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Let $( I )$ be an ideal in the ring $( R )$ of all holomorphic functions of a single complex variable on the complex plane. I am interested in understanding whether it is possible for $( I )$ to be non-finitely generated.

Has there been any prior research on this specific question, or are there known conditions under which an ideal in $( R )$ would necessarily be finitely generated or not? Any insights, references, or approaches would be greatly appreciated.

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    $\begingroup$ The ideal of the holomorphic functions that vanish on all but finitely many points in $\mathbb{Z}$ can not be finitely generated. I think every non-principal ultrafilter $\mathcal{F}$ on $\mathbb{Z}$ also gives such an example (the ideal is given by the holomorphic functions that vanish on some set in $\mathcal{F}$). $\endgroup$
    – Zerox
    Commented Oct 29 at 18:39
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    $\begingroup$ thanks for your insight. appreciate it! $\endgroup$
    – Haze
    Commented Oct 29 at 18:48
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    $\begingroup$ In my first example, any function in the ideal generated by a finite set of holomorphic functions that vanish on all but finitely many points in $\mathbb{Z}$ can only obtain nonzero values on the union $S$ (which is finite) of the nonzero-valued points of the generators. So this finite generated ideal will not contain the holomorphic functions vanishing on $S$ but nonzero elsewhere. $\endgroup$
    – Zerox
    Commented Oct 29 at 18:50
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    $\begingroup$ Concretely, the ideal that Zerox gives is generated by $\{ \sin \pi z, \tfrac{\sin \pi z}{z}, \tfrac{\sin \pi z}{(z-1)z(z+1)}, \tfrac{\sin \pi z}{(z-2)(z-1)z(z+1)(z+2)}, \cdots \}$. Each of these functions individually has a Weierstrass factorization, but that doesn't address whether the ideal is finitely generated. $\endgroup$ Commented Oct 29 at 20:58
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    $\begingroup$ This ring is not noetherian. As in David Speyer's comment, there is a sequence of ring elements $f_1,f_2,\dots$ such that $f_n$ is a multiple of $f_{n+1}$ but not conversely. The ideals $(f_n)$ form a strictly increasing sequence whose union cannot be finitely generated. Another example would be $sin\frac{z}{2^n}$. $\endgroup$ Commented Oct 29 at 23:11

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