# invertible ideals are projective

If $R$ is a ring and $f:M\to N$ is a homomorphism^{} of $R$-modules, then a right inverse^{} of $f$ is a homomorphism $g:N\to M$ such that $f\circ g$ is the identity map on $N$. For a right inverse to exist, it is clear that $f$ must be an epimorphism^{}. If a right inverse exists for every such epimorphism and all modules $M$, then $N$ is said to be a projective module^{}.

For fractional ideals^{} over an integral domain^{} $R$, the property of being projective as an $R$-module is equivalent^{} to being an invertible ideal.

###### Theorem.

Let $R$ be an integral domain. Then a fractional ideal over $R$ is invertible if and only if it is projective as an $R$-module.

In particular, every fractional ideal over a Dedekind domain^{} is invertible, and is therefore projective.

Title | invertible ideals are projective |
---|---|

Canonical name | InvertibleIdealsAreProjective |

Date of creation | 2013-03-22 18:35:47 |

Last modified on | 2013-03-22 18:35:47 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 5 |

Author | gel (22282) |

Entry type | Theorem |

Classification | msc 16D40 |

Classification | msc 13A15 |

Related topic | ProjectiveModule |

Related topic | FractionalIdeal |