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Jun 5, 2013 at 15:09 vote accept Sean Tilson
Jun 5, 2013 at 15:07 vote accept Sean Tilson
Jun 5, 2013 at 15:08
Mar 17, 2013 at 10:31 comment added Markus Land I also believe that it is "well-known" that stable operations of degree $1$ are just given by Ext-groups of the two coefficient groups between which you want to calculate operations. Of course this implies that $H\mathbb{Z}^1H\mathbb{Z} = 0$.
Dec 28, 2010 at 13:54 history edited Tom Goodwillie CC BY-SA 2.5
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Dec 28, 2010 at 9:08 comment added Torsten Ekedahl The composite of reduction mod p and the integral Bockstein is zero by construction as the integral Bockstein is the boundary map coming from the sequence $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/p\to0$.
Dec 28, 2010 at 6:59 comment added Sean Tilson I do not understand how those groups can be finite for all n larger than 0. It seems that there should be an operation $\beta_p$ for each prime, a composition of the reduction mod p and then the integral bockstein, in $H\mathbb{Z}^1 H\mathbb{Z}$.
Dec 28, 2010 at 4:31 history answered Tom Goodwillie CC BY-SA 2.5