# has a rank

Let $R$ be a Noetherian ring^{} with total quotient ring $\mathrm{Quot}(R)$, and $M$ a finitely generated^{} $R$-module. We say $M$ has a rank if $M{\otimes}_{R}\mathrm{Quot}(R)\cong \mathrm{Quot}{(R)}^{n}$ for some non-negative integer $n$. And in this situation, we say $M$ has rank $n$.

Title | has a rank |
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Canonical name | HasARank |

Date of creation | 2013-03-22 18:13:41 |

Last modified on | 2013-03-22 18:13:41 |

Owner | yshen (21076) |

Last modified by | yshen (21076) |

Numerical id | 6 |

Author | yshen (21076) |

Entry type | Definition |

Classification | msc 13C99 |