Timeline for Homological smoothness implies projectivity?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2013 at 10:17 | comment | added | Pierre-Guy Plamondon | @Mariano Wouldn't the resolution, being free as a graded algebra, stop being perfect over $\mathbb{Z}$? | |
| Oct 31, 2013 at 3:25 | comment | added | Mariano Suárez-Álvarez | Can't you just take a semi free dg resolution of this example? HH will not notice the change. | |
| Oct 30, 2013 at 23:57 | comment | added | John Salvatierrez | I think cofibrant here is the sense of dg-algebras (or simplicial algebras). Instead of the derived category of an abelian category, you have a model category (or $\infty$-category) of "derived" algebras, where derived tensor product makes sense. | |
| Oct 30, 2013 at 21:13 | comment | added | Pierre-Guy Plamondon | @Sasha I think I'm confused now by the requirement for $A$ to be cofibrant. As an object of which category? Unless I am mistaken, for $A \otimes_R A^{op}$ to be an algebra, it makes no sense to take the left derived tensor product (as this gives an object of a derived category, and not an algebra). Thus I don't see why $A$ being cofibrant in any sense would change the question. Am I missing something? | |
| Oct 30, 2013 at 17:08 | comment | added | Sasha | Is $A$ cofibrant? | |
| Oct 30, 2013 at 16:46 | history | answered | Pierre-Guy Plamondon | CC BY-SA 3.0 |