Suppose $xy\in J$ and $x\notin J$. Then the <colon ideal> $(J:x)$ properly contains $J$ because it contains $y$. If also $y\notin J$, then $J+(x)$ properly contains $J$. The maximality of $J$ in the family makes both larger ideals lie outside the family. Applying the stated closure condition with $I=J$ and $a=x$ would imply $J$ lies outside the family, a contradiction. Hence $x\in J$ or $y\in J$, and $J$ is a <prime ideal>. This argument is the <Prime ideal principle for an Oka family>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 4 / a / Solution

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Every ideal maximal outside an Oka family is prime. If $xy\in I$ while $x,y\notin I$, both $I+(x)$ and $(I:x)$ properly contain $I$, contradicting the Oka condition.


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 Incoming links: Prime ideal principle for an Oka family