proof that a domain is Dedekind if its ideals are invertible
Every fractional ideal is invertible.
If every fractional ideal is invertible, then is Dedekind.
First, every invertible ideal is finitely generated, so is Noetherian.
Now, let be a prime ideal, and be a maximal ideal containing . As is invertible, there exists an ideal such that . That is a prime ideal implies or . The first case gives and, by cancelling the invertible ideal implies that , a contradiction. So, the second case must be true and, by maximality of , , showing that all prime ideals are maximal.
Now let be an element of the field of fractions and be integral over . Then, we can write for coefficients . Letting be the fractional ideal
gives , so . As is invertible, it can be cancelled to give , showing that is integrally closed. ∎
It only remains to show the converse, that is if is Dedekind then every nonzero ideal is invertible. We start with the following lemmas.
Every nonzero ideal contains a product of prime ideals. That is, for some nonzero prime ideals .
We use proof by contradiction, so suppose this is not the case. As is Noetherian, the set of nonzero ideals which do not contain a product of nonzero primes has a maximal element (w.r.t. the partial order of set inclusion) say, .
In particular cannot be prime itself, so there exist such that and . Therefore is strictly contained in and and, by the choice of , these ideals must contain a product of primes. So,
contains a product of primes, which is the required contradiction. ∎
Let be a maximal ideal containing and be a nonzero element of . By the previous lemma there are prime ideals satisfying
We choose as small as possible. As is prime, this gives for some and, as every prime ideal is maximal, this is an equality. Without loss of generality we may take . As was assumed to be as small as possible, is not a subset of , so there exists . Then, gives and
as required. ∎
We finally show that every nonzero ideal is invertible. If its inverse exists then it should be the largest fractional ideal satisfying , so we set
Choosing any nonzero gives so is indeed a fractional ideal. It only remains to be shown that , for which we use proof by contradiction. If this were not the case then the previous lemma gives an such that . By the definition of , this gives and therefore is an -module. Furthermore, as is Noetherian, will be finitely generated as an -module. This implies that is integral over the integrally closed ring , so , giving the required contradiction.
|Title||proof that a domain is Dedekind if its ideals are invertible|
|Date of creation||2013-03-22 18:34:54|
|Last modified on||2013-03-22 18:34:54|
|Last modified by||gel (22282)|