Suppose $R$ is a commutative ring with identity, $P$ is a prime ideal of $R$. Can we prove that there exist an ideal $p\subset P$ such that $p$ is a minimal prime ideal? (I think it might be used Axiom of Choice?) If $R$ is Noetherian, can we avoid to use Axiom of Choice?
2026-03-30 12:02:10.1774872130
Minimal prime ideals
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An equivalent characterization of $P$ being a prime ideal is that its complement $R\setminus P$ is a multiplicative system, ie. is closed under multiplication and containing the $1$. Let $\mathcal{S}$ denote the set of multiplicative systems, which contain the complement of $P$ as subsystem. It is partially ordered by inclusion, obviously nonempty and the union of ascending chains of multiplicative systems yields a multiplicative system (that is to say: ascending chains have an upper bound). By Zorn there is a maximal element, which corresponds to a minimal prime ideal.
I am not sure, how noetherianess affects this argument. Maybe there is another argument, which can be simplified by it.