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For an integral projective curve on a smooth projective surface, blowing up a point of multiplicity $m$ gives $C^{\prime }={\pi }^{\ast }C-mE$. The intersection formula for blowing up a surface and canonical divisor formula for a surface blowup yield $C^{\prime 2}=C^2-m^2$ and $K_{X^{\prime }}\cdot C^{\prime }=K_X\cdot C+m$. Substitute both in the arithmetic adjunction formula on a smooth surface. This does not require the transformed curve to be smooth.

Blow up the distinct singular points of the integral projective curve, once at each original point, and let $C^{\prime }$ be its strict transform of an algebraic subvariety. Write $H$ for the pullback of a projective line and $E_i$ for the algebraic exceptional divisors. The multiplicity of a plane curve at a point and the intersection formula for blowing up a surface giveThe canonical divisor formula for a surface blowup gives $K=-3H+{\sum }_i{E}_i$. Applying the arithmetic adjunction formula on a smooth surface, rather than a formula requiring a smooth strict transform, yieldsand henceThe strict transform of an algebraic subvariety is still an integral projective curve. Its arithmetic genus is $h^1(C^{\prime },{\mathcal{O}}_{C^{\prime }})\ge 0$, even if some singularities remain. Therefore the plane curve singularity multiplicity bound isThe argument uses only one blowup of a smooth algebraic surface at each original singular point; it does not assume these blowups resolve every singularity.

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.