# $V(I)=\mathrm{\varnothing}$ implies $I=R$

Note that most of the notation used here is defined in the entry prime spectrum.

###### Theorem.

If $R$ is a commutative ring with identity^{} and $I$ is an ideal of $R$ with $V\mathit{}\mathrm{(}I\mathrm{)}\mathrm{=}\mathrm{\varnothing}$, then $I\mathrm{=}R$.

###### Proof.

Let $R$ be a commutative ring with identity and $I$ be an ideal of $R$ with $I\ne R$. Then, by this theorem (http://planetmath.org/EveryRingHasAMaximalIdeal), there exists a maximal ideal^{} $M$ of $R$ containing $I$. Since $M$ is , then $M$ is a proper prime ideal^{} of $R$. Thus, $M\in V(I)$. The theorem follows.
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Title | $V(I)=\mathrm{\varnothing}$ implies $I=R$ |
---|---|

Canonical name | VIemptysetImpliesIR |

Date of creation | 2013-03-22 16:07:43 |

Last modified on | 2013-03-22 16:07:43 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 10 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 14A15 |

Related topic | ProofThatOperatornameSpecRIsQuasiCompact |