Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 134 / 2 / iv

By Kodaira's lemma for the big divisor $K_X$, choose an integer $q>0$, an ample Cartier divisor $A$, and an effective divisor $E$ withIf an integral curve $C$ is not a component of $E$, nonnegative intersections of distinct curves giveThus every curve with $K_X\cdot C<0$ is one of the finitely many components of $E$. In particular,This is the surface case of negative curves of a big real divisor lie in finitely many divisors. The proof does not assert that every negative curve on $X$ belongs to this finite set; it concerns curves negative against this particular big canonical class.

Let $C\in \mathrm{Neg}(K_X)$. The arithmetic adjunction formula on a smooth surface givesWe first exclude $C^2\ge 0$, following the PDF's hint and the finiteness proved in (a).

If $C^2=0$, adjunction and $K_X\cdot C<0$ force $p_a(C)=0$ and $K_X\cdot C=-2$. An integral projective curve of arithmetic genus zero is a smooth rational curve: normalization and the nonnegative singularity-length correction show both normalization genus and singularity correction vanish. Part (iii) makes $C$ semiample. Choose a basepoint-free $|mC|$. Over the infinite algebraically closed field, a general section avoids containing any of the finitely many curves of $\mathrm{Neg}(K_X)$ as a component. Its effective divisor $G$ then has $K_X\cdot G\ge 0$, but $K_X\cdot G=m{K}_X\cdot C<0$, a contradiction.

If $C^2>0$, Riemann–Roch theorem for algebraic surfaces and Serre duality show $h^0(X,mC)$ is unbounded. Indeed $h^2(X,mC)=h^0(X,K_X-mC)=0$ for $m\gg 0$, because the latter divisor has negative intersection with a fixed ample divisor. HenceThere is therefore some $m$ with $h^0(X,mC)>h^0(X,(m-1)C)$, giving a section whose divisor does not contain $C$. The canonical section of $mC$ does not vanish identically along any other curve. A general linear combination of these two sections consequently contains none of the finite set $\mathrm{Neg}(K_X)$, again contradicting its negative canonical intersection. Thus $C^2<0$.

Now put $a=-C^2\ge 1$ and $b=-K_X\cdot C\ge 1$. Adjunction givesIt follows that $a=b=1$ and $p_a(C)=0$. HenceThe general-section argument only avoids finitely many proper linear subspaces, so it works over any algebraically closed field, including positive characteristic.

The Castelnuovo contraction criterion says that a smooth rational curve $C$ on a smooth projective surface with $C^2=-1$ can be contracted by a birational morphism to a smooth point of a smooth projective surface. The contraction is an isomorphism away from $C$, and its inverse is the blowup of a smooth algebraic surface at that point. This is the algebraic contraction criterion, valid over the algebraically closed field here; it is not the rationality criterion bearing Castelnuovo's name.

By (b), every $C\in \mathrm{Neg}(K_X)$ satisfies the hypothesis. Therefore the required morphism exists:The PDF calls $f_1$ a birational map; the conclusion is stronger, since this map is everywhere defined and is a birational morphism.