# maximal ideal is prime (general case)

Theorem. In a ring (not necessarily commutative^{}) with unity, any maximal ideal^{} is a prime ideal^{}.

Proof. Let $\U0001d52a$ be a maximal ideal of such a ring $R$ and suppose $R$ has ideals $\U0001d51e$ and $\U0001d51f$ with $\U0001d51e\U0001d51f\subseteq \U0001d52a$, but $\U0001d51e\u2288\U0001d52a$. Since $\U0001d52a$ is maximal, we must have $\U0001d51e+\U0001d52a=R$. Then,

$$\U0001d51f=R\U0001d51f=(\U0001d51e+\U0001d52a)\U0001d51f=\U0001d51e\U0001d51f+\U0001d52a\U0001d51f\subseteq \U0001d52a+\U0001d52a=\U0001d52a.$$ |

Thus, either $\U0001d51e\subseteq \U0001d52a$ or $\U0001d51f\subseteq \U0001d52a$. This demonstrates that $\U0001d52a$ is prime.

Note that the condition that $R$ has an identity element^{} is essential. For otherwise, we may take $R$ to be a finite zero ring^{}. Such rings contain no proper prime ideals. As long as the number of elements of $R$ is not prime, $R$ will have a non-zero maximal ideal.

Title | maximal ideal is prime (general case) |
---|---|

Canonical name | MaximalIdealIsPrimegeneralCase |

Date of creation | 2013-03-22 17:38:02 |

Last modified on | 2013-03-22 17:38:02 |

Owner | mclase (549) |

Last modified by | mclase (549) |

Numerical id | 8 |

Author | mclase (549) |

Entry type | Theorem |

Classification | msc 16D25 |

Related topic | MaximalIdealIsPrime |