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I was reminded of this problem by a recent question of Regenbogen. It is not too difficult to prove that since every number field has finite class group, that every ideal in the ring of algebraic integers is either principal or infinitely generated. I've been unable to prove that these statements are actually equivalent, and this is what I'd be interested to know.

In particular, if they are equivalent, then this may provide a different way of proving finiteness of class number.

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    $\begingroup$ Indeed, aren't all ideals two generator? Take an ideal I and choose ideals J and K so gcd(J,K) = (1) for which IJ=(a) and IK=(b) are principal. Then I=gcd(IJ, IK) = gcd((a), (b)) = (a,b). There are lots of missing details, e.g. invertibility, but I think this is the basic idea. $\endgroup$
    – Matthew Stover
    Mar 22, 2010 at 16:32
  • $\begingroup$ BTW, the `indeed' referred to a now-deleted comment. $\endgroup$
    – Matthew Stover
    Mar 22, 2010 at 16:34
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    $\begingroup$ Any Dedekind domain is Noetherian, so all its ideals are finitely generated. In fact, any ideal $I$ can be generated by two elements, and the first element can be chosen as any non-zero $a\in I$. This is a straightforward Chinese remainder theorem argument using unique factorization of ideals. $\endgroup$
    – Keenan Kidwell
    Mar 22, 2010 at 16:41

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In the statement of the question you meant "every ideal is finitely generated".

In any event, these two types of finiteness (of ideals as modules and of the ideal class group) are not at all equivalent. For example, you can replace integral closures of $\mathbf Z$ in finite extensions of the rationals with integral closures of $F[x]$ in finite extensions of $F(x)$, where $F$ is a field, and thus speak about ideal class groups in this other setting. There the ideal class groups can be infinite, even though the ideals are still finitely generated. To take a concrete example, consider the integral closure of $\mathbf C[x]$ in $\mathbf C(x,y)$ where $y^2 = x^3 - x$. That integral closure is a Dedekind domain and its class group is infinite; in fact the ideal class group is isomorphic to the group of complex points on the elliptic curve $y^2 = x^3 - x$, which as a group looks like $\mathbf C/\mathbf Z[i]$.

Finiteness of class groups is somewhat special (not unique to it, but still special) to the setting of rings of algebraic integers and not valid in general Dedekind domains. In a general Dedekind domain, any ideal has at most 2 generators and this is unrelated to the size of the class group.

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    $\begingroup$ Is there a general property that implies finiteness of class group (or perhaps some nice equivalent conditions)? $\endgroup$
    – Harry Gindi
    Mar 22, 2010 at 17:15
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    $\begingroup$ A "reason" integral closures of $\mathbf Z$ and $\mathbf F_q[x]$ have finite class numbers is the finiteness of the residue rings of $\mathbf Z$ and $\mathbf F_q[x]$, which (through a Jordan-Holder filtration) basically comes down to the fact that $\mathbf Z$ and $\mathbf F_q[x]$ have finite residue fields. For a contrast, $\mathbf C[x]$ has infinite residue fields and its integral closures in extensions of $\mathbf C(x)$ typically have infinite class groups. $\endgroup$
    – KConrad
    Mar 22, 2010 at 18:19
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    $\begingroup$ @fpqc. Good question. I think the answer is no -- not only can the class group of a Dedekind domain be anything it wants to be (a theorem of Luther Claborn), but it can vary wildly even with the choice of Dedekind ring inside a given fraction field. As KConrad points out, to get a finiteness out, you have to put some finiteness in, e.g. of residue fields. I have been wondering for some time whether a Dedekind domain such that $R/I$ is finite for any nontrivial ideal $I$ must have finite class group. The proofs I have seen in the classical cases don't seem to work so abstractly... $\endgroup$
    – Pete L. Clark
    Mar 22, 2010 at 19:58
  • $\begingroup$ Sorry, haven't checked into this site in a couple years. I actually meant infinitely generated. For example the ideal that is generated by all the nth roots of 2 for all positive n. This is an ideal in the ring of all algebraic integers. In that ring, all ideals are principle or infinitely generated. $\endgroup$
    – Ben Weiss
    Jun 25, 2013 at 23:49
  • $\begingroup$ Oh, so the context was the ring of algebraic integers in an alg. closure of the rationals, not in a number field. An integral domain in which all finitely generated ideals are principal (not principle) is called a Bezout domain. I don't see why you think this Bezout property of the ring of all algebraic integers should imply number fields have finite class groups. And do you have a proof that the ring of all algebraic integers is a Bezout domain that doesn't rely on finiteness of class groups of number fields? I'll check for your answer in 3 years. :) $\endgroup$
    – KConrad
    Jun 26, 2013 at 0:37

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