4
$\begingroup$

In my answer to Do there exist a non-PIR in which every countably generated prime ideal is principal?, I gave an example of an integral domain in which no nonzero prime ideal is countably generated. However, my example is rather complicated, and so I have the following question:

What are some other (hopefully simpler) examples of integral domains in which no nonzero prime ideal is countably generated (besides the trivial examples of fields)?

$\endgroup$
  • $\begingroup$ what about rings of real analytic functions? $\endgroup$ – vidyarthi Nov 27 '16 at 23:13
  • $\begingroup$ If you take an uncountably generated minimal prime and localize at it, can anything be said about the minimal generating set of the maximal ideal in the localization? Is it hard to control that? I don't know enough about the topic. $\endgroup$ – rschwieb Nov 28 '16 at 3:14
  • $\begingroup$ @vidyarthi: Well, those have some countably generated primes (the maximal ideals corresponding to points, which are principal). But it seems plausible that if you localize at a maximal ideal which does not correspond to a point you will have no countably generated nonzero primes. $\endgroup$ – Eric Wofsey Nov 28 '16 at 23:01
  • $\begingroup$ @rschwieb: Assuming you mean minimal nonzero prime, it seems plausible that would be a source of examples. But it's not obvious to me how to even find a domain with a minimal nonzero prime that is not countably generated. $\endgroup$ – Eric Wofsey Nov 28 '16 at 23:04
  • $\begingroup$ I think rings of functions and localizations at minimal primes are almost never domains unless they're fields. I found this reference for domains with every nonzero prime of infinite height http://www.tandfonline.com/doi/abs/10.1080/00927879808826338#preview and was hoping something like this would work $\endgroup$ – Jay Nov 29 '16 at 0:11
1
$\begingroup$

One natural example of such a ring is the ring $R$ of meromorphic functions on $\mathbb{C}$ (or any connected open subset of $\mathbb{C}$) which have only finitely many poles. This ring is clearly a domain and not a field. For $f\in R$, we will write $Z(f)$ for the set of zeroes of $f$. The key fact we use is that if $f,g\in R$, then there exists $h\in (f,g)$ such that $Z(h)=Z(f)\cap Z(g)$ (see A problem about generalization of Bezout equation to entire functions).

Now suppose $I\subset R$ is a countably generated nonzero proper ideal; we will prove $I$ is not prime. To prove this, let $f\in I$ be nonzero. Let $C$ be a countable set of generators of $I$. For each finite subset $F\subseteq C$, let $h_F\in I$ be such that $Z(h_F)=Z(f)\cap\bigcap_{g\in F}Z(g)$. Since $I$ is a proper ideal, $h_F$ is not a unit, so $Z(h_F)$ is infinite. Since $f$ is nonzero, $Z(f)$ is countably infinite. Now since there are only countably many finite subsets $F\subseteq C$, we can diagonalize to partition $Z(f)$ into two subsets $S$ and $T$ such that $S\cap Z(h_F)$ and $T\cap Z(h_F)$ are both infinite for each $F$. Let $f_0\in R$ be a function whose zero set is $S$ with zeroes of the same multiplicity as $f$ and let $f_1$ be a function whose zero set is $T$ with zeroes of the same multiplicity as $f$. Then $f_0f_1$ is divisible by $f$, so $f_0f_1\in I$.

However, I claim that neither $f_0$ nor $f_1$ is in $I$, so $I$ is not prime. Indeed, if $f_0$ were in $I$, it would be in the ideal generated by some finite subset $F\subseteq C$. But then $f_0$ would be $0$ at all but finitely many points of $\bigcap_{g\in F}Z(g)$ (the finitely many exceptions coming from the fact that elements of $R$ can have finitely many poles to cancel the zeroes of the elements of $F$). Since $T\cap Z(h_F)\subseteq T\cap \bigcap_{g\in F}Z(g)$ is infinite and $f_0$ has no zeroes on $T$, this is a contradiction. Thus $f_0\not\in I$, and by a similar argument $f_1\not\in I$ as well.

(More conceptually, what is going on in this proof is that if $I$ were prime, then the collection of zero sets of its elements would give a nonprincipal ultrafilter on $Z(f)$ for any nonzero $f\in I$. If $I$ were countably generated, this ultrafilter would be countably generated, but a nonprincipal ultrafilter cannot be countably generated. For some related discussion and another neat place this ring pops up as a counterexample, see my answer to Is the axiom of choice necessary to prove that closed points in the Zariski topology are maximal ideals?.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.