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In ZF without choice, can a commutative ring $R$ have a prime ideal $P$ that is maximal among prime ideals but not among proper ideals?

By replacing $R$ with the quotient ring $R/P$, we may assume WLOG that $P$ is the zero ideal, so that $R$ would then be an integral domain that is not a field but has the zero ideal as its only prime ideal.

Note that it is possible for an integral domain (in ZFC or otherwise) to have exactly one nonzero prime ideal (e.g. any discrete valuation ring).

Geoffrey Trang
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    Take a ring without prime ideals, and take some product with some other ring, so you get a prime ideal, but not a maximal ideal. – Asaf Karagila Aug 06 '19 at 14:58
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    @AsafKaragila I upvoted your comment, but now I realize that I don't understand it. The prime ideals of $R\times S$ are in bijection with the disjoint union of the prime ideas of $R$ and the prime ideals of $S$, so if $R$ has no prime ideals, every prime ideal of $R\times S$ will have the form $R\times \mathfrak{p}$, where $\mathfrak{p}$ is prime in $S$. Right? – Alex Kruckman Aug 06 '19 at 15:42
  • @Alex: That's the plan. – Asaf Karagila Aug 06 '19 at 15:45
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    @AsafKaragila: But $R\times\mathfrak{p}$ will not be maximal among prime ideals of $R\times S$ unless $\mathfrak{p}$ is maximal among prime ideals of $S$, and so you are just reduced back to the original question for $S$. – Eric Wofsey Aug 06 '19 at 16:17
  • @Eric: Yes, you're right. I was thinking "diagonally" and it made sense in my head. – Asaf Karagila Aug 06 '19 at 16:22

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