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Timeline for Dedekind spectra

Current License: CC BY-SA 4.0

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Apr 2, 2020 at 7:25 history edited YCor CC BY-SA 4.0
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Oct 2, 2013 at 23:34 comment added David White Hovey's paper mentioned in my earlier comment is now available online: mhovey.web.wesleyan.edu/smithideals.pdf
Dec 5, 2011 at 17:27 vote accept Jonathan Beardsley
Dec 4, 2011 at 15:21 history edited Tyler Lawson
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Dec 4, 2011 at 15:11 answer added Tyler Lawson timeline score: 17
Dec 4, 2011 at 3:56 comment added Sean Tilson Will: I think that is where the difficulty comes in, what a subsepctrum is does not work the way homotopy theorists might want to. That is, it is not really a homotopical notion. Also, I am not sure what you mean by manifolds either.
Dec 2, 2011 at 22:13 comment added David White Let me comment that I thought a bit about how to do things like principal ideals and even that was not at all obvious. Ideals act very nicely in some regards, but they are objects in a different category now and a lot of things become very hard to define or compute. Rings of integers and ideal class groups seem wildly out of reach right now.
Dec 2, 2011 at 22:11 comment added David White Actually, I think it's much more complicated than Will's comment suggests. Every map is homotopic to an inclusion, so "subobjects" are really not the way to go if you want to define ideals. Jeff Smith had a clever idea to do this by looking in the arrow category Arr$(C$) where objects are morphisms in $C$ and morphisms are commutative squares. My advisor, Mark Hovey, has written an extensive unpublished paper on this subject and has been giving talks on the subject for the past year. He found the requisite model structure and discussed a homotopy theory of ideals. I guess you could email him
Dec 2, 2011 at 20:51 comment added Jonathan Beardsley @Will I mean, okay yes there is the notion of modules over ring spectra in homotopy theory ($E$ is a module over $R$ if there is some map $E\wedge R\to E$ satisfying some diagrams, etc.). So for ideal I'm guessing you just use subspectra because we can always localize at that subspectrum? How is the definition of a principal ideal obvious?
Dec 2, 2011 at 20:48 comment added Jonathan Beardsley I'm not sure I follow your use of manifolds?
Dec 2, 2011 at 20:27 comment added Will Sawin They seem like you could just pretty much directly try to port the definitions. The definition of module seems clear. For an ideal you just need a concept of what counts as an inclusion, which might just be a set-theoretic inclusion (up to homotopy) but might be stranger. The definition of a principal ideal is obvious, and this gets you the Picard group. Prime ideal seems a bit subtle. I would say that, for any two submanifolds not in the ideal, there product is homotopic to a mnaifold not in the ideal. I doubt a ring of integers would work because defining the field of fractions seems hard.
Dec 2, 2011 at 16:51 history asked Jonathan Beardsley CC BY-SA 3.0