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Is there a class of ring spectra that corresponds to and/or extends the class of Dedekind rings from traditional algebra? Is there a notion of "ring of integers" of a ring spectrum which is somehow "over" either $H\mathbb{Z}$ or $\mathbb{S}$? Additionally, is there a notion of an ideal class group of a Dedekind ring spectrum?

Thanks

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    I don't know about homotopy theory, but in algebraic geometry, the generalisation of Dedekind ring is (essentially) a non-singular algebraic curve, and the corresponding notion of ideal class group is the (Weil) divisor class group.2011-12-01
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    Thankyou. What you say may have some correlation to homotopy theory, at some level. I keep finding out that things do...2011-12-01
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    I guess you could try to look at $\Sigma^\infty X$ where $X$ is such a curve, and investigate the properties there.2011-12-01
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    I think that's probably not the right generalization -- algebraic varieties are by their very nature meant to be understood "smoothly" (i.e. I think that they're studied by working in local charts and then patching to recover global information -- not that I know anything about varieties!). At this point, people are still working hard to generalize commutative algebra to spectra. On the other hand, "derived algebraic geometry" is where one does algebraic geometry but for every affine scheme Spec(R) hangs on to a particular ring spectrum whose homotopy is R (or maybe just 0th homotopy?), ...2011-12-02
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    which sounds to me much more likely to be the 'right' way to homotopically generalize the notion of a non-singular algebraic curve. As for your question, I have no idea. I attended a talk a while ago that I understood almost nothing of where they discussed (among other things) a generalization of Azumaya algebras to ring spectra. I texed notes, and they're nowhere near fit to be made available online but I'd be happy to email them to you if you'd like to get a sense of how these things go.2011-12-02
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    A ring spectrum over HZ is, more or less, the same thing as DGA; and any spectrum is a spectrum over S.2011-12-02
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    @GrigoryM Thanks. I guess I knew that, hahah, or I should have (although I have heard that a ring spectrum is significantly more useful and flexible than a DGA, though I cannot support this statement other than saying I think Jack Morava said something along those lines...haha).2011-12-02
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    @AaronMazel-Gee that would be great. I'm always looking for more info on this stuff. I'm especially interested in this idea of generalizing number theory and various areas of algebra to homotopy theory. I believe my e-mail is in my profile.2011-12-02
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    @GrigoryM: Sorry if I'm being dense -- why is $\mathbb{S}$ terminal?2011-12-02
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    @AaronMazel-Gee It's not terminal, it's initial (terminology is somewhat misleading here -- like "an algebra over ring R" is, actually, an object under R, not over R)2011-12-02
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    (Reference for the first half of my comment, btw: http://arxiv.org/abs/math/0209215)2011-12-02
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    For more info see http://mathoverflow.net/questions/82485/dedekind-spectra2011-12-03
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    @GrigoryM: That's a pretty awful abuse. I just ran into the same issue when I was thinking that $f_*M$ should push a module $M$ forward along a ring homomorphism $f$, when really it's the other way around for algebro-geometric reasons. But anyways, isn't the 1-point spectrum $\Sigma^\infty \mbox{pt}$ the initial object?2011-12-05
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    In the symmetric monoidal structure on the category of spectra (i.e. the EKMM or Symmetric Spectra constructions), the sphere spectrum is the "unit" object meaning there is a map from it to everything (similarly to the way that the unique map from the trivial group chooses the identity in a group), and that $S\wedge E=E$, but the former is the sense in which it is initial, I believe.2011-12-05
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    Which actually is $\Sigma^\infty$ of two points (which sort of morally makes sense since things are based, I guess).2011-12-05
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    Erm, $S\wedge E\simeq E$.2011-12-05
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    @AaronMazel-Gee Algebraic geometers (working in the opposite category, of affine schemes) are to blame here ;-) And yes, S is initial in the category of *ring* spectra (it's analogue of $\mathbb Z$, not of a point).2011-12-05

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