3 corrected spelling of "cannot"

The same thing is true in positive characteristic, the degree of $c_n$ is equal to the Euler characteristic (except if you consider de Rham cohomology where it only is the Euler characteristic mod $p$). The proof of course can not cannot use the standard proof in the complex case, using Hopf's theorem that says that the degree of the Euler class is the Euler characteristic and the identification of the Euler class with the top Chern class). One can instead use the Riemann-Roch theorem and the identification of the Euler characteristic with the de Rham Euler characteristic. Given the latter the rest is just to verify that the seemingly complicated expresssion of the Riemann-Roch simplifies by a calculation using the splitting principle to just $c_n$. Alternatively, if I remember correctly one can use a Lefschetz pencil and induction over the dimension.

Addendum: It's all coming back to me, a third possibility is to use that the self-intersection of the diagonal is on the one hand the Euler characteristic (as it is gives the trace of the identitity map), on the other hand that self-intersection is given by the top Chern class of the normal bundle of the diagonal which is exactly the cotangent bundle.

2 added 364 characters in body

The same thing is true in positive characteristic, the degree of $c_n$ is equal to the Euler characteristic (except if you consider de Rham cohomology where it only is the Euler characteristic mod $p$). The proof of course can not use the standard proof in the complex case, using Hopf's theorem that says that the degree of the Euler class is the Euler characteristic and the identification of the Euler class with the top Chern class). One can instead use the Riemann-Roch theorem and the identification of the Euler characteristic with the de Rham Euler characteristic. Given the latter the rest is just to verify that the seemingly complicated expresssion of the Riemann-Roch simplifies by a calculation using the splitting principle to just $c_n$. Alternatively, if I remember correctly one can use a Lefschetz pencil and induction over the dimension.

Addendum: It's all coming back to me, a third possibility is to use that the self-intersection of the diagonal is on the one hand the Euler characteristic (as it is gives the trace of the identitity map), on the other hand that self-intersection is given by the top Chern class of the normal bundle of the diagonal which is exactly the cotangent bundle.

1

The same thing is true in positive characteristic, the degree of $c_n$ is equal to the Euler characteristic (except if you consider de Rham cohomology where it only is the Euler characteristic mod $p$). The proof of course can not use the standard proof in the complex case, using Hopf's theorem that says that the degree of the Euler class is the Euler characteristic and the identification of the Euler class with the top Chern class). One can instead use the Riemann-Roch theorem and the identification of the Euler characteristic with the de Rham Euler characteristic. Given the latter the rest is just to verify that the seemingly complicated expresssion of the Riemann-Roch simplifies by a calculation using the splitting principle to just $c_n$. Alternatively, if I remember correctly one can use a Lefschetz pencil and induction over the dimension.