Dimension of $\mathcal{O}_X(X-\{P\})$.

Question

Let $X$ be a smooth projective and irreducible curve and $P\in X$. I am asked to show that the dimension (as a $k$-vector space where $k$ is a algebraically closed field) of $\mathcal{O}_X(X-P)$ is infinite using the Riemann-Roch theorem. A curve over $k$ is defined as a quasi-projective variety all of whose irreducible components are of dimension one and the Riemann-Roch theorem states that $$\dim_kH^0(X,D)-\dim_kH^1(X,D)=1-g+\deg(D).$$ Where if we write $X=U\cup V$ for $U,V$ affine opens, $W:=U\cap V$, $D$ a divisor on $X$, and a map $\varphi:\mathcal{O}_X(D)(U)\oplus\mathcal{O}_X(D)(V)\to\mathcal{O}_X(D)(W)$ given by $\varphi(f,g)=f|_W-g|_W$. Then $H^0(X,D):=\text{ker}(\varphi)=\mathcal{O}_X(D)(X)$, $H^1(X,D):=\text{coker}(\varphi)$ and $\dim_kH^1(X,0)=g$.

I do not know where to start with this question. I suspect it has something to do with the canonical divisor since I probably have to recover $\mathcal{O}_X(X-P)$ using some divisor but even then I am completely lost.

Answer 1

By Riemann-Roch it suffices to show that for every $n$ the restriction map $H^0(X, \mathcal{O}(nP)) \to H^0(X \setminus \{P\}, \mathcal{O}_{X \setminus \{P\}})$ is injective.

This is well defined: rational functions in $H^0(X, \mathcal{O}(nP))$ only have poles at $P$ so their restriction to $X \setminus \{P\}$ are regular. Injectivity is just the fact that if a rational function vanishes on a nonempty open set, it vanishes everywhere.

Answer 2

I'm not really sure if this answers your question because I'm also confused, as in the comments, so let me treat this as an extended comment:

A projective curve minus any finite number of points is affine. To see this, let the points $p_1, ..., p_n$ be enumerated and consider the divisor $D = \sum p_i$. This divisor has positive degree, so some multiple of it is very ample, say, $rD$, and thus you get a map into projective space $\varphi: C \to \mathbb{P}^N$ for some $N$ and some $\varphi$. Furthermore, the points $p_i$ are all exactly the intersection of $\varphi(C)$ with a hyperplane $H$, since very ample implies that the pullback of $\mathcal{O}_{\mathbb{P}^N}(1)$ is exactly $\mathcal{O}_C(D)$. Thus, removing this hyperplane from $\mathbb{P}^N$, we obtain an affine space, and $\varphi(C) - H = \varphi(C) - \{ p_i \}$ is an affine variety.

Now we can abuse notation and write $\varphi(C - \{ p_i \}) = C - \{p_i\}$, identifying the curve with its image under this map. In this case $H^0(C - \{p_i\} , \mathcal{O})$ can now be identified with the coordinate ring of the affine variety. As a vector space over $k$, this is an infinite dimensional space. This follows because $C - \{ p_i \}$ is a curve, so the transcendence degree of the coordinate ring over $k$ is $1$. Therefore, there is a trancendental element, and powers of this element are a basis as a $k$-vector space.