Class number measuring the failure of unique factorization

Question

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?

Answer 1

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://alpha.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!

Answer 2

Here is a partial answer.

In a UFD, the following statement is true : "either an element is prime or you can write it as a nontrivial product". In a Dedekind ring with finite class number h, it is not true, but you have the following "quantitative" statement : "either an element is prime or you can write its h-th power as a nontrivial product".

Answer 3

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.

Answer 4

I know I'm late to the party, but it seems to me the other answers don't directly answer the last part of the question:

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?

No. Of course it's not necessarily true that a given integer in two fields $K_1 \cap K_2$ will have more distinct factorizations in $K_1$ than $K_2$ if $h(K_1) \ge h(K_2)$. But even in simple situations where you could hope for this to be true, the opposite behavior can happen.

In this paper, I gave a combinatorial expression for the number $\eta(x)$ of distinct factorizations of $x \in \mathcal O_K$ ($K$ a number field) in terms of the ideal classes of $\mathfrak p_i$ and the exponents $e_i$, where $x \mathcal O_K$ has prime ideal factorization $\prod \mathfrak p_i^{e_i}$. There is also a precise description of the structure of distinct factorizations, which makes it clear that the structure of the distinct factorizations can get more complicated with more complicated class groups. (In particular, you can use this description to get Carlitz's theorem and related results.)

But to return to your question, here is an example where you can describe $\eta(x)$ simply. Suppose $K$ is quadratic, $x=p$ is a rational prime and $\mathfrak p$ is the prime ideal of $\mathcal O_K$ above $p$, then for any $n \in \mathbb N$,

$$ \eta(p^n) = \begin{cases} 1 & \mathfrak p \text{ principal or $p$ ramifies} \\ \lfloor \frac nm \rfloor + 1 & \text{else}, \end{cases} $$

where $m$ is the order of $\mathfrak p$ in the class group.

In particular, suppose $K_1$ and $K_2$ are quadratic fields with class numbers 3 and 2 respectively, and $p$ is rational prime unramified in both $K_1$ and $K_2$. Let $\mathfrak p_1$ and $\mathfrak p_2$ be primes of $K_1$ and $K_2$ above $p$. Then $p^n$ (for any $n > 3$) has more distinct factorizations in $K_2$ (a class number 2 field) than $K_1$ (a class number 3 field) if and only if $\mathfrak p_2$ is not principal.

Answer 5

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 $p^h = 1$.