Timeline for Structure of Hopf algebras - trouble understanding an old paper
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2015 at 9:50 | history | edited | Neil Strickland | CC BY-SA 3.0 |
added 24 characters in body
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| Sep 19, 2015 at 9:44 | history | edited | Neil Strickland | CC BY-SA 3.0 |
added 4 characters in body
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| Sep 18, 2015 at 6:04 | vote | accept | Neil Strickland | ||
| Sep 18, 2015 at 4:04 | answer | added | Peter May | timeline score: 20 | |
| Sep 15, 2015 at 21:40 | history | edited | Neil Strickland | CC BY-SA 3.0 |
I have changed the notation and logical structure slightly, and added some further details.
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| Sep 14, 2015 at 16:26 | answer | added | Peter May | timeline score: 13 | |
| Sep 14, 2015 at 16:25 | comment | added | Neil Strickland | I need to adapt these results to a different context, where $A$ is not graded and connected, but I have other assumptions that may do the same job. May's approach looks promising for that purpose, if the issue that I described can be resolved. I should probably try to adapt the Milnor-Moore argument as well, but for the moment I am looking at May's paper. I do not have an antipode, which makes an approach via Dieudonne modules less promising. | |
| Sep 14, 2015 at 16:12 | comment | added | user43326 | If you aren't happy with Milnor-Moore Theorem 7.11, you can always use Dieudonne modules. The injectivity of the $p$-th power implies that Dieudonne module is torsion-free, and the connectedness hypothesis implies that there is no infinitely divisible elements. | |
| Sep 14, 2015 at 15:58 | comment | added | user43326 | Isn't this in Milnor-Moore? | |
| Sep 14, 2015 at 15:56 | comment | added | Neil Strickland | @user43326 How do you know that $A$ is polynomial? That is a theorem of Borel, but a large part of the point of May's paper is to give a sharpened version and alternative proof of Borel's result, so I am not happy to quote it. | |
| Sep 14, 2015 at 15:47 | comment | added | user43326 | Since you are happy to suppose the $p$-th power is injective, $A$ is polynomial. So if $\overline{f}$ is surjective, it is split epi as a map of algebras. So if $f$ is injective on $Q(A)$ it is injective. | |
| Sep 14, 2015 at 12:07 | history | asked | Neil Strickland | CC BY-SA 3.0 |