Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 22F (Riemann Surfaces)

Choose coordinates $z$ at $p$ and $w$ at $f(p)$ vanishing at those points. The local expansion is $w\circ f\circ z^{-1}(\zeta )=a{\zeta }^v+O({\zeta }^{v+1})$, $a\ne 0$. Define the branching order $v_f(p)=v$. A change of either coordinate has nonzero linear coefficient, so it preserves the leading exponent, proving well-definedness.

Ramification means $v>1$, equivalently vanishing of the derivative in local coordinates. These zeros are isolated unless the local derivative is identically zero, which by analytic continuation would make $f$ constant. An infinite set of ramification points on a compact surface would have an accumulation point, contradicting isolation. Thus that set is finite.

A nonconstant holomorphic map of compact connected Riemann surfaces is onto, since its image is open by the local form and closed by compactness. It has a finite degree $d\ge 1$. The Riemann-Hurwitz formula isIf both genera equal $g>1$, positivity of $2g-2$ forces $d=1$ and every ramification term zero. The map is bijective and has nonzero derivative everywhere. The holomorphic inverse function theorem says that a holomorphic function with nonzero derivative has a local holomorphic inverse; these inverses patch, so $f$ is biholomorphic. For $g=1$, multiplication by two on $\mathbb{C}/\Lambda$ is a degree-four holomorphic map of an elliptic curve to itself and is not biholomorphic.

For a nonsingular plane curve $F(s,t)=0$, at a point with $F_t\ne 0$ the holomorphic implicit function theorem gives $t=t(s)$ and the chart coordinate $s$. Where $F_s\ne 0$, use $t$ and express $s=s(t)$. Nonsingularity ensures at least one choice is available, and the coordinate transition functions and both restricted projections are holomorphic.

Here $F=t^4-P(s)$ with $P(s)=(s^2-1)(s-4)$. A singular point would require $t=0$, $P(s)=0$ and $P^{\prime }(s)=0$. But $P$ has three distinct roots $-1,1,4$, so this is impossible. Hence the curve is nonsingular.

If $t\ne 0$, $F_t=4t^3\ne 0$, so $s$ is a local coordinate and the projection to $s$ is unramified. At $(s_0,0)$ for $s_0\in \{-1,1,4\}$, use $t$ as coordinate. Since $P^{\prime }(s_0)\ne 0$, solving $P(s)=t^4$ gives $s-s_0=t^4/P^{\prime }(s_0)+O(t^8)$. ThusThese are all the ramification points on the stated affine curve; no points at infinity belong to $Y\subset {\mathbb{C}}^2$.