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Let $\mathcal{O}(\mathbb{C})$ be the ring of entire functions, that is, those functions $f : \mathbb{C} \to \mathbb{C}$ which are holomorphic for all $z \in \mathbb{C}.$ For each $z_0 \in \mathbb{C}$.

Are there any other maximal ideals in $\mathcal{O}(\mathbb{C})$ besides these obvious ones?

If anyone can give a concise description of $\text{Spec }\mathcal{O}(\mathbb{C})$, that would be extremely helpful. I'm trying to understand wether or not knowing the closed subset $V(f)$ of $\text{Spec }\mathcal{O}(\mathbb{C})$ of ideals containing $f$ gives you more information about $f$ than simply knowing the vanishing set of $f$ in the classical sense.

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5 Answers 5

up vote 8 down vote accepted

Here's a more analytic description of exactly what knowing $V(f)$ tells you. Let us say $f$ ~ $g$ if their vanishing sets are the same, and moreover there exist positive constants $c,C$ such that $c\cdot ord_g(z)< ord_f(z)< C\cdot ord_g(z)$ as $z$ ranges over the vanishing set. Then knowing $V(f)$ is the same is as knowing the equivalence class of $f$. Indeed,

$V(f) = V(g)$ $\iff$ there exists $n$ such that $f^n\in (g)$ and $g^n\in (f)$ $\iff$ $f$ ~ $g$.

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By $V(f)$ do you mean the set of all zeros of $f$ for the set of all prime ideals containing $f$? –  Jamie Weigandt Jul 31 '10 at 4:37
The first equivalence indicates that $V(f)$ is meant to be the closed subset of the prime spectrum. –  Martin Brandenburg Jul 31 '10 at 6:42
Right, I interpreted Spec to mean prime spectrum. –  Kevin Ventullo Aug 2 '10 at 7:06
Having thought about it, this is an incredibly useful criterion for what I've been thinking about. Thanks Kevin! –  Jamie Weigandt Aug 3 '10 at 18:28

Pick an ultrafilter on the integers. The functions whose set of zeros contain some element of this ultrafilter form a maximal ideal $I$. (Proof: if $f$ is not in $I$ then it does not vanish on any element of the ultrafilter, so it is nonzero at all points of some element $S$ of the ultrafilter, so we can find some holomorphic $g$ so that $fg$ is 1 on $S$, so that $1-fg$ is 0 on $S$ and therefore in the ideal $I$. So $1$ is in the ideal generated by $f$ and $I$, so $I$ is maximal.)

Ultrafilters are somewhat weird, so there is no easy description of all maximal ideals.

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Might there be a reasonable description of the maximal ideals in terms of the Stone-Cech compactification of C? –  Qiaochu Yuan Jul 30 '10 at 23:16

An addendum to Richard Borcherds's answer: Of course, instead of the integers, you could use any closed, discrete subset $A$ of $\mathbb C$. By taking different such $A$'s and different ultrafilters $U$ on them, the ideals $I$ that you get are all of the maximal ideals in the ring of entire functions.

More generally, with any $A$ and $U$ as above, suppose $S$ is a nonempty, proper initial segment of the ultrapower $\mathbb N^A/U$ that is closed under addition. Let $P$ be the collection those entire functions $f$ such that either $f$ is identically 0 or the function $v_f:A\to \mathbb N$ sending $a$ to the order-of-vanishing of $f$ at $a$ represents, in the ultrapower, an element not in $S$. Then $P$ is a prime ideal in the ring of entire functions, and every prime ideal arises in this way. (The maximal ideals correspond to the case where $S=\{0\}$.)

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This is probably helpful:



By the way, what where you thinking about? I have been also thinking abouy some related stuf, what about $Spec \; \mathcal{O}_0$, the spectrum of the ring of convergent power series? Does it have any geometric interpretetation?

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I think was pawing around for Nevanlinna's five value theorem. –  Jamie Weigandt May 22 '12 at 17:59

I beleive the set of all enitre functions vanishing on all but finitely many integers is a maximal ideal.

edit: by the comments below - not maximal, but prime, and not contained in any of the "close points".

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yeah, you are right. But anyway, this proves that there are other maximal ideals, since it is not contained in any off the "obvious ones". –  the L Jul 30 '10 at 22:17
Wait... nevermind, that's silly, its the same ideal. Certainly is lives in some maximal ideal other than the obvious ones... –  Jamie Weigandt Jul 30 '10 at 22:19
well, you can just take the set of all functions vanishing on all but finitely many even integers - stricly containing it. –  the L Jul 30 '10 at 22:22

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