= Solution
We prove <prime avoidance> by contradiction. Remove any <prime ideal> contained in another one on the list; this leaves the same union. If the <ideal> $I$ is contained in none of the remaining <prime ideals>, choose $a_i\in I\setminus P_i$. For every $j\ne i$, incomparability lets us choose $b_{ij}\in P_j\setminus P_i$. Set
$$
b_i=\prod_{j\ne i}b_{ij},\qquad c=\sum_i a_i b_i.
$$
For a one-element list, use the empty product $b_1=1$. By the defining property of a <prime ideal>, $b_i\notin P_i$, and also $a_i b_i\notin P_i$. But $b_j\in P_i$ for every $j\ne i$. Thus all summands except $a_i b_i$ lie in $P_i$, and consequently $c\notin P_i$ for every $i$. On the other hand, $c\in I$ because each $a_i\in I$. This contradicts the assumed containment in the union. Therefore
$$
\boxed{I\subseteq P_i\text{ for at least one }i.}
$$
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