What is the spectrum of the ring of entire functions? 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.
 A: 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 $S=\{0\}$.)
A: 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. 
A: 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 $g^n\in (f)$ $\iff$ $f$ ~ $g$.
A: This is probably helpful:
http://www.math.hmc.edu/~henriksen/publications/1951_Henriksen_On_the_ideal_structure_of_the_ring_of_entire_functions.pdf
http://scholarship.claremont.edu/cgi/viewcontent.cgi?article=1373&context=hmc_fac_pub
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?
A: 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".
