Principal ideal ring analytic functions

Question

Could someone sketch a proof and explain me in words, why the set of analytic functions on $\mathbb{C}$ does not form form a principal ideal ring?

Thank you!

Answer 1

While I am not very familiar with analytic functions themselves, I would imagine that one could construct an infinite ascending chain of ideals, showing that the ring is not Noetherian (and hence, definitely not principal.)

So say for example $I_j=\{f \mid \forall n\in\mathbb{N}, n\geq j\ \ f(n)=0\}$ would probably constitute an infinite ascending chain of ideals (as it does in the ring of continuous functions.)

Added: Experts have added useful examples in the comments to show that there is such an ascending sequence:

Ragib Zaman: $f_j(x)=\prod_{k=j+1}^{\infty} (1-z^2/k^2).$

Georges Elencwajg: $f_j(z)=\frac {sin(\pi z)}{z(z-1)...(z-j) }$

Thank you for contributing these: I too have learned something, now :)