 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 $b_{ij}\in P_j\setminus P_i$. SetFor 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