In (Rick Miranda, page 129) Divisor on a Riemann Surface is defined as :
A divisor on Riemann surface $X$ is a function $D: X \rightarrow \mathbb{Z}$ whose support is a discrete subset of $X$.
Author states that The divisors on $X$ form a group under pointwise addition.
I'm having difficulty in verifying that divisors form a group under pointwise addition. I tried to check the above fact by taking $X = \mathbb{C}$ as Riemann Surface. Let $f, g$ be two divisors on $X$ with supports $\{\frac{1}{n}\}_{i=1}^\infty$ and $\{0\}$ respectively. f and g are defined as follows
$$f(x) = \begin{cases} n \text{ ,if } x = 1/n\\ 0 \text{ ,otherwise}\\
\end{cases}
$$
$$g(x) = \begin{cases} 1 \text{ , if } x = 0 \\ 0 \text{ ,otherwise}\end{cases}$$
Consider $f+g:X \rightarrow \mathbb{C}$, clearly $(f+g)(x) \ne 0 $ when $x \in \{\frac{1}{n}\}_{i=1}^\infty$ or $x = 0$, Hence support of $f+g$ has a limit. So, by definition $f+g$ is not a divisor.
Is there something that I don't understand?
Best Answer
The definition of divisor in the book is wrong; I believe this is in the errata. For a map $D:X\rightarrow\mathbb{Z}$, the support should be $\overline{\{p\in X:D(p)\neq0\}}$ (the closure of what Miranda calls the support) and a divisor is such a map with discrete support. So, with this definition, your $f$ isn't a divisor and there is no issue.