Suppose $\mathcal{L}$ is a base-point free ample invertible sheaf on a complex surface $S$ inducing a finite morphism $S\to \mathbb{P}^2$ (so $\mathcal{L}$ has 3 global sections). I am wondering if there is a connection between the degree of this map and the Hilbert polynomial of $\mathcal{L}$? I have some other hypotheses on $S$ if needed.

Is it correct that the degree of the map will be ($2!$ times) the leading coefficient of the Hilbert polynomial?

If this is not true, can you say anything else about the Hilbert polynomial of $\mathcal{L}$ from the existence of such a map?

Any suggestions/hints are greatly appreciated!

  • 1
    $\begingroup$ The degree of the map is $L^2$, the self intersection number and you can easily see its relation to the Hilbert polynomial. $\endgroup$ – Mohan Aug 29 '18 at 2:23
  • $\begingroup$ thanks, and then my claim about the leading coefficient is true by Riemann-Roch, right $\endgroup$ – anon Aug 29 '18 at 2:50
  • $\begingroup$ also can you say that $S$ must be isomorphic to $\mathbb{P}^2$? my surface is also rational and normal. $\endgroup$ – anon Aug 29 '18 at 2:54
  • $\begingroup$ $S$ may not be isomorphic to the plane. For example, take a double cover of the plane ramified along a smooth quartic. Then, it is rational, but not isomorphic to the plane. The pull back of the hyperplane section has precisely 3-dimensional sections. $\endgroup$ – Mohan Aug 29 '18 at 15:19

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