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 $I⊈R_i$. Then for any $i\in \{1,\mathrm{\dots },n\}$ there exists $P_i\in V(R_i)$ such that $I⊈P_i$ (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). $\mathrm{\square }$