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I keep reading the term "rational (ring-)spectrum" but can't find a definition.

My original motivation for researching this was to understand some basic examples of complex oriented cohomology theories. One of these reads:

Every rational ring spectrum $E$ is complex orientable since we have a morphism $S^0_\mathbb{Q} \simeq H\mathbb{Q} \to E$ (and we know $H\mathbb{Q}$ to be).

Unfortunately I wasn't very rigorous when taking these notes, so I can't reconstruct what $S^0_\mathbb{Q}$ is supposed to be. $H\mathbb{Q}$ is the Eilenberg-Mac Lane spectrum for the rationals.

So can anybody clarify this? What is a rational spectrum and where does the asserted map come from?

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I hope this rough answer helps. A rational spectrum is a spectrum whose homotopy groups are rational vector spaces and $S^0_\mathbb Q$ is the localization of the sphere spectrum at the rational homotopy equivalences. Both $E$ and $H\mathbb Q$ are rationally local, and rationalization is smashing. Hence you get unit maps $S^0_\mathbb Q\to E$ and $S^0_\mathbb Q\to H\mathbb Q$ and you can check that the latter is a homotopy equivalence.

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  • $\begingroup$ The answer definitely helps, thanks! But can you clarify what you mean by "Both $E$ and $H\mathbb{Q}$ are rationally local, and rationalization is smashing."? $\endgroup$
    – Julian Kniephoff
    Sep 9, 2013 at 14:05
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    $\begingroup$ Please take a look at Chapter VIII in the book math.uchicago.edu/~may/BOOKS/EKMM.pdf where these notions and much more are explained. Your case corresponds to $R=S^0$ and $E=H\mathbb Q$. (If you are interested in other models than $S$-modules, like orthogonal or symmetric spectra, you need to look at sources dicussing those, of course.) $\endgroup$
    – Rasmus
    Sep 9, 2013 at 19:02
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    $\begingroup$ I thought I'd add another source on localization of spectra (which includes rationalization): math.uwo.ca/~mfrankla/Bousfield_LocalnSpectraHomol.pdf. It's much older and less technical than EKMM, but it's pretty readable and most of the useful fundamental properties are there. Cheers! $\endgroup$
    – Cary
    Dec 29, 2013 at 23:58

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