In Lurie's "A Survey of Elliptic Cohomology", he writes on page 14 that if $A$ is an ordinary commutative ring considered as an $E_\infty$-ring, then $A$-module spectra are the same thing as objects of the derived category of $A$-modules. This is mysterious to me. On the one hand, to an $A$-module spectrum $M$ we might associate the $A$-modules $\pi_n(M)$, but I don't know of any interesting maps between these; perhaps this will just end up being the homology of any representative chain complex of $A$-modules. But then, I certainly don't see a natural way of getting from an object of $\mathcal{D}(\mbox{Mod}_A)$ to an $A$-module spectrum.

Incidentally, what does this induce on the level of categories? The obvious first guess is that $A$-module spectra actually form a topological category and that passing to $\mathcal{D}(\mbox{Mod}_A)$ applies $\pi_0$.

  • 1
    $\begingroup$ Alternatively you can use the universal property of the derived category (for $\infty$-categories). Also you can look at the proof of this claim in Lurie's Higher Algebra p.681 (Prop. $\endgroup$ – Dylan Wilson Jan 29 '12 at 20:47
  • 7
    $\begingroup$ The answer to your question is rather technical, but you can find all details in the following nice paper by Shipley: math.uic.edu/~bshipley/zdga17.pdf $\endgroup$ – Fernando Muro Jan 29 '12 at 20:48
  • 1
    $\begingroup$ @Dylan: My previous comment was in response to your first. In response to your second comment: titcr (urbandictionary.com/define.php?term=titcr) $\endgroup$ – Aaron Mazel-Gee Jan 29 '12 at 20:49
  • 2
    $\begingroup$ So what he means by module spectra over a discrete ring $A$ are module spectra over $HA$. There are interesting maps between these as spectra, but not as $HA$ modules. $\endgroup$ – Sean Tilson Jan 29 '12 at 23:00
  • 6
    $\begingroup$ I dare say that the only way of regarding a ring as a ring spectrum is via its Eilenberg-MacLane ring spectrum $\endgroup$ – Fernando Muro Jan 30 '12 at 0:23

This question already has been answered in the comments.

(Tilson) We regard a commutative ring as an $E_\infty$ spectrum via the EM functor $H$. This is definitely what Jacob is doing. One could also use associative rings and $A_\infty$ spectra for what follows.

(Wilson) Many of the correspondences between algebra and stable homotopy theory are described in Chapter 7 in Lurie's higher algebra book.

(Muro) The correspondence between algebras/modules and the associated EM-spectra is laid out in math.uic.edu/~bshipley/zdga17.pdf (Cor 2.15) which depends on her paper with Stefan Schwede "Equivalences of monoidal model categories."

It is a bit technical, but is easier to work out the correspondence if you restrict to non-negatively graded $\mathbb{Z}$-chain complexes and connective $H\mathbb{Z}$-modules. The correspondence can be spread into two stages: 1) Use the Dold-Kan correspondence to move between chain complexes and simplicial abelian groups. 2) Take the geometric realization of your simplicial abelian group which is a topological abelian group and hence an infinite loop space, so we can take its associated connective spectrum (by repeatedly applying the bar construction). The fact that geometric realization preserves products can then be used to see that this spectrum is an $H\mathbb{Z}$-module.

Now given an $H\mathbb{Z}$-module $M$ we can form the associated simplicial abelian group $H\mathbb{Z}-mod(H\mathbb{Z}\wedge\Sigma^\infty_+ \Delta^i, M)$ to go back.

This equivalence induces an equivalence their associated stable infinity categories.

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks! This is very nice (and much more believable than my initial naive attempt). I should've guessed that Dold-Kan would be involved. $\endgroup$ – Aaron Mazel-Gee Jan 31 '12 at 2:52
  • $\begingroup$ Wow Justin! that is very clear. $\endgroup$ – Sean Tilson Feb 2 '12 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.