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

Proposition. 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). $\square$