## Class number measuring the failure of unique factorization

The statement that the class number measures the failure of the ring of integers to be a ufd is very common in books. ufd iff class number is 1. This inspires the following question:

Is there a quantitative statement relating the class number of a number field to the failure of unique factorization in the maximal order - other than $h = 1$ iff $R$ is a ufd?

In what sense does a maximal order of class number 3 "fail more" to be a ufd than a maximal order of class number 2?

Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?

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I'd say that the group tells you in what different ways does UFD fail in your ring, rather than that its size measures how bad the failure is. – Mariano Suárez-Alvarez Jan 6 2010 at 17:09
@Mariano: it sounds like you are hinting at the group structure of the class group as a measure of failure. This sounds as interesting as the original question. Can you elaborate on such a relationship? – Dror Speiser Jan 6 2010 at 17:24

Theorem (Carlitz, 1960): The ring of integers $\mathbb{Z}_F$ of an algebraic number field $F$ has class number at most $2$ iff for all nonzero nonunits $x \in \mathbb{Z}_F$, any two factorizations of $x$ into irreducibles have the same number of factors.

A proof of this (and a 1990 generalization of Valenza) can be found in $\S 22.3$ of my commutative algebra notes.

This paper has spawned a lot of research by ring theorists on half-factorial domains: these are rings in which every nonzero nonunit factors into irreducibles and such that the number of irreducible factors is independent of the factorization.

To be honest though, I think there are plenty of number theorists who think of the class number as measuring the failure of unique factorization who don't know Carlitz's theorem (or who know it but are not thinking of it when they make that kind of statement).

Here is another try [edit: this is essentially the same as Olivier's response, but said differently; I think it is worthwhile to have both]: when trying to solve certain Diophantine problems (over the integers), one often gets nice results if the class number of a certain number field is prime to a certain quantity. The most famous example of this is Fermat's Last Theorem, which is easy to prove for an odd prime $p$ for which the class number of $\mathbb{Q}(\zeta_p)$ is prime to $p$: a so-called "regular" prime.

For an application to Mordell equations $y^2 + k = x^3$, see

http://math.uga.edu/~pete/4400MordellEquation.pdf

Especially see Section 4, where the class of rings "of class number prime to 3" is defined axiomatically and applied to the Mordell equation. (N.B.: These notes are written for an advanced undergraduate / first year grad student audience.)

The Mordell equation is probably a better example than the Fermat equation because:

(i) the argument in the "regular" case is more elementary than FLT in the regular case (the latter is too involved to be done in a first course), and

(ii) when the "regularity" hypothesis is dropped, it is not just harder to prove that there are no nontrivial solutions, it is actually very often false!

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there is a theorem: we know that for every number field $K$ (with class number not necessarily 1) there exists a unique factorization domain $\mathcal R_K$, $\mathcal O_K\subset \mathcal R_K\subset K$ such that it's group of invertible elements is finitely generated.

what is the reference for this result?

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This follows immediately from proposition 11.6 in Algebraic Number Theory, Neukirch, English first edition, by taking any $X$ such that the map $\bigoplus_{\mathfrak{p}\not\in X}K^*/O_\mathfrak{p}^*\rightarrow Cl(O_K)$ is surjective, which is possible since $Cl(O_K)$ is finite. – Dror Speiser Jul 21 at 13:04
You can take for $R_K$ any ring of $S$-integers ${\mathcal O}_{K,S}$ where $S$ is a finite set of primes from ${\mathcal O}_K$ (plus the archimedean places) whose ideal classes generate the class group of $K$. – KConrad Jul 22 at 3:48
thank you...... – quaestion Jul 23 at 19:17

Class number $h(K)$ is exactly the quantitative measure of the failure of unique factorization: by its definition it measures "how many more ideas are there compared to numbers".

To clarify: decomposition is always unique for ideals, so if the only ideals you have are numbers (that is, $h = 1$), then you don't have any problem decomposing numbers (so, you have PID). Furthermore, the more "leftover" ideals you have (ideal class group), the more possibilities of writing different decompositions of numbers exist.

This vague statement can be turned into some precise ones. If you have different factorizations of number $x$, this means the prime ideals in the decomposition $x = \mathfrak p_1\mathfrak p_2\dots\mathfrak p_n$ are grouped in a different way.You can establish from here the bound on the number of possible different factorizations; may be (not sure here) it can be shown to be no more than $C(h)$.

Another theorem that follows (mentioned by Olivier): $x^h$ must always have a decomposition into true numbers rather then ideals. Indeed, $x^n = \mathfrak p_1^h\mathfrak p_2^h\dots\mathfrak p_n^h$ and you need to use the fact that any element $p$ in abelian group of size $h$ has the property $p^h = 1$.

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For Dedekind domains, like the integers of a number field, PID iff UFD. There's definitely a quantitative statement relating the class number to failure of PIDness: the higher the class number, the smaller the density of principal prime ideals amongst the prime ideals; this is just Cebotarev plus standard facts about the Hilbert class field.

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