Chinese remainder theorem for rings

Statement

Let $R$ be a commutative ring and let $I_1, \dots, I_n$ be ideals of $R$, with $n \ge 1$. Suppose that $I_j$ and $I_k$ are coprime for all $j \neq k$. Then the map $$R/(I_1 \cap \dots \cap I_n) \to R/I_1 \times \dots \times R/I_n, \qquad a + (I_1 \cap \dots \cap I_n) \mapsto (a + I_1, \dots, a + I_n)$$ is a ring isomorphism.

Remarks

Under the same hypothesis, $I_1 \cap \dots \cap I_n = I_1 I_2 \cdots I_n$. For $R = \mathbb{Z}$ and pairwise coprime integers $m_1, \dots, m_n$, the theorem gives $\mathbb{Z}/(m_1 \cdots m_n)\mathbb{Z} \cong \mathbb{Z}/m_1\mathbb{Z} \times \dots \times \mathbb{Z}/m_n\mathbb{Z}$. For example, $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.

Explanations

L1 human

For integers, the theorem says that a number modulo $15$ is the same information as the number modulo $3$ and modulo $5$. For each pair of remainders, exactly one class modulo $15$ has them. The ring version says the same for any family of pairwise coprime ideals.