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…
Best Answer
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.