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Suppose $D$ is a region (connected open set) in complex plane, and $f$ is a meromorphic function on $D$.

Question: Does there always exist two holomorphic function $g$ and $h$ such that $f=\frac{g}{h}$? When $D$ is the whole complex plane I know it is true thank to Weierstrass infinite multiply formula, and I don't know whether it is true for any region $D$.

Any comments are welcome.

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Yes, the Weierstrass product theorem holds for arbitrary regions $D$ in $\mathbb{C}$ (and is not too hard to deduce from the Weierstrass theorem in $\mathbb{C}$, and this implies that every meromorphic function on $D$ can be written as a fraction $f = g/h$ with $g$ and $h$ holomorphic in $g$. Details can be found e.g. in Remmert, Classical Topics in Complex Function Theory, see chapter 4. – t.b. Apr 18 '12 at 4:41
up vote 9 down vote accepted

1) Yes, every meromorphic function in $f\in \mathcal O(D)$ can be written $f=\frac{g}{h}$ with $g,h\in \mathcal O(D)$.
Proof: The divisor associated to $f$ has a canonical minimal decomposition $div(f)=D^+-D^-$ with $D^+-D^-\geq 0$ with $D^+, D^-$ effective (=positive)
[Explicitly: $D^+(z) =max(0, ord_zf), \; D^-(z) =max(0, -ord_zf)$ ]
The fundamental result (Rudin, Real and Complex Analysis, Theorem 15.11 page 295) is that there exists a function $h\in \mathcal O(D)$ with $div(h)=D^-$. If we then put $g=fh\in \mathcal O(D)$, we have our required decomposition as quotient $f=\frac{g}{h}$.

2) As a consequence, note the pleasant algebraic result $\mathcal M(D)=Frac(\mathcal O(D))$ characterizing the fraction field of the domain $\mathcal O(D)$.

3) The decomposition $f=\frac{g}{h}$ is also valid on any non-compact Riemann surface $X$ and again we have $\mathcal M(X)=Frac(\mathcal O(X))$.
The exact same proof as above works since any divisor $E\in Div(X)$ is the divisor of some meromorphic function $ m\in \mathcal M(X)^*$, that is $E=div(m)$: that result is proved in Theorem 26.5 of Forster's Lectures on Riemann Surfaces, the ultimate reference on the subject and a book I can't praise highly enough.

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How to prove the fact that every holomorphic line bundle on a non-compact Riemann surface $X$ is trivial? – Yuchen Liu Apr 18 '12 at 7:05
I have added an explanation and a reference for the generalization to non-compact Riemann surfaces in part 3) of my answer. – Georges Elencwajg Apr 18 '12 at 10:01
Thanks for your reply! – Yuchen Liu Apr 18 '12 at 11:20

Yes; $M(D)$ is the quotient field of $H(D)$. Conway, "Functions of one complex variable", Corollary 5.20.

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