Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 24F / a

Let $H=V(L)$ for a nonzero homogeneous linear polynomial $L$. Because $H$ does not contain $X$, the restriction $L{|}_X$ is a nonzero section of ${\mathcal{O}}_X(1)$. Its zero divisor is the hyperplane-section divisor of a projective plane curvewhere $I_p(X,H)$ is the local intersection multiplicity; equivalently, it is the order of vanishing of $L{|}_X$ at $p$.

The curves $X$ and $H$ have no common component and have degrees $d$ and $1$. The Bézout theorem therefore gives

The Adjunction formula gives $K_X\cong {\mathcal{O}}_X(d-3)$. Since a hyperplane section has degree $d$, the canonical degree of a smooth plane curve is

For any smooth projective curve of genus $g$, a canonical divisor has degree $2g-2$. If this plane curve had genus two, part (ii) would implyso $d^2-3d-2=0$. The quadratic discriminant is $17$, which is not a perfect square, so the equation has no integral solution. Thus a smooth projective plane curve cannot have genus two, as recorded by no smooth plane curve has genus two.