Proposition 1   Let $I$ be an ideal of the ring $R$ and $P_1, P_2, ... ,P_n$ be prime ideals of $R$ . If $I\not\subseteq P_i$ , for all $i$ , then $I \not\subseteq \cup P_i$ .

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 $s_i \not\in \bigcup_{j\ne i}P_j$ . If for some $i$ , $s_i\not\in P_i$ then we are done. Thus, we may consider only the case $s_i \in P_i$ , for all $i$ .
Let $a_i = r_1...r_{i-1}r_{i+1}...r_n$ . Since $P_i$ is prime then $a_i\not\in P_i$ , 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 $a \not\in P_i$ , 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 $P_1,P_2,...,P_n$ be prime ideals of $R$ . If $I \subseteq \cup P_i$ , then $I\subseteq P_i$ , for some $i$ .