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I have an $E_\infty$ ring spectrum $R$. I suspect it is trivial, but I'm not sure. What I do know is that there is an $R$-linear equivalence $R \simeq \Sigma R$. Unless I am very confused, this already implies, for example, that $\pi_* R$ is a $\mathbb{F}_2$-algebra.

What else does a condition like $R \simeq \Sigma R$ imply? For example, does it imply statements about, say, the action of Steenrod operators?

I am also interested in statements that require extra niceness conditions on this equivalence, because in my case I understand the equivalence reasonably well and can try to check for specific niceness conditions.

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  • $\begingroup$ Sorry can you share the argument that it is linear over F2? I can only see that 2 can't be invertible $\endgroup$
    – Andy Jiang
    Commented Mar 10, 2022 at 14:03
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    $\begingroup$ MO is an E_{infinity}-algebra with homotopy groups a polynomial ring over F_2. If you invert a polynomial generator in odd degree, I think you get a non-trivial ring of the sort you are describing. What you can say if 2=0 in pi_0 of an E_{infinity} ring is that the ring receives an E_2-ring map from F_2, and in particular is an F_2 module spectrum. This means the cohomology with Steenrod action is determined from the homotopy groups, and vice versa. $\endgroup$
    – Jeremy Hahn
    Commented Mar 10, 2022 at 14:30
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    $\begingroup$ @davik : if $x$ is invertible in odd degree, then $x^2$ is invertible, and also $2$-torsion ($x^2 = (-1)^{|x|} x^2 = -x^2$), therefore $2 = 0$ $\endgroup$
    – Maxime Ramzi
    Commented Mar 10, 2022 at 18:00
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    $\begingroup$ @kiran : this seems to require that it receives a map from its $\pi_0$, and I'm not sure how that follows - am I missing some other argument ? $\endgroup$
    – Maxime Ramzi
    Commented Mar 11, 2022 at 11:03
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    $\begingroup$ @MaximeRamzi good point thanks, I guess my statement is just at the level of $H\mathbb{F}_2$-modules. $\endgroup$
    – kiran
    Commented Mar 11, 2022 at 21:29

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