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Feb 2, 2019 at 19:40 comment added Sasha @EricPeterson: Thank you! I will have to think about it.
Feb 2, 2019 at 19:39 vote accept Sasha
Feb 2, 2019 at 13:26 comment added Eric Peterson Denis's answer is 100% the reason why the specific object S/p appears in the theory of localization. You could still ask whether you could use HZ/p, and here the answer is somewhat unintuitive: for connective spectra, localization at S/p and at HZ/p agree, and so you cannot tell the difference before passing to nonconnective spectra, which are somewhat exotic objects. The spectrum $KU := \Sigma^\infty_+ \mathbb CP^\infty[\beta^{-1}]$ is a concrete example of a spectrum with $H_*(KU; Z)$ rational, hence $H_*(KU; Z/p)$ and $L_{HZ/p} KU$ null, but $L_{S/p} KU$ nontrivial.
Feb 2, 2019 at 11:07 answer added Denis Nardin timeline score: 7
Feb 2, 2019 at 8:38 history asked Sasha CC BY-SA 4.0