Talk:Ideal class group
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Great article, thanks! One question:
- Do the ideal classes form a group for every integral domain, or just for Dedekind domains?
AxelBoldt 01:34 Nov 25, 2002 (UTC)
It depends whether you are including fractional ideals. If not, then no, never, as you have pointed out: the ideal group of a Dedekind domain is a free monoid with the prime ideals as generators. If we include fractional ideals, then it turns out that (every fractional ideal is invertible) implies (the ring is Dedekind). However, the relation on nonzero ideals is still an equivalence relation for integral domains that aren't Dedekind. I'll try to make this clear in the article.
This sentence
- the ideal group of a Dedekind domain is a free monoid with the prime ideals as generators
seems to contradict this passage from the article:
- If R is a ring of algebraic integers, or more generally a Dedekind domain, the multiplication defined above turns the set of ideal classes into an abelian group, the ideal class group of R.
I believe this latter sentence comes from me, and I would appreciate if you could correct it. So it is not possible to define the ideal class group without recourse to fractional ideals? AxelBoldt 03:56 Nov 27, 2002 (UTC) AxelBoldt 03:56 Nov 27, 2002 (UTC)
Oops! I misread your question. The ideals never form a group that I know of (at least in non-pathological cases (</handwaving>)). The ideals are a free monoid on the prime ideals exactly when the ring is Dedekind exactly when the fractional ideals form a group. The first sentence you have above should read: "The nonzero ideals of a Dedekind domain form a free monoid with the nonzero prime ideals as generators." But I have to think about whether the ideal classes always form a group. I'm not sure anymore. The question is, given an ideal class [I], can we find a class [J] such that the product class is the principal ideals? I'll need a day or so to think it over.
I think I have resolved this issue in the article, but there may be circularity in the article still: i'd appreciate a once-over! So here's the beef in plain old hand-waving english: the integral (i.e., honest) ideals are just a monoid, because they don't have inverses under ideal multiplication. But you defined multiplication of ideal classes in exactly the right way - to multiply two classes pick representatives, multiply them, and take the class of their product to be the product of the classes. The ideal classes still form a group, with R = (1) as the identity. The problem now is to show that a class is invertible. I don't know any good way to do it without using fractional ideals, in which case we might as well just represent the whole thing as a quotient group. It seems to be a matter (in the literature at least) of how dirty you want to get your hands. Most of the modern writers use fractional ideals right from the start: it seems to be the "right" way to look at things. But Marcus, for example, never mentions them that I remember (haven't looked at his book in a while though), and you can't even read that book (Number Fields) without getting your hands absolutely filthy.
I tried to find all the places where I forgot to say "nonzero" but I probably missed some. But the zero ideal isn't really an ideal anyway. ;)
Here we go: To show the ideal classes form a group in a Dedekind domain, you have to show that given any ideal I there is an ideal L such that IL is principal. This relies on the fact that in Dedekind domains, a containment of ideals I ⊆ J is equivalent to a divisibility relation J | I. So in general the ideal classes don't form a group in any natural way. So I've changed the example involving the infinite class number accordingly. Any other holes?
K0 - I think there is a point here. In general this is generated by projective modules of any rank, so that there is
- K0(R) → Z
given by the rank; and the ideal class group is actually the kernel of this?
Charles Matthews 08:30, 16 Oct 2004 (UTC)
