proof that a domain is Dedekind if its ideals are products of maximals
Let be an integral domain. We show that it is a Dedekind domain if and only if every nonzero proper ideal can be expressed as a product of maximal ideals. To do this, we make use of the characterization of Dedekind domains as integral domains in which every nonzero integral ideal is invertible (proof that a domain is Dedekind if its ideals are invertible).
First, let us suppose that every nonzero proper ideal in is a product of maximal ideals. Let be a maximal ideal and choose a nonzero . Then, by assumption,
for some and maximal ideals . As is a principal ideal, each of the factors is invertible. Also,
As is prime, this gives for some . However, is maximal so must equal , showing that is indeed invertible. Then, every nonzero proper ideal is a product of maximal, and hence invertible, ideals and so is invertible, and it follows that is Dedekind.
We now show the reverse direction, so suppose that is Dedekind. Proof by contradiction will be used to show that every nonzero ideal is a product of maximals, so suppose that this is not the case. Then, as is defined to be Noetherian (http://planetmath.org/Noetherian), there is an ideal maximal (http://planetmath.org/MaximalElement) (w.r.t. the partial order of set inclusion) among those proper ideals which are not a product of maximal ideals. Then cannot be a maximal ideal itself, so is strictly contained in a maximal ideal and, as is invertible, we can write for an ideal .
Therefore and we cannot have equality, otherwise cancelling from would give . So, is strictly larger than and, by the choice of , is therefore a product of maximal ideals. Finally, is then also a product of maximal ideals.
|Title||proof that a domain is Dedekind if its ideals are products of maximals|
|Date of creation||2013-03-22 18:35:04|
|Last modified on||2013-03-22 18:35:04|
|Last modified by||gel (22282)|