Lemma 10.8 (Chinese remainder theorem). Assuming that:

$R$ a ring
$I_{1}, \dots, I_{n} \subseteq R$ ideals
$I_{i} + I_{j} = R$ for all $i \neq j$

Then

(i) $⋂_{i = 1}^{n} = \prod_{i = 1}^{n} I_{i}$ ( $= I$ say).
(ii) $R ∕ I ≅ \prod_{I = 1}^{n} R ∕ I_{i}$ .