Dedekind Domains

Duals of Fractional Ideals and Projectivity Criteria

Recall that $I$ is a fractional ideal, if $I\subseteq F=\mathrm{Frac}(R)$ is an $R$-submodule, and there exists $\lambda \in R$ such that $\lambda I\subseteq R$. If $(e_i{)}_{1⩽i⩽r}$ is an $F$-basis of $V$ and $(I_i{)}_{1⩽i⩽r}$ is a family of fractional ideals, then $M={\oplus }_{i=1}^r{I}_i{e}_i$ is a fractional module in $V$. By definition, it is easy to check the following lemma.

Lemma

Let $I_1,I_2$ be fractional ideals.

• $I_1+I_2$, which is the $R$-submodule of $F$ generated by $I_1\cup I_2$
• $I_1\cap I_2$
• $I_1{I}_2$, which is the $R$-submodule of $F$ generated by all $x_1{x}_2$ with $x_1\in I_1$, $x_2\in I_2$

are all fractional ideals.

The following proposition introduces the concept of the inverse of a fractional ideal, $I^{\ast }$, which is the key to showing that the set of nonzero fractional ideals forms an abelian group. The main result provides a module-theoretic criterion for invertibility: $I^{\ast }I=R$ if and only if $I$ is a projective $R$-module.

Proposition

• For any fractional ideal $I$, there is a natural identification between the dual of $I$ and a set of $F$: $I^{\ast }=\{x\in F:xI\subseteq R\}$.
• Let $V$ be an $r$-dimensional $F$-space with a basis $(e_i{)}_{1⩽i⩽r}$, and let $(I_i{)}_{1⩽i⩽r}$ be a family of fractional ideals. Set $M={\oplus }_{i=1}^r{I}_i{e}_i$. Then
  • $M^{\ast }={\oplus }_{i=1}^r{I}_i^{\ast }{e}_i^{\ast }$, where $e_i^{\ast }\in V^{\ast }$.
  • the fractional module $M$ is a finitely generated projective $R$-module iff for all $1⩽i⩽r$, we have $I_i^{\ast }{I}_i=R$. In particular, a fractional ideal $I$ is finitely generated projective $R$-module iff $I^{\ast }I=R$.

\begin{proof} i) Let $0\ne \lambda \in R$ such that $\lambda I\subseteq R$. Take $\phi \in {\mathrm{Hom}}_R(I,R)=I^{\ast }$. For $x,y\in I$ and

$$\phi (\lambda xy)=\lambda \phi (xy)=\lambda x\phi (y)=\lambda y\phi (x).$$

Since $R$ is an integral domain, one has

$$\phi (xy)=x\phi (y)=y\phi (x).$$

If $x,y\ne 0$, there is $y^{-1}\phi (y)=x^{-1}\phi (x)$. So we can define $h_{\phi }\in F$ by $h_{\phi }=x^{-1}\phi (x)$ for any $x\in I\setminus \{0\}$. For any $i\in I$, $h_{\phi }$ can be written as $h_{\phi }=i^{-1}\phi (i)$ and so $h_{\phi }i=\phi (i)\in R$. We can check the map $\phi \mapsto h_{\phi }$ is an isomorphism, that is, ${\mathrm{Hom}}_R(I,R)\mapsto I^{\ast }=\{x\in F:xI\subseteq R\}$. Moreover, $I^{\ast }$ is a fractional ideal.

ii) a) is easy.

ii) b) It suffices to show, a fractional ideal $I$ is finitely generated projective iff $I^{\ast }I=R$. Recall that $I$ is a finitely generated projective module iff $\mathrm{id}\in {\tau }_{I,I}$, i.e., there exists ${\sum }_i{b}_i\otimes a_i\in I^{\ast }{\otimes }_{R}I$ such that for any $x\in I$, ${\sum }_{i}x{b}_i{a}_i=x$, i.e. ${\sum }_i{b}_i{a}_i=1$ iff $I^{\ast }I=R$. \end{proof}

Definition

A finitely generated projective fractional ideal is called invertible. That is, for a fractional ideal $I$, $I$ is invertible iff $I^{\ast }I=R$.

Dedekind Domain

Definition

A principal ideal domain is a integral domain $R$ all ideals of which are free.

Remark. Here the definition of principal ideal domain is equivalent to the standard definition (an integral domain whose ideals are principal). One can get principal ideal is free easily; conversely, if a free ideal $I$ has rank more than $1$, assume $a$ and $b$ are two distinct basis elements of $I$ and then $(-b)\cdot a+a\cdot b=0$ leads to a contradiction.

As a generalization of PID, we define Dedekind Domain.

Definition

A Dedekind domain is a integral domain $R$ all ideals of which are finitely generated projective.

Remark. In fact, an equivalent definition of Dedekind domain is a $1$-dimensional integral closed Noetherian ring.

• A ring $R$ is $1$-dimensional integral closed Noetherian ring iff $R_p$ is DVR for each nonzero prime ideal $p$. See ^rmtvh7.

• For an ideal $I\subseteq R$, $I$ is projective iff $I_p$ is projective for each prime ideal $p$. See ^766550

• It suffices to show $R_p$ is DVR iff $I_p$ is projective for each ideal for each prime ideal $p$. Since a DVR is a PID, $I_p$ is a principal ideal and so is free and projective. Furthermore, $I$ finitely generated is guaranteed by Noetherian. Now we finish the proof.

• an integral domain is a Dedekind domain if and only if it is a hereditary ring

• Local PID and local Dedekind domain are equivalent. See ^wzfm6a.

Local+Dedekind = PID

Recall definition and properties of local rings:

We say a ring $R$ (possibly non-commutative) is local if the set of non-units of $R$ form an ideal of $R$.

TFAE for a ring R.

• R is a local ring
• the set of nonunits of R coincides with J(R), that is, J(R) is a unique maximal left (or right) ideal of R
• R/J(R) is a division ring.

Proposition

Let $R$ be a commutative ring. Assume $R$ is local with unique maximal ideal $I$. Then every finitely generated projective $R$-module is free.

\begin{proof} Let $X$ be a finitely generated projective $R$-module. Then the module $X/IX$ is a finite-dimensional $(R/I)$-vector space with dimension $r$. The isomorphism $(R/I{)}^r\stackrel{\sim }{\to }X/IX$ can be lifted to an $R$-morphism $\phi :R^r\to X$.

Note that $\phi$ is an epimorphism as $\phi (R^r)+IX=X$ by Nakayama lemma. Define $X^{\prime }=\ker \phi$. Since

$$0\to X^{\prime }\to R^r\to X\to 0$$

is exact and $X$ is projective, one have $X^{\prime }$ is a direct summand of $R^r$ by ^3rkmd6 and so it is finitely generated. Since $R^r\simeq X^{\prime }\oplus X$, there is $(R/I{)}^r\simeq (X^{\prime }/I{X}^{\prime })\oplus (X/IX)$. So $X^{\prime }/I{X}^{\prime }=0$ and then $X^{\prime }=0$. Hence $X=R^r$ is free. \end{proof}

Corollary

Local Dedekind domain = Local PID

\begin{proof} Recall that PID is always Dedekind. Conversely, for a local Dedekind domain, all its ideals are finitely generated projective and so free by ^spgbvh. Hence it is PID. \end{proof}

Localization of PIDs and Dedekind Rings

Proposition

Let $R$ be an integral domain. Let $S$ be a nonempty multiplicatively closed subset of $R\setminus \{0\}$.

• If $R$ is a PID, so is $S^{-1}R$.
• If $R$ is a Dedekind domain, so is $S^{-1}R$.

\begin{proof} Define $j:R\hookrightarrow S^{-1}R,r\mapsto r/1$, which is an embedding. For any ideal $J\subseteq S^{-1}R$, we have that

$$I=j^{-1}(J)=\{r\in R:j(r)\in J\}$$

is an ideal of $R$. For any $r/s\in J$, there is $r/1\in J$ and so $r\in I$. Thus $S^{-1}I\supseteq J$. On the other hand, for any $x\in S^{-1}I$, $x$ can be written as $x=r^{\prime }/s^{\prime }$ with $r^{\prime }\in I,s^{\prime }\in S$. Then $r^{\prime }/s^{\prime }\in J$ by $r^{\prime }\in I$ and $r^{\prime }/s^{\prime }=(r^{\prime }/1)(1/s^{\prime })$. Now we proved $S^{-1}I=J$, that is, each ideal $J$ of $S^{-1}R$ can be written as $J=S^{-1}I$ for some ideal $I\subseteq R$.

i) For any ideal $J\subseteq S^{-1}R$, $J=S^{-1}I$ for some ideal $I$ of $R$. Since $R$ is a principal ideal domain, we know $I$ is free. Note that $xy-yx=0$ for any $x,y\in I$, so $\mathrm{rank}I⩽1$. If $I=(0)$, then $J=S^{-1}I=(0)$ is also free. If $I\ne (0)$, then $I=(a)$ for some $a\in I$. In this case, $S^{-1}J=(a/1)$ is also free. By the arbitrary of $J$, $S^{-1}R$ is a principal ideal domain.

ii) For any ideal $J\subseteq S^{-1}R$, $J=S^{-1}I$ for some ideal $I$ of $R$. Since $R$ is a Dedekind ring, $I$ is finitely generated projective, that is, $I$ is a direct summand of $R^n$ for some $n$. Assume that $R^n=I\oplus I^{\prime }$. Note that $J=S^{-1}R\otimes I$ and $S^{-1}R$ is flat, so we have

$$(S^{-1}R{)}^n=S^{-1}{R}^n=S^{-1}R\otimes R^n=(S^{-1}R\otimes I)\oplus (S^{-1}R\otimes I^{\prime })=J\oplus J^{\prime }$$

and $J$ is a direct summand of $(S^{-1}R{)}^n$. So $J$ is also a finitely generated projective ideal. Now we finish the proof. \end{proof}

Ideal Theory in Dedekind Domains

Recall that we have defined invertible fractional ideal before in ^iazl34.

Lemma

An integral domain $R$ is a Dedekind domain iff all fractional ideals are invertible.

\begin{proof} By ^19yf5a and ^iazl34. \end{proof}

For Dedekind domain, the dual $I^{\ast }$ of a fractional ideal $I$ is denoted by $I^{-1}$.

Lemma

The set of fractional ideals of a Dedekind domain $R$ is an abelian group. The identity element is $R$, the inverse of a fractional ideal $I$ is $I^{-1}$.

Definition

A fractional ideal $I$ of $R$ is called integral ideal if $I\subseteq R$.

Let $I$ and $J$ be fractional ideal of $R$, we say $J$ divides $I$, that is, $J\mid I$, if there is an integral ideal such that $I=JK$.

Lemma

Let $I$ be fractional ideals in a Dedekind domain $R$. TFAE:

• $I\subseteq J$;
• $J\mid I$.

\begin{proof} ii)→i) is easy.

i)→ii). Let $K=J^{-1}I$. Then $I=JK$. Note that $J^{-1}I\subseteq J^{-1}J=R$ and so $J^{-1}I=K$ is an integral ideal. \end{proof}

Corollary

Every nonzero prime ideal of a Dedekind domain $R$ is maximal.

\begin{proof} Assume that $p$ is a nonzero prime ideal and $p\subseteq I$ for some ideal $I$ of $R$. We aim to show $I=p$ or $I=R$. By ^8e8e1e, there exists an ideal $I^{\prime }$ such that $p=I^{\prime }I$. Since $p$ is prime, one have either $I\subseteq p$ or $I^{\prime }\subseteq p$.

• exercise: Since $p$ is prime, one have either $I\subseteq p$ or $I^{\prime }\subseteq p$.

If $I\subseteq p$, we have done. If $I^{\prime }\subseteq p$, then $p=pI$ and $I=R$. \end{proof}

Corollary

A Dedekind domain is integral closed.

\begin{proof} Let $R$ be a Dedekind ring, and let $F$ be the fraction field of $R$. Let $x\in F$ be integral over $R$, i.e., $x^n+a_{n-1}{x}^{n-1}+\cdots +a_0=0$. Then $x$ belongs to the fractional ideal $I$ generated by $1,x,\cdots ,x^{n-1}$. We can check $I^2=I$. So $I=I^2\cdot I^{-1}=I{I}^{-1}=R$ and so $x\in R$.

Alternating proof. See ^26zbyp. \end{proof}

unique factorization of ideals in Dedekind domain

Let $R$ be a Dedekind domain. Every nonzero proper ideal of $R$ can be written in a unique way as a product of prime ideals.

\begin{proof} Step 1. Prove any proper ideal is a product of prime ideals.

If not, then there exists an ideal $I$ which is maximal among proper ideals which are not product of prime ideals. In particular, $I$ is not prime and so not maximal. So there exists a maximal ideal $J$ such that $I⊊J⊊R$. By ^8e8e1e, $I=JK$ for some ideal $K$ of $R$.

Note that $I\ne K$, otherwise $J=R$. Hence $I⊊K⊊R$. By maximality of $I$, $J$ and $K$ are products of prime ideals. So $I$ can be written as a product of prime ideal, which is a contradiction.

Step 2. Uniqueness.

Assume that $I=p_1\cdots p_m=q_1\cdots q_n$ where $p_i$, $q_j$ are prime ideals. We do induction on $m$. Note that $q_1\cdots q_n\subseteq p_1$. Since $p_1$ is prime, there exists $j$ such that $q_j\subseteq p_1$. WLOG $j=1$ and $q_1\subseteq p_1$. By ^ib3ruw, $p_1=q_j$ and then $p_2\cdots p_m=q_2\cdots q_n$. Apply inductive hypothesis and we finish the proof. \end{proof}

Lemma

Let $R$ be a Dedekind domain. For any fractional ideal $I$, there exists integral ideals $J$ and $K$ such that $I=J{K}^{-1}$.

\begin{proof} Since $I$ is a fractional ideal, there exists $\lambda \in R$ such that $\lambda I\subseteq R$. Let $J=\lambda I$ and $K=\lambda R$. Then $K^{-1}=(1/\lambda )R$ and $J{K}^{-1}=\lambda I\cdot (1/\lambda )R=I$. Now we finish the proof. \end{proof}

Definition

For every fractional ideal $I$, we write $I={\prod }_p{p}^{{\nu }_p(I)}$ with ${\nu }_p(I)\in \mathbb{Z}$ and $p$ runs over all prime ideals. We call ${\nu }_p(I)$ the valuation of $I$ at $p$.