Timeline for Examples of the moduli space of X giving facts about a certain X
Current License: CC BY-SA 4.0
7 events
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| Oct 22, 2022 at 15:34 | history | edited | LSpice | CC BY-SA 4.0 |
`\DeclareMathOperator`, while this is on the front page
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| May 30, 2010 at 9:42 | comment | added | Torsten Ekedahl | @BCnrd: Characteristic $0$ comes in when one uses the fact that $exp(r)$ and $log(1+r)$ are defined when $r$ is nilpotent (this is actually the basis for the fact that group schemes are smooth so they are equally algebraic). This implies that $mR$ as an additive group and $1+mR$ as multiplicative group are isomorphic when $m$ is a nilpotent ideal in $R$. This is false when we are not in characteristic $0$, see for instance the case of $1+tF[[t]]$ and $tF[[t]]$ when $F$ is a finite field. Mumford in his book "Curves on surfaces" has a nice description in these terms of what happens for $p>0$. | |
| May 28, 2010 at 16:21 | comment | added | BCnrd | @Torsten: I put emphasis on a "purely algebraic" argument (to rule out complex-analytic methods). What is it about char. 0 that you use in a purely algebraic argument if not the smoothness of lft groups? | |
| May 28, 2010 at 6:18 | comment | added | Torsten Ekedahl | @BCnrd: It is easy to see directly using the cohomological description of the Picard group that infinitesimal lifting is true in characteristic $0$ (with or without rational points) so this is (unfortunately) not an example of the kind the poser wants. | |
| May 28, 2010 at 5:56 | history | edited | Lars | CC BY-SA 2.5 |
added 33 characters in body
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| May 28, 2010 at 1:00 | comment | added | BCnrd | If $X(k)$ is empty then the natural map ${\rm{Pic}}(X) \rightarrow {\rm{Pic}}_ {X/k}(k)$ may not be an isomorphism (unless $k$ has trivial Brauer group). This already arises for $k = \mathbf{R}$ and $k = \mathbf{Q}$. Logically, it is tied up with the actual definition of the relative Picard functor. A nice application is that if $k$ has characteristic 0 then by Cartier's theorem the Picard scheme is then smooth, so there is no obstruction to infinitesimal deformation of line bundles if $X(k)$ is non-empty (false in char. > 0). Could otherwise be hard to see purely algebraically. | |
| May 27, 2010 at 22:59 | history | answered | Lars | CC BY-SA 2.5 |