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Say $f_1,...,f_k$ are given holomorphic functions, on, say, $\mathbb{C}$ (I'm working in a noncompact Riemann surface but it doesn't matter, $\mathbb{C}$ will do). Let $f$ be a fixed holomorphic function such that the order of $f$ at any point $z \in \mathbb{C}$ is equal to the minimum of the orders of $f_1,..f_n$, in other words, $f_i/f$ is always holomorphic for any $i = 1, \dots, n$. Now let $g$ be any holomorphic function. I want to show that there exist holomorphic functions $g_1,...,g_n$ such that $g_1f_1 + \cdots + g_nf_n = gf$. That's all I want, and I would love a bit of guidance please.

This is part of a bigger problem I'm working on and it seems true but for some reason I'm not seeing it...? Is the argument simple?

Acton
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1 Answers1

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Since the order of $f$ is the minimum of the $f_i$'s, you can write $$ f=f_iu $$ with $u$ a unit in $\mathcal{O}_z$ for some $i$. But then the ideal $(f)\subset(f_1,f_2,...,f_n)$. Hence for any holomorphic function $g$, there are $g_j$'s holomorphic so that near $z$ $$gf=g_1f_1+g_2f_2+...g_nf_n\;.$$

  • I don't understand. The OP's question requires $g$ to be aribitrary. –  May 09 '18 at 15:24
  • Isn't it possible that the $i$ changes as the point changes. In that case $f/f_{i}$ need not be a unit. –  May 09 '18 at 15:35
  • @harmonicuser Yes, you're right. I misread the question -- this answer only shows that this is true near the point $z$, of course. –  May 09 '18 at 15:43
  • Sorry, what do you mean by a unit in $\mathcal{O}_z$? Also, is that your notation for the stalk at $z$ of the sheaf of holomorphic functions or something? – Acton May 09 '18 at 15:45
  • I do want this to hold globally :) – Acton May 09 '18 at 15:46
  • @ActonGrey Yep. That's the stalk. As harmonicuser pointed out, this doesn't work globally. –  May 09 '18 at 15:46