Genus of a smooth plane curve

A smooth plane curve of degree $d$ has genus

If a smooth degree-$d$ projective plane curve had genus two, the genus-degree formula would give $(d-1)(d-2)=4$, equivalently $d^2-3d-2=0$. Its discriminant is $17$, so it has no integral solution for the degree $d$.

A smooth plane quartic is a smooth degree-four curve in ${\mathbb{P}}^2$ and has genus three.

The projective closure of the superelliptic curve $y^3=x^4+1$ is the smooth plane quarticIts unique point at infinity is $P_{\infty }=[0:1:0]$. The rational function $x=X/Z$ defines a degree-three morphism to ${\mathbb{P}}^1$ ramified with index three at $P_{\infty }$ and at the four affine points $(\alpha ,0)$ with ${\alpha }^4=-1$. Its genus is three, and

The Klein quartic has the smooth projective plane modelIts affine chart $Z=1$ is $x^{3}y+y^3+x=0$. If $\zeta$ is a primitive seventh root of unity, the projective transformationis an automorphism of order seven.

The projective completion ofis the smooth plane quarticThe coordinate relabelling $[A:B:C]=[Y:X:Z]$ turns this equation into the standard Klein quartic equation $A^{3}B+B^{3}C+C^{3}A=0$.

The coordinate function $x$ has degree three on the Klein quartic. It has ramification index three at $(0,0)$, index two at the seven affine points satisfying $y^7=-3/8$ and $x=2y^3$, and index two at $[0:1:0]$. The total ramification is ten, so the Riemann-Hurwitz formula gives genus three.

For a smooth plane quartic $C\subset {\mathbb{P}}^2$, the Adjunction formula gives $K_C\cong {\mathcal{O}}_C(1)$. Hence every line section $C\cap \ell$ is a canonical divisor.

Projection from a point of a smooth plane quartic gives a degree-three map to the projective line. A degree-two map would make the curve hyperelliptic, but its canonical map is its plane embedding, whereas a hyperelliptic canonical map is not an embedding. Its gonality is therefore three.