Spectrum of infinite product of rings

$\def\Z{{\mathbb{Z}}\,} \def\Spec{{\rm Spec}\,}$ Suppose $R$ a ring and consider $\Spec(\prod_{i \in \mathbb{Z}} R)$. Now for the finite case, I know that holds $\Spec(R \times R) = \Spec(R) \coprod \Spec(R)$.

My intutition says that this does not extend to the infinite case. Maybe $\Spec(\oplus_{i \in \Z} R) = \coprod_{i \in \Z}\Spec(R)$ holds, but I am not sure. Can anybody give a proof or counter example for both the infinite direct sum and direct product?

Solution 1:

The spectrum of an infinite direct product is complicated. For example, the spectrum of an infinite direct product $\prod_{i \in I} F_i$ of fields can canonically be identified with the space $\beta I$ of ultrafilters on $I$, also known as the Stone-Čech compactification of $I$. For more on this in the special case that each $F_i$ is $\mathbb{F}_2$ see this blog post.

In general the spectrum of an arbitrary infinite direct product $\prod_{i \in I} R_i$ fibers over $\beta I$, where the fiber over an ultrafilter $U \in \beta I$ is the spectrum of the ultraproduct $\prod_{i \in I} R_i / U$. For more on this see Eric Wofsey's excellent answer here and this blog post.