Let $\{I_{\alpha}\}_{\alpha \in A}$ be an arbitrary family of ideals of a commutative ring $R$. Suppose that $I_{\alpha} + I_{\beta} = R$ for all $\alpha \neq \beta$.
Then is there a corresponding situation if we take the infinite product to be $\prod_{\alpha \in A} R/I_{\alpha}$ or can you only apply it to finitely many ideals?
Define a directed system to be for any $B \subset A$, $M_{B} = \prod_{b \in B} R/I_{b}$. Then for $B \subset C$, clearly we have an embedding $f_{BC}:B \hookrightarrow C$, and with all maps $(M_B, f_{BC})$ forms a directed system.
So can't we take the direct limit of $M_B$?
No, the canonical map $R\to \prod R/I_\alpha$ will typically not be surjective. For instance, let $R=\mathbb{Z}$ and let the $I_\alpha$ be the ideals $(p)$ for all primes $p$. Then the product $\prod_p\mathbb{Z}/(p)$ is uncountable, so there cannot be a surjection from $\mathbb{Z}$ to it.