# a characterization of the radical of an ideal

###### Proposition 1.

Let $I$ be an ideal in a ring $R$, and $\sqrt{I}$ be its radical^{}. Then $\sqrt{I}$ is the intersection^{} of all prime ideals^{} containing $I$.

###### Proof.

Suppose $x\in \sqrt{I}$, and $P$ is a prime ideal containing $I$. Then $R-P$ is an $m$-system (http://planetmath.org/MSystem). If $x\in R-P$, then $(R-P)\cap I\ne \mathrm{\varnothing}$, contradicting the assumption^{} that $I\subseteq P$. Therefore $x\notin R-P$. In other words, $x\in P$, and we have one of the inclusions.

Conversely, suppose $x\notin \sqrt{I}$. Then there is an $m$-system $S$ containing $x$ such that $S\cap I=\mathrm{\varnothing}$. Enlarge $I$ to a prime ideal $P$ disjoint from $S$, so that $x\notin P$ (we can do this; for a proof, see the second remark in this entry (http://planetmath.org/MSystem)). By contrapositivity, we have the other inclusion. ∎

Remark. This shows that every prime ideal is a radical ideal: for $\sqrt{P}$ is the intersection of all prime ideals containing $P$, and if $P$ is itself prime, then $P=\sqrt{P}$.

Title | a characterization of the radical of an ideal |
---|---|

Canonical name | ACharacterizationOfTheRadicalOfAnIdeal |

Date of creation | 2013-03-22 18:04:50 |

Last modified on | 2013-03-22 18:04:50 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 7 |

Author | CWoo (3771) |

Entry type | Derivation |

Classification | msc 16N40 |

Classification | msc 13-00 |

Classification | msc 14A05 |