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Suppose that $X\subset\mathbb P^n$ is a $d$-dimentional smooth projective variety (not a linear subspace) over an algebraically closed field. If $\gamma\colon X\to\mathrm{Gr}(d,\mathbb P^n)$ is Gauss mapping that attaches to each point $x\in X$ the embedded Zariski tangent space to $X$ at $x$, then it is known that $\gamma$ is finite.

If characteristic is zero, it is known that $\gamma$ is not just finite but birational onto its image. My question is whether $\gamma$ is generically one to one in finite characteristic.

Edit: removed the question about birationality in finite characteristic, thanks to the example given by Felipe Voloch.

Thanks in advance,

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You can start reading here: link.springer.com/article/10.1007%2Fs10711-008-9334-1#page-1 –  M P Apr 25 '13 at 10:05
@MP: Great! Thanks for the reference. –  Serge Lvovski Apr 25 '13 at 15:33
There are papers by Kleiman-Piene that discuss this question. My best recollection is that they tend to be inseparable, but finite. –  aginensky Apr 25 '13 at 15:34
MP's reference provides counter-examples that are smooth space curves. On the other hand, all smooth plane curves have generically one-to-one Gauss map, by [Hajime Kaji : On the Gauss maps of space curves in characteristic p, Corollary 4.5]. –  Olivier Benoist May 12 '13 at 14:06
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1 Answer

No. The plane curve $x^{p+1}+y^{p+1}=1$ has an inseparable Gauss map.

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@Felipe: Thank yuo for the example (I will edit the question accordingly). However, this Gauss mapping is 1-1. Does there exist an example where it is not generically one to one? –  Serge Lvovski Apr 25 '13 at 15:26
See MP's reference. –  Felipe Voloch Apr 25 '13 at 18:12
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