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The Picard group $\mathrm{Pic}(X)$ is the set of isomorphism classes of line bundles on $X$, with tensor product as addition, the trivial bundle as zero and the dual bundle as inverse. On a smooth projective surface, line bundles can be represented by divisors. The intersection pairing on the Picard group of a surface is the symmetric bilinear mapobtained by moving the divisors into proper position and counting their intersections with multiplicity.

The surfaceis the first Hirzebruch surface ${\mathbb{F}}_1$. If $S$ is its negative section and $F$ a fiber of the ruling, its Picard lattice of a Hirzebruch surface is

Now let $\pi :X\to {\mathbb{P}}^2$ be the given nonisomorphic birational morphism. A birational morphism between smooth projective surfaces factors as a nonempty sequence of point blowups. Let $H={\pi }^{\ast }[\text{line}]$, and take the total transform $E$ on $X$ of the exceptional curve of the first blowup. The intersection formula for blowing up a surface givesTherefore the nonzero Picard group element $H+E$ satisfiesThis is the isotropic divisor from a nontrivial birational morphism to the projective plane.

For a morphism $\varphi :{\mathbb{P}}^2\to {\mathbb{P}}^n$, the pullback of the hyperplane bundle has the formfor an integer $d\ge 0$. If a line $\ell$ is contracted to a point, this bundle restricts trivially to $\ell$, whereasIts degree is therefore zero, so $d=0$. The homogeneous sections defining $\varphi$ are then constants, and $\varphi$ is constant. This proves the morphism from the projective plane contracting a line criterion.

Finally choose an integer $r>C^2$ and blow up $r$ distinct points of the smooth curve $C$. For the resulting morphism $\pi :X^{\prime }\to X$, let $E_1,\dots ,E_r$ be the exceptional curves. The strict transform isand the self-intersection after blowing up points on a smooth curve formula givesBlowing up a smooth point of a smooth curve does not change that curve itself, so $\pi {|}_{C^{\prime }}:C^{\prime }\to C$ is an isomorphism.