ideal of holomorphic function that vanish on all but finite number of integers is not finite generated

Question

Let $\mathcal{O}$ be the ring of holomorphic function on $\Bbb{C}^1$. I want to show that the ideal of functions that vanish on all but a finite number of integers is not finitely generated, therefore the ring above is non-Noetherian.

My attempt if it's generated by $f_1,...,f_n$, my feeling is that I need to use something like the identity principle in some place, sorry I don't have idea.

I believe this is done in Berenstein & Gay's Complex Variables in Chapter 3. — mathematics2x2life, Feb 25 '23 at 05:03
any finitely generated ideal in $\mathcal{O}$ is principal (not hard to prove by induction using Weierstrass theorem about the existence of functions that vanish at precisely given discrete set and some manipulations with $\bar \partial$ — Conrad, Feb 25 '23 at 05:21

score 5 · Accepted Answer · answered Feb 25 '23 at 06:16

5

Let $I$ denote that ideal and say $f_1, \ldots, f_n$ generate $I$. Note that by definition, there exists an integer $N$ large enough such that each $f_i$ vanishes on $N$. But then every function in $I$ must vanish on $N$. However, you can cook up a function which vanishes on $\Bbb Z$ and not on $N$, which would finish the job.

answered Feb 25 '23 at 06:16

Aryaman Maithani

15,771

2

Nice answer! Can we recommend $\frac{\sin \pi z}{z-N}$ for the example? – orangeskid Feb 25 '23 at 06:43
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Nice answer and nice comment thank you ! – yi li Feb 25 '23 at 07:09
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@orangeskid: That's exactly what I had in mind! :) – Aryaman Maithani Feb 25 '23 at 20:06

ideal of holomorphic function that vanish on all but finite number of integers is not finite generated

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