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I haven't learned that much about primary decomposition, but from I understand about Dedekind domains, we have that all fractional ideals are invertible and all (plain old) ideals factor uniquely into a product of prime ideals, so that Dedekind domains should satisfy this condition. Are these the only such rings, or is there a weaker condition we can give?

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It is proved in Jacobson, Basic Algebra, Vol 2., that unique decomposition into prime ideals is equivalent to being a dedekind domain. –  Anweshi Jan 19 '10 at 0:01
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If I remember correctly then the definition of fractional ideal (that I know) requires the ring to be Dedekind. Can a more general definition be given? –  Tran Chieu Minh Jan 19 '10 at 0:03
    
And fractional ideals are invertible anyway, since they are finitely generated. –  Anweshi Jan 19 '10 at 0:04
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If our ideals don't have a unique prime factorization, then either associativity fails or invertibility fails. There's a non-noetherian generalization of a dedekind domain, i.e. all finitely generated ideals have a unique factorization into prime ideals. Then we could take the fractional ideals of the finitely generated ones, but this is a pretty useless generalization in my opinion. I would place my bet on there being no nontrivial generalization. –  Harry Gindi Jan 19 '10 at 0:05
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Unless you're in a dedekind domain, there is no reason to believe that the primes have any interesting properties with respect to factorization, so taking the free abelian group on the primes won't really help. –  Harry Gindi Jan 19 '10 at 0:12

3 Answers 3

up vote 13 down vote accepted

First some responses to comments above:

One can (and should) define fractional ideals for any integral domain $R$. A fractional $R$-ideal is a nonzero $R$-submodule $I$ of the fraction field $K$ such that there exists $x \in K \setminus \{0\}$ such that $xI \subset R$. The product of two fractional ideals is again a fractional ideal, and the multiplication is associative and has $R$ itself as an identity: in other words, the set $\operatorname{Frac}(R)$ of fractional ideals of $R$ forms a commutative monoid.

We have the notion of a principal fractional ideal: this is a submodule of the form $xR$ for $x \in K^{\times}$. The set of principal fractional ideals forms a subgroup $\operatorname{Prin}(R)$ of $R$. One could take the quotient $\operatorname{Frac(R)}/\operatorname{Prin(R)}$, but this is a bit of a false step, as it is not a group in general.

In the case of Dedekind domains there is nothing to worry about:

Theorem: For an integral domain $R$, the following are equivalent:
(i) $R$ is Noetherian, integrally closed of dimension at most one (a Dedekind domain).
(ii) Every nonzero integral ideal of $R$ factors into a product of prime ideals.
(ii)' Every nonzero integral ideal of $R$ factors uniquely into a product of prime ideals.
(iii) The fractional ideals of $R$ form a group.

So for a Dedekind domain, certainly $\operatorname{Frac}(R)/\operatorname{Prin(R)}$ is a group, called the ideal class group of $R$.

In general, there is an easy way to remedy the problem that $\operatorname{Frac}(R)/\operatorname{Prin(R)}$ need not be a group. Namely, instead of taking the full monoid of fractional ideals, we restrict to the unit group $I(R)$, the invertible fractional ideals. For any domain $R$, we may define the Picard group

$\operatorname{Pic}(R) = I(R)/\operatorname{Prin}(R)$.

Because of the theorem above, for a non-Dedekind domain the Picard group isn't capturing any information about the prime ideals in particular. There is however a different construction -- coinciding with $\operatorname{Pic}(R)$ when $R$ is a Dedekind domain -- which does just this. For (a tiny bit of) motivation: even in the case of a Dedekind domain we don't take the free abelian group on all the prime ideals: we omit $(0)$.

Now let $R$ be any Noetherian domain. One can define the divisor class group $\operatorname{Cl}(R)$ as follows: let $\operatorname{Div}(R)$ be the free abelian group generated by the height one prime ideals $\mathfrak{p}$ (these are ideals so that there is no prime ideal $\mathfrak{q}$ properly in between $(0)$ and $\mathfrak{p}$). One can also define, for each $f \in K^{\times}$, a principal divisor $\operatorname{div}(f)$. (I don't want to give the exact recipe in the general case: it involves lengths of modules. If $R$ happens to be integrally closed, then the localization $R_{\mathfrak{p}}$ at a height one prime $\mathfrak{p}$ is a DVR, say with valuation $v_{\mathfrak{p}}$, and then one takes $\operatorname{div}(f) = \sum_{\mathfrak{p}} v_{\mathfrak{p}}(f) [\mathfrak{p}]$.) Again the principal divisors $\operatorname{Prin}(R)$ form a subgroup of $\operatorname{Div}(R)$ and the quotient $\operatorname{Div}(R)/\operatorname{Prin}(R)$ is the divisor class group.

There is a canonical homomorphism $\operatorname{Pic}(R) \rightarrow \operatorname{Cl}(R)$, which is in general neither injective nor surjective. However, the map is an isomorphism in the case that $R$ is a regular ring.

These constructions are the affine versions of more familiar constructions in classical algebraic geometry: they are, respectively, Cartier divisors and Weil divisors, which agree on a nonsingular variety but not in general.

Finally, one can also define analogues of these groups for certain non-Noetherian domains (the Noetherianity is used to ensure that $v_{\mathfrak{p}}(f) = 0$ except for finitely many primes $\mathfrak{p}$), e.g. Krull domains and Prufer domains. The latter is a domain in which each finitely generated nonzero ideal is invertible. Both are natural and interesting classes of rings.

For more details on this material, see e.g. the (rather rough and incomplete) notes

http://www.math.uga.edu/~pete/classgroup.pdf

[Addendum: also see Section 11 of http://math.uga.edu/~pete/factorization2010.pdf.]

For much more detail see the references cited therein, especially Larsen and McCarthy's Multiplicative Ideal Theory.

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This is a great answer. –  Harry Gindi Jan 19 '10 at 1:28
    
Wow - thanks for another amazing answer! This really brings together a lot of things I've been thinking about (and also a lot of things I keep hearing the basic premise of, but hadn't really seen defined yet). I'll definitely have to check out Larsen and McCarthy - what else is a good reference for just straight-up theory of commutative rings (obviously using some algebraic geometry tools is unavoidable, but I think I like this kind of structural analysis of rings a bit more)? This is really the math that grabs me - I can't wait to work on it when I "grow up" :) –  Zev Chonoles Jan 19 '10 at 4:30
    
Notwithstanding the title, Eisenbud's Commutative Algebra: with a View Toward Algebraic Geometry is an excellent text even if you just completely ignore all the geometric stuff. In particular, he has a very nice treatment of this topic in Chapter 11. The best commutative algebra text that I know that (so far as I can tell!) is not secretly thinking about things geometrically is Kaplansky's Commutative Rings. –  Pete L. Clark Jan 19 '10 at 4:59
    
Bourbaki is actually wonderful also. I would say that they're not secretly teaching you geometry either. –  Harry Gindi Jan 19 '10 at 7:29
    
Great - off to get both of those! –  Zev Chonoles Jan 22 '10 at 16:34

When one considers the ring of algebraic integers R of a number field, the Picard group Pic(R)=Inv(R)/Prin(R) of invertible fractional ideals modulo principal fractional ideals is a measure of how far the ring is from being a UFD (the finiteness of the group is a classical theorem of algebraic number theory). In particular, Pic(R)=0 precisely when R is a UFD. If the cardinality of Pic(R) is h>1, then the hth power of every invertible fractional ideal factors uniquely into the product of prime ideals.

Perhaps unfortunately, there exist rings R which are not UFDs but for which the Picard group is trivial. An example of such a ring is $\mathbb{Q}+x\mathbb{R}[x]$. This is not to say that the Picard group does not convey important information when one leaves the world of Dedekind Domains.

A ring is a Bezout domain if it is an integral domain in which every finitely generated ideal is principal. For example, if the ring is Noetherian, then being a Bezout domain is equivalent to being a PID.

Now define a ring R to be a Prufer domain if it is an integral domain in which every non-zero finitely generated ideal is invertible. It is a theorem that for a Prufer ring R, Pic(R) is trivial if and only if R is a Bezout domain.

Let T be a non-empty set of indeterminates and R be an integral domain. Mathematicians have long been interested in exploring the relationshipa between (1) Pic(R) and Pic(R[T]) and (2) Cl(R) and Cl(R[T]). It was proven by Gabelli that Cl(R)=C(R[T]) if and only if R is integrally closed. Call an arbitrary commutative ring seminormal if $b^2=c^3$ implies that there exists an element $a$ such that $a^3=b$ and $a^2=c$. Then Pic(R)=Pic(R[T]) precisely when R is seminormal.

Finally, it makes sense - and can in fact be incredibly fruitful - to define the Picard group of a noncommutative ring. The most common set up in which this group is defined is as follows. Let R be an integral domain with quotient field K, let A be a finite-dimensional semisimple algebra over K and O be an order of A. Then one can define a group $Pic_R(O)$, which is a certain quotient of the group of invertible fractional ideals, and a homomorphism $\varphi: Aut_R(O)\rightarrow Pic_R(O)$. One can adapt techniques from K-theory in order to study $Pic_R(O)$, and hence the automorphism group of the order O as well.

An excellent reference for this is the follwoing paper of Frolich.


A. Frolich The Picard Group of Noncommutative Rings, in Particular of Orders, Transactions of the AMS, Vol 180 (Jun 1973), pp. 1-45


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I figured I might as well copy my comments into an answer.

If our ideals don't have a unique prime factorization, then either associativity fails or invertibility fails. There's a non-noetherian generalization of a dedekind domain, i.e. all finitely generated ideals have a unique factorization into prime ideals. Then we could take the fractional ideals of the finitely generated ones, but this is a pretty useless generalization in my opinion. I would place my bet on there being no nontrivial generalization.

Unless you're in a dedekind domain, there is no reason to believe that the primes have any interesting properties with respect to factorization, so taking the free abelian group on the primes won't really help.

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Thanks again for your help - I have voted you up. –  Zev Chonoles Jan 19 '10 at 6:36

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