Friday, February 29, 2008

More concepts from ring theory

We are at an important juncture in our discussion of algebraic number theory. From here on out, the path starts to go uphill more steeply, with quite a bit more abstraction and technical complexity. I hope you'll follow along anyhow. Don't feel like you need to grasp all the fine points immediately.

We're now going to cover some concepts of ring theory that are essential for talking about rings of algebraic integers. We introduced algebraic integers themselves here. The previous discussion of rings is here. Mathematics being what it is, you sort of need to have some exposure to these preliminaries in order to go further. All preceding installments in this series are listed here.

In the previous three installments we spent of lot of time on the concept of "unique factorization". What we are about to do is formalize the concept in terms of ideals of commutative rings. This discussion needs to be somewhat lengthy, but at the end we will be able to state some general properties that rings of algebraic integers have, and some conditions that are equivalent to uniqueness of factorization. Don't get too stressed by this. Detailed proofs won't be given. The level of difficulty here is comparable to what's in an introductory college course in abstract algebra or linear algebra.

Because of the importance of unique factorization, an integral domain which has unique factorization is called a unique factorization domain, or UFD. A little more precisely, an integral domain R is a UFD if there is a set P of irreducible elements such that every nonzero element α of R can be written in a unique way as a finite product α=u∏1≤k≤n pkek, where u is a unit, n is a nonnegative integer, pk are distinct elements of P, and exponents ek are nonnegative integers (all of which depend on α).

The defining property of a UFD is pretty clear and understandable, but it is not expressed in the language of ideals. As we shall see, a great deal of what we want to know about algebraic numbers can be expressed in terms of ideals, so we'd like to know how unique factorization fits in. To this end, we have this centrally important concept: An integral domain R is called a principal ideal domain (PID for short) in case every ideal I⊆R is equal to an ideal of the form (α)=αR for some α∈R.

It is a fact, which is not difficult to prove, but not immediately obvious either, that every principal ideal domain is a unique factorization domain. So if we want to show that a given integral domain R has unique factorization, it is sufficient to show that R is a PID. Unfortunately, among all integral domains there are some which are UFDs but not PIDs. So R can be a UFD even if it is not a PID – being a PID is not a necessary condition. The class of UFDs contains the class of PIDs, but is strictly larger. For instance, if F is a field, the ring of polynomials in one variable F[x] is a PID (and a UFD). (The reason this is true will be mentioned in a moment.) However, the ring of polynomials in two variables F[x,y] is a UFD but not a PID.

Obviously, it would be convenient to have a simple criterion to determine whether a domain R is a PID (and hence a UFD). It turns out that the the process of division-with-remainder that can be performed in ℤ and in polynomial rings in one variable fills the bill. All that's needed is an integer valued function on the domain R with certain properties. For a∈R, let this function be written as |a| (since it is much like an absolute value). This function should have three properties:
  1. |a|≥0 for all a∈R, and |a|=0 if and only if a=0
  2. |ab| = |a|⋅|b| for all a,b∈R
  3. if a,b∈R and b≠0, then there exist q,r∈R such that a=qb+r, with 0≤|r|<|b|
A domain R with such a function is called a Euclidean domain, because one has a Euclidean algorithm that works just as it does in ℤ. If I⊆R is a nonzero ideal, it has an element b∈I, with b≠0, of smallest nonzero "norm" |b|. If a∈I we can write a=qb+r for q,r∈R, and |r|<|b|. Yet r=a-qb is in I, so by the assumption of minimality of |b| we have |r|=0 and therefore r=0. Hence a=qb, I⊆(b), and finally I=(b) is principal. In other words, every Euclidean domain is a PID, and therefore a UFD.

Unfortunately, even among rings of integers of quadratic fields, only finitely many are known to be Eucldean. If F=ℚ(√d) and OF is the ring of integers, it is known that for d<0 the ring is Euclidean only when d=-1, -2, -3, -7, or -11. If d>0, the number of rings which are Euclidean is larger. What is known is that only a finite number of these are Euclidean using the norm function. At least one other is Euclidean using a function other than the norm function, but so far it's not known whether there are only a finite number like that.

This may seem rather disappointing, but in fact the quadratic fields where OF is a PID are also quite scarce. If d<0, then in addition to the Euclidean cases, the only other values are d=-19, -43, -67, and -163. The proof that this is a complete list for d<0 is quite difficult and was not satisfactorily done until 1966.

The situation with d>0 is even more difficult. It is not actually known whether there are only finitely many d>0 such that Oℚ(√d) is a PID. Gauss himself conjectured that there are infinitely many, but this is still an important open question.

Returning to concepts, we recall that among all integral domains, the class of PIDs is strictly smaller than the class of UFDs. It turns out that there is a subclass of all integral domains in which the notions of UFD and PID are equivalent. In fact, this is an important class, because it includes all rings of algebraic integers. This class itself can be defined by a number of equivalent conditions. But to explain this, we need to discuss the group of fractional ideals of an integral domain.

Fractional ideals



If A and B are ideals of any commutative ring R, it's easy to define the product of two ideals as a set of finite sums: A⋅B = {∑1≤k≤n akbk | ak∈A, bk∈B, n∈ℤ, n>0}. By definition of an ideal, A⋅B⊆A and A⋅B⊆B, hence A⋅B⊆A∩B. If R has a unit (as we usually assume), then clearly A⋅R=A. So in the set of ideals of R there is a commutative binary operation of multiplication, and it has an identity. Multiplication of ideals is associative since R multiplication in R is associative. (A set with an associative multiplication is called a semigroup, and if an identity exists, it's a monoid. In neither case is multiplication necessarily commutative.)

In a situation like this, it's natural (for a mathematician anyhow) to wonder what conditions on R would make the set of its ideals into a full group -- that is, how the inverse of an ideal might be defined. It turns out that for integral domains the conditions are beautiful and everything one could hope for.

To get an idea of where to start, consider the principal ideal domain ℤ. For any nonzero n∈ℤ, the obvious thing to consider is (1/n) = {m/n | m∈ℤ}. That's sort of like an ideal, since it's a commutative group under addition, and m(1/n)⊆(1/n) for all m∈ℤ. Also, under the obvious definition, (n)(1/n) = ℤ (since the product contains 1). So (1/n) surely acts like the "inverse" of the ideal (n) of ℤ for any n.

Suppose R is any commutative ring with an identity and M is any set at all (not necessarily related directly to R) where one can define an operation of multiplication rm=mr for r∈R and m∈M. Suppose further that:
  1. M is a commutative group under addition.
  2. rm∈M for all r∈R and m∈M.
  3. 1m = m for all m∈M.
  4. r(m1 + m2) = rm1 + rm2 for all r∈R, m1,m2∈M.
Then M is said to be an R-module. (If R isn't commutative, one can define R-modules by being a little more picky about the definition.) So from the example above, (1/n) is a ℤ-module, and also any ideal of a ring R is an R-module.

Given all that, if R is an integral domain, whatever a fractional ideal of R might be, it certainly should be an R-module. Indeed, we can formally define a fractional ideal of R as an R-module M such that:
  1. M⊆F, if F is the field of quotients of R.
  2. The multiplication rm for r∈R and m∈M is just the normal multiplication in F.
  3. There is some nonzero r∈R such that rM⊆R.
(As a reminder, the field of quotients of an integral domain R is defined as follows. Consider the set of pairs (m,n), with m,n∈R, n≠0 Consider two pairs (m,n) and (m′,n′) to be equivalent just in case mn′=m′n. (Think of this as fractions, where m/n = m′/n&prime when mn′=m′n.) The underlying set of the field of quotients is the set of equivalence classes of pairs under this relation. Define addition on this set by (m,n)+(m′,n′) = (mn′+m′n,nn′) and multiplication by (m,n)(m′,n′) = (mm′,nn′). Then it can be shown that addition and multiplication are well-defined, and the set of equivalence classes of pairs is indeed a field.)

Multiplication of fractional ideals is defined just like multiplication of ideals: M⋅M′= {∑1≤k≤n mkm′k | mk∈M, m′k∈M′, n∈ℤ, n>0}. With this definition, the set of fractional ideals of R is a monoid. The question is: what conditions on R will guarantee that fractional ideals form a group? This is not just a matter of idle curiosity, because it turns out that for rings with the right properties, one has unique factorization in the group of fractional ideals, and in the set of integral (i. e. ordinary) ideals as well. For rings of algebraic integers, which just happen to have the right properties, this unique factorization of ideals is almost as good, for many purposes, as having unique factorization of ring elements themselves.

We need just a few more concepts before we can state the necessary conditions. A proper ideal of R is an ideal I that is not equal to R, i. e. a proper subset. One writes I⊂R. A prime ideal P is a proper ideal such that for all a,b∈R, ab∈P only if either a∈P or b∈P (or both). In ℤ, for example, (6) isn't a prime ideal, since 2⋅3∈(6), but 2∉(6) and 3∉(6). A maximal ideal is a proper ideal P that is not properly contained in some other proper ideal P′. The integral domain R, with field of fractions F, is said to be integrally closed if every α∈F that is integral (i. e. an algebraic integer of F) over R is actually an element of R. For example, if R=ℤ, then F=ℚ, and to say α∈R is integral over F means f(α)=0 for some monic polynomial f(x) with coefficients in R, i. e. f(x)∈ℤ[x]. If α=a/b for a,b∈ℤ, then when you "clear fractions" in f(a/b)=0, you find b|a. This argument applies in any UFD, so in fact any UFD is integrally closed.

Lastly, we need a type of finiteness condition. An ideal I is said to be finitely generated if it has the form I = {∑x∈S axx | ax∈R for all x∈S}, where S⊆R is a finte set of generators. It is much like a principal ideal, except for having n=#(S) generators instead of 1. If all ideals of a ring are finitely generated, the ring is said to be Noetherian, after Emmy Noether (1882-1935). There are several equivalent characterizations of Noetherian rings. For instance, R is Noetherian if and only if every nonempty family of ideals of R has a maximal element (which contains all the other members of the family) with respect to inclusion.

Finally we can state the crucial result: If R is an integral domain, then the following are equivalent:
  1. R is Noetherian, integrally closed, and every nonzero prime ideal is maximal.
  2. Every nonzero ideal of R is uniquely expressible as a product of prime ideals.
  3. Every nonzero ideal of R is a product of prime ideals.
  4. The set of nonzero fractional ideals of R forms a group under multiplication.
The first item on this list characterizes R in terms of several ring-theoretic properties. The second item is unique factorization into prime ideals, and it is in fact equivalent to the apparently weaker third item. The fourth item is the key fact, which answers our earlier quesion about when the fractional ideals of R form a group.

Any integral domain R that has one of these properties has all of them, and is called a Dedekind domain, after Richard Dedekind (1831-1916). It isn't hard to show that if F⊇ℚ is a finite field extension, then the ring OF of algebraic integers of F has the properties listed in the first item, and so OF is a Dedekind domain and has the other properties also.

As rings, Dedekind domains have some very nice properties. For example, if R is a Dedekind domain:
  1. P is a prime ideal of R if and only if it is indecomposable, i. e. P ≠ I⋅I′ where I and I′ are ideals other than P or R.
  2. If P is a prime ideal and P divides the product I⋅I′ of two ideals (P|I⋅I′), then P|I or P|I′.
  3. Divisibility between fractional ideals is equivalent to inclusion, i. e. if M and M′ are nonzero fractional ideals, then M divides M′ if and only if M⊇M′. (Multiplication of ideals yields a result that is smaller.)
  4. If R is a unique factorization domain, it is a principal ideal domain.
  5. Every fractional ideal M of R can be generated by at most two elements, and one of these elements of M can be chosen arbitrarily.


Dedekind, Emmy Noether, and a few others were the main developers of ideal theory in this form, and Dedekind was a leading figure in the theory of algebraic numbers in general. Ernst Kummer (1810-1893) somewhat earlier had a more primitive theory of "ideal numbers" which provided a kind of unique factorization of algebraic numbers, but Dedekind made the theory much simpler and more general.

In the next installment we'll look at simple examples of how ideals that are prime in a ring of integers may split into factors in the ring of integers of an extension field. This is a very key issue in the overall theory. Eventually we will see how this abstract point of view generalizes some important ideas, called "reciprocity laws" from classical number theory.

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2 Comments:

Anonymous Anonymous said...

"...even among rings of integers of quadratic fields, only finitely many are Eucldean"--actually, this is still unknown (and it's one of the bizarre things about algebraic number theory that such an apparently simple question can still be unresolved).

For some rings, the norm function (described in one of your previous articles) happens to be a Euclidean function--this is an easy way to prove that a particular ring is Euclidean. Such rings are referred to as "norm-Euclidean", and there are indeed only finitely many of them amongst quadratic number fields.

But is it possible for a ring of integers to be Euclidean without being norm-Euclidean? The first example was found only a few years ago (2004? sorry, I don't have the reference handy). We still don't know how many more such things are out there.

8/22/2009 03:38:00 AM  
Blogger Charles Daney said...

But is it possible for a ring of integers to be Euclidean without being norm-Euclidean? The first example was found only a few years ago (2004? sorry, I don't have the reference handy). We still don't know how many more such things are out there.

Thanks for the information. I was obviously unaware that there was a possibility of such rings being Euclidean but not norm-Euclidean.

I've made an appropriate correction.

8/24/2009 03:38:00 PM  

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