# exact sequences for modules with finite projective dimension

Proposition^{}. Let $R$ be a ring and $M$ be a (left) $R$-module, such that $$. If

$$0\to K\to {P}_{n}\to \mathrm{\cdots}\to {P}_{0}\to M\to 0$$ |

is an exact sequence^{} of $R$-modules, such that each ${P}_{i}$ is projective, then $K$ is projective.

Proof. Since $$, then there exists exact sequence of $R$-modules

$$0\to {P}_{n}^{\prime}\to \mathrm{\cdots}\to {P}_{0}^{\prime}\to M\to 0,$$ |

Note that sequences^{}

$${P}_{n}\to \mathrm{\cdots}\to {P}_{0}\to M\to 0;$$ |

$${P}_{n}^{\prime}\to \mathrm{\cdots}\to {P}_{0}\to M\to 0,$$ |

are projective resolutions of $M$. Let $\delta :{P}_{n}\to {P}_{n-1}$ and $\beta :{P}_{n}^{\prime}\to {P}_{n-1}^{\prime}$ be maps take from these resolutions. Then generalized Schanuel’s lemma implies that $\mathrm{ker}\delta $ and $\mathrm{ker}\beta $ are projectively equivalent. But $\mathrm{ker}\delta \simeq K$ and $\mathrm{ker}\beta =0$. This means, that there are projective modules^{} $P,Q$ such that

$$K\oplus P\simeq Q.$$ |

Therefore $K$ is a direct summand of a free module^{} (since $Q$ is), which completes^{} the proof. $\mathrm{\square}$

Title | exact sequences for modules with finite projective dimension |
---|---|

Canonical name | ExactSequencesForModulesWithFiniteProjectiveDimension |

Date of creation | 2013-03-22 19:04:55 |

Last modified on | 2013-03-22 19:04:55 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Corollary |

Classification | msc 16E10 |

Classification | msc 18G20 |

Classification | msc 18G10 |