Proposition 10.28.8. If $\mathcal{F}$ is an Oka family of ideals, then any maximal element of the complement of $\mathcal{F}$ is prime.

Proof. Suppose $I \not\in \mathcal{F}$ is maximal with respect to not being in $\mathcal{F}$ but $I$ is not prime. Note that $I \not= R$ because $R \in \mathcal{F}$. Since $I$ is not prime we can find $a, b \in R - I$ with $ab \in I$. It follows that $(I, a) \neq I$ and $(I : a)$ contains $b \not\in I$ so also $(I : a) \neq I$. Thus $(I : a), (I, a)$ both strictly contain $I$, so they must belong to $\mathcal{F}$. By the Oka condition, we have $I \in \mathcal{F}$, a contradiction. $\square$

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