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.