# maximal ideal is prime

Theorem. In a commutative ring with non-zero unity, any maximal ideal^{} is a prime ideal^{}.

Proof.β Let $\mathrm{\pi \x9d\x94\u037a}$ be a maximal ideal of such a ring $R$ and let the ring product $r\beta \x81\u2019s$ belong to $\mathrm{\pi \x9d\x94\u037a}$ but e.g. β$r\beta \x88\x89\mathrm{\pi \x9d\x94\u037a}$. The maximality of $\mathrm{\pi \x9d\x94\u037a}$ implies thatβ $\mathrm{\pi \x9d\x94\u037a}+(r)=R=(1)$.β Thus there exists an element β$m\beta \x88\x88\mathrm{\pi \x9d\x94\u037a}$β and an elementβ $x\beta \x88\x88R$β such thatβ $m+x\beta \x81\u2019r=1$.β Now $m$ and $r\beta \x81\u2019s$ belong to $\mathrm{\pi \x9d\x94\u037a}$, whence

$$s=1\beta \x81\u2019s=(m+x\beta \x81\u2019r)\beta \x81\u2019s=s\beta \x81\u2019m+x\beta \x81\u2019(r\beta \x81\u2019s)\beta \x88\x88\mathrm{\pi \x9d\x94\u037a}.$$ |

So we can say that along with $r\beta \x81\u2019s$, at least one of its factors (http://planetmath.org/Product) belongs to $\mathrm{\pi \x9d\x94\u037a}$, and therefore $\mathrm{\pi \x9d\x94\u037a}$ is a prime ideal of $R$.

Title | maximal ideal is prime |
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Canonical name | MaximalIdealIsPrime |

Date of creation | 2013-03-22 17:37:59 |

Last modified on | 2013-03-22 17:37:59 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 16D25 |

Classification | msc 13A15 |

Related topic | SumOfIdeals |

Related topic | MaximumIdealIsPrimeGeneralCase |

Related topic | CriterionForMaximalIdeal |