proof that a domain is Dedekind if its ideals are invertible

Let $R$ be an integral domain with field of fractions $k$ . We show that the following are equivalent.

1.

$R$ is Dedekind. That is, it is Noetherian (http://planetmath.org/Noetherian), integrally closed, and every prime ideal is maximal (http://planetmath.org/MaximalIdeal).
2.

Every nonzero (integral) ideal is invertible.
3.

Every fractional ideal is invertible.

As every fractional ideal is the product of an element of $k$ and an integral ideal, statements (2) and (3) are equivalent. We start by proving that (3) implies $R$ is Dedekind.

Lemma.

If every fractional ideal is invertible, then $R$ is Dedekind.

Proof.

First, every invertible ideal is finitely generated, so $R$ is Noetherian.

Now, let $\mathfrak{p}$ be a prime ideal, and $\mathfrak{m}$ be a maximal ideal containing $\mathfrak{p}$ . As $\mathfrak{m}$ is invertible, there exists an ideal $\mathfrak{a}$ such that $\mathfrak{p}=\mathfrak{m}\mathfrak{a}$ . That $\mathfrak{p}$ is a prime ideal implies $\mathfrak{a}\subseteq\mathfrak{p}$ or $\mathfrak{m}\subseteq\mathfrak{p}$ . The first case gives $\mathfrak{p}\subseteq\mathfrak{m}\mathfrak{p}$ and, by cancelling the invertible ideal $\mathfrak{p}$ implies that $\mathfrak{m}=R$ , a contradiction. So, the second case must be true and, by maximality of $\mathfrak{m}$ , $\mathfrak{p}=\mathfrak{m}$ , showing that all prime ideals are maximal.

Now let $x$ be an element of the field of fractions $k$ and be integral over $R$ . Then, we can write $x^{n}=c_{0}+c_{1}x+\cdots+c_{n-1}x^{n-1}$ for coefficients $c_{k}\in R$ . Letting $\mathfrak{a}$ be the fractional ideal

\mathfrak{a}=(1,x,x^{2},\ldots,x^{n-1})

gives $x^{n}\in\mathfrak{a}$ , so $x\mathfrak{a}\subseteq\mathfrak{a}$ . As $\mathfrak{a}$ is invertible, it can be cancelled to give $x\in R$ , showing that $R$ is integrally closed. ∎

It only remains to show the converse, that is if $R$ is Dedekind then every nonzero ideal is invertible. We start with the following lemmas.

Lemma.

Every nonzero ideal $\mathfrak{a}$ contains a product of prime ideals. That is, $\mathfrak{p}_{1}\cdots\mathfrak{p}_{n}\subseteq\mathfrak{a}$ for some nonzero prime ideals $\mathfrak{p}_{k}$ .

Proof.

We use proof by contradiction, so suppose this is not the case. As $R$ 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, $\mathfrak{a}$ .

In particular $\mathfrak{a}$ cannot be prime itself, so there exist $x,y\in R$ such that $xy\in\mathfrak{a}$ and $x,y\not\in\mathfrak{a}$ . Therefore $\mathfrak{a}$ is strictly contained in $\mathfrak{a}+(x)$ and $\mathfrak{a}+(y)$ and, by the choice of $\mathfrak{a}$ , these ideals must contain a product of primes. So,

(\mathfrak{a}+(x))(\mathfrak{a}+(y))=\mathfrak{a}^{2}+x\mathfrak{a}+y\mathfrak% {a}+(xy)\subseteq\mathfrak{a}

contains a product of primes, which is the required contradiction. ∎

Lemma.

For any nonzero proper ideal $\mathfrak{a}$ there is an element $x\in k\setminus R$ such that $x\mathfrak{a}\subseteq R$ .

Proof.

Let $\mathfrak{p}$ be a maximal ideal containing $\mathfrak{a}$ and $a$ be a nonzero element of $\mathfrak{a}$ . By the previous lemma there are prime ideals $\mathfrak{p}_{1},\ldots,\mathfrak{p}_{n}$ satisfying

\mathfrak{p}_{1}\cdots\mathfrak{p}_{n}\subseteq(a)\subseteq\mathfrak{a}% \subseteq\mathfrak{p}.

We choose $n$ as small as possible. As $\mathfrak{p}$ is prime, this gives $\mathfrak{p}_{k}\subseteq\mathfrak{p}$ for some $k$ and, as every prime ideal is maximal, this is an equality. Without loss of generality we may take $\mathfrak{p}=\mathfrak{p}_{n}$ . As $n$ was assumed to be as small as possible, $\mathfrak{p}_{1}\cdots\mathfrak{p}_{n-1}$ is not a subset of $(a)$ , so there exists $b\in\mathfrak{p}_{1}\cdots\mathfrak{p}_{n-1}\setminus(a)$ . Then, $b\not\in(a)$ gives $x\equiv a^{-1}b\not\in R$ and

x\mathfrak{a}\subseteq a^{-1}b\mathfrak{p}\subseteq a^{-1}\mathfrak{p}_{1}% \cdots\mathfrak{p}_{n-1}\mathfrak{p}\subseteq a^{-1}(a)=R

as required. ∎

We finally show that every nonzero ideal $\mathfrak{a}$ is invertible. If its inverse exists then it should be the largest fractional ideal satisfying $\mathfrak{ba}\subseteq R$ , so we set

\mathfrak{b}=\left\{x\in k:x\mathfrak{a}\subseteq R\right\}.

Choosing any nonzero $a\in\mathfrak{a}$ gives $a\mathfrak{b}\subseteq\mathfrak{ba}\subseteq R$ so $\mathfrak{b}$ is indeed a fractional ideal. It only remains to be shown that $\mathfrak{ba}=R$ , for which we use proof by contradiction. If this were not the case then the previous lemma gives an $x\in k\setminus R$ such that $x\mathfrak{ba}\subseteq R$ . By the definition of $\mathfrak{b}$ , this gives $x\mathfrak{b}\subseteq\mathfrak{b}$ and therefore $\mathfrak{b}$ is an $R[x]$ -module. Furthermore, as $R$ is Noetherian, $\mathfrak{b}$ will be finitely generated as an $R$ -module. This implies that $x$ is integral over the integrally closed ring $R$ , so $x\in R$ , giving the required contradiction.

Title	proof that a domain is Dedekind if its ideals are invertible
Canonical name	ProofThatADomainIsDedekindIfItsIdealsAreInvertible
Date of creation	2013-03-22 18:34:54
Last modified on	2013-03-22 18:34:54
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 13A15
Classification	msc 13F05
Related topic	DedekindDomain
Related topic	FractionalIdeal