Timeline for Abelianized fundamental group of a curve over a finite field
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 29, 2012 at 5:53 | answer | added | Will Sawin | timeline score: 5 | |
| Nov 29, 2012 at 5:20 | history | edited | Justin Campbell | CC BY-SA 3.0 |
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| Nov 29, 2012 at 4:53 | comment | added | user28172 | Justin, for your final question you are tacitly assuming that $\pi_1({\rm{Pic}}^1_{X/k})$ is abelian, but is false. More specifically, with $k := \mathbf{F}_q$, although ${\rm{Pic}}^1_{X/k}$ admits a structure of abelian variety (since $k$ is finite and ${\rm{Pic}}^1_{X/k}$ is a torsor for the abelian variety ${\rm{Pic}}^0_{X/k}$), for an abelian variety $A \ne 0$ over a finite field $k$ the group $\pi_1(A)$ is never abelian. Indeed, for $n$ coprime to char($k$) and $|A(k)|$, $[n]:A \rightarrow A$ is a nontrivial connected finite etale cover with no nontrivial automorphisms! | |
| Nov 29, 2012 at 4:19 | comment | added | Emerton | ... compares etale cohomology of $\overline{X}$ to that of $X$. Regards, | |
| Nov 29, 2012 at 4:00 | comment | added | Emerton | Dear Justin, You have an exact sequence of groups with the right hand term being abelian, in fact procyclic, and you are trying to understand the associated right exact sequence of abelianizations. This is just an elementary exercise in group theory; if you do it, you will find that the map $\pi_1(\overline{X})^{ab} \to \pi_1(X)^{ab}$ has image equal to the Galois coinvariants of the source. Incidentally, you might find it profitable to rewrite your abelianized sequence in terms of etale (co)homology; you will then find that it is part of the Hochschild--Serre spectral sequence that ... | |
| Nov 29, 2012 at 3:55 | history | edited | Justin Campbell | CC BY-SA 3.0 |
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| Nov 29, 2012 at 3:29 | history | asked | Justin Campbell | CC BY-SA 3.0 |