Timeline for Comparison between singular and etale cohomology in Batyrev's paper on Birational Calabi-Yau varieties
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7 events
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| Mar 12, 2012 at 14:22 | comment | added | Tom Lovering | Thanks for the quick reply. I think it's helped focus in on my confusion. If $R$ were constructed as you say, I'm suspicious that its function field might have some transcendence degree over $\mathbb{Q}_p$. On the other hand, the paper seems to say that $R$ is "the maximal compact subring in a local $p$-adic field," which I took to mean an algebraic extension of $\mathbb{Q}_p$, and I think actually being a local field is necessary to make the rest of the argument work. So I guess my question now is: is $Frac(R)$ (as you define it) obviously algebraic $/\mathbb{Q}_p$? | |
| Mar 12, 2012 at 13:30 | comment | added | Donu Arapura | Unless I misunderstood your notation, $\mathcal{R}$ is an integral domain. You first localize $\mathcal{R}\to \mathcal{R}_{J(\pi)}=S$ and then complete to get $S\to R=\hat S$. Both maps are injective. The unnatural embedding is $R\to \mathbb{C}$, and here you do have to play around. If you prefer, another way to do this is to observe that the Betti numbers are the same for any two complex embeddings, because they both coincide with dimension of etale cohomology groups. | |
| Mar 12, 2012 at 12:43 | comment | added | Tom Lovering | I don't think there's any kind of natural embedding of $\mathcal{R}$ into $R$ :the paper seems to go via $\mathcal{R} \rightarrow \mathcal{R}\otimes_\mathbb{Z} \mathbb{Z}_p \rightarrow R$ which feels unlikely to be injective. For example, if $\mathcal{R}$ is something like $\mathbb{Z}[x_1,...,x_n]$. So do you mean I should artificially construct some kind of injection $\mathcal{R} \hookrightarrow R$,using the fact $\mathbb{Z}_p$ has lots of transcendental elements,and then play around to make it compatible with the inclusion of $\mathcal{R}$ in $\mathbb{C}$? | |
| Mar 10, 2012 at 18:44 | comment | added | Donu Arapura | I should add that you should choose $K\subset \mathbb{C}$ to be compatible with $\mathcal{K}\subset \mathbb{C}$. I leave it to you to see that this is possible. | |
| Mar 10, 2012 at 18:34 | comment | added | Donu Arapura | Form a commutative square, which is hard to do in the comments. The bottom edge is $\mathcal{R}\subset \mathcal{K}$ where $\mathcal{K}$ is the fraction field; top $R\subset K$ also fraction field; left $\mathcal{R}\subset R$, and right $\mathcal{K}\subset K$. Now include $K\subset \mathbb{C}$. So all is well | |
| Mar 10, 2012 at 18:05 | history | edited | Tom Lovering | CC BY-SA 3.0 |
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| Mar 10, 2012 at 17:59 | history | asked | Tom Lovering | CC BY-SA 3.0 |