \mathbb{N}} D_i$ of connected rings is $\beta \mathbb{N}$.