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I see two "obvious" classes of nonzero finite $E_\infty$ ring spectra $R$:

  • $R = \Sigma^\infty_+ (S^1)^{\times n}$

  • $R = D\Sigma^\infty_+ X$ ($X$ a finite space)

Questions:

  1. Are there any others?

  2. In all the above examples, the unit map $\mathbb S \to R$ splits off. Is this always the case?

  3. In the second class of examples, I believe all elements of $\pi_\ast R$ not in the image of the unit $\pi_\ast \mathbb S \to \pi_\ast R$ are nilpotent. How generally is this true? Is it true for all examples not in the first class of examples?

  4. How does the answer change if we localize at a prime, or perform some more drastic localization?

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    $\begingroup$ Well, as an obvious generalization of your first example there's $R=\mathbb{S}[G]$ for $G$ a compact Lie group (or more generally, a stably dualizable topological group). $\endgroup$
    – Denis Nardin
    Commented Jun 11, 2019 at 6:41
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    $\begingroup$ G has to be abelian to get an E_infty thing (as opposed to an E_1 thing). $\endgroup$
    – Dylan Wilson
    Commented Jun 11, 2019 at 10:20
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    $\begingroup$ Another class of examples are square zero extensions by finite complexes. $\endgroup$
    – Dylan Wilson
    Commented Jun 11, 2019 at 10:21
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    $\begingroup$ @DylanWilson Duh... sorry. But at least this includes also finite commutative groups :) $\endgroup$
    – Denis Nardin
    Commented Jun 11, 2019 at 11:31
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    $\begingroup$ Good point! Also, regarding nilpotence: by Mathew-Naumann-Noel, nilpotence is detected in integral homology; if R is finite then every element in nonzero degree is automatically nilpotent in homology so the same is true in homotopy. So the only thing that could go wrong is stuff in degree zero- and indeed that can happen. For example, take X to be disconnected and use a component to define an idempotent. (Note your claim in Question 3 is not quite right because of this example- the unit won’t hit this idempotent). $\endgroup$
    – Dylan Wilson
    Commented Jun 11, 2019 at 12:14

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