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Taken from Miles Reid "Undergraduate Commutative Algebra" p.35 ex. 1.12 b)

Let $I,J_1,J_2 \subset A$ be ideals of a commutative ring $A$. Let $P$ be a prime ideal, then if $I \subset J_1 \cup J_2 \cup P$ then $I \subset J_1$ or $J_2$ or $P$.

I've handled the case with two ideals, but I can't manage to generalize the reasoning. Moreover, I don't know how to use the hypothesis of P prime ideal. Can someone provide me some advices?

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  • $\begingroup$ What do you mean when you say "the case with two ideals"? You mean the same statement with $J_1$ and $J_2$ but no $P$, right? $\endgroup$
    – rschwieb
    Feb 19, 2014 at 13:32
  • $\begingroup$ Yes :) exactly that case $\endgroup$
    – Riccardo
    Feb 19, 2014 at 13:36

1 Answer 1

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This is a special case of the Prime Avoidance lemma, and proofs are abundant in books and on the net. For example, here is one.

I hope you still want to try out the problem before reading the proof, though. Here's a hint for that:

Without loss of generality, you can assume that $I$ isn't contained in $P\cup J_1$,$P\cup J_2$, or $J_1\cup J_2$, for in that case you'd fall back on your two-ideal version.

So, we can choose $a\in I\setminus (J_1\cup J_2)$,$b\in I\setminus (P\cup J_1)$, and $c\in I\setminus (P\cup J_2)$.

Hints:

  • Can you say anything about what sets $a,b,c$ do belong to, other than $I$?
  • Look at $a+bc$.
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