Dedekind domain

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

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 the following

Theorem: All fractional ideals of $R$ are invertible.

Proof: 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 $\gamma\in K\setminus R$ 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 $b\in P_1\cdots P_{n-1}\setminus (a)$ 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 $\gamma\in K\setminus R$ 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.

This result has a number of important applications.

First, as we clearly have $RI = I$ for all fractional ideals $I$, the set of fractional ideals, $G$, of $R$ becomes an abelian group under multiplication. The identity element is $R$ and the existence of inverses is guaranteed by the above result. The set of principal ideal fractional ideals, $H$ forms a subgroup of this group. The quotient group is $G/H$ is then the ideal class group of $R$, a object of great importance in algebraic number theory.

We may define divisibility of ideals in the "obvious" way. Namely for fractional ideals $A,B$ we say that $A|B$ iff there is some integral ideal $C$ for which $B = AC$. The above result gives a useful characterization of divisibility.

Theorem: For fractional ideals $A,B$ of $R$, $A|B$ iff $B\subseteq A$.

Proof: First, if $A|B$, then there is some $C\subseteq R$ for which $B = AC$. But this gives $B = AC\subseteq AR = A$. Conversely assume that $B\subseteq A$. Consider the fractional ideal $A^{-1}$. We have $A^{-1}B\subseteq A^{-1}A = R$. Thus $A^{-1}B$ is a fractional ideal and $A^{-1}B\subseteq R$. It follows that $A^{-1}B$ is an integral ideal of $R$. But now $A(A^{-1}B) = AA^{-1}B = RB = B$ and so $A|B$. $\blacksquare$

As an application of this, we get that $\gcd(I,J) = I+J$ and $\text{lcm}(I,J) = I\cap J$ for all ideals $I$ and $J$.