Class group and factorizations

I agree that some catchy phrases in that context are not really to the point, especially if generalized to contexts for which they were not conceived.

However, if you stick to the original context then it is in a certain sense completely justified to say that the class group measures, indeed not only measures but determines, the failure of uniqueness of factorization.

It seems there is agreement we can ignore invertible elements and we care about multiplication, so here is a result.

Let $K,K'$ be number fields with class groups $G,G'$ then $\mathcal{O}_K/\mathcal{O}_K^{\times}$ and $\mathcal{O}_{K'}/\mathcal{O}_{K'}^{\times}$ are isomorphic (as multiplicative semigroups) if and only if $G$ and $G'$ are isomorphic.

So, $\mathcal{O}_K/\mathcal{O}_K^{\times}$ is completely determined by the class group only.

Added: This result is in 'Über Ringe mit gemeinsamer multiplikativer Halbgruppe' [On rings with common multiplicative semigroup] Rédei and Steinfeld (1952).

But they only state it somewhat in passing and it seems do not give a proof even. Here is an argument.

First observe that the semi-groups of ideals are isomorphic no matter what. Both are free commutative semigroups with countable set of generators (the prime ideals). We need to find an isomorphism that preserves principal ideals.

Suppose the class groups are isomorphic. Let $f$ denote an isopmorphism. Recall that each of the ideal classes contains a countable number of prime ideals. For $C$ and ideal class of $K_1$, let $F_C$ be a bijection from the set of prime ideals of $K_1$ in $C$ to that of prime ideals of $K_2$ in $f(C)$. The maps $F_C$ combined yield a bijection $F$ from the prime ideals to the prime ideals. This extends to a semigroup isomorphism of the semigroups of all ideals that we also denote by $F$.

An ideal $I=P_1 \dots P_n$ is principal if and only if the combination of the classes of the $[P_i]$ in the class group is the neutral element, that is $[P_1] \dots [P_n]= 1_G$. Yet $F(I) = F(P_1) \dots F(P_n)$ and $[F(P_i)]= f([P_i])$. Finally, recall that $f$ was an isomorphism of groups, so $[P_1] \dots [P_n]= 1_G$ if and only if $f([P_1]) \dots f([P_n])= 1_{G'}$ So, $I$ is principal if and only if $F(I)$ is principal. And the claim is proved.

A key-word to look for, though this does not exactly yield this is 'block monoid' but see the reference I at the end.

Thus, if you care about multiplication only, which is a possible point of view regarding factorizations, then the class group really tells you all there ever is to know, for the structure in its entirety. For individual elements it is true that passing to idea classes group can loose information.

For your specific question.

1. Here is one example, possibly somewhat relating to your idea idea on relations, but there are numerous results (you could look for example at the relevant chapter of Narkiewicz's book on algebraic number theory to find many references):

   Let us consider two irreducible elements $a,b$. The product $ab$ might have different factorizations. Say, $ab= c_1 \dots c_n$. What is a global (over all $a,b \in \mathcal{O}_K$ irreducible) upper bound for $n$?

   A rough upper bound is the cardinality of the class group (which is sharp if and only if the group is cyclic). A precise bound is the the Davenport constant of the class group.

2. The result I mentioned above gives one justification in a certain context. In more general context less is true and if one wishes to be precise one should not say that the class group of a Dedekind domain measures the deviation from unique factorization, as Dedekind domains with the same class group can have hugely different properties regarding factorizations. It still provides a rough upper bound for the complexity that could occur though.

   If one goes beyond domains where the ideals have unique factorizations into prime ideals (in a suitable sense) the link gets even still vaguer.

3. Every property of the class group is relevant in some way. If the class group is changed the structure of $\mathcal{O}_K/\mathcal{O}_K^{\times}$ changes by the above result. There are a lot of specific phenomena where such things show, like the Davenport constant example above.

   A property of factorizations of an individual element that in general cannot be seen in the ideal classes is the number of distinct factorizations. Let $P,Q$ be two ideals in the same non-trivial ideal class $C$ and assume the class has order $n$. Then $a,b$ such that $ (a)=P^{2n}$ and $(b) = (PQ)^{n}$ are indistinguishable as regards ideal classes, both being principla ideals (that is in the trivial class) and each being the product of $2n$ prime ideals from the class $C$, yet the former has an essentially unique factorization while the latter has several different ones, namely $P^iQ^{n-i}$ times $P^{n-i}Q^{i}$ for $i= 0 , \dots , \lfloor n/2 \rfloor $. (For an explicit example see below)

   For number of factorizations also the frequency of different types in the same class is relevant; it is irrelevant which prime ideals precisely one has, but if they are different or equal is relevant.

I hope this answer goes a bit in the direction you envisioned. The class group is very relevant, but your ideal of relations is also good and indeed something like this is used in recent investigations on factorizations. See for example Semigroup-theoretical characterizations of arithmetical invariants with applications to numerical monoids and Krull monoids

Supplement to 2. There are various phenomena to consider but let us stick with the one we considered; in how many irreducibles factors can the product of two irreducibles decompose.

There are Dedekind domains with class group isomorphic to the integers for which the product of two irreducibles stays always the product of two irreducibles (so really simple, just like class group one or two elements for this phenomenon).

And there are Dedekind domains with isomorphic class group where however for every integer $n \ge 2$ there are two irreducible elements that are also the product of $n$ irreducible elements. Even more if you take this into account, for whatever finite set $I$ of integers at least $2$ you can find two irreducibles that are also product of $n$ irreducible for each $n$ in $I$.

This is about as different at it can get. And the class groups are the same.

Supplement to 3. Let us consider an explicit example. The ring of integers of $\mathbb{Q} [\sqrt{-5}]$ is $\mathbb{Z} [\sqrt{-5}]$. The class group has two elements.

We convince ourselves that $4$ can (up to equivalence) only be factored as $2^2$, yet $6$ is $2 \ 3$ and $(1 + \sqrt{-5}) (1 - \sqrt{-5})$. So $4$ and $6$ behave differently regarding factorizations.

This is easily understood looking at factorizations into prime ideals.
Let $P= (2, 1+ \sqrt{-5})$ , $Q_1 = (3, 1+ \sqrt{-5})$ and $Q_2 = (3, 1- \sqrt{-5})$. All these ideals are prime and not principal. They are thus lying in the nontrivial class of the class group.

We have $(2)=P^2$, so $(4) = P^4$. Further $Q_1Q_2 = (3)$ and $PQ_1 = (1+ \sqrt{-5}) $ and $PQ_2 = (1- \sqrt{-5})$. So $(6) = P^2Q_1Q_2$ and depending on how we group the ideals together we get either $(6) = (P^2) (Q_1Q_2) = (2) (3)$ or $(6) = (PQ_1)(PQ_2) =(1+ \sqrt{-5}) (1- \sqrt{-5})$. While for $(4)= P^4$ we cannot group in different, we just have $P^2 \ P^2$.

But if you pass to the class group you loose that information, as both $(4)$ and $(6)$ are product of $4$ prime ideals in the same class. So not all elements that are product of four ideals from the same class have the same factorization properties. So not all the information is in the classes.

If you want to know more about this in general you might be interested in reading Baginski and Chapman 'Factorizations of Algebraic Integers, Block Monoids, and Additive Number Theory' American Math Monthly 2011