Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 24F (Algebraic Geometry)

Homogenizing the affine equation gives the projective closureIts partial derivatives areIf all three vanished at a point of $X$, then $X=0$ and either $Y=0$ or $Z=0$. If $Y=0$, the equation $F=0$ forces $Z=0$; if $Z=0$, the equation $F_Z=0$ forces $Y=0$. Either conclusion gives the forbidden zero triple of homogeneous coordinates. The Jacobian criterion therefore shows that $X$ is a smooth projective curve; more specifically, it is the projective closure of y cubed equals x to the fourth plus one, a smooth plane quartic.

The formula $[X:Y:Z]\mapsto [X:Z]$ is initially undefined at the unique point $P_{\infty }=[0:1:0]$ of $X$ at infinity. Since $X$ is a smooth projective curve, the extension of a rational map from a smooth projective curve extends it uniquely to a morphismOn the affine chart it is the rational function $x$, and the equation $y^3=x^4+1$ shows that $\phi$ has degree three.

For each of the four roots $\alpha$ of $x^4+1$, let $P_{\alpha }=(\alpha ,0)$. Since the root is simple, $t=y$ is a uniformizer at $P_{\alpha }$, and the equation givesThus the ramification index of a holomorphic map at every $P_{\alpha }$ is three.

At $P_{\infty }$, use the chart $Y=1$ and put $u=X/Y$, $v=Z/Y$. The equation becomesThe implicit function theorem makes $u$ a uniformizer and gives $v=u^4$ times a local unit. Henceso $P_{\infty }$ also has ramification index three. These are all the ramification points of a holomorphic map because the resulting ramification degree is $5(3-1)=10$. Indeed, the Riemann-Hurwitz formula givesTherefore $g_X=3$, in agreement with the genus of a smooth plane curve, and the Euler characteristic of the underlying compact orientable surface is

LetAt a finite ramification point $P_{\alpha }=(\alpha ,0)$, the uniformizer $t=y$ from part b satisfies $x-\alpha =c_{\alpha }{t}^3+O(t^6)$. ConsequentlyAt every other affine point, $x-x(p)$ is a uniformizer and $y$ is a unit, so again ${\nu }_p(\omega )=0$.

At $P_{\infty }$, take the uniformizer $t=u=X/Y$ used above. Since $v=Z/Y=t^4$ times a local unit,It follows that ${\nu }_{P_{\infty }}(dx)=-4$ and ${\nu }_{P_{\infty }}(y^2)=-8$, so the valuation of a rational differential isThuswhose degree four agrees with $2g_X-2$. For the resulting canonical divisor $K_X=4P_{\infty }$, the functions $1,x,y$ have pole orders $0,3,4$ at $P_{\infty }$ and no other poles. They therefore belong to the canonical Riemann-Roch space. Since $\ell (K_X)=g_X=3$, they form a basis:Equivalently, multiplying by $\omega$ gives the basisof holomorphic differential forms, also matching the general description of holomorphic differentials on a smooth plane curve.