For commutative rings $A_1,...,A_n$ we have that $$ \operatorname{Spec} (A_1 \times \cdots \times A_n) \simeq \operatorname{Spec} A_1 \sqcup \cdots \sqcup \operatorname{Spec} A_n .$$ My question is, whether we have a similar identification for an infinite product of rings.
My guess is that the answer is no. I would argue like this:
For $n \in \mathbb N$, let $k_n$ be a field and consider the scheme $X = \bigsqcup_{n \in \mathbb N} \operatorname{Spec}k_n$. Topologically, this is a disjoint union of points, and it should hence carry the discrete topology. But then, $X$ cannot be affine, as affine schemes are quasi-compact, but the open covering $X = \bigcup_{n \in \mathbb N} \operatorname{Spec} k_n$ does not admit a finite subcover.
Is this ok? If so, can one generalize the statement for finite products to something similar for infinite products?
