Let $S$ be a finite set and let the ring $R=\prod_S\mathbb{Z}$ and let $B$ be a ring. I know that $$ \operatorname{Spec}R=\coprod_S\operatorname{Spec} \mathbb{Z}$$ Let $S$ be a finite set. Why is $$ \operatorname{Hom}( \operatorname{Spec}B, \coprod_S \operatorname{Spec}\mathbb{Z})=S^{\pi_0( \operatorname{Spec}B)}$$ where $S^{\pi_0( \operatorname{Spec}B)}$ is the sets of maps (of sets) from $\pi_0(\operatorname{Spec}B) \to S$, and $\pi_0( \operatorname{Spec}B)$ is the set of connected components on $ \operatorname{Spec}B$. Equivalently why is $$ \operatorname{Hom}_\text{ Rings}(\prod_S\mathbb{Z},B) = \operatorname{Hom}_\text{ Sets}(\pi_0( \operatorname{Spec}B), S)$$ EDIT: I think I figured it out. Decompose $\operatorname{Spec} B = \coprod \operatorname{Spec} (B_i)$ into its connected components, then $B=\prod_{\pi_0(Spec B)} B_i$ and so we need to compute $$Hom(\prod_S \mathbb{Z}, \prod_{\pi_0(Spec B)} B_i) = \prod_{i \in \pi_0(Spec B)} Hom(\prod_S \mathbb{Z}, B_i)=\prod_{i \in \pi_0(Spec B)} \operatorname{Hom}(Spec B_i, \coprod_S Spec \mathbb{Z}) $$ Since $Spec B_i$ is connected, the image of any map $Spec B_i \to \coprod_S Spec \mathbb{Z}$ must land in one of the $|S|$ copies of $Spec \mathbb{Z}$, and there is only one map $Spec B_i \to \operatorname{Spec} \mathbb{Z}$. So $\operatorname{Hom}(Spec B_i, \coprod_S Spec \mathbb{Z}) \cong S$ and the previous displayed equation line continues as
$$= \prod_{i \in \pi_0(Spec B)}S = S^{\pi_0(Spec B)}$$
The motivation for my question is that it seems to be being claimed in the second highlighted inequality from Hida's book Geometry Modular Forms:
