Infinite dimensional space of regular functions over a punctured curve

Question

Let $X$ be a smooth projective curve, $P\in X$ and $U=X\setminus\{P\}$. I want to show that $\mathcal O_X(U)$ is an infinite dimensional vector space. A hint tells me to use Riemann-Roch.

I was thinking about considering the sheaf $\mathcal O_X|_U$ and using RR on this sheaf for some divisors to show that $\operatorname{dim}H^0(U,\mathcal O_X|_U)$ diverges in some way. However the most obvious divisor we have is $P$ which is not in $U$. So using RR on $X$ we get that $\dim H^0(X,\mathcal O_X(nP))=\dim H^1(X,\mathcal O_X(nP))+1-g(X)$ which will diverge when increasing $n$. Also $nP$ will be very ample, but I don't know how to proceed.

Any ideas ?

Answer 1

I will give a further hint rather than a full answer:

A global section of $\mathcal{O}_X(nP)$ is a function regular on $U = X\setminus \{P\}$ that has a pole of order $\le n$ at $P$. In particular, $\Gamma(X,\mathcal{O}_X(nP)) \subset \Gamma(U,\mathcal{O}_X|_U)$ for all $n\ge 0$.

This implies $\dim \Gamma(X,\mathcal{O}_X(nP))\le \dim \Gamma(U,\mathcal{O}_X|_U)$.

Now, apply Riemann-Roch to compute $\dim \Gamma(X,\mathcal{O}_X(nP))$ (after increasing $n$ a bit) and you will get the result.

Answer 2

Alternate approach: $U = X \setminus P$ is affine since $P$ is ample, so $U$ is isomorphic to $\operatorname{Spec} \mathcal{O}_X(U)$. If this ring were finite-dimensional, it would be of Krull dimension $0$. But of course it has Krull dimension $1$.