---
tag: 0W97J8QED7
kind: statement
role: theorem
name: Chinese remainder theorem for rings
msc: [13A15]
uses: [0MDJP82TEH, 0VFW4Y69BT, 0MNZCN4BQK, 0B6QC0C0G1]
sources:
  - url: https://en.wikipedia.org/wiki/Chinese_remainder_theorem
    locator: Generalization to arbitrary rings
---

## Statement

Let $R$ be a [[commutative ring|0GYGHKX834]] and let $I_1, \dots, I_n$ be ideals of $R$, with $n \ge 1$.
Suppose that $I_j$ and $I_k$ are [[coprime|0MDJP82TEH]] 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|0B6QC0C0G1]].

## 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.