proof that a Noetherian domain is Dedekind if it is locally a PID
For a given maximal ideal and ideal of , we shall write for the ideal generated by in , which consists of the elements of the form for and . It is then easily seen that gives a bijection between the prime ideals of contained in and the prime ideals of , with inverse . In particular is the unique maximal ideal of , which is therefore a local ring.
Now suppose that is Dedekind, then the localization will be a Dedekind domain (localizations of Dedekind domains are Dedekind) with a unique maximal ideal, so it is a principal ideal domain (Dedekind domains with finitely many primes are PIDs).
Only the converse remains to be shown, so suppose that is a Noetherian domain such that is a principal ideal domain for every maximal ideal . In particular, is integrally closed and every nonzero prime ideal is maximal, so it contains a unique nonzero prime ideal .
We start by showing that every nonzero prime ideal of is maximal. Choose a maximal ideal containing . Then, is a nonzero prime ideal, so and therefore is maximal.
We finally show that is integrally closed. So, choose any integral over and lying in its field of fractions. Let be the ideal
We use proof by contradiction to show that is the whole of . So, supposing that this is not the case, there exists a maximal ideal containing . Then will be integral over the integrally closed ring and therefore . So, for some and . Then, so , which is the required contradiction. Therefore, and, in particular, and , showing that is integrally closed.
|Title||proof that a Noetherian domain is Dedekind if it is locally a PID|
|Date of creation||2013-03-22 18:35:27|
|Last modified on||2013-03-22 18:35:27|
|Last modified by||gel (22282)|