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.

There is a common characterization of the class group ${\rm Cl}(R)$ as a kind of measure of how badly factorization fails to be unique. The most obvious justification for this sentiment is that the order of the class group of $R$ determines if it has unique factorization or not:

$$h=1\quad\Leftrightarrow\quad {\rm PID}\quad \Rightarrow \quad {\rm UFD}. \tag{$\star$}$$

This is very unsatisfying though because the exact size of $h$ (not even to mention all of the group structure that ${\rm Cl}(R)$ has in general) is not used, and "UFD / not UFD" doesn't in any sense measure the extent to which $R$ fails to be a UFD, only if it does. Plus, the converse, UFD $\Rightarrow$ PID, is specific to when $R$ is Dedekind, so this is only a partial justification.

There are a number of answers exposing specific ideas in How does a class group measure the failure of unique factorization? (which is basically my question here, but I am resurrecting it because I am not satisfied) and Class number measuring failure of unique factorization at MO.

PLC and BD cite Carlitz ($h=2$ iff factorization lengths are invariant), PLC notes that $h$ can yield arithmetic obstructions to certain paths like in FLT, and BD cites a theorem of Kaczorowski which goes from factorization information to an exact isomorphism class characterization of the class group (which is reverse of the desired order: going from class group to factorization information).

Kevin has the strongest skeptical vibe among the responses in his conclusion:

Unfortunately, you can see we lose a lot of information in passing to the class group, and in particular it doesn't tell us anything at all about which elements are obstacles to unique factorization. The intuition, rather, is that a more complicated class group implies we're further from unique factorization.

I am not sure how the mentioned intuition translates into concrete facts, but I do get the impression that $\rm Cl$ doesn't have the right "type" of information to talk about factorizations. Rather, I think the more direct description for $\rm Cl$ is as a measure of how ideals fail to act like numbers.

Here is my own crack at "measuring" factorization's failure to be unique. Let $\Gamma(R)$ be the set of all associates classes of irreducible elements. (Suppose $R$ is a factorization domain, so all elements have some factorization if not a unique one, and $K$ is $R$'s fraction field.) The group of principal fractional ideals is essentially $K^\times/R^\times$, and it is generated by irreducible elements. Thus there is a surjective map $\Bbb Z^{\oplus \Gamma(R)}\to K^\times/R^\times$, and the kernel is comprised of all relations satisfied between irreducibles under multiplication. These relations are precisely the inequivalent factorizations into irreducibles that occur in $R$. Our knowledge so far can be put into a diagram:

$$\{{\rm relations}\}\to\Bbb Z^{\oplus \Gamma(R)}\to K^\times/R^\times\to I(R)\to{\rm Cl}(R).$$

Observe this has two subsequences which are short exact, the first ending and the second beginning with $K^\times/R^\times$. (Can we determine if $\rm relations$ is infinitely generated or not?) One thing to notice is that in such a sequence $A\to B\to C\to D\to E$, the exact sequence $C\to D\to E$ generically is expected to have little control over the exact sequence $A\to B\to C$, which at face value seems like good evidence that ${\rm Cl}(R)$ does not influence factorization very much, but in fact $(\star)$ says $|{\rm Cl}|=1$ implies $\{\rm relations\}=0$, so this diagram does not fully capture the situation.

So, my questions:

  • What direct, specific relationships exist between elemental factorizations and ideal classes? (In particular, beyond Kaczorowski's theorem.)
  • Why is it useful to think of ${\rm Cl}(R)$ as measuring the failure of unique factorization, as opposed to being only indirectly related (i.e. by measuring the failure of ideals to act like numbers)?
  • What properties of ${\rm Cl}(R)$ definitively don't say anything specific about factorizations, and conversely what properties of factorizations are definitively not captured by ideal classes?

Sorry if I have been rambling, or not justified posting about this again, or my questions too vague.

share|improve this question
Let me make a few remarks, but first let me say that this is a great question, one which I have asked myself (and have asked several number theorists, to no avail). The class group definitely does tell you other things. For example, in terms of arithmetic of a number field $K$, one object you would like to study is the maximal unramified extension of $K$. This is called the Hilbert Class Field of $K$, and satisfies the awesome relation that $K/\mathbb{Q}$ is Galois, and $\text{Gal}(K/\mathbb{Q})\cong \text{Cl}(K)$. Another motivation, is a geometric one. Namely, $\text{Cl}(K)$ is precisely the –  Alex Youcis Oct 25 '13 at 8:50
Picard group $\text{Pic}(\text{Spec }(\mathcal{O}_K))$, and there we see that the group structure is precisely telling us not only about non-trivial line bundles, but how their interactions can "fix" or "make worse" the triviality. Now, onto more practical things. For the sake of pure, basic, interesting results coming from the class group, there are only two things that I have seen used to great effect: the order (the class number) and the exponent (the minimal $m$ such that $m\text{Cl}(K)=0$). The former is used in many places, as you know, and the latter is used often times when trying –  Alex Youcis Oct 25 '13 at 8:52
to solve Diophantine equations. Now, both of these statistics are things easily deducible from, and in fact part of the statistics characterizing something more important. Namely, the class group (of a global field) is always a finite group. So, it's isomorphism type is really just determined by its invariant factor decomposition--a list of numbers. There is no fancy group theory, in some sense, going on here. Thus, really you're asking what this list of numbers, tells you. The order and exponent tell you two statistics of this list--the sum of the tuples, and the greatest common multiple of –  Alex Youcis Oct 25 '13 at 8:54
the tuples (interpreted correctly). So, maybe a more natural question, before jumping straight to the whole list of numbers (the isomorphism type) would be what other, natural statistics of the list tell you? For example, if instead of invariant factor, we decomposed maximally into $p$-parts, what does the number of $p$-parts tell you (i.e. what does $\dim_{\mathbb{F}_p}(\text{Cl}(K)\otimes_\mathbb{Z}\mathbb{F}_p)$)? What about the minimal number of generators of $\text{Cl}(K)$ (i.e. the total number of factors in the $p$-part decomposition) etc. These are just some observations. Good luck! –  Alex Youcis Oct 25 '13 at 8:57
@AlexYoucis Mike is referring to your earlier comment about what you'd do if a professional didn't weigh in. –  blue Mar 2 at 7:28

1 Answer 1

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

share|improve this answer
Thanks for your answer. I have a few questions. Is there a name for the result that the class group determines the monoid ${\cal O}_K/{\cal O}_K^\times$, or keywords that can be searched? In (2) what kind of "hugely different properties" are we talking about? Finally, I don't understand your middle paragraph in (3) at all. Since $(a)$ and $(b)$ are equal classes, isn't every ideal factorization of one also an ideal factorization of the other? What does "having $2n$ in $C$" mean? And what "property of factorizations of an individual element" is being discussed there? –  blue Jun 2 at 19:22
Sorry it took a while. Finally I update the answer to address your questions. –  quid Jun 6 at 23:19

Your Answer


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.