If is an integral domain, the * of is
the field of all elements , where . The field of
fractions of is the smallest field that contains . For
example, the field of fractions of
is
and of
is
.
*

Suppose is integral over . Then since is integrally closed, is an element of , so , as required.

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

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

To see that this notion of divides is sensible, suppose , so . Then and for some integer and , and divides means that , i.e., that there exists an integer such that , which exactly means that divides , as expected.

We are now ready to prove the theorem.

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 either or . If , we are done since then is an inverse of . Thus suppose that . Our strategy is to show that there is some not in ; such a would leave invariant (i.e., ), so since is an -module it will follow that , a contradiction.

By Lemma 6.1.8, we can choose a product , with minimal, such that

So far we have proved that if is a prime ideal of , then is the inverse of in the monoid of nonzero fractional ideals of . As mentioned after Definition 6.1.5 [on Tuesday], every nonzero fractional ideal is of the form for and an integral ideal, so since has inverse , it suffices to show that every integral ideal has an inverse. If not, then there is a nonzero integral ideal that is maximal among all nonzero integral ideals that do not have an inverse. Every ideal is contained in a maximal ideal, so there is a nonzero prime ideal such that . Multiplying both sides of this inclusion by and using that , we see that . If , then arguing as in the proof that is the inverse of , we see that each element of preserves the finitely generated -module and is hence integral. But then , which implies that , a contradiction. Thus . Because is maximal among ideals that do not have an inverse, the ideal does have an inverse, call it . Then is the inverse of , since .

We can finally deduce the crucial Theorem 6.1.10, 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 rings of integers of number fields 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 the , which is a subset of , which contradicts the fact that is a prime ideal. Thus for some . We can thus cancel and from both sides of the equation. Repeating this argument finishes the proof of uniqueness.

> R<x> := PolynomialRing(RationalField()); > K := NumberField(x^2+6); > OK := MaximalOrder(K); > [K!b : b in Basis(OK)]; [ 1, K.1 // this is sqrt(-6) ] > Factorization(6*OK); [ <Prime Ideal of OK Two element generators: [2, 0] [2, 1], 2>, <Prime Ideal of OK Two element generators: [3, 0] [3, 1], 2> ]The output means that

William Stein 2004-05-06