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
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| Jan 24, 2023 at 23:26 | comment | added | Doron Grossman-Naples | Interesting. It was pointed out to me by Kiran Luecke that Ravenel's flatness condition is equivalent to $E*E$ being free (not flat) over $E_*$, so it makes sense that he might not care to distinguish them. | |
| Jan 24, 2023 at 23:09 | history | answered | Nicholas Kuhn | CC BY-SA 4.0 |