I need to show ord$_p(fg)$=ord$_p(f)$+ord$_p(g)$ where $f,g$ are meromorphic functions on some Riemann Surface.
So I just need help clarifying how to set this up.
Could I start by:
$\space \space \space \space \space $ Let $f(z) = \sum_{m=-\infty}^\infty a_m(z-p)^m$, $g(z) = \sum_{n=-\infty}^\infty b_n(z-p)^n$
Then since order is the smallest $m$ or $n$ such that the coefficient is nonzero, am I allowed to say:
$\space \space \space \space \space $ From here we can denote ord$_p(f)$ by $m_0$ and ord$_p(g)$ by $n_0$
$\space \space \space \space \space $ Then we have $f(z)g(z) = \sum_{m=-\infty}^\infty \sum_{n=-\infty}^\infty a_mb_n(z-p)^{m+n}$ Then does ord of this $= a_0+b_0$
My indexing gymnastics is not up to par. I know if I do the RHS it is easier, we have
$\space \space \space \space \space $ ord$_p(f)$ $+$ ord$_p(g)$ = $m_0+n_0$
Am I on the right track?
Let $f,g$ be meromorphic functions ($\neq 0$) with orders $\operatorname{ord}_p(f) = m_0$ and $\operatorname{ord}_p(g) = n_0$ First note that $m_0, n_0\in \mathbb{Z}$ since meromorphic functions have at most poles as singularities. Then we can write $$ f(z) = (z-p)^{m_0} \sum_{n=0}^\infty a_n (z-p)^n , \quad g(z) = (z-p)^{n_0} \sum_{n=0}^\infty b_n (z-p)^n $$ where $a_0, b_0 \neq 0$. It follows that $$ f(z)g(z) = (z-p)^{m_0+n_0}\left(\sum_{n=0}^\infty a_n (z-p)^n \right)\left(\sum_{n=0}^\infty b_n (z-p)^n \right) $$ i.e.$f(z)g(z) = (z-p)^{m_0+n_0}h(z)$ where $h$ is holomorphic and $h(p) \neq 0$. Hence $\operatorname{ord}_p(fg) = m_0+n_0$.