# ideals contained in a union of radical ideals

Let $R$ be a commutative ring and $I\subseteq R$ an ideal. Recall that $I$ 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

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

Of course $V(I)\neq\emptyset$ (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 $I=r(I)$.

Let $I,R_{1},\ldots,R_{n}$ be ideals in $R$, such that each $R_{i}$ is radical. If

 $I\subseteq R_{1}\cup\cdots\cup R_{n},$

then there exists $i\in\{1,\ldots,n\}$ such that $I\subseteq R_{i}$.

Proof. Assume that this not true, i.e. for every $i$ we have $I\not\subseteq R_{i}$. Then for any $i\in\{1,\ldots,n\}$ there exists $P_{i}\in V(R_{i})$ such that $I\not\subseteq P_{i}$ (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

 $I\subseteq P_{1}\cup\cdots\cup P_{n}.$

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

Title ideals contained in a union of radical ideals IdealsContainedInAUnionOfRadicalIdeals 2013-03-22 19:04:23 2013-03-22 19:04:23 joking (16130) joking (16130) 4 joking (16130) Corollary msc 13A15