If is an integral domain, the *field of fractions*
of is the field of all equivalence classes of formal quotients
, where with , and
if .
For example, the field of fractions of
is (canonically isomorphic
to)
and the field of fractions of
is
. The field of fractions of the ring of integers
of a number field is just the number field .

sage: Frac(ZZ) Rational FieldIn Sage the

sage: K.<a> = QuadraticField(5) sage: OK = K.ring_of_integers(); OK Order with module basis 1/2*a + 1/2, a in Number Field in a with defining polynomial x^2 - 5 sage: Frac(OK) Number Field in a with defining polynomial x^2 - 5

The fraction field of an *order* - i.e., a subring of of
finite index - is also the number field again.

sage: O2 = K.order(2*a); O2 Order with module basis 1, 2*a in Number Field in a with defining polynomial x^2 - 5 sage: Frac(O2) Number Field in a with defining polynomial x^2 - 5

Without loss we may assume that , so that . Suppose is integral over . Then since is integrally closed, is an element of , so , as required.

The ring is not a Dedekind domain because it is not an integral domain. The ring is not a Dedekind domain because it is not integrally closed in its field of fractions, as is integrally over and lies in , but not in . The ring is a Dedekind domain, as is any ring of integers of a number field, as we will see below. Also, any field is a Dedekind domain, since it is an integral domain, it is trivially integrally closed in itself, and there are no nonzero prime ideals so the condition that they be maximal is empty. The ring is not noetherian, but it is integrally closed in its field of fraction, and every nonzero prime ideal is maximal.

If and are ideals in a ring , the product is the ideal generated by all products of elements in with elements in :

sage: K.<a> = NumberField(x^2 + 23) sage: I = K.fractional_ideal(2, 1/2*a - 1/2) sage: J = I^2 sage: I Fractional ideal (2, 1/2*a - 1/2) of Number Field ... sage: J Fractional ideal (4, 1/2*a + 3/2) of Number Field ... sage: I*J Fractional ideal (-1/2*a - 3/2) of Number Field ...

Since fractional ideals are finitely generated, we can clear denominators of a generating set to see that there is some nonzero such that

For example, the set of rational numbers with denominator or is a fractional ideal of .

The product of two fractional ideals is again finitely generated, so it is a fractional ideal, and for any ideal , so to prove that the set of fractional ideals under multiplication is a group it suffices to show the existence of inverses. We will first prove that if is a prime ideal, then has an inverse, then we will prove that all nonzero integral ideals have inverses, and finally observe that every fractional ideal has an inverse. (Note: Once we know that the set of fractional ideals is a group, it will follows that inverses are unique; until then we will be careful to write ``an'' instead of ``the''.)

Suppose is a nonzero prime ideal of . We will show that the -module

For the rest of the proof, fix a nonzero element . Since is an -module, is an ideal, hence is a fractional ideal. Since we have , hence since is maximal, either or . If , we are done since then is an inverse of . Thus suppose that . Our strategy is to show that there is some , with . Since , such a would leave invariant, i.e., . Since is an -module we will see that it will follow that , a contradiction.

By Lemma 3.1.10, we can choose a product , with minimal, with

To finish the proof that has an inverse, we observe that preserves the -module , and is hence in , a contradiction. More precisely, if is a basis for as a -module, then the action of on is given by a matrix with entries in , so the minimal polynomial of has coefficients in (because satisfies the minimal polynomial of , by the Cayley-Hamilton theorem - here we also use that , since is a finite set). This implies that is integral over , so , since is integrally closed by Proposition 3.1.3. (Note how this argument depends strongly on the fact that is integrally closed!)

So far we have proved that if is a prime ideal of , then

We can finally deduce the crucial Theorem 3.1.11, which will allow us to show that any nonzero ideal of a Dedekind domain can be expressed uniquely as a product of primes (up to order). Thus unique factorization holds for ideals in a Dedekind domain, and it is this unique factorization that initially motivated the introduction of ideals to mathematics over a century ago.

Suppose . If no is contained in , then for each there is an such that . But the product of the is in , which is a subset of , which contradicts that is a prime ideal. Thus for some . We can thus cancel and from both sides of the equation by multiplying both sides by the inverse. Repeating this argument finishes the proof of uniqueness.

where each of the ideals and is prime. We will discuss algorithms for computing such a decomposition in detail in Chapter 4. The first idea is to write , and hence reduce to the case of writing the , for prime, as a product of primes. Next one decomposes the finite (as a set) ring .

The factorization (3.1.1) can be compute using
*SAGE* as follows:

sage: K.<a> = NumberField(x^2 + 6); K Number Field in a with defining polynomial x^2 + 6 sage: K.factor_integer(6) (Fractional ideal (2, a) of Number Field ...)^2 * (Fractional ideal (3, a) of Number Field ...)^2

William Stein 2012-09-24