# How to prove $\mathcal{l}(D+P) \leq \mathcal l{(D)} + 1$

Let $X$ be an irreducible curve, and define $\mathcal{L}(D)$ as usual for $D \in \mathrm{Div}(X)$. Define $l(D) = \mathrm{dim} \ \mathcal{L}(D)$. I'd like to show that for any divisor $D$ and point $P$, $\mathcal{l}(D+P) \leq \mathcal l{(D)} + 1$.

Say $D = \sum n_i P_i$. I can prove this provided $P$ is not any of the $P_i$, by considering the map $\lambda : \mathcal{L}(D) \to k$, $f \mapsto f(P)$. This map has kernel $\mathcal{L}(D-P)$, and rank-nullity gives the result.

But if $P$ is one of the $P_i$, say $P=P_j$ then I'm struggling. Any help would be appreciated.

Thanks

-

Let $X$ be a projective smooth connected curve over $\mathbb{C}$, let $D \in \mathrm{Div}(X)$ be a divisor and let $p$ be a point of $X$. Consider the exact sequence of sheaves over $X$ $$0 \to \mathcal{O}_X(-p) \to \mathcal{O}_X \to \mathcal{O}_p \to 0,$$ where $\mathcal{O}_p$ is the skyscraper sheaf that is $\mathbb{C}$ on the point $p$. Tensor with the invertible sheaf $\mathcal{O}_X(D)$ and you get another exact sequence of sheaves over $X$: $$0 \to \mathcal{O}_X(D - p) \to \mathcal{O}_X(D) \to \mathcal{O}_p \to 0.$$ Taking global sections, you get an exact sequence of $\mathbb{C}$-vector spaces $$0 \to \mathcal{L}(D-p) \to \mathcal{L}(D) \to \mathbb{C}$$ that implies $1 \geq l(D) - l(D-p)$, as you want.

The only difficult point is to understand the (not canonical) isomorphism $\mathcal{O}_p \otimes_{\mathcal{O}_X} \mathcal{O}_X(D) \simeq \mathcal{O}_p$ in terms of rational functions.

EDIT. Now I am giving a proof without mentioning sheaves, as required by algeom. This is substantially a translation of what I wrote above. Let $K$ be the field of rational functions of $X$. It is well known that Weil divisors are locally principal (this follows from the fact that $X$ is locally factorial), hence there exist an open neighborhood $U \subseteq X$ of the point $p$ and a rational function $\phi \in K^*$ such that $D \vert_U = \mathrm{div}_U(\phi)$ as divisors over $U$. Now consider the the map $\alpha \colon \mathcal{L}(D) \to \mathbb{C}$ defined by $$\alpha \colon f \mapsto (f \phi)(p), \quad \forall f \in \mathcal{L}(D).$$ This is well posed because the rational function $f \phi$ is regular in $p$. You should be able to prove that $\mathcal{L}(D-p) = \ker \alpha$. Conclude applying rank-nullity theorem to $\alpha$.

-
Now's the point when I realise I should have mentioned I don't know what a sheaf is. If we assume $X$ is a smooth projective curve, is there a less general proof? –  algeom May 18 '12 at 14:48

Here is an elementary formulation, without sheaves.

Let $t\in Rat(X)$ be a uniformizing parameter at $P$ (that is, $t$ vanishes with order $1$ at $P$) and let $n_P\in \mathbb Z$ be the coefficient of $D=\sum n_QQ$ at $P$.

You then have en evaluation map $$\lambda: \mathcal L(D+P)\to k:f\mapsto (t^{n_P +1}\cdot f)(P)$$ and you can conclude with the reank-nullity theorem or in more sophisticated terminology with the exact sequence of $k$-vector spaces
$$0\to \mathcal L(D)\to \mathcal L(D+P)\stackrel {\lambda}{\to} k$$

-