I am trying to prove Generalized Chinese Remainder Theorem based on the following hint.
The statement is: $\{Q_i\}$ is a set of coprime ideals of a commutative ring $R$ and $\phi : R \to \prod_{i} R/Q_i$ is a mapping obtained from projections. Prove that $\phi$ is surjective.
Hint: If $I$ is a maximal ideal than it can contain at most one of the $Q_i$.
I thought of trying to prove it by contradiction: like taking an element that has no preimage and then prove that preimage of the ideal generated by this element is maximal (and also contains the $ker(\phi) = \cap_{i} Q_i$). Something like this. But failed actually :)
If the set is finite, the hint tells you to localize at any maximal ideal. Then use that finite products are the same as finite direct sums and localization commutes with arbitrary direct sums.
After localizing at a maximal ideal, the surjecitivity is trivial, hence the result follows, since we can test surjectivity locally.
Just another comment: This is a proof for the fact that localization does not commute with arbitrary products. Because if it would, we could use the above proof to show surjecitivity of $\mathbb Z \to \prod_{p \in \mathbb P} \mathbb Z/p\mathbb Z$, which is false.