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I would like to prove the following variation of 1.11 in AM.

If $\mathfrak{p_1},\ldots,\mathfrak{p_n}$ are prime ideals and $\mathfrak{a}$ an ideal such that $\mathfrak{a}\neq\mathfrak{p_i}$ for each $i$. Then $\mathfrak{a}\neq\bigcup \mathfrak{p}_i$.

Any ideas how to go about this? Or possibly provide a counter example.

coolpenguin
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  • A union of ideals is very rarely an ideal. Including a union of prime ideals. – Arthur Oct 06 '21 at 14:33
  • This is a version of what is known as prime avoidance. It holds under more general conditions: if the ring contains an infinite field, or if at most two of the ideals are not prime. There is a proof on page 91 of Eisenbud, Commutative Algebra. – Chris Leary Oct 06 '21 at 14:42
  • That's the contrapositive of the prime avoidance lemma. – Bernard Oct 06 '21 at 15:36

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