= Solution
The <Picard group> $\operatorname{Pic}(X)$ is the set of isomorphism classes of <line bundle>[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 map
$$
\operatorname{Pic}(X)\times\operatorname{Pic}(X)\longrightarrow\mathbb Z,
\qquad ([D],[E])\longmapsto D\cdot E,
$$
obtained by moving the divisors into proper position and counting their intersections with multiplicity.
The surface
$$
\mathbb P_{\mathbb P^1}(\mathcal O\oplus\mathcal O(1))
$$
is the first <Hirzebruch surface> $\mathbb F_1$. If $S$ is its <negative section of a Hirzebruch surface>[negative section] and $F$ a fiber of the ruling, its <Picard lattice of a Hirzebruch surface> is
$$
\operatorname{Pic}(\mathbb F_1)=\mathbb ZS\oplus\mathbb ZF,
\qquad
S^2=-1,quad S\cdot F=1,quad F^2=0.
$$
Now let $\pi:X\to\mathbb P^2$ be the given nonisomorphic <birational variety>[birational] morphism. A birational morphism between smooth projective surfaces factors as a nonempty sequence of point blowups. Let $H=\pi^*[\text{line}]$, and take the total transform $E$ on $X$ of the exceptional curve of the first blowup. The <intersection formula for blowing up a surface> gives
$$
H^2=1,qquad E^2=-1,qquad H\cdot E=0.
$$
Therefore the nonzero <Picard group> element $H+E$ satisfies
$$
(H+E)^2=1-1=0.
$$
This is the <isotropic divisor from a nontrivial birational morphism to the projective plane>.
For a morphism $\phi:\mathbb P^2\to\mathbb P^n$, the pullback of the hyperplane bundle has the form
$$
\phi^*\mathcal O_{\mathbb P^n}(1)\cong\mathcal O_{\mathbb P^2}(d)
$$
for an integer $d\geq0$. If a line $\ell$ is contracted to a point, this bundle restricts trivially to $\ell$, whereas
$$
\mathcal O_{\mathbb P^2}(d)|_\ell\cong\mathcal O_{\mathbb P^1}(d).
$$
Its <degree of a divisor>[degree] is therefore zero, so $d=0$. The homogeneous sections defining $\phi$ are then constants, and $\phi$ 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 $\pi:X'\to X$, let $E_1,\ldots,E_r$ be the <exceptional divisor>[exceptional curves]. The <strict transform> is
$$
C'=\pi^*C-\sum_{i=1}^rE_i,
$$
and the <self-intersection after blowing up points on a smooth curve> formula gives
$$
(C')^2=C^2-r<0.
$$
Blowing up a smooth point of a smooth curve does not change that curve itself, so $\pi|_{C'}:C'\to C$ is an isomorphism.
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