Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 23G (Algebraic Geometry)

Use the usual hypothesis that the ground field has characteristic different from two, and work over its algebraic closure for the geometric points. This hypothesis is necessary for the smooth elliptic-curve calculation: in characteristic two, the displayed cubic model can be singular even when its three roots are distinct.

Homogenization givesAt $Z=0$ this says $X^3=0$, soTo compute the pole orders on a smooth cubic in Weierstrass form, work in the chart $Y=1$, writing $u=X/Y$ and $v=Z/Y$. The equation is $v={\prod }_j(u-{\lambda }_{j}v)$. Its derivative in $v$ at $(0,0)$ is one on the left after bringing the right side across, so $u$ is a local parameter. The equation gives $v=u^3+O(u^5)$. Consequently $x=u/v$ and $y=1/v$ have poles of orders two and three at $P_{\infty }$.

Let $P_j=({\lambda }_j,0)$, and let $Q_{\pm }=(0,\pm \sqrt{-{\lambda }_1{\lambda }_2{\lambda }_3})$. The latter are distinct because the product is nonzero and the characteristic is not two. At each $Q_{\pm }$, the derivative of the curve equation with respect to $y$ is nonzero, so $x$ is a local parameter and has a simple zero. At $P_j$, the derivative of the cubic in $x$ is nonzero, so $y$ is a local parameter with a simple zero; $x-{\lambda }_j$ has order two there.

The only poles occur at the point at infinity, and their orders were computed in part (i). Thus the principal divisors areOver a field where the $Q_{\pm }$ are not rational, their sum is interpreted as the degree-two zero divisor, or computed after scalar extension and then descended.

A rational function with no poles away from $P_{\infty }$ is regular on the smooth affine curve. Its ring of regular functions is the coordinate ringso it has a unique expression $A(x)+yB(x)$ with polynomials $A,B$. This regular-function assertion uses smoothness, hence normality, of the affine curve, not Riemann-Roch theorem.

The nonconstant leading term of $A$ has even pole order $2\mathrm{deg}A$ at infinity, while $yB$ has odd pole order $3+2\mathrm{deg}B$. These different parities prevent cancellation. Thus no nonconstant such function has a pole of order at most one. Therefore

The same pole-order argument provides a basisFor $n\ge 0$ the first set contains $\lfloor n/2\rfloor +1$ elements, and the second contains $\max (0,\lfloor (n-3)/2\rfloor +1)$. Distinct pole orders imply linear independence, and the coordinate-ring expression proves spanning. HenceFor $n<0$ a function in the space would have no poles anywhere and would vanish at infinity; a regular function on a projective integral curve is constant, so it must be zero. Equivalently the coordinate-ring pole argument first makes it constant, then the vanishing condition excludes nonzero constants.