Does the ring of analytic functions have zero divisors? Question I have to show that the ring of complex analytic functions on open unit disk has no zero divisors. 
My attempt let suppose $fg≡0$ such that $f≢0$ and $g≢0$ on open unit disk $U$ then $f$ and $g$ have finitely many zeros on $U$ and so that $fg$ have finitely many zeros on $U$ and hence $fg≢0$. Hence we must have either $f≡0$ or $g≡0$. Hence given ring has no zero divisors. 
I am not that good in complex analysis. However i am familiar with abstract algebra. 
So please give details. Is my attempt correct? I didnt know, why $f$ and $g$ have finitely many zeros on $U$?  please elaborate this point too. 
Please help...
 A: It is possible to find analytic functions defined on the open unit disc having infinitely many zeros. For example, consider the Blaschke product. Another example, given in Conway's Functions of One Complex Variable, is $$f(z)=\cos\left(\frac{1+z}{1-z}\right),\;|z|\lt 1.$$ But if $f$ is a nonzero analytic function on the closed unit disc, it can only have finitely many zeros. To see this, note that every infinite subset of a compact set has a limit point.
As I mentioned in the comment section, the standard approach to this problem is to use the identity theorem and a solution can be found here.
A: Your idea that $f$ and $g$ have a finite numbers of $0$ is not true. However, there's still a proof based on the number of zeros, a more "algebraic"-flavored proof:
If $fg\equiv 0$; $\{f=0\}\cup \{g=0\}=\mathbb{D}$. However, if the two functions are not identically zero they can have at maximum a countable number of zeros, which gives a contradiction.
To prove that the set of zeros is countable, note that
$\mathbb{D}=\cup_{\mathbb{N}} \{|z|\le 1-\frac1n\}$, so if one of the two functions (lets say $f$) has more than a countable number of zeros, at least a countable number of them is contained in one of those sets $\{|z|\le 1-\frac1n\}$. Since they are compact, this implies that the set of zeros has a cluster point, which by the identity theorem implies that $f\equiv 0$, a contradiction.
Another interesting thing about the set of holomorphic functions on $\mathbb{D}$ is that, while it is an integral domain, it is not a UFD: the irreducible elements are the Blaschke factors, but not every element can be written as a finite product of them, as $$\sin\left(\frac{1}{z-1}\right)$$ shows.
This also implies that this set is not a Noetherian ring.
