# ideal included in union of prime ideals

In the following $R$ is a commutative ring with unity.

###### Proposition 1.

Let $I$ be an ideal of the ring $R$ and ${P}_{\mathrm{1}}\mathrm{,}{P}_{\mathrm{2}}\mathrm{,}\mathrm{\dots}\mathrm{,}{P}_{n}$ be prime ideals^{} of $R$. If $I\mathrm{\u2288}{P}_{i}$, for all $i$, then $I\mathrm{\u2288}\mathrm{\cup}{P}_{i}$.

###### Proof.

We will prove by induction^{} on $n$. For $n=1$ the proof is trivial. Assume now that the result is true for $n-1$. That implies the existence, for each $i$, of an element ${s}_{i}$ such that ${s}_{i}\in I$ and ${s}_{i}\notin {\bigcup}_{j\ne i}{P}_{j}$. If for some $i$, ${s}_{i}\notin {P}_{i}$ then we are done. Thus, we may consider only the case ${s}_{i}\in {P}_{i}$, for all $i$.

Let ${a}_{i}={r}_{1}\mathrm{\dots}{r}_{i-1}{r}_{i+1}\mathrm{\dots}{r}_{n}$. Since ${P}_{i}$ is prime then ${a}_{i}\notin {P}_{i}$, for all $i$. Moreover, for $j\ne i$, the element ${a}_{i}\in {P}_{j}$. Consider the element $a=\sum {a}_{j}\in I$. Since ${a}_{i}=a-{\sum}_{i\ne j}{a}_{j}$ and ${\sum}_{i\ne j}{a}_{j}\in {P}_{i}$, it follows that $a\notin {P}_{i}$, otherwise ${a}_{i}\in {P}_{i}$, contradiction^{}. The existence of the element $a$ proves the proposition^{}.∎

###### Corollary 1.

Let $I$ be an ideal of the ring $R$ and ${P}_{\mathrm{1}}\mathrm{,}{P}_{\mathrm{2}}\mathrm{,}\mathrm{\dots}\mathrm{,}{P}_{n}$ be prime ideals of $R$. If $I\mathrm{\subseteq}\mathrm{\cup}{P}_{i}$, then $I\mathrm{\subseteq}{P}_{i}$, for some $i$.

Title | ideal included in union of prime ideals |
---|---|

Canonical name | IdealIncludedInUnionOfPrimeIdeals |

Date of creation | 2013-03-22 16:53:14 |

Last modified on | 2013-03-22 16:53:14 |

Owner | polarbear (3475) |

Last modified by | polarbear (3475) |

Numerical id | 10 |

Author | polarbear (3475) |

Entry type | Result |

Classification | msc 16D99 |

Classification | msc 13C99 |

Synonym | prime avoidance lemma |

Related topic | IdealsContainedInAUnionOfIdeals |