# set of all holomorphic functions is a integral domain

To show $H(G)=$ set of holomorphic functions on $G$ is a integral domain.

i would like to know whether my proof of $H(G)$ does not have divisors of zero correct or not?

to show if $fg\equiv 0 \implies f\equiv0$ or $g \equiv 0$

let $fg=0$ $\forall z\in G$ and $a \in G$ if $f(a) \neq 0 \implies f(z) \neq 0$ $\forall z \in B(a,R)$ for some $R$ this shows $g(z) = 0$ $\forall z \in B(a,R)$

$g^n(a)=0$ $\forall n \implies g(z)=0$ on $G$ .

• What is $G$? Presumably connected? Mar 5, 2013 at 13:53
• @GEdgar I was thinking the same, is it necessary that G be connected? Oct 29, 2017 at 17:27
• If not connected, take $f \equiv 0$ on part of the space, $f\equiv 1$ on the rest of the space, and $g=1-f$. Oct 29, 2017 at 19:30
• Where is this question from? Oct 2 at 17:59

The idea of the proof is right, but you should mention why the implication of $f(a) \neq 0 \implies f(z)\neq 0 \forall z \in B(a,R)$ for some $R$ holds.
As we have holomorphic functions you don't need to make the assumption $f(a)\neq 0$ it is enough to say $f\not \equiv 0$, as zeroes of holomorphic functions can't have an accumulation point, when the function is not constant zero.
And why from $g^n(a)=0$ the implication $g(z)=0$ on $G$ holds.
And you should mention somewhere that $\mathbb{C}$ is a field and hence $f(a)\cdot g(a)=0\implies f(a)=0 \vee g(a)=0$