Dedekind Domains

Recall (Corollary 5.2.5) that we proved that the ring of integers $\O _K$ of a number field is Noetherian. 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 of $R$ is the field of all elements $a/b$, where $a,b \in R$. The field of fractions of $R$ is the smallest field that contains $R$. For example, the field of fractions of $\mathbf{Z}$ is $\mathbf{Q}$ and of $\mathbf{Z}[(1+\sqrt{5})/2]$ is $\mathbf{Q}(\sqrt{5})$.

Definition 6.1.1 (Integrally Closed)   An integral domain $R$ is 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 6.1.2   If $K$ is any number field, then $\O _K$ is integrally closed. Also, the ring $\overline{\mathbf{Z}}$ of all algebraic integers is integrally closed.

Proof. We first prove that  $\overline{\mathbf{Z}}$ is integrally closed. Suppose $c\in\overline{\mathbf{Q}}$ 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 $f(c)=0$. 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 $f(c)=0$, we can write $c^n$ as a $\mathbf{Z}[a_0,\ldots, a_{n-1}]$-linear combination of $c^i$ for $i<n$, so the ring $\mathbf{Z}[a_0,\ldots, a_{n-1},c]$ is also finitely generated as a $\mathbf{Z}$-module. Thus $\mathbf{Z}[c]$ is finitely generated as $\mathbf{Z}$-module because it is a submodule of a finitely generated $\mathbf{Z}$-module, which implies that $c$ is integral over  $\mathbf{Z}$.

Suppose $c\in K$ is integral over $\O _K$. Then since $\overline{\mathbf{Z}}$ is integrally closed, $c$ is an element of $\overline{\mathbf{Z}}$, so $c\in K \cap \overline{\mathbf{Z}}=\O _K$, as required. $\qedsymbol$

Definition 6.1.3 (Dedekind Domain)   An integral domain $R$ is a if it is Noetherian, integrally closed in its field of fractions, and every nonzero prime ideal of $R$ is maximal.

The ring $\mathbf{Q}\oplus \mathbf{Q}$ is Noetherian, integrally closed in its field of fractions, and the two prime ideals are maximal. However, it 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 a domain, it is trivially integrally closed in itself, and there are no nonzero prime ideals so that condition that they be maximal is empty.

Proposition 6.1.4   The ring of integers $\O _K$ of a number field is a Dedekind domain.

Proof. By Proposition 6.1.2, the ring $\O _K$ is integrally closed, and by Proposition 5.2.5 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, 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 the homework problems for examples.)

Definition 6.1.5 (Fractional Ideal)   A is an $\O _K$-submodule of $I\subset K$ that is finitely generated as an $\O _K$-module.

To avoid confusion, we will sometimes call a genuine ideal $I\subset \O _K$ an . Also, since fractional ideals are finitely generated, we can clear denominators of a generating set to see that every fractional ideal is of the form $a I = \{a b : b \in I\}$ for some $a\in K$ and ideal $I\subset \O _K$.

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

Theorem 6.1.6   The set of nonzero fractional ideals of a Dedekind domain $R$ is an abelian group under ideal multiplication.

Before proving Theorem 6.1.6 we prove a lemma. For the rest of this section $\O _K$ is the ring of integers of a number field $K$.

Definition 6.1.7 (Divides for Ideals)   Suppose that $I,J$ are ideals of $\O _K$. Then $I$  $J$ 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 6.1.8   Suppose $I$ is an ideal of $\O _K$. Then there exist prime ideals $\mathfrak{p}_1,\ldots, \mathfrak{p}_n$ such that $\mathfrak{p}_1\cdot \mathfrak{p}_2\cdots \mathfrak{p}_n \subset{}I$. In other words, $I$ divides a product of prime ideals. (By convention the empty product is the unit ideal. Also, if $I=0$, then we take $\mathfrak{p}_1=(0)$, which is a prime ideal.)

Proof. The key idea is to use that $\O _K$ is Noetherian to deduce that the set $S$ of ideals that do not satisfy the lemma is empty. If $S$ is nonempty, then because $\O _K$ 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 $I$ is not prime. By definition of prime ideal, 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. Thus so does $I$, since

$$J_1 J_2 = I^2 + I(b)+ (a)I+ (ab) \subset I,$$

which is a contradiction. Thus $S$ is empty, which completes the proof. $\qedsymbol$

We are now ready to prove the theorem.

Proof. [Proof of Theorem 6.1.6] The product of two fractional ideals is again finitely generated, so it is a fractional ideal, and $I\O _K=\O _K$ for any nonzero 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 nonzero integral ideals have inverses, and finally observe that every fractional ideal has an inverse.

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 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 $d\in I$ not in $\O _K$; such a $d$ would leave  $\mathfrak{p}$ invariant (i.e., $d \mathfrak{p}\subset \mathfrak{p}$), so since $\mathfrak{p}$ is an $\O _K$-module it will follow that $d\in \O _K$, a contradiction.

By Lemma 6.1.8, we can choose a product $\mathfrak{p}_1,\ldots, \mathfrak{p}_m$, with $m$ minimal, such that

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

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 $\mathbf{Z}$. This implies that $d$ is integral over $\mathbf{Z}$, so $d\in \O _K$, since $\O _K$ is integrally closed by Proposition 6.1.2. (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 $\mathfrak{p}^{-1} = \{a \in {\mathbf K}: a\mathfrak{p}\subset \O _K\}$ is the inverse of $\mathfrak{p}$ in the monoid of nonzero fractional ideals of $\O _K$. As mentioned after Definition 6.1.5 [on Tuesday], 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 $I \subset \mathfrak{p}^{-1} I \subset \O _K$. If $I = \mathfrak{p}^{-1} I$, then arguing as in the proof that $\mathfrak{p}^{-1}$ is the 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 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, call it $J$. Then $\mathfrak{p}{}J$ is the inverse of $I$, since $\O _K = (\mathfrak{p}{}J)(\mathfrak{p}^{-1}I) = JI$. $\qedsymbol$

We can finally deduce the crucial Theorem 6.1.10, 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 rings of integers of number fields over a century ago.

Theorem 6.1.9   Suppose $I$ is an 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. (Exception: If $I=0$, then the representation is not unique.)

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 6.1.6 we can cancel $I$ from both sides of this equation to see that $\mathfrak{p}^{-1}=\O _K$, a contradiction. Thus $I$ is strictly contained in $I\mathfrak{p}^{-1}$, so 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 the $\mathfrak{p}_1\cdots \mathfrak{p}_n$, which is a subset of $\mathfrak{p}_1$, which contradicts the fact 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. Repeating this argument finishes the proof of uniqueness. $\qedsymbol$

Corollary 6.1.10   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 6.1.10 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 6.1.10 $\qedsymbol$

Example 6.1.11   The ring of integers of $K=\mathbf{Q}(\sqrt{-6})$ is $\O _K=\mathbf{Z}[\sqrt{-6}]$. In $\O _K$, 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 6.1.9, however, implies that the principal ideal $(6)$ can, however, be written uniquely as a product of prime ideals. Using we find such a decomposition:

> R<x> := PolynomialRing(RationalField());
> K := NumberField(x^2+6);
> OK := MaximalOrder(K);
> [K!b : b in Basis(OK)];
[
    1,
    K.1    // this is sqrt(-6)
]   
> Factorization(6*OK);
[
    <Prime Ideal of OK
    Two element generators:
        [2, 0]
        [2, 1], 2>,
    <Prime Ideal of OK
    Two element generators:
        [3, 0]
        [3, 1], 2>
]

The output means that

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

where each of the ideals $(2, 2+\sqrt{-6})$ and $(3, 3+\sqrt{-6})$ is prime. I will discuss algorithms for computing such a decomposition in detail, probably next week. 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 Artinian ring $\O _K\otimes \mathbf{F}_p$ as a product of local Artinian rings.