Ideal class group

From Wikipedia, the free encyclopedia

In mathematics, the extent to which unique factorization fails in the ring of integers of an algebraic number field (or more generally any Dedekind domain) can be described by a certain abelian group known as an ideal class group (or class group). If this group is finite, (as it is in the case of the ring of integers of a number field) then the order of the group is called the class number. The multiplicative theory of a Dedekind domain is intimately tied to the structure of its class group. For example, the class group of a Dedekind domain is trivial if and only if the ring is a unique factorization domain.

Contents

[edit] History and origin of the ideal class group

The first ideal class groups encountered in mathematics were part of the theory of quadratic forms: in the case of binary integral quadratic forms, as put into something like a final form by Gauss, a composition law was defined on certain equivalence classes of forms. This gave a finite abelian group, as was recognised at the time.

Later Kummer was working towards a theory of cyclotomic fields. It had been realised (probably by several people) that failure to complete proofs in the general case of Fermat's last theorem by factorisation using the roots of unity was for a very good reason: a failure of the fundamental theorem of arithmetic to hold, in the rings generated by those roots of unity, was a major obstacle. Out of Kummer's work for the first time came a study of the obstruction to the factorisation. We now recognise this as part of the ideal class group: in fact Kummer had isolated the p-torsion in that group for the field of p-roots of unity, for any prime number p, as the reason for the failure of the standard method of attack on the Fermat problem (see regular prime).

Somewhat later again Dedekind formulated the concept of ideal, Kummer having worked in a different way. At this point the existing examples could be unified. It was shown that while rings of algebraic integers do not always have unique factorization into primes (because they need not be principal ideal domains), they do have the property that every proper ideal admits a unique factorization as a product of prime ideals (that is, every ring of algebraic integers is a Dedekind domain). The ideal class group gives some answer to the question: which ideals are principal ideals? The answer comes in the form all of them, if and only if the ideal class group (which is a finite group) has just one element.

[edit] Technical development

If R is an integral domain, define a relation ~ on nonzero fractional(!) ideals of R by I ~ J whenever there exist nonzero elements a and b of R such that (a)I = (b)J. (Here the notation (a) means the principal ideal of R consisting of all the multiples of a.) It is easily shown that this is an equivalence relation. The equivalence classes are called the ideal classes of R. Ideal classes can be multiplied: if [I] denotes the equivalence class of the ideal I, then the multiplication [I][J] = [IJ] is well-defined and commutative. The principal ideals form the ideal class [R] which serves as an identity element for this multiplication. Thus a class [I] has an inverse [J] if and only if there is an ideal J such that IJ is a principal ideal. In general, such a J may not exist and consequently the set of ideal classes of R may only be a monoid.

However, if R is the ring of algebraic integers in an algebraic number field, or more generally a Dedekind domain, the multiplication defined above turns the set of fractional ideal classes into an abelian group, the ideal class group of R. The group property of existence of inverse elements follows easily from the fact that, in a Dedekind domain, every non-zero ideal (except R) is a product of prime ideals.

The ideal class group is trivial (i.e. has only one element) if and only if all ideals of R are principal. In this sense, the ideal class group measures how far R is from being a principal ideal domain, and hence from satisfying unique prime factorization (Dedekind domains are unique factorization domains if and only if they are principal ideal domains).

The number of ideal classes (the class number of R) may be infinite in general. But if R is in fact a ring of algebraic integers, then this number is always finite. This is one of the main results of classical algebraic number theory.

Computation of the class group is hard, in general; it can be done by hand for the ring of integers in an algebraic number field of small discriminant, using a theorem of Minkowski. This result gives a bound, depending on the ring, such that every ideal class contains an ideal of norm less than the bound. In general the bound is not sharp enough to make the calculation practical for fields with large discriminant, but computers are well suited to the task.

It was remarked above that the ideal class group provides part of the answer to the question of how much ideals in a Dedekind domain behave like elements. The other part of the answer is provided by the multiplicative group of units of the Dedekind domain, since passage from principal ideals to their generators requires the use of units (and this is the rest of the reason for introducing the concept of fractional ideal, as well).

Define a map from K* to the set of all nonzero fractional ideals of R by sending every element to the principal (fractional) ideal it generates. This is a group homomorphism; its kernel is the group of units of R, and its cokernel is the ideal class group of R. The failure of these groups to be trivial is a precise measure of the failure of the map to be an isomorphism: that is the failure of ideals to act like ring elements, that is to say, like numbers.

The mapping from rings of integers R to their corresponding class groups is functorial, and the class group can be subsumed under the heading of algebraic K-theory, with K0(R) being the functor assigning to R its ideal class group; more precisely, K0(R) = Z×C(R), where C(R) is the class group. Higher K groups can also be employed and interpreted arithmetically in connection to rings of integers.

[edit] Examples of ideal class groups

The rings Z, Z[i], and Z[ω], (where i is a square root of -1 and ω is a cube root of 1) are all principal ideal domains, and so have class number 1: that is, they have trivial ideal class groups. If k is a field, then the polynomial ring k[X1, X2, X3, ...] is an integral domain. It has a countably infinite set of ideal classes.

If d is a square-free integer (in other words, a product of distinct primes) other than 1, then Q(√d) is a finite extension of Q. In particular it is a 2-dimensional vector space over Q, called a quadratic field. If d < 0, then the class number of the ring R of algebraic integers of Q(√d) is equal to 1 for precisely the following values of d: d = -1, -2, -3, -7, -11, -19, -43, -67, and -163. This result was first conjectured by Gauss and proven by Kurt Heegner, although Heegner's proof was not believed until Harold Stark gave a later proof in 1967, which Stark showed was actually equivalent to Heegner's. (See Stark-Heegner theorem.) This is a special case of the famous class number problem.

If, on the other hand, d > 0, then it is unknown whether there are infinitely many fields Q(√d) with class number 1. Computational results indicate that there are a great many such fields to say the least.

[edit] Example of a non-trivial class group

The ring R = Z [√−5] is the ring of integers of Q(√−5). It does not possess unique factorisation; in fact the class group of R is cyclic of order 2. Indeed, the ideal

J = (2, 1 + √−5)

is not principal, which can be proved by contradiction as follows. If J were generated by an element x of R, then x would divide both 2 and 1 + √−5. Then the norm N(x) of x would divide both N(2) = 4 and N(1 + √−5) = 6, so N(x) would divide 2. We are assuming that x is not a unit of R, so N(x) cannot be 1. It cannot be 2 either, because R has no elements of norm 2, that is, the equation b2 + 5c2 = 2 has no solutions in integers.

One also computes that J2 = (2), which is principal, so the class of J in the ideal class group has order two. Showing that there aren't any other ideal classes requires more effort.

The fact that this J is not principal is also related to the fact that the element 6 has two distinct factorisations into irreducibles:

6 = 2 × 3 = (1 + √−5) × (1 − √−5).

[edit] Connections to class field theory

Class field theory is a branch of algebraic number theory which seeks to classify all the abelian extensions of a given algebraic number field, meaning Galois extensions with abelian Galois group. A particularly beautiful example is found in the Hilbert class field of a number field, which can be defined as the maximal unramified abelian extension of such a field. The Hilbert class field L of a number field K is unique and has the following properties:

  • Every ideal of the ring of integers of K becomes principal in L, i.e., if I is an integral ideal of K then the image of I is a principal ideal in L.
  • L is a Galois extension of K with Galois group isomorphic to the ideal class group of K.

Neither property is particularly easy to prove.

[edit] See also