4.5 The Ideal Class Group of a Dedekind Domain

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We continue to assume that R is a Dedekind domain, with fraction field

E. An R-submodule B _ E is called a fractional ideal if it is a finitely

generated module.

Lemma 4.5.1 Let B be a fractional ideal. Then there exist prime ideals

P1, P2, . . . , Pr,Q1,Q2, . . . ,Qs such that B = RP1P2 · · · PrQ1Q2 · · ·Qs. (It is

possible for either r = 0 or s = 0.)

Corollary 4.5.1.1 The set of fractional ideals in E forms an abelian group

under multiplication.

A fractional ideal B _ E is called a principal fractional ideal if it is of

the form R_, for some _ 2 E. Note that in this case, B−1 = R( 1

_). It is easy

to show that if R is a principal ideal domain, then every fractional ideal is

principal (Exercise 1).

If F is the set of fractional ideals in E we have seen that F is an abelian

group under multiplication, with identity R. If we denote by P the set of

prinicpal fractional ideals, then it is easy to see that P is a subgroup of

F; the quotient group C = F/P is called the ideal class group of R; it is

trivial precisely when R is a principal ideal domain. If R = OE for a finite

extension E _ Q, then it is known that C is a finite group. The order h = |C|

is called the class number of R (or of E) and is a fundamental invariant in

algebraic number theory.

Exercises

1. If R is a p.i.d., prove that every fractional ideal of E is principal.

2. Let R be a Dedekind domain with fraction field E. Prove that E itself

is not a fractional ideal (except in the trivial case in which case R is a

field to be begin with).

3. Let R be a Dedekind domain with ideal class group C. Let P _ R

be a prime ideal and assume that the order of the element [P] 2 C is

k > 1. If Pk = (_), for some _ 2 R, show that _ is irreducible but not

prime.

4.6. A CHARACTERIZATION OF DEDEKIND DOMAINS 101

4. Let R be a Dedekind domain with ideal class group of order at most

2. Prove that the number of irreducible factors in a factorization

of an element a 2 R depends only on a.1 (Hint: Note first that

by Exercise 6 of Section 3.3, any non-unit of R can be factored into

irreducibles. By induction on the minimal length of a factorization of

a 2 R into irreducibles, we may assume that a has no prime factors.

Next assume that _ 2 R is a non-prime irreducible element. If we

factor the principal ideal into prime ideals: (_) = Q1Q2 · · ·Qr then

the assumption guarantees that Q1Q2 = (_), for some _ 2 R. If

r > 2, then (_) is properly contained in Q1Q2 = (_) and so _ is

a proper divisor of _, a contradiction. Therefore, it follows that a

principal ideal generated by a non-prime irreducible element factors

into the product of two prime ideals. Now what?)

5. Let R be as above, i.e., a Dedekind domain with ideal class group

of order at most 2. Let _1, _2 2 R be irreducible elements. As we

seen in Exercise 4 above, any factorization of _1_2 will involve exactly

two irreducibles. Show that, up to associates, there can be at most

three distinct factorizations of _1_2 into irreducibles. (As a simple

illustration, it turns out that the Dedekind domain Z[

p

−5] has class

group of order 2; correspondingly we have distinct factorizations: 21 =

3 · 7 = (1 + 2

p

−5)(1 − 2

p

−5) = (4 +

p

−5)(4 −

p

−5).)

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