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$\def\Z{{\mathbb{Z}}\,} \def\Spec{{\rm Spec}\,}$ 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?

Kenta S
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Rico
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    Note that the spectrum of any ring is quasi-compact. However, $\coprod_{i\in\mathbb{Z}} X$ is never quasi-compact if $X$ is not empty. So $\coprod_{i\in\mathbb{Z}} X$ is never the spectrum of a ring. – Roland Nov 17 '15 at 12:43
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    And $\bigoplus_{i \in I} R$ is not a ring if $I$ is infinite. In general, it's bad practice to write direct sum for rings (even if $I$ is finite), because it only satisfies the universal property for the direct product (and not that of the coproduct). – Remy Nov 18 '15 at 02:28
  • Ah yeah you are right, I forgot it had no 1! Thnx :) – Rico Nov 18 '15 at 18:02

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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.

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Qiaochu Yuan
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