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Jun 3, 2012 at 15:00 comment added Arkandias Thank you for you answer. I agree that we only have to show that $$\mathcal{C}^{sm}(\mathbb{Z}_p,A)_{\mathbb{Z}_p} = 0$$ but I do not see the isomorphism $$\mathcal{C}^{sm}(\mathbb{Z}_p,A) = \varprojlim A[\mathbb{Z}_p/p^n \mathbb{Z}_p]$$ Moreover the Iwasawa algebra of $\mathbb{Z}_p$ over $A$ is the continuous dual of $\mathcal{C}^{sm}(\mathbb{Z}_p,A)$ (as David said) but why this latter should be isomorphic to $\mathcal{C}^{sm}(\mathbb{Z}_p,A)$ ?...
Jun 3, 2012 at 10:50 vote accept Arkandias
Jun 3, 2012 at 14:32
Jun 2, 2012 at 8:57 history edited Filippo Alberto Edoardo CC BY-SA 3.0
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Jun 2, 2012 at 8:41 history edited Filippo Alberto Edoardo CC BY-SA 3.0
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Jun 2, 2012 at 3:26 history answered Filippo Alberto Edoardo CC BY-SA 3.0