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!

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    $\begingroup$ A very similar question were asked a few days ago. Specifically, any PID is an UFD, and $\mathcal{O}(\mathbb{C})$ is not an UFD, as shown in the answer to this question: math.stackexchange.com/questions/153877/… $\endgroup$ – Fredrik Meyer Jun 7 '12 at 14:11

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 :)

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    $\begingroup$ Your argument is correct, but it is not trivial that the $I_j$'s form a strictly increasing sequence. A possible proof is through Weierstrass products. $\endgroup$ – Georges Elencwajg Jun 7 '12 at 14:52
  • $\begingroup$ @GeorgesElencwajg isn't it as simple as producing an analytic function in one that is not in the next? It sounds simple anyway... but then again I have already admitted not knowing analytic functions well :) $\endgroup$ – rschwieb Jun 7 '12 at 14:56
  • $\begingroup$ Dear rschwieb, what I wrote means that there exists $g\in I_{j+1}\setminus I_j$. (So it would be a function in one that is not in the preceding one !) I agree that it is simple, but still an argument is required. $\endgroup$ – Georges Elencwajg Jun 7 '12 at 15:15
  • $\begingroup$ An entire function which has roots for all $i>j$ but not at $j$ is $\prod_{k=j+1}^{\infty} (1-z^2/k^2).$ $\endgroup$ – Ragib Zaman Jun 7 '12 at 15:24
  • $\begingroup$ @GeorgesElencwajg I was betting it would be trivial for anyone who knew about analytic functions. Is Ragib's suggestion enough? $\endgroup$ – rschwieb Jun 7 '12 at 15:32

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