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Over an algebraically closed field, the blowup of a smooth algebraic surface at a closed point replaces the point by an exceptional smooth rational curve $E\cong {\mathbb{P}}^1$, with $E^2=-1$. Locally it is the blowup of the affine plane at the origin. It is a proper birational morphism, an isomorphism off the exceptional curve, and the blown-up surface is again smooth and projective when the original is projective.

A smooth rational curve with self-intersection $-1$ on a smooth projective surface over an algebraically closed field contracts to a smooth point of another smooth projective surface. The contraction is a birational morphism and an isomorphism off the curve, with inverse a blowup of a smooth algebraic surface. This algebraic criterion is valid in arbitrary characteristic and should not be confused with Castelnuovo's rationality criterion.

A primitive effective nef class of square zero on a complex K3 surface defines a basepoint-free pencil with connected genus-one fibres. Its general fibre is a smooth elliptic curve. A section is a smooth rational curve $\Gamma$ with $E\cdot \Gamma =1$ and ${\Gamma }^2=-2$.

Let $X\subset {\mathbb{P}}^3$ have degree $d\ge 3$ and hyperplane class $H$. The adjunction formula gives $K_X=(d-4)H$. For a line $L\subset X$, the same formula on the smooth rational curve $L$ yields $(K_X+L)\cdot L=-2$, henceThus all lines are negative curves, and part (b) makes their set countable. The Fano scheme of lines is a closed projective scheme of $\mathrm{Gr}(2,4)$. A positive-dimensional complex algebraic component would have uncountably many points, so this scheme is zero-dimensional. A projective scheme of finite type has only finitely many zero-dimensional points. This proves the finiteness of lines on a smooth surface of degree at least three.

The adjunction formula on the degree-$d$ projective hypersurface gives $K_X=(d-4)H$. Since $C$ is a smooth rational curve, adjunction on $C$ gives $(K_X+C)\cdot C=2g(C)-2=-2$. Its degree is $H\cdot C=a$, so the smooth rational-curve self-intersection on a projective hypersurface formula is

The Noether–Lefschetz theorem says that a very general smooth degree-$d$ surface in ${\mathbb{P}}_{\mathbb{C}}^3$, for $d\ge 4$, has Picard group freely generated by its hyperplane class $H$. Here very general means outside a countable union of proper closed loci in the parameter space of surfaces.

If such a surface contained a smooth rational curve $C$, its divisor class would be $mH$ for an integer $m$. Since $H\cdot C=md>0$, one has $m>0$ and therefore $C^2=m^{2}d>0$. Part (a), however, gives $C^2=-2-(d-4)\mathrm{deg}C\le -2$, a contradiction. Thus a very general surface of degree at least four has no smooth rational curve. The statement concerns smooth embedded curves; a singular curve can still have rational normalization.

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.

A smooth rational curve $C$ with $C^2=0$ on a smooth projective surface is semiample and has Iitaka dimension one. Its normal bundle is trivial, so the restriction sequences for $mC$ have quotient ${\mathcal{O}}_C$ and zero quotient $H^1$. The finite dimensions of $H^1(X,mC)$ decrease and stabilize. Restriction of sections to $C$ is then surjective; a lift of $1$ and the canonical section generate globally. Exactness also gives $h^0(X,mC)-h^0(X,(m-1)C)=1$ eventually.