Timeline for Why did Ravenel define a ring spectrum to be flat if its smash-square splits into copies of itself?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 31, 2023 at 20:23 | comment | added | Drew Heard | Lazard's theorem for connective modules over connective ring spectra is equivalent to Definition (2) (Theorem 7.2.2.15 of Higher Algebra) | |
| Jan 27, 2023 at 19:08 | comment | added | Tim Campion | The statement about modules in algebra is known as Lazard's theorem (and I think this is considered a pretty standard fact?). Apparently it's a different Lazard than the Lazard formal group law. | |
| Jan 25, 2023 at 1:12 | comment | added | kiran | Is this equivalent to definition (3)? One direction seems to follow from pi_* commuting with filtered colims. For the other direction if D is the filtered diagram of free E_-modules whose colimit is E_*E, then E_*( )(x)_{E_}D is a diagram of homology theories which I think lifts to a diagram of spectra (the nodes are free E-modules) whose colimit has a map to E(x)E by universality, and by another pi_*-commuting with filtered colims that map is an equivalence I think. | |
| Jan 24, 2023 at 22:37 | comment | added | Tom Goodwillie | No, I am saying exactly the opposite. And I have also added to that paragraph, and corrected something higher up. | |
| Jan 24, 2023 at 22:32 | history | edited | Tom Goodwillie | CC BY-SA 4.0 |
added 344 characters in body
|
| Jan 24, 2023 at 22:22 | comment | added | Z. M | I am not following the last paragraph. You seem to say that, if $E\wedge E$ is a filtered colimit of shifts of free $E$-modules, then $E$ is a filtered colimit of shifts of free $\mathbb S$-modules, and you say that its converse is false. However, it seems to me that its converse holds (basically a base change along $\mathbb S\to E$), but I don't know how to see the statement itself. | |
| Jan 24, 2023 at 22:06 | history | answered | Tom Goodwillie | CC BY-SA 4.0 |