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Aug 9, 2022 at 19:54 comment added Onur Oktay Perhaps paraphrasing didn't lower the level of difficulty, but I believe an expert's eyes can catch what amateurs like myself may miss. I value your opinion and input as before.
Aug 9, 2022 at 19:54 comment added Onur Oktay After pondering a while, I'm convinced that the problem is difficult in this generality. In a naive attempt to find the subsets of $\mathcal{S}$ that are closed under addition, let $E_m = E\cap maxspec(R)$ for $E\subseteq spec(R)$ and let $\mathcal{S}^* = \{E\in spec(R): k(E) = k(E_m)\}$ where $k(E)$ is the intersection of the prime ideals in $E$. $\mathcal{S}^*$ is closed under unions, so a slightly weaker version of Q1 above is: classify the maximal subsets of $\mathcal{S}^*$ that are closed under intersection.
Aug 9, 2022 at 16:11 comment added Keith Kearnes @OnurOktay: About the questions in the body (For which classes of rings $\ldots$), I think that these questions are too hard if you want to consider arbitrary classes. If $\mathcal C$ is a class of commutative unital rings that (i) is closed under the formation of products and subrings and (ii) contains a field that has a valuation subring that is not a subfield, then I think you can mimic the above construction within $\mathcal C$. This rules out a lot of classes.
Aug 9, 2022 at 16:06 vote accept Onur Oktay
Aug 9, 2022 at 16:06 history edited Keith Kearnes CC BY-SA 4.0
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Aug 9, 2022 at 15:53 history edited Keith Kearnes CC BY-SA 4.0
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Aug 9, 2022 at 15:39 history edited Keith Kearnes CC BY-SA 4.0
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Aug 9, 2022 at 15:29 comment added Onur Oktay I'd like to note for the readers like myself that the construction has immediate (made obvious by the stages) generalization if one starts with an integral domain and a prime ideal different from $D=\mathbb{Z}$ and $P=2\mathbb{Z}$ respectively. $L$ is the localization $D_P$. $S=c_{00}(\mathbb{N}\to D_{{0}})\oplus\mathbf{1}L$, where $\mathbf{1}=(1,1,\dots)$. $c_{00}$ could be substituted with another ring of sequences without $\mathbf{1}$ in it, e.g., $\ell^p(\mathbb{N}\to\mathbb{Q})$ for $D=\mathbb{Z}$.
Aug 9, 2022 at 15:29 comment added Onur Oktay Professor @KeithKearnes, thank you for your helpful approach and extensive reply. Your construction is work of art in my humble opinion. Thanks also for all the details and explanation - words are like salad, help to digest the main course, clearly the more the merrier. $R$ is indeed a counterexample to the question in the title. May I ask what's your insight for the two questions in the body? Perhaps I should post these two in a separate question.
Aug 9, 2022 at 7:46 history edited Keith Kearnes CC BY-SA 4.0
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Aug 9, 2022 at 7:40 history edited Keith Kearnes CC BY-SA 4.0
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Aug 9, 2022 at 7:14 history edited Keith Kearnes CC BY-SA 4.0
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Aug 9, 2022 at 5:16 history edited Keith Kearnes CC BY-SA 4.0
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Aug 9, 2022 at 4:52 history answered Keith Kearnes CC BY-SA 4.0