# Can there be a point on a Riemann surface such that every rational function is ramified at this point?

Let $X$ be a compact connected Riemann surface, and let $S\subset X$ be a finite subset.

Does there exist a morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ which is unramified at the points of $S$?

I'm interested in the case where $X$ is of genus at least $2$. (The genus zero case is trivial: take $f$ to be the identity.)

The answer is trivial when $S$ is empty. (Any morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ will do.)

Let $h:X\to \mathbf{P}^1(\mathbf{C})$ be a morphism with ramification locus $R(h)$. Then, if $S\subset X\backslash R(h)$, the answer is yes.

How effective can our answer be? That is, suppose that there exists such an $f$. Then, can we bound its degree?

The title is a special case of the above question: take $S=\{\textrm{pt}\}$.

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The answer to your first question is yes by a (a weak version of) Riemann-Roch. Pick a point $P \notin S$ and write $\ell(nP)$ for the dimension of the space of meromorphic functions unramified outside of $P$ and with a pole of order at most $n$ at $P$. Then $\ell(nP) \underset{n\to +\infty}{\longrightarrow} +\infty$. Regarding your second question, the Riemann-Hurwitz formula relates the degree to the local ramification indexes. –  YBL Dec 25 '11 at 0:45
@YBL: the condition "unramified outside of $P$" is not linear - how does Riemann-Roch say anything about such a space of functions? –  user8268 Dec 25 '11 at 21:20

If $S=\lbrace P\rbrace$, put an algebraic structure $X^{alg}$ on $X$ and take a uniformizing parameter $t\in \mathcal O_{X^{alg},P}$ at $P$.
Since $\mathcal O_{X^{alg},P}\subset Rat(X^{alg})=\mathcal Mer(X)$, the meromorphic function $t$ seen as a map $X\to \mathbb P^1(\mathbb C)$ solves your problem.
You can generalize that to the case where $S$ is an arbitrary finite set, again by putting an algebraic structure on the Riemann surface $X$: look up Corollary 1.16 in Chapter VI of Miranda's book, which solves your problem.

Edit: Let me however give a self-contained proof using Riemann-Roch.

Fix an arbitrary point $x_0\in X\setminus S$ outside of $S=\lbrace x_1,...,x_n \rbrace$.
Consider the divisors (where $N$ will be determined later)
$$D_1=(-2)\cdot x_1+...+(-2)\cdot x_n+N\cdot x_0\quad \text {and} \quad D_2=(-1)\cdot x_1+...+(-1)\cdot x_n+N\cdot x_0$$ and their associated sheaves (=line bundles) $\mathcal O(D_1), \mathcal O(D_2)$.
They give rise to a short exact sequence of sheaves $$0\to \mathcal O(D_1)\to \mathcal O(D_2)\to \mathcal Q\to 0$$ where the quotient sheaf $\mathcal Q$ is a finite sum of sky-scraper sheaves.
Taking cohomology we get $$...\to H^0(X, D_2)\to H^0(X,\mathcal Q) \to H^1(X,D_1)\to ...$$
Now $H^1(X,D_1)$ is dual to $H^0(X,K_X(2\cdot x_1+...+2\cdot x_n-N\cdot x_0))$ (by Serre duality) and is thus zero for $N\gt 2n+2g-2$.
This choice of $N$ implies that the morphism $\gamma : H^0(X, D_2)\to H^0(X,\mathcal Q)$ is surjective.

And what has this got to do with your problem? It solves it!
Indeed, if you choose a coordinate $z_i$ near $x_i$, the stalk $\mathcal Q_{x_i}$ is identified with the complex line $\mathbb C\cdot z_i$ and by choosing a section $s\in H^0(X, \mathcal O(D_2))$ mapping to a $\gamma(s)\in H^0(X,\mathcal Q)$ non-zero at every $x_i$ you obtain the required meromorphic function $s$: its only pole is at $x_0$ and it has zeros of order exactly $1$ at the $x_i$'s.

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Dear Georges, I'm trying to see the geometry in your answer. Correct me if I'm wrong, but are we saying something along the lines of: We embed $X$ in $\mathbb P^3$ (i.e. put an algebraic structure on $X$). Given a line $L \simeq \mathbb P^1$ in $\mathbb P^3$ we consider a linear projection $f : \mathbb P^3 \to L$. The claim is now that given a point $p$ on $X$, we can choose $L$ in such a way that $f|_X$ is non-ramified at $p$? –  Gunnar Magnusson Feb 7 '12 at 13:34
Dear @Gunnar, I might be misunderstanding you but in principle you must project from a point of projective space onto a plane. A variant of what you suggest would be to choose a plane $F$ with linear eqution $f$ not going through $p$ and a plane $G$ with linear eqution $g$ going through $p$ but transversal to $X$ at $p$ (that is not containing the tangent projective line $\mathbb T_p(X)$ to $X$ at $p$) . The required rational function would then be the restriction $(g/f)|X$. –  Georges Elencwajg Feb 7 '12 at 14:22

You can always get an $f$ of degree $\max(g+1,n)$, where $g$ is the genus and $n$ the number of points in $S$. I don't think it is a good estimate (the problem is that I construct $f$ so that every point of $S$ is a (simple) pole; there should be $f$'s with lower degrees if the values at points of $S$ are different)

Here is how to see it. Let $P_1,\dots,P_n$ be the points in $S$ and let $z_i$ be a local coordinate around $P_i$ (s.t. $z_i(P_i)=0$) on some disc $D_i\subset \Sigma$. Suppose the discs don't overlap. The function $1/z_i$ on $D_i\cap (\Sigma -P_i)$ gives a class $\alpha_i\in H^1(\Sigma,\mathcal{O})\cong\Omega^1(\Sigma)^*$ (use $D_i$ and $\Sigma-P_i$ as an open cover of $\Sigma$; $1/z_i$ is a function on $D_i\cap(\Sigma-P-i)$). We know that $\alpha_i\neq0$ (if $\alpha_i$ were a coboundary then $1/z_i=h-k$, where $h$ is holomorphic on $\Sigma-P_i$ and $k$ on $D_i$, but that means that $h$ is meromorphic on $\Sigma$ with a unique pole at $P_i$, which is impossible if $g>0$).

We can suppose $n>g$ (if not then add some more points to $S$). As $\dim H^1(\Sigma,\mathcal{O})=g$ and all $\alpha_i$'s are non-$0$, there are $c_1,\dots,c_n\in\mathbb{C}$, all non-$0$, such that $\sum c_i\alpha_i= 0\in H^1(\Sigma,\mathcal{O})$. If you write $\sum c_i\alpha_i$ as a coboundary for the open cover $\Sigma-S,D_1,D_2,\dots,D_n$ of $\Sigma$, the holomorphic function on $\Sigma-S$ is a meromorphic function on $\Sigma$ with a simple pole at every point of $S$, hence non-ramified at $S$, and its degree is the number of poles (i.e. $n$).

edit I changed my answer completely as it contained a mortal gap

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