Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 24F (Algebraic Geometry)

Let the smooth projective curve be the projective completion of $f(x,y)=0$, of degree $d$, and suppose after interchanging $x$ and $y$ if necessary that $f_y=\partial f/\partial y$ is nonzero. A basis of its holomorphic differential forms isEquivalently one may use $-x^i{y}^{j}dy/f_x$. These expressions agree on their common domain because differentiating $f(x,y)=0$ gives $f_{x}dx+f_{y}dy=0$. The allowed monomials are precisely the homogeneous polynomials of degree at most $d-3$, in accordance with the Adjunction formula $K_C\cong {\mathcal{O}}_C(d-3)$.

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

The intersection with a line $\ell$ is a hyperplane section:The canonical bundle of a smooth projective hypersurface formula, obtained from the Adjunction formula, gives for a smooth degree-four curve in ${\mathbb{P}}^2$Thus ${\mathcal{O}}_C(D)\cong K_C$, so the divisor class of $D$ is the canonical divisor class. Therefore $D$ is a canonical divisor. Its degree is also $4=2g-2$, consistently with $g=3$.

Choose a point $p\in C$ and project from $p$ to a line not containing $p$. Every line through $p$ meets the quartic in four points counted with intersection multiplicity by Bézout theorem; removing the fixed intersection at $p$ leaves a generic fibre of three points. This projection from a point on a plane curve therefore defines a nonconstant morphismof degree three, so the gonality is at most three.

It cannot be one, because a degree-one morphism between smooth projective curves is an isomorphism of algebraic varieties, whereas $C$ has genus three and the projective line has genus zero. It cannot be two either: that would make $C$ a hyperelliptic curve. The canonical map of a hyperelliptic curve is not a closed embedding, but for a smooth plane quartic the canonical map of a smooth plane curve is exactly its given embedding $C\hookrightarrow {\mathbb{P}}^2$, since $K_C\cong {\mathcal{O}}_C(1)$. This contradiction proves