# Chinese remainder theorem

Let $R$ be a commutative ring with identity^{}. If ${I}_{1},\mathrm{\dots},{I}_{n}$ are ideals of $R$ such that ${I}_{i}+{I}_{j}=R$ whenever $i\ne j$, then let

$$I={\cap}_{i=1}^{n}{I}_{i}=\prod _{i=1}^{n}{I}_{i}.$$ |

The sum of quotient maps $R/I\to R/{I}_{i}$ gives an isomorphism^{}

$$R/I\cong \prod _{i=1}^{n}R/{I}_{i}.$$ |

This has the slightly weaker consequence that given a system of congruences^{} $x\cong {a}_{i}\phantom{\rule{veryverythickmathspace}{0ex}}(mod{I}_{i})$, there is a solution in $R$ which is unique mod $I$, as the theorem is usually stated for the integers.

Title | Chinese remainder theorem^{} |
---|---|

Canonical name | ChineseRemainderTheorem1 |

Date of creation | 2013-03-22 12:16:43 |

Last modified on | 2013-03-22 12:16:43 |

Owner | bwebste (988) |

Last modified by | bwebste (988) |

Numerical id | 7 |

Author | bwebste (988) |

Entry type | Theorem |

Classification | msc 11N99 |

Classification | msc 11A05 |

Classification | msc 13A15 |

Related topic | ChineseRemainderTheoremInTermsOfDivisorTheory |