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Question: What are some ring spectra $E$ satisfying the following conditions?

  1. The coefficients $E_\ast = \pi_\ast(E)$ are graded-commutative;

  2. There is a Kunneth spectral sequence $E_\ast(X) \otimes_{E_\ast} E_\ast(Y) \Rightarrow E_\ast(X \wedge Y)$ for finite spectra $X,Y$;

  3. The coefficients $E_\ast$ form a graded-artinian ring (i.e. satisfies the descending chain condition for homogeneous ideals).

Clearly there is a canonical sequence of examples $E = H\mathbb Q, K(1), K(2), \dots, H \mathbb F_p$. More generally, if $E_\ast$ is a graded field then $E$ is an example -- but this essentially reduces to the examples listed.

I'm particularly curious if the telescope $T(n)$ is (can be chosen to) satisfy (1,2,3).

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    $\begingroup$ I'm not sure of what you mean with "Künneth SS", but there's a Künneth SS of the form $\mathrm{Tor}_{E_\ast}(E_\ast X,E_\ast Y)\Rightarrow E_\ast(X\wedge Y)$ for all spectra $X,Y$ and commutative ring spectrum $E$ (type of convergence may vary). $\endgroup$ Jun 14, 2020 at 16:31
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    $\begingroup$ @DenisNardin I want to be a bit loose with what "commutative ring spectrum" means -- in order to have $Tor_{E_\ast}(E_\ast X, E_\ast Y) \Rightarrow E_\ast(X \wedge Y)$, does it suffice for $E$ to be a spectrum with maps $E \wedge E \to E \leftarrow \mathbb S$ which endow $E_\ast$ with the structure of a graded commutative ring? Does one need to assume the maps are $E_1$? $E_\infty$? $\endgroup$
    – Tim Campion
    Jun 14, 2020 at 16:37
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    $\begingroup$ For the version I know you need $E_\infty$. There's another variant due to Adams with different restrictions (you need $E$ to be written as a colimit of finite spectra with special properties) which might be more suited to your goals. Regardless, I don't think you can discard the higher Tor in the signature of the spectral sequence in general. $\endgroup$ Jun 14, 2020 at 16:40
  • $\begingroup$ @DenisNardin Of course you're right. That kind of sinks my original motivation for this question, but maybe it's still of some interest with an amended "Kunneth" requirement... $\endgroup$
    – Tim Campion
    Jun 14, 2020 at 19:18
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    $\begingroup$ Perhaps an observation that might be useful is that the Künneth formula holds "on the nose" (i.e. $E_\ast X\otimes_{E_\ast} E_\ast Y\cong E_\ast(X\wedge Y)$) whenever $E_\ast X$ is a flat $E_\ast$-module (this doesn't even require the Künneth spectral sequence to exist!). $\endgroup$ Jun 14, 2020 at 20:22

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