Dedekind Domains

Recall (Corollary 2.4.6) that we proved that the ring of integers $\O_K$ of a number field is noetherian, as follows. As we saw before using norms, the ring $\O_K$ is finitely generated as a module over  $\mathbf {Z}$, so it is certainly finitely generated as a ring over  $\mathbf {Z}$. By the Hilbert Basis Theorem, $\O_K$ is noetherian.

If $R$ is an integral domain, the field of fractions $\Frac (R)$ of $R$ is the field of all equivalence classes of formal quotients $a/b$, where $a,b \in R$ with $b\neq 0$, and $a/b\sim c/d$ if $ad=bc$. For example, the field of fractions of  $\mathbf {Z}$ is (canonically isomorphic to) $\mathbf {Q}$ and the field of fractions of $\mathbf{Z}[(1+\sqrt{5})/2]$ is $\mathbf{Q}(\sqrt{5})$. The field of fractions of the ring $\O_K$ of integers of a number field $K$ is just the number field $K$.

Example 3.1.1   We compute the fraction fields mentioned above.

sage: Frac(ZZ)
Rational Field

In Sage the Frac command usually returns a field canonically isomorphic to the fraction field (not a formal construction).

sage: K.<a> = QuadraticField(5)
sage: OK = K.ring_of_integers(); OK
Order with module basis 1/2*a + 1/2, a in Number Field 
in a with defining polynomial x^2 - 5
sage: Frac(OK)
Number Field in a with defining polynomial x^2 - 5

The fraction field of an order - i.e., a subring of $\O_K$ of finite index - is also the number field again.

sage: O2 = K.order(2*a); O2
Order with module basis 1, 2*a in Number Field 
in a with defining polynomial x^2 - 5
sage: Frac(O2)
Number Field in a with defining polynomial x^2 - 5

Definition 3.1.2 (Integrally Closed)   An integral domain $R$ is integrally closed in its field of fractions if whenever $\alpha$ is in the field of fractions of $R$ and $\alpha$ satisfies a monic polynomial $f\in R[x]$, then $\alpha \in R$.

Proposition 3.1.3   If $K$ is any number field, then $\O_K$ is integrally closed. Also, the ring $\overline{\mathbf{Z}}$ of all algebraic integers (in a fixed choice of $\overline{\mathbf{Q}}$) is integrally closed.

Proof. We first prove that  $\overline{\mathbf{Z}}$ is integrally closed. Suppose $\alpha\in\overline{\mathbf{Q}}$ is integral over  $\overline{\mathbf{Z}}$, so there is a monic polynomial $f(x)=x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ with $a_i\in\overline{\mathbf{Z}}$ and $f(\alpha)=0$. The $a_i$ all lie in the ring of integers $\O_K$ of the number field $K=\mathbf{Q}(a_0,a_1,\ldots a_{n-1})$, and $\O_K$ is finitely generated as a $\mathbf {Z}$-module, so $\mathbf{Z}[a_0,\ldots, a_{n-1}]$ is finitely generated as a $\mathbf {Z}$-module. Since $f(\alpha)=0$, we can write $\alpha^n$ as a $\mathbf{Z}[a_0,\ldots, a_{n-1}]$-linear combination of $\alpha^i$ for $i<n$, so the ring $\mathbf{Z}[a_0,\ldots, a_{n-1},\alpha]$ is also finitely generated as a $\mathbf {Z}$-module. Thus $\mathbf{Z}[\alpha]$ is finitely generated as $\mathbf {Z}$-module because it is a submodule of a finitely generated $\mathbf {Z}$-module, which implies that $\alpha$ is integral over  $\mathbf {Z}$.

Without loss we may assume that $K\subset \overline{\mathbf{Q}}$, so that $\O_K = \overline{\mathbf{Z}}\cap K$. Suppose $\alpha \in K$ is integral over $\O_K$. Then since $\overline{\mathbf{Z}}$ is integrally closed, $\alpha$ is an element of $\overline{\mathbf{Z}}$, so $\alpha\in K \cap \overline{\mathbf{Z}}=\O_K$, as required. $\qedsymbol$

Definition 3.1.4 (Dedekind Domain)   An integral domain $R$ is a Dedekind domain if it is noetherian, integrally closed in its field of fractions, and every nonzero prime ideal of $R$ is maximal.

The ring $\mathbf{Z}\oplus \mathbf{Z}$ is not a Dedekind domain because it is not an integral domain. The ring $\mathbf{Z}[\sqrt{5}]$ is not a Dedekind domain because it is not integrally closed in its field of fractions, as $(1+\sqrt{5})/2$ is integrally over $\mathbf {Z}$ and lies in $\mathbf{Q}(\sqrt{5})$, but not in $\mathbf{Z}[\sqrt{5}]$. The ring  $\mathbf {Z}$ is a Dedekind domain, as is any ring of integers $\O_K$ of a number field, as we will see below. Also, any field $K$ is a Dedekind domain, since it is an integral domain, it is trivially integrally closed in itself, and there are no nonzero prime ideals so the condition that they be maximal is empty. The ring $\overline{\mathbf{Z}}$ is not noetherian, but it is integrally closed in its field of fraction, and every nonzero prime ideal is maximal.

Proposition 3.1.5   The ring of integers $\O_K$ of a number field is a Dedekind domain.

Proof. By Proposition 3.1.3, the ring $\O_K$ is integrally closed, and by Proposition 2.4.6 it is noetherian. Suppose that  $\mathfrak{p}$ is a nonzero prime ideal of $\O_K$. Let $\alpha\in \mathfrak{p}$ be a nonzero element, and let $f(x) \in \mathbf{Z}[x]$ be the minimal polynomial of $\alpha$. Then

$$f(\alpha)=\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_1\alpha+a_0=0,$$

so $a_0 = -(\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_1\alpha)\in \mathfrak{p}$. Since $f$ is irreducible, $a_0$ is a nonzero element of $\mathbf {Z}$ that lies in  $\mathfrak{p}$. Every element of the finitely generated abelian group $\O_K/\mathfrak{p}$ is killed by $a_0$, so $\O_K/\mathfrak{p}$ is a finite set. Since  $\mathfrak{p}$ is prime, $\O_K/\mathfrak{p}$ is an integral domain. Every finite integral domain is a field (see Exercise 10), so $\mathfrak{p}$ is maximal, which completes the proof. $\qedsymbol$

If $I$ and $J$ are ideals in a ring $R$, the product $IJ$ is the ideal generated by all products of elements in $I$ with elements in $J$:

$$IJ = (ab : a\in I, b\in J) \subset R.$$

Note that the set of all products $ab$, with $a\in I$ and $b\in J$, need not be an ideal, so it is important to take the ideal generated by that set (see Exercise 11).

Definition 3.1.6 (Fractional Ideal)   A fractional ideal is a nonzero $\O_K$-submodule $I$ of $K$ that is finitely generated as an $\O_K$-module.

We will sometimes call a genuine ideal $I\subset \O_K$ an integral ideal. The notion of fractional ideal makes sense for an arbitrary Dedekind domain $R$ - it is an $R$-module $I\subset K=\Frac (R)$ that is finitely generated as an $R$-module.

Example 3.1.7   We multiply two fractional ideals in SAGE:

sage: K.<a> = NumberField(x^2 + 23)
sage: I = K.fractional_ideal(2, 1/2*a - 1/2)
sage: J = I^2
sage: I
Fractional ideal (2, 1/2*a - 1/2) of Number Field ...
sage: J
Fractional ideal (4, 1/2*a + 3/2) of Number Field ...
sage: I*J
Fractional ideal (-1/2*a - 3/2) of Number Field ...

Since fractional ideals $I$ are finitely generated, we can clear denominators of a generating set to see that there is some nonzero $\alpha \in K$ such that

$$\alpha I = J \subset \O_K$$

with $J$ an integral ideal. Thus dividing by $\alpha$, we see that every fractional ideal is of the form

$$a J = \{a b : b \in J\}$$

for some $a\in K$ and integral ideal $J\subset \O_K$.

For example, the set $\frac{1}{2}\mathbf{Z}$ of rational numbers with denominator $1$ or $2$ is a fractional ideal of $\mathbf {Z}$.

Theorem 3.1.8   The set of fractional ideals of a Dedekind domain $R$ is an abelian group under ideal multiplication with identity element $R$.

Note that fractional ideals are nonzero by definition, so it is not necessary to write ``nonzero fractional ideals'' in the statement of the theorem. We will only prove Theorem 3.1.8 in the case when $R=\O_K$ is the ring of integers of a number field $K$. Before proving Theorem 3.1.8 we prove a lemma. For the rest of this section $\O_K$ is the ring of integers of a number field $K$.

Definition 3.1.9 (Divides for Ideals)   Suppose that $I,J$ are ideals of $\O_K$. Then we say that $I$ divides $J$ if $I\supset J$.

To see that this notion of divides is sensible, suppose $K=\mathbf{Q}$, so $\O_K=\mathbf{Z}$. Then $I=(n)$ and $J=(m)$ for some integer $n$ and $m$, and $I$ divides $J$ means that $(n)\supset (m)$, i.e., that there exists an integer $c$ such that $m=cn$, which exactly means that $n$ divides $m$, as expected.

Lemma 3.1.10   Suppose $I$ is a nonzero ideal of $\O_K$. Then there exist prime ideals $\mathfrak{p}_1,\ldots, \mathfrak{p}_n$ such that $\mathfrak{p}_1\cdot \mathfrak{p}_2\cdots \mathfrak{p}_n \subset{}I$, i.e., $I$ divides a product of prime ideals.

Proof. Let $S$ be the set of nonzero ideals of $\O_K$ that do not satisfy the conclusion of the lemma. The key idea is to use that $\O_K$ is noetherian to show that $S$ is the empty set. If $S$ is nonempty, then since $\O_K$ is noetherian, there is an ideal $I\in S$ that is maximal as an element of $S$. If $I$ were prime, then $I$ would trivially contain a product of primes, so we may assume that $I$ is not prime. Thus there exists $a,b\in \O_K$ such that $ab\in I$ but $a\not\in I$ and $b\not\in I$. Let $J_1 = I+(a)$ and $J_2=I+(b)$. Then neither $J_1$ nor $J_2$ is in $S$, since $I$ is maximal, so both $J_1$ and $J_2$ contain a product of prime ideals, say $\mathfrak{p}_1 \cdots \mathfrak{p}_r \subset J_1$ and $\mathfrak{q}_1\cdots \mathfrak{q}_s \subset J_2$. Then

$$\mathfrak{p}_1 \cdots \mathfrak{p}_r \cdot \mathfrak{q}_1\cdots \mathfrak{q}_s \subset J_1 J_2 = I^2 + I(b)+ (a)I+ (ab) \subset I,$$

so $I$ contains a product of primes. This is a contradiction, since we assumed $I\in S$. Thus $S$ is empty, which completes the proof. $\qedsymbol$

We are now ready to prove the theorem.

Proof. [Proof of Theorem 3.1.8] Note that we will only prove Theorem 3.1.8 in the case when $R=\O_K$ is the ring of integers of a number field $K$.

The product of two fractional ideals is again finitely generated, so it is a fractional ideal, and $I\O_K=I$ for any ideal $I$, so to prove that the set of fractional ideals under multiplication is a group it suffices to show the existence of inverses. We will first prove that if  $\mathfrak{p}$ is a prime ideal, then  $\mathfrak{p}$ has an inverse, then we will prove that all nonzero integral ideals have inverses, and finally observe that every fractional ideal has an inverse. (Note: Once we know that the set of fractional ideals is a group, it will follows that inverses are unique; until then we will be careful to write ``an'' instead of ``the''.)

Suppose $\mathfrak{p}$ is a nonzero prime ideal of $\O_K$. We will show that the $\O_K$-module

$$I = \{a \in K : a\mathfrak{p}\subset \O_K \}$$

is a fractional ideal of $\O_K$ such that $I\mathfrak{p}= \O_K$, so that $I$ is an inverse of $\mathfrak{p}$.

For the rest of the proof, fix a nonzero element $b\in \mathfrak{p}$. Since $I$ is an $\O_K$-module, $bI\subset \O_K$ is an $\O_K$ ideal, hence $I$ is a fractional ideal. Since $\O_K \subset I$ we have $\mathfrak{p}\subset I \mathfrak{p} \subset \O_K$, hence since  $\mathfrak{p}$ is maximal, either $\mathfrak{p}= I\mathfrak{p}$ or $I\mathfrak{p}= \O_K$. If $I\mathfrak{p}= \O_K$, we are done since then $I$ is an inverse of  $\mathfrak{p}$. Thus suppose that $I\mathfrak{p}=\mathfrak{p}$. Our strategy is to show that there is some $d\in I$, with $d\not\in\O_K$. Since $I\mathfrak{p}=\mathfrak{p}$, such a $d$ would leave  $\mathfrak{p}$ invariant, i.e., $d \mathfrak{p}\subset \mathfrak{p}$. Since $\mathfrak{p}$ is an $\O_K$-module we will see that it will follow that $d\in \O_K$, a contradiction.

By Lemma 3.1.10, we can choose a product $\mathfrak{p}_1,\ldots, \mathfrak{p}_m$, with $m$ minimal, with

$$\mathfrak{p}_1\mathfrak{p}_2\cdots \mathfrak{p}_m \subset (b) \subset \mathfrak{p}.$$

If no $\mathfrak{p}_i$ is contained in $\mathfrak{p}$, then we can choose for each $i$ an $a_i \in \mathfrak{p}_i$ with $a_i\not\in \mathfrak{p}$; but then $\prod a_i\in \mathfrak{p}$, which contradicts that $\mathfrak{p}$ is a prime ideal. Thus some $\mathfrak{p}_i$, say $\mathfrak{p}_1$, is contained in $\mathfrak{p}$, which implies that $\mathfrak{p}_1 = \mathfrak{p}$ since every nonzero prime ideal is maximal. Because $m$ is minimal, $\mathfrak{p}_2\cdots \mathfrak{p}_m$ is not a subset of $(b)$, so there exists $c\in \mathfrak{p}_2\cdots \mathfrak{p}_m$ that does not lie in $(b)$. Then $\mathfrak{p}(c) \subset (b)$, so by definition of $I$ we have $d=c/b\in I$. However, $d\not\in\O_K$, since if it were then $c$ would be in $(b)$. We have thus found our element $d\in I$ that does not lie in $\O_K$.

To finish the proof that $\mathfrak{p}$ has an inverse, we observe that $d$ preserves the $\O_K$-module $\mathfrak{p}$, and is hence in $\O_K$, a contradiction. More precisely, if $b_1,\ldots,b_n$ is a basis for $\mathfrak{p}$ as a $\mathbf {Z}$-module, then the action of $d$ on  $\mathfrak{p}$ is given by a matrix with entries in  $\mathbf {Z}$, so the minimal polynomial of $d$ has coefficients in  $\mathbf {Z}$ (because $d$ satisfies the minimal polynomial of $\ell_d$, by the Cayley-Hamilton theorem - here we also use that $\mathbf{Q}\otimes \mathfrak{p}= K$, since $\O_K/\mathfrak{p}$ is a finite set). This implies that $d$ is integral over  $\mathbf {Z}$, so $d\in \O_K$, since $\O_K$ is integrally closed by Proposition 3.1.3. (Note how this argument depends strongly on the fact that $\O_K$ is integrally closed!)

So far we have proved that if $\mathfrak{p}$ is a prime ideal of $\O_K$, then

$$\mathfrak{p}^{-1} = \{a \in K : a\mathfrak{p}\subset \O_K\}$$

is the inverse of $\mathfrak{p}$ in the monoid of nonzero fractional ideals of $\O_K$. As mentioned after Definition 3.1.6, every nonzero fractional ideal is of the form $aI$ for $a\in K$ and $I$ an integral ideal, so since $(a)$ has inverse $(1/a)$, it suffices to show that every integral ideal $I$ has an inverse. If not, then there is a nonzero integral ideal $I$ that is maximal among all nonzero integral ideals that do not have an inverse. Every ideal is contained in a maximal ideal, so there is a nonzero prime ideal $\mathfrak{p}$ such that $I\subset \mathfrak{p}$. Multiplying both sides of this inclusion by $\mathfrak{p}^{-1}$ and using that $\O_K\subset \mathfrak{p}^{-1}$, we see that

$$I \subset \mathfrak{p}^{-1} I \subset \mathfrak{p}^{-1}\mathfrak{p}= \O_K.$$

If $I = \mathfrak{p}^{-1} I$, then arguing as in the proof that $\mathfrak{p}^{-1}$ is an inverse of  $\mathfrak{p}$, we see that each element of $\mathfrak{p}^{-1}$ preserves the finitely generated $\mathbf {Z}$-module $I$ and is hence integral. But then $\mathfrak{p}^{-1}\subset \O_K$, which, upon multiplying both sides by $\mathfrak{p}$, implies that $\O_K = \mathfrak{p}\mathfrak{p}^{-1} \subset \mathfrak{p}$, a contradiction. Thus $I \neq \mathfrak{p}^{-1} I$. Because $I$ is maximal among ideals that do not have an inverse, the ideal $\mathfrak{p}^{-1} I$ does have an inverse $J$. Then $\mathfrak{p}{}^{-1}J$ is an inverse of $I$, since $(J\mathfrak{p}^{-1})I = J(\mathfrak{p}^{-1}I) = \O_K.$ $\qedsymbol$

We can finally deduce the crucial Theorem 3.1.11, which will allow us to show that any nonzero ideal of a Dedekind domain can be expressed uniquely as a product of primes (up to order). Thus unique factorization holds for ideals in a Dedekind domain, and it is this unique factorization that initially motivated the introduction of ideals to mathematics over a century ago.

Theorem 3.1.11   Suppose $I$ is a nonzero integral ideal of $\O_K$. Then $I$ can be written as a product

$$I = \mathfrak{p}_1\cdots \mathfrak{p}_n$$

of prime ideals of $\O_K$, and this representation is unique up to order.

Proof. Suppose $I$ is an ideal that is maximal among the set of all ideals in $\O_K$ that can not be written as a product of primes. Every ideal is contained in a maximal ideal, so $I$ is contained in a nonzero prime ideal $\mathfrak{p}$. If $I\mathfrak{p}^{-1} = I$, then by Theorem 3.1.8 we can cancel $I$ from both sides of this equation to see that $\mathfrak{p}^{-1}=\O_K$, a contradiction. Since $\O_K\subset \mathfrak{p}^{-1}$, we have $I\subset I\mathfrak{p}^{-1}$, and by the above observation $I$ is strictly contained in $I\mathfrak{p}^{-1}$. By our maximality assumption on $I$, there are maximal ideals $\mathfrak{p}_1,\ldots, \mathfrak{p}_n$ such that $I\mathfrak{p}^{-1} = \mathfrak{p}_1\cdots \mathfrak{p}_n$. Then $I=\mathfrak{p}\cdot \mathfrak{p}_1\cdots \mathfrak{p}_n$, a contradiction. Thus every ideal can be written as a product of primes.

Suppose $\mathfrak{p}_1\cdots \mathfrak{p}_n=\mathfrak{q}_1\cdots \mathfrak{q}_m$. If no $\mathfrak{q}_i$ is contained in $\mathfrak{p}_1$, then for each $i$ there is an $a_i\in \mathfrak{q}_i$ such that $a_i\not\in\mathfrak{p}_1$. But the product of the $a_i$ is in $\mathfrak{p}_1\cdots \mathfrak{p}_n$, which is a subset of $\mathfrak{p}_1$, which contradicts that $\mathfrak{p}_1$ is a prime ideal. Thus $\mathfrak{q}_i=\mathfrak{p}_1$ for some $i$. We can thus cancel $\mathfrak{q}_i$ and $\mathfrak{p}_1$ from both sides of the equation by multiplying both sides by the inverse. Repeating this argument finishes the proof of uniqueness. $\qedsymbol$

Theorem 3.1.12   If $I$ is a fractional ideal of $\O_K$ then there exists prime ideals $\mathfrak{p}_1,\ldots, \mathfrak{p}_n$ and $\mathfrak{q}_1,\ldots, \mathfrak{q}_m$, unique up to order, such that

$$I = (\mathfrak{p}_1\cdots \mathfrak{p}_n)(\mathfrak{q}_1\cdots \mathfrak{q}_m)^{-1}.$$

Proof. We have $I=(a/b)J$ for some $a,b\in \O_K$ and integral ideal $J$. Applying Theorem 3.1.11 to $(a)$, $(b)$, and $J$ gives an expression as claimed. For uniqueness, if one has two such product expressions, multiply through by the denominators and use the uniqueness part of Theorem 3.1.11 $\qedsymbol$

Example 3.1.13   The ring of integers of $K=\mathbf{Q}(\sqrt{-6})$ is $\O_K=\mathbf{Z}[\sqrt{-6}]$. We have

$$6 = -\sqrt{-6}\sqrt{-6} = 2 \cdot 3.$$

If $ab=\sqrt{-6}$, with $a,b\in \O_K$ and neither a unit, then $\Norm (a)\Norm (b) = 6$, so without loss $\Norm (a)=2$ and $\Norm (b)=3$. If $a=c + d\sqrt{-6}$, then $\Norm (a) = c^2 + 6d^2$; since the equation $c^2 + 6d^2 = 2$ has no solution with $c,d\in\mathbf{Z}$, there is no element in $\O_K$ with norm $2$, so $\sqrt{-6}$ is irreducible. Also, $\sqrt{-6}$ is not a unit times $2$ or times $3$, since again the norms would not match up. Thus $6$ can not be written uniquely as a product of irreducibles in $\O_K$. Theorem 3.1.12, however, implies that the principal ideal $(6)$ can, however, be written uniquely as a product of prime ideals. An explicit decomposition is

$$(6) = (2, 2+\sqrt{-6})^2 \cdot (3,3+\sqrt{-6})^2,$$ (3.1)

where each of the ideals $(2, 2+\sqrt{-6})$ and $(3, 3+\sqrt{-6})$ is prime. We will discuss algorithms for computing such a decomposition in detail in Chapter 4. The first idea is to write $(6)=(2)(3)$, and hence reduce to the case of writing the $(p)$, for $p\in\mathbf{Z}$ prime, as a product of primes. Next one decomposes the finite (as a set) ring $\O_K/p\O_K$.

The factorization (3.1.1) can be compute using SAGE as follows:

sage: K.<a> = NumberField(x^2 + 6); K
Number Field in a with defining polynomial x^2 + 6
sage: K.factor_integer(6)
(Fractional ideal (2, a) of Number Field ...)^2 * 
(Fractional ideal (3, a) of Number Field ...)^2