# Why is $H^0(C,\mathcal O(D))$ a vector space?

Given a divisor $D$ on a smooth curve, one can define the sheaf $\mathcal O(D)$ by the prescription $\Gamma(U,\mathcal O(D) :=$ $\{$meromorphic functions on $U$ that satisfy $(f) + D \ge 0\}$. Then, one can define a line bundle as $\mathcal L(D) = H^0(C,\mathcal O(D))$.

So, line bundles are supposed to be rank one $\mathcal O_X$-modules. But how does this definition above guarantee that you can add two global sections (that's what I take the $H^0$ to mean) of $\mathcal O(D)$ and still get a section in $\mathcal O(D)$?

It seems to me that one can get no control over the divisor $(f + g)$, for $f,G \in \mathcal O(D)$, other than that its degree is zero (since the sum of two meromorphic functions is meromorphic).

## 1 Answer

Let $D_+$, $D_-$ denote respectively the positive and negative parts of a divisor.

Just notice that $f + g$ has poles exactly at the points either $f$ or $g$ has poles and with order equal to the maximum of the order of the two. Hence $(f + g)_-$ plus $D_+$ is still $\ge 0$. On the other hand, $D_-$ dictates where $f$ and $g$ must have zeros (and to what order). Since each one of $f$ and $g$ have zeros at those points to at least the right order, so will $(f+g)_+ + D_- \ge 0$.

• This is a nice explanation, although I don't think I believe your claim from the question that you don't see how to get control over $(f+g)$, since you answered yourself immediately! – Kevin Carlson Nov 19 '13 at 2:29