Parametrization of Rational Surfaces

The parametric degree of a rational surface

“…It follows that C i K < 0. By Lemma A.1 in [13], each C i is exceptional, and C 2 i = C i K = −1 for all i. Hence C K = −r , and C D is an integral multiple of r .…”

“…Proof Lemma A.8 from [13] -which is also true in the case K is non-closed -says that we can conclude from D a 2 > 0 that either 3D a + K a = 0, or 3D a + 2K a = 0, or (2D a +K a ) 2 = 0. Using Corollary 1, we get level(D) = level(D a )+a = a+i/3 in the i-th case, for i = 1, 2.…”

“…Proof The proof of Lemma A.7 in [13] generalizes without difficulty to non-closed fields. This shows the existence of P with the desired properties.…”

“…There is only one step in the proofs of [13] that uses the assumption that K is algebraically closed: in this case, any prime divisor C of dimension 0 with C K < 0 is exceptional, and we have C 2 = C K = −1 (Lemma A.1 in [13]). It follows that if D is nef and D + K is effective but not nef, then there exists an exceptional divisor orthogonal to D.…”

“…We can therefore say that the parametric degree depends "just slightly" on the choice of the ground field, as this choice can change by a factor of at most 2. Of course, this is a bit misleading because there are surfaces for which rationality depends on the choice of the field, for instance surfaces that are rational over C but not rational over R. We note that this result implies that the question of rationality is decidable over any perfect field with decidable first order theory: first, we compute a parametrization over the algebraic closure, using [13]. If it is of degree d, then, by the result in this paper, the surface is rational iff there is a parametrization of degree at most 2d.…”