Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I premise that I do not know anything about fractionary ideals and class groups of Dedekind domains. But I know the theory of divisors on regular schemes, as treated in Hartshorne. What I would like to know is if there exist some geometric approaches to calculate the class group of a Dedekind domain.

In fact, for an algebraic variety $X$, the usual method consists of choosing an open subset $U$ of $X$ such that it is easy to prove that $Cl \ U = 0$ and then finding $1$-codimension points of $X \setminus U$. These points are generators of the group $Cl \ X$ and, with rational functions, one finds the relations among them.

But, what must I do to calculate the class group of $\mathbb{Z}[\sqrt{-5}]$, for example? To me it is not easy to choose an affine subset that is the spectrum of a UFD. (I know only $\mathbb{Z}$ and $\mathbb{Z}[i]$.)

share|improve this question

2 Answers 2

up vote 10 down vote accepted

I guess the analogue for number fields is the following. The Minkowski bound furnishes, for every number field $K$, an explicit number

$$M_K = \sqrt{|D_K|} \left( \frac{4}{\pi} \right)^s \frac{n^n}{n!}$$

such that $\text{Cl}(\mathcal{O}_K)$ is generated by prime ideals of norm at most $M_K$, where

In particular, it follows that one can write down an explicit list $S$ of primes (the primes less than or equal to $M_K$) such that $S^{-1} \mathcal{O}_K$ has trivial class group (recall that the localization of a Dedekind domain is a Dedekind domain). The inclusion $\text{Spec } S^{-1} \mathcal{O}_K \to \text{Spec } \mathcal{O}_K$ is the analogue of the inclusion of the open subset $U$ in the function field case. Then, as you say, finding rational functions is how one finds relations among the generators.

For $K = \mathbb{Q}(\sqrt{-5})$ we have $|D_K| = 20, s = 1, n = 2$, so the Minkowski bound is $\frac{4 \sqrt{5}}{\pi} < 3$. It follows that the class group is generated by ideals of norm at most $2$, so we only have to consider prime ideals above $2$. Since $x^2 + 5 \equiv (x + 1)^2 \bmod 2$, we have

$$(2) = P^2$$

for some prime ideal $P$ of norm $2$, hence the class group is either trivial or the cyclic group $C_2$ generated by $P$, and by inspection $\mathcal{O}_K$ does not have unique factorization so it is the latter. (A more generalizable way to end this argument is that $\mathcal{O}_K$ does not contain an element of norm $2$.)

share|improve this answer
    
I don't know what one can say about abstract Dedekind domains. It's known that every abelian group is the class group of a Dedekind domain... I would bet that computing the class group of a Dedekind domain is undecidable in general. –  Qiaochu Yuan Aug 8 '11 at 23:54
    
Very useful! But the Minkowski bound is an algebraic tool, although I have understood your geometric explanation. Thank you! –  Andrea Aug 9 '11 at 12:29

One point to bear in mind is that it is not typically possible to choose an open subset $U$ of Spec $A$, for a Dedekind domain $A$, so that $Cl U$ is trivial. Indeed, this is possible if and only if $Cl A$ is finitely generated. (The complement of $U$ in Spec $A$ will then be a finite set of generators of $Cl A$.)

For example, if $A$ is the affine ring of a smooth curve over $\mathbb C$ whose completion has positive genus, e.g. $A = \mathbb C[x,y]/(y^2 - x^3 - x)$, then $Cl A$ is infinitely generated, and so no such $U$ exists.

If $Cl A$ is finitely generated, then the problem of finding $U$ is equivalent (as you already note in your question) to the problem of finding generators for $Cl A$, and I think it is fairly standard to do this via height bounds/geometry of numbers, as in the approach via Minkowski that Qiaochu suggests.

For a related, but different, arithmetic context, one can consider the proof of the Mordell--Weil theorem and the problem of finding explicitly generators for the group of rational points on an elliptic curve or abelian varieties --- here one uses height arguments. For a geometric analogue, one can consider the Neron--Severi Theorem of the Base, about finite generation of the Neron--Severi group. In one of Lang's book, he presents a unified account of these various theorems.

Note that the connection with the Theorem of the Base is more than superficial: if $X$ is a smooth and projective variety, then Pic $X$ will be finitely generated if and only if its connected component is trivial (so that Pic $X$ coincides with the Neron--Severi group of $X$). Then the problem of computing a finite set of generators (which is the same as computing an open $U$ such that Pic $U$ is trivial) is the problem of making the Theorem of the Base effective for $X$. I don't know how to do this in practice, but I'd be surprised if it's easy to find $U$ just by inspection in general.

E.g. if $X$ is a K$3$ surface, then we know that Pic $X = NS(X)$, but the free rank of Pic $X$ can be as high as $20$. Is it possible just by inspection to find a $U$ inside $X$ with trivial Pic? I would guess that in practice one would use some kind of geometric analogue of a Minkowski bound to find generators for Pic $X$ --- i.e. some effective version of the Theorem of the Base --- and, having done this, one could then compute $U$ if one was so inclined. (In other words, computing Pic $X$ would come first, and computing $U$ would come second.)

I guess I can summarize this post by asking a question of my own: are there really that many $U$ for which it's easy to prove that $Cl U = 0$?

share|improve this answer
    
Great answer! Now I'm realizing that I have computed only Picard groups of rational varieties; so in this case it is easy to find an open subset $U$ for which $Cl U = 0$. –  Andrea Aug 24 '11 at 8:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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