# ideals contained in a union of radical ideals

Let $R$ be a commutative ring and $I\subseteq R$ an ideal. Recall that the radical^{} of $I$ is defined as

$$r(I)=\{x\in R|{\exists}_{n\in \mathbb{N}}{x}^{n}\in I\}.$$ |

It can be shown, that $r(I)$ is again an ideal and $I\subseteq r(I)$. Let

$$V(I)=\{P\subseteq R|P\text{is a prime ideal and}I\subseteq P\}.$$ |

Of course $V(I)\ne \mathrm{\varnothing}$ (because $I$ is contained in at least one maximal ideal^{}) and it can be shown, that

$$r(I)=\bigcap _{P\in V(I)}P.$$ |

Finaly, recall that an ideal $I$ is called radical, if $I=r(I)$.

Proposition^{}. Let $I,{R}_{1},\mathrm{\dots},{R}_{n}$ be ideals in $R$, such that each ${R}_{i}$ is radical. If

$$I\subseteq {R}_{1}\cup \mathrm{\cdots}\cup {R}_{n},$$ |

then there exists $i\in \{1,\mathrm{\dots},n\}$ such that $I\subseteq {R}_{i}$.

Proof. Assume that this not true, i.e. for every $i$ we have $I\u2288{R}_{i}$. Then for any $i\in \{1,\mathrm{\dots},n\}$ there exists ${P}_{i}\in V({R}_{i})$ such that $I\u2288{P}_{i}$ (this follows from the fact, that ${R}_{i}=r({R}_{i})$ and the characterization^{} of radicals via prime ideals^{}). But for any $i$ we have ${R}_{i}\subseteq {P}_{i}$ and thus

$$I\subseteq {P}_{1}\cup \mathrm{\cdots}\cup {P}_{n}.$$ |

Contradiction^{}, since each ${P}_{i}$ is prime (see the parent object for details). $\mathrm{\square}$

Title | ideals contained in a union of radical ideals |
---|---|

Canonical name | IdealsContainedInAUnionOfRadicalIdeals |

Date of creation | 2013-03-22 19:04:23 |

Last modified on | 2013-03-22 19:04:23 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Corollary |

Classification | msc 13A15 |