8
$\begingroup$

A finite quotient Dedekind domain is a Dedekind domain $A$ such that $|A/I|$ is finite for every nonzero integral ideal $I$. Can it happen that, for some $d$, $|A/I|\leq d$ for infinitely many $I$? (Equivalently, infinitely many primes.)

I am wondering that how far a finite quotient Dedekind domain could be different from "global rings", i.e. rings of integers of number fields and function fields over finite fields. For example, there are such rings that have only finitely many maximal ideals but are not DVR, like $\mathbb Z$ localized at all integers coprime to 6. The example I am asking in this question is more or less the opposite: whether there is a finite quotient Dedekind domain with "too many" primes.

$\endgroup$
0

2 Answers 2

4
$\begingroup$
  • If $char(A)=0$ : $N(I) \le d$ implies $A/I$ is a quotient of $A/(d!)$, your assumption is that it is a finite ring with finitely many quotients thus only finitely many ideals with $N(I) \le d$

  • If $char(A) = p$ : look at the ideal $J_m = \sum_{a \in A} (a^{p^m}-a) A$,

    assume this time $I$ is a prime ideal,

    $N(I) \le p^k$ implies $A/I$ is a quotient of $A/J_{k!}$, again your assumption is that it is a finite ring with finitely many quotients thus only finitely many ideals with $N(I) \le p^k$

That is to say the next step is to ask what happens if $A/(d!)$ or $A/J_{k!}$ is not a finite ring

$\endgroup$
1
$\begingroup$

More generally one has :

Theorem (Samuel, 1971): Let $R$ be a Noetherian ring, and let $n$ be a positive integer. The set of ideals $I$ of $R$ such that $R/I$ is finite of cardinality $n$ is finite.

For a proof see e.g. Theorem 22.3 of my commutative algebra notes.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .