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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?

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    $\begingroup$ For the general case maybe you want to read this. $\endgroup$
    – user26857
    Jun 1, 2015 at 12:14
  • $\begingroup$ 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. $\endgroup$
    – user 1
    Jun 1, 2015 at 13:12
  • $\begingroup$ one can just google "Prime avoidance theorem" + the comment says how one can reach the proof. $\endgroup$
    – user 1
    Jun 1, 2015 at 13:23
  • $\begingroup$ @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. $\endgroup$
    – user26857
    Jun 1, 2015 at 16:57
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    $\begingroup$ I am more than content with the answer. Thank you^^ $\endgroup$
    – Marc
    Jun 1, 2015 at 19:14

1 Answer 1

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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 $x_1x_2+x_2x_3+x_3x_1$.)

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