# proof that a domain is Dedekind if its ideals are 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.

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 ProofThatADomainIsDedekindIfItsIdealsAreInvertible 2013-03-22 18:34:54 2013-03-22 18:34:54 gel (22282) gel (22282) 5 gel (22282) Proof msc 13A15 msc 13F05 DedekindDomain FractionalIdeal