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Apr 28, 2017 at 11:35 history edited AAK CC BY-SA 3.0
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Apr 28, 2017 at 11:25 comment added AAK Ah, good point. That simplifies the proof a bit as well, thanks!
Apr 28, 2017 at 5:01 comment added A Rock and a Hard Place I do think that should be the case: if $\pi_0R$ is a $\mathbb Q$-algebra, then all $\pi_nR$ are rational vector spaces, so $R$ is $\mathbb Q$-local. Since rationalization is a smashing equivalence and $L_\mathbb QS\simeq \mathbb Q,$ then $R\simeq L_\mathbb QR\simeq \mathbb Q\otimes R$ which is clearly an $\mathbb E_\infty$-$\mathbb Q$-algebra.
Apr 28, 2017 at 0:29 comment added A Rock and a Hard Place Wonderful, thank you very much! Does $\pi_0R$ being a $\mathbb Q$-algebra suffice for $R$ (maybe connective?) to be a $\mathbb Q$-algebra, as an $\mathbb E_\infty$-ring? Is that something that has to do with the motto: "the rational sphere is the rational Eilenberg-MacLane"?
Apr 28, 2017 at 0:11 vote accept A Rock and a Hard Place
Apr 27, 2017 at 23:13 history answered AAK CC BY-SA 3.0