# Union of specific prime ideals is not an ideal

Let $R$ be a commutative ring with $1$ with three prime ideals $P_1,P_2,P_3$ such that $P_i\subseteq P_j$ if and only if $i=j$. I want to show that the union of these prime ideals, which I denote $J$, is not an ideal of $R$.

Somehow I need to find two elements $x,y\in J$ such that $x+y\notin J$. Can someone help me out?

• For the general case maybe you want to read this. Jun 1, 2015 at 12:14
• By Prime avoidance theorem, (3.61 of Sharp's book: Steps in Commutative Algebra), if $J$ is an ideal, then it will be one of $P_i$ s. and from its proof you can take the proof for your question. Jun 1, 2015 at 13:12
• one can just google "Prime avoidance theorem" + the comment says how one can reach the proof. Jun 1, 2015 at 13:23
• @user1 If I understand well the OP wants a direct proof (which is more or less similar to the proof of the prime avoidance lemma), not to use some other result for proving the property. Jun 1, 2015 at 16:57
• I am more than content with the answer. Thank you^^
– Marc
Jun 1, 2015 at 19:14

Let $x_i\in P_i-(P_j\cup P_k)$. (Why there is such an element?) Then $$x_1x_2+x_2x_3+x_3x_1\notin P_1\cup P_2\cup P_3.$$ (If you like can use $x_1x_2+x_3$ instead of $x_1x_2+x_2x_3+x_3x_1$.)