proof that a domain is Dedekind if its ideals are products of maximals

Let $R$ 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 $R$ is a product of maximal ideals. Let $\mathfrak{m}$ be a maximal ideal and choose a nonzero $x\in\mathfrak{m}$ . Then, by assumption,

(x)=\mathfrak{m}_{1}\cdots\mathfrak{m_{n}}

for some $n\geq 0$ and maximal ideals $\mathfrak{m}_{k}$ . As $(x)$ is a principal ideal, each of the factors $\mathfrak{m}_{k}$ is invertible. Also,

\mathfrak{m}_{1}\cdots\mathfrak{m_{n}}\subseteq\mathfrak{m}.

As $\mathfrak{m}$ is prime, this gives $\mathfrak{m}_{k}\subseteq\mathfrak{m}$ for some $k$ . However, $\mathfrak{m}_{k}$ is maximal so must equal $\mathfrak{m}$ , showing that $\mathfrak{m}$ 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 $R$ is Dedekind.

We now show the reverse direction, so suppose that $R$ 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 $R$ is defined to be Noetherian (http://planetmath.org/Noetherian), there is an ideal $\mathfrak{a}$ 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 $\mathfrak{a}$ cannot be a maximal ideal itself, so is strictly contained in a maximal ideal $\mathfrak{m}$ and, as $\mathfrak{m}$ is invertible, we can write $\mathfrak{a}=\mathfrak{mb}$ for an ideal $\mathfrak{b}$ .

Therefore $\mathfrak{a}\subseteq\mathfrak{b}$ and we cannot have equality, otherwise cancelling $\mathfrak{a}$ from $\mathfrak{a}=\mathfrak{ma}$ would give $\mathfrak{m}=R$ . So, $\mathfrak{b}$ is strictly larger than $\mathfrak{a}$ and, by the choice of $\mathfrak{a}$ , is therefore a product of maximal ideals. Finally, $\mathfrak{a}=\mathfrak{mb}$ is then also a product of maximal ideals.

Title	proof that a domain is Dedekind if its ideals are products of maximals
Canonical name	ProofThatADomainIsDedekindIfItsIdealsAreProductsOfMaximals
Date of creation	2013-03-22 18:35:04
Last modified on	2013-03-22 18:35:04
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 13F05
Classification	msc 13A15
Related topic	DedekindDomain
Related topic	MaximalIdeal
Related topic	FractionalIdeal