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The arithmetic genus of a proper curve $C$ is $p_a(C)=1-\chi (C,{\mathcal{O}}_C)$. For an integral projective curve over an algebraically closed field, $H^0(C,{\mathcal{O}}_C)=k$, so $p_a=h^1\ge 0$. Its finite normalization has $p_a(C)=g(C^{\nu })+\mathrm{length}({\nu }_{\ast }{\mathcal{O}}_{C^{\nu }}/{\mathcal{O}}_C)$. Hence arithmetic genus zero forces the curve to be smooth and rational.

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.

Over an algebraically closed field, a smooth projective rational curve is isomorphic to ${\mathbb{P}}^1$. Its line bundles are determined by degree, its degree-zero line bundle is trivial, and $H^1({\mathbb{P}}^1,\mathcal{O})=0$. An integral projective curve of arithmetic genus zero is automatically smooth and rational by the normalization genus formula.