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

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

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$:

$\displaystyle 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

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

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

$\displaystyle 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

$\displaystyle \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

$\displaystyle 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

$\displaystyle 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

$\displaystyle 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)];
    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

$\displaystyle (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.

William Stein 2004-05-06