Dedekind domain

A Dedekind domain is a integral domain $R$ satisfying the following properties:

$R$ is a noetherian ring.
Every prime ideal of $R$ is a maximal ideal.
$R$ is integrally closed in its field of fractions.

Dedekind domains are very important in abstract algebra and number theory. For example, the ring of integers of any number field is a Dedekind domain.

There are several very nice properties of Dedekind domains:

Dedekind domains have unique prime factorizations of ideals (but not necessarily of elements).

There are also various properties of homological importance that Dedekind domains satisfy.

Invertibility of Ideals

Let $R$ be a Dedekind domain with field of fractions $K$ , and let $I$ be any nonzero fractional ideal of $R$ . We call $I$ invertible if there is a fractional ideal $I^{-1}$ such that $II^{-1}=R$ . We shall show that all fractional ideals of $R$ are invertible.

Given any nonzero fractional ideal $I$ of $R$ define $I^{-1} = \{\beta\in K|\beta I\subseteq R\}$ . $I^{-1}$ is clearly an $R$ -module. Moreover, for any nonzero $\alpha \in I\cap R$ (such an alpha clearly exists, if $x/y\in I$ for $x,y\in R$ then $x\in I$ ) we have $\alpha I^{-1}\subseteq R$ by the definition of $I^{-1}$ , and so $I^{-1}$ must be a fractional ideal of $R$ . It follows that $II^{-1}$ is a fractional ideal of $R$ as well, let $II^{-1} = A$ . By definition, $A\subseteq R$ , and so $A$ is an integral ideal. We claim that in fact $A = R$ , and so $I$ is invertible.

We will need the following lemmas.

Lemma 1: Every nonzero integral ideal $J$ of $R$ contains a product of prime ideals (counting $R$ as the empty product).

Proof: Assume that this is not the case. Let $\mathcal S$ be the collection of integral ideals of $R$ not containing a product of prime ideals, so $\mathcal S$ is nonempty and $R\not\in \mathcal S$ . As $R$ is noetherian, $\mathcal S$ must have a maximal element, say $M$ . Clearly $M$ cannot be prime (otherwise it would contain itself), so there must be $x,y\in R$ with $xy\in M$ but $x,y\not\in M$ . But then $M\subsetneq M+(x),M+(y)$ , and so $M+(x)$ and $M+(y)$ contain products of prime ideals. But then $(M+(x))(M+(y)) = M+(xy)\subseteq M$ also contains a product of prime ideals, contradicting the choice of $M$ . $\square$

Lemma 2: For any proper integral ideal $J$ , there is some $\gamma\in K\sm R$ (Error compiling LaTeX. Unknown error_msg) for which $\gamma J\subseteq R$ .

Proof: Take any nonzero $a\in J$ . By Lemma 1, $(a)$ contains a product of prime ideals, say $(a)\supseteq P_1P_2\cdots P_n$ with $n$ minimal (i.e. $J$ does not contain a product of $n-1$ prime ideals). As $R\not\subseteq (a)$ , $n\ge 1$ . As $J$ is a proper ideal, it must be contained in some maximal ideal, $P$ . Since maximal ideals are prime in commutative rings, $P$ is prime. But now $P_1P_2\cdots P_n\subseteq P$ . Thus as $P$ is prime, $P_i\subseteq P$ for some $i$ (if $P$ is prime and $A,B$ are ideals with $AB\subseteq P$ then either $A\subseteq P$ or $B\subseteq P$ ). But as $R$ is a Dedekind domain, $P_i$ must be maximal, so $P = P_i$ . Now assume WLOG that $i = n$ . By the minimality of $n$ , $(P_1\cdots P_{n-1})\not\subseteq (a)$ . Take any $b\in P_1\cdots P_{n-1}\sm (a)$ (Error compiling LaTeX. Unknown error_msg) let $\gamma = b/a\in K$ . We claim that this is the desired $\gamma$ .

First if $\gamma\in R$ then $b = \gamma a\in (a)$ , a contradiction, so $\gamma\not\in R$ . Now for any $x\in J$ , $bx\in P_1\cdots P_{n-1}J\subseteq P_1\cdots P_n\subseteq (a)$ , and so $bx = ar$ for some $r\in R$ . But now $\gamma x = \frac{bx}{a} = r\in R$ , and so $\gamma J\subseteq R$ , as required. $\square$

Now we return to the main proof. Assume that $A\ne R$ . Then by Lemma 2, there is some $\gamma\in K\sm R$ (Error compiling LaTeX. Unknown error_msg) for which $\gamma A\subseteq R$ . By the definition of $I^{-1}$ , for any $\beta\in I^{-1}$ we have $(\gamma\beta)I = \gamma(\beta I) \subseteq \gamma II^{-1} = \gamma A \subseteq R,$ and so $\gamma\beta\in I^{-1}$ . It follows that $\gamma I^{-1}\subseteq I^{-1}$ . We claim that this implies $\gamma\in R$ (contradicting the choice of $\gamma$ ).

Indeed, the map $f:x\mapsto \gamma x$ is an $R$ -linear map from $I^{-1}\to I^{-1}$ . As $R$ is noetherian, $I^{-1}$ must be a finitely generated $R$ -module. Indeed, for some nonzero $r\in R$ , $rI^{-1}$ must be an integral ideal of $R$ , which is finitely generated by the definition of noetherian rings. But if $rI^{-1} = Ry_1+\cdots + Ry_m$ then $I^{-1} = R(y_1/r)+\cdots+R(y_m/r)$ , so $I^{-1}$ is finitely generated as well. Now take $I^{-1} = Rx_1+\cdots +Rx_m$ , and let $M_f$ be the matrix representation of $f$ with respect to $x_1,\ldots,x_m$ . Then $M_f$ is an $m\times m$ matrix with coefficients in $R$ and we have: $M_f \left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_m \end{array} \right) = \gamma \left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_m \end{array} \right),$ and so $\gamma$ is an eigenvalue of $M_f$ . But then $\gamma$ is a root of the characteristic polynomial, $g(t) = |I_m(t)-M|$ of $M_f$ . But as $M_f$ has all of its entries in $R$ , $g(x)$ is a monic polynomial in $R[x]$ . Thus as $R$ is integrally closed in $K$ , $\gamma\in R$ .

This is a contradiction, and so we must have $A = II^{-1} = R$ , and so $I$ is indeed invertible. $\blacksquare$

In fact, the converse is true as well: if all nonzero ideals are invertible, then $R$ is a Dedekind domain. This is sometimes used as a definition of Dedekind domains.

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Dedekind domain

Invertibility of Ideals