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How to classify K3 surfaces over an arbitrary field k?

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Although I completely agree with David Lehavi's answer, if you work over $\mathbb{Z}_p$, probably you can also make sense of rigid analytic K3 surfaces that are not necessarily algebraic schemes over $\mathbb{Z}_p$ (just as there are complex analytic K3 surfaces that are not algebraic schemes over $\mathbb{C}$). –  Jason Starr Feb 13 '13 at 22:12

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up vote 8 down vote accepted

The "standard" definition of a K3 surface is field independent (unless you are a physicist):

$p_g=1, q=0$, and trivial canonical class.

Some results:

  • Mumford and Bombieri showed that you get (just as in the complex case) a 19 dimensional family of K3 surfaces for any degree (the 19 dimensional thingy is a deformation theory argument which is completely algebraic).
  • Deligne showed that all the K3 surfaces in finite characteristics are reductions mod p.

What you obviously don't get is the fact that all these spaces sit together in a nice 20 dimensional complex ball. I also don't know if you can carry over any of the recent Kodaira dimension computation of these moduli (which are very analytic in nature).

Reference: Complex algebraic surfaces (Beauville): Chapter VIII and Appendix A.

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Field independent with the small twist that when people refer to K3 over k they usually mean algebraic surface, of which there are 19 parameters, while over C they mean a member of 20-dimensional family. –  Ilya Nikokoshev Oct 20 '09 at 4:11
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@ilya - well, it depends if these people are algebraic or differential geometers I guess... –  David Lehavi Oct 20 '09 at 5:46
    
Or mathematicians and physicists. –  Ilya Nikokoshev Oct 22 '09 at 17:56
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it should be $p_g=1$ and $q=0$... –  Yoyontzin Feb 12 '13 at 1:45
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+ Yoyontzin: oops (and it stayed like this for the last 3.5 years) –  David Lehavi Feb 12 '13 at 20:55

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