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**3.1 The Definition and Some Basic Properties**

We identify the natural class of integral domains in which unique factorization of ideals

is possible.

**3.1.1 Definition**

A Dedekind domain is an integral domain A satisfying the following three conditions:

(1) A is a Noetherian ring;

(2) A is integrally closed;

(3) Every nonzero prime ideal of A is maximal.

A principal ideal domain satisfies all three conditions, and is therefore a Dedekind

domain. We are going to show that in the AKLB setup, if A is a Dedekind domain, then

so is B, a result that provides many more examples and already suggests that Dedekind

domains are important in algebraic number theory.

**3.1.2 Proposition**

In the AKLB setup, B is integrally closed, regardless of A. If A is an integrally closed

Noetherian ring, then B is also a Noetherian ring, as well as a finitely generated A-module.

Proof. By (1.1.6), B is integrally closed in L, which is the fraction field of B by (2.2.8).

Therefore B is integrally closed. If A is integrally closed, then by (2.3.8), B is a submodule

of a free A-module M of rank n. If A is Noetherian, then M, which is isomorphic to the

direct sum of n copies of A, is a Noetherian A-module, hence so is the submodule B. An

ideal of B is, in particular, an A-submodule of B, hence is finitely generated over A and

therefore over B. It follows that B is a Noetherian ring. ♣

**3.1.3 Theorem**

In the AKLB setup, if A is a Dedekind domain, then so is B. In particular, the ring of

algebraic integers in a number field is a Dedekind domain.Proof. In view of (3.1.2), it suffices to show that every nonzero prime ideal Q of B is

maximal. Choose any nonzero element x of Q. Since x ∈ B, x satisfies a polynomial

equation

xm + am−1xm−1 + · · · + a1x + a0 = 0

with the ai ∈ A. If we take the positive integer m as small as possible, then a0 = 0 by

minimality of m. Solving for a0, we see that a0 ∈ Bx ∩ A ⊆ Q ∩ A, so the prime ideal

P = Q ∩ A is nonzero, hence maximal by hypothesis. By Section 1.1, Problem 6, Q is

maximal. ♣

Problems For Section 3.1

This problem set will give the proof of a result to be used later. Let P1, P2, . . . , Ps, s ≥ 2,

be ideals in a ring R, with P1 and P2 not necessarily prime, but P3, . . . , Ps prime (if

s ≥ 3). Let I be any ideal of R. The idea is that if we can avoid the Pj individually, in

other words, for each j we can find an element in I but not in Pj , then we can avoid all

the Pj simultaneously, that is, we can find a single element in I that is in none of the Pj .

The usual statement is the contrapositive of this assertion.

Prime Avoidance Lemma

With I and the Pi as above, if I ⊆ ∪si

=1Pi, then for some i we have I ⊆ Pi.

1. Suppose that the result is false. Show that without loss of generality, we can assume

the existence of elements ai ∈ I with ai ∈ Pi but ai /∈

P1 ∪ ·· ·∪Pi−1 ∪ Pi+1 ∪ ·· ·∪Ps.

2. Prove the result for s = 2.

3. Now assume s > 2, and observe that a1a2 · · · as−1 ∈ P1 ∩ ·· · ∩ Ps−1, but as /∈

P1 ∪ ·· ·∪Ps−1. Let a = (a1 · · · as−1) + as, which does not belong to P1 ∪· · ·∪Ps−1, else

as would belong to this set. Show that a ∈ I and a /∈ P1 ∪ ·· · ∪ Ps, contradicting the

hypothesis.

**3.2 Fractional Ideals**

Our goal is to establish unique factorization of ideals in a Dedekind domain, and to do

this we will need to generalize the notion of ideal. First, some preliminaries.

**3.2.1 Products of Ideals**

Recall that if I1, . . . , In are ideals, then their product I1 · · · In is the set of all finite sums

i a1ia2i · · · ani, where aki ∈ Ik, k = 1, . . . , n. It follows from the definition that I1 · · · In

is an ideal contained in each Ij . Moreover, if a prime ideal P contains a product I1 · · · In

of ideals, then P contains Ij for some j.3.2.2 Proposition

If I is a nonzero ideal of the Noetherian integral domain R, then I contains a product of

nonzero prime ideals.

Proof. Assume the contrary. If S is the collection of all nonzero ideals that do not contain

a product of nonzero prime ideals, then, as R is Noetherian, S has a maximal element J,

and J cannot be prime because it belongs to S. Thus there are elements a, b ∈ R such

that a /∈ J, b /∈

J, and ab ∈ J. By maximality of J, the ideals J + Ra and J + Rb each

contain a product of nonzero prime ideals, hence so does (J+Ra)(J+Rb) ⊆ J+Rab = J.

This is a contradiction. (Notice that we must use the fact that a product of nonzero ideals

is nonzero, and this is where the hypothesis that R is an integral domain comes in.) ♣

**3.2.3 Corollary**

If I is an ideal of the Noetherian ring R (not necessarily an integral domain), then I

contains a product of prime ideals.

Proof. Repeat the proof of (3.2.2), with the word “nonzero” deleted. ♣

Ideals in the ring of integers are of the form nZ, the set of multiples of n. A set of

the form (3/2)Z is not an ideal because it is not a subset of Z, yet it behaves in a similar

manner. The set is closed under addition and multiplication by an integer, and it becomes

an ideal of Z if we simply multiply all the elements by 2. It will be profitable to study

sets of this type.

**3.2.4 Definitions**

Let R be an integral domain with fraction field K, and let I be an R-submodule of K.

We say that I is a fractional ideal of R if rI ⊆ R for some nonzero r ∈ R. We call r a

denominator of I. An ordinary ideal of R is a fractional ideal (take r = 1), and will often

be referred to as an integral ideal.

**3.2.5 Lemma**

(i) If I is a finitely generated R-submodule of K, then I is a fractional ideal.

(ii) If R is Noetherian and I is a fractional ideal of R, then I is a finitely generated

R-submodule of K.

(iii) If I and J are fractional ideals with denominators r and s respectively, then I∩J, I+J

and IJ are fractional ideals with respective denominators r (or s), rs and rs. [The product

of fractional ideals is defined exactly as in (3.2.1).]

Proof.

(i) If x1 = a1/b1, . . . , xn = an/bn generate I and b = b1 · · · bn, then bI ⊆ R.

(ii) If rI ⊆ R, then I ⊆ r−1R. As an R-module, r−1R is isomorphic to R and is therefore

Noetherian. Consequently, I is finitely generated.

(iii) It follows from the definition (3.2.4) that the intersection, sum and product of fractional

ideals are fractional ideals. The assertions about denominators are proved by noting

that r(I ∩ J) ⊆ rI ⊆ R, rs(I + J) ⊆ rI + sJ ⊆ R, and rsIJ = (rI)(sJ) ⊆ R. ♣

The product of two nonzero fractional ideals is a nonzero fractional ideal, and the

multiplication is associative because multiplication in R is associative. There is an identity

element, namely R, since RI ⊆ I = 1I ⊆ RI. We will show that if R is a Dedekind domain,

then every nonzero fractional ideal has a multiplicative inverse, so the nonzero fractional

ideals form a group.

**3.2.6 Lemma**

Let I be a nonzero prime ideal of the Dedekind domain R, and let J be the set of all

elements x ∈ K such that xI ⊆ R. Then R ⊂ J.

Proof. Since RI ⊆ R, it follows that R is a subset of J. Pick a nonzero element a ∈ I,

so that I contains the principal ideal Ra. Let n be the smallest positive integer such

that Ra contains a product P1 · · · Pn of n nonzero prime ideals. Since R is Noetherian,

there is such an n by (3.2.2), and by (3.2.1), I contains one of the Pi, say P1. But in a

Dedekind domain, every nonzero prime ideal is maximal, so I = P1. Assuming n ≥ 2, set

I1 = P2 · · · Pn, so that Ra ⊇ I1 by minimality of n. Choose b ∈ I1 with b /∈ Ra. Now

II1 = P1 · · · Pn ⊆ Ra, in particular, Ib ⊆ Ra, hence Iba−1 ⊆ R. (Note that a has an

inverse in K but not necessarily in R.) Thus ba−1 ∈ J, but ba−1 /∈

R, for if so, b ∈ Ra,

contradicting the choice of b.

The case n = 1 must be handled separately. In this case, P1 = I ⊇ Ra ⊇ P1, so

I = Ra. Thus Ra is a proper ideal, and we can choose b ∈ R with b /∈ Ra. Then

ba−1 /∈

R, but ba−1I = ba−1Ra = bR ⊆ R, so ba−1 ∈ J. ♣

We now prove that in (3.2.6), J is the inverse of I.

**3.2.7 Proposition**

Let I be a nonzero prime ideal of the Dedekind domain R, and let J = {x ∈ K : xI ⊆ R}.

Then J is a fractional ideal and IJ = R.

Proof. If r is a nonzero element of I and x ∈ J, then rx ∈ R, so rJ ⊆ R and J is a

fractional ideal. Now IJ ⊆ R by definition of J, so IJ is an integral ideal. Using (3.2.6),

we have I = IR ⊆ IJ ⊆ R, and maximality of I implies that either IJ = I or IJ = R.

In the latter case, we are finished, so assume IJ = I.

If x ∈ J, then xI ⊆ IJ = I, and by induction, xnI ⊆ I for all n = 1, 2, . . . . Let r be

any nonzero element of I. Then rxn ∈ xnI ⊆ I ⊆ R, so R[x] is a fractional ideal. Since

R is Noetherian, part (ii) of (3.2.5) implies that R[x] is a finitely generated R-submodule

of K. By (1.1.2), x is integral over R. But R, a Dedekind domain, is integrally closed, so

x ∈ R. Therefore J ⊆ R, contradicting (3.2.6). ♣

The following basic property of Dedekind domains can be proved directly from the

definition, without waiting for the unique factorization of ideals.

**3.2.8 Theorem**

If R is a Dedekind domain, then R is a UFD if and only if R is a PID.

Proof. Recall from basic algebra that a (commutative) ring R is a PID iff R is a UFD

and every nonzero prime ideal of R is maximal. ♣

Problems For Section 3.2

1. If I and J are relatively prime ideals (I + J = R), show that IJ = I ∩ J. More

generally, if I1, . . . , In are relatively prime in pairs, show that I1 · · · In = ∩ni

=1Ii.

2. Let P1 and P2 be relatively prime ideals in the ring R. Show that Pr

1 and Ps

2 are

relatively prime for arbitrary positive integers r and s.

3. Let R be an integral domain with fraction field K. If K is a fractional ideal of R, show

that R = K.

**3.3 Unique Factorization of Ideals**

In the previous section, we inverted nonzero prime ideals in a Dedekind domain. We now

extend this result to nonzero fractional ideals.

**3.3.1 Theorem**

If I is a nonzero fractional ideal of the Dedekind domain R, then I can be factored uniquely

as Pn1

1 Pn2

2

· · · Pnr

r , where the ni are integers. Consequently, the nonzero fractional ideals

form a group under multiplication.

Proof. First consider the existence of such a factorization. Without loss of generality, we

can restrict to integral ideals. [Note that if r = 0 and rI ⊆ R, then I = (rR)−1(rI).] By

convention, we regard R as the product of the empty collection of prime ideals, so let S

be the set of all nonzero proper ideals of R that cannot be factored in the given form, with

all ni positive integers. (This trick will yield the useful result that the factorization of

integral ideals only involves positive exponents.) Since R is Noetherian, S, if nonempty,

has a maximal element I0, which is contained in a maximal ideal I. By (3.2.7), I has an

inverse fractional ideal J. Thus by (3.2.6) and (3.2.7),

I0 = I0R ⊆ I0J ⊆ IJ = R.

Therefore I0J is an integral ideal, and we claim that I0 ⊂ I0J. For if I0 = I0J, then the

last paragraph of the proof of (3.2.7) can be reproduced with I replaced by I0 to reach a

contradiction. By maximality of I0, I0J is a product of prime ideals, say I0J = P1 · · · Pr

(with repetition allowed). Multiply both sides by the prime ideal I to conclude that I0 is

a product of prime ideals, contradicting I0 ∈ S. Thus S must be empty, and the existence

of the desired factorization is established.

To prove uniqueness, suppose that we have two prime factorizations

Pn1

1

· · · Pnr

r = Qt1

1

· · ·Qts

s

where again we may assume without loss of generality that all exponents are positive.

(If P−n appears, multiply both sides by Pn.) Now P1 contains the product of the Pni

i ,

so by (3.2.1), P1 contains Qj for some j. By maximality of Qj , P1 = Qj , and we may

renumber so that P1 = Q1. Multiply by the inverse of P1 (a fractional ideal, but there is

no problem), and continue inductively to complete the proof. ♣

**3.3.2 Corollary**

A nonzero fractional ideal I is an integral ideal if and only if all exponents in the prime

factorization of I are nonnegative.

Proof. The “only if” part was noted in the proof of (3.3.1). The “if” part follows because

a power of an integral ideal is still an integral ideal. ♣

**3.3.3 Corollary**

Denote by nP (I) the exponent of the prime ideal P in the factorization of I. (If P does

not appear, take nP (I) = 0.) If I1 and I2 are nonzero fractional ideals, then I1 ⊇ I2 if

and only if for every prime ideal P of R, nP (I1) ≤ nP (I2).

Proof. We have I2 ⊆ I1 iff I2I

−1

1

⊆ R, and by (3.3.2), this happens iff for every P,

nP (I2) − nP (I1) ≥ 0. ♣

**3.3.4 Definition**

Let I1 and I2 be nonzero integral ideals. We say that I1divides I2 if I2 = JI1 for some

integral ideal J. Just as with integers, an equivalent statement is that each prime factor

of I1 is a factor of I2.

**3.3.5 Corollary**

If I1 and I2 are nonzero integral ideals, then I1 divides I2 if and only if I1 ⊇ I2. In other

words, for these ideals,

DIV IDES MEANS CONTAINS.

Proof. By (3.3.4), I1 divides I2 iff nP (I1) ≤ nP (I2) for every prime ideal P. By (3.3.3),

this is equivalent to I1 ⊇ I2. ♣

**3.3.6 GCD’s and LCM’s**

As a nice application of the principle that divides means contains, we can use the prime

factorization of ideals in a Dedekind domain to compute the greatest common divisor

and least common multiple of two nonzero ideals I and J, exactly as with integers. The

greatest common divisor is the smallest ideal containing both I and J, that is, I +J. The

least common multiple is the largest ideal contained in both I and J, which is I ∩ J.

A Dedekind domain comes close to being a principal ideal domain in the sense that

every nonzero integral ideal, in fact every nonzero fractional ideal, divides some principal

ideal.

fractional ideal of the Dedekind domain R. Then there is a nonzero

integral ideal J such that IJ is a principal ideal of R.

Proof. By (3.3.1), there is a nonzero fractional ideal I such that II = R. By definition

of fractional ideal, there is a nonzero element r ∈ R such that rI is an integral ideal. If

J = rI , then IJ = Rr, a principal ideal of R. ♣

Problems For Section 3.3

By (2.3.11), the ring B of algebraic integers in Q(

√

−5) is Z[

√

−5]. In Problems 1-3, we will

show that Z[

√

−5] is not a unique factorization domain by considering the factorization

(1 +

√

−5)(1 −

√

−5) = 2 × 3.

1. By computing norms, verify that all four of the above factors are irreducible.

2. Show that the only units of B are ±1.

3. Show that no factor on one side of the above equation is an associate of a factor on

the other side, so unique factorization fails.

4. Show that the ring of algebraic integers in Q(

√

−17) is not a unique factorization

domain.

5. In Z[

√

−5] and Z

√

−17], the only algebraic integers of norm 1 are ±1. Show that this

property does not hold for the algebraic integers in Q(

√

−3).

**3.4 Some Arithmetic in Dedekind Domains**

Unique factorization of ideals in a Dedekind domain permits calculations that are analogous

to familiar manipulations involving ordinary integers. In this section, we illustrate

some of the ideas.

Let P1, . . . , Pn be distinct nonzero prime ideals of the Dedekind domain R, and let

J = P1 · · · Pn. Let Qi be the product of the Pj with Pi omitted, that is,

Qi = P1 · · · Pi−1Pi+1 · · · Pn.

(If n = 1, we take Q1 = R.) If I is any nonzero ideal of R, then by unique factorization,

IQi ⊃ IJ. For each i = 1, . . . ,n, choose an element ai belonging to IQi but not to IJ,

and let a = n

i=1 ai.

**3.4.1 Lemma**

The element a belongs to I, but for each i, a /∈

IPi. (In particular, a = 0.)

Proof. Since each ai belongs to IQi ⊆ I, we have a ∈ I. Now ai cannot belong to IPi,

for if so, ai ∈ IPi ∩ IQi, which is the least common multiple of IPi and IQi [see (3.3.6)].

But by definition of Qi, the least common multiple is simply IJ, which contradicts the

choice of ai. We break up the sum defining a as follows:

a = (a1 + · · · + ai−1) + ai + (ai+1 + · · · + an).

If j = i, then aj ∈ IQj ⊆ IPi, so the first and third terms of the right side of (1) belong

to IPi. Since ai /∈

IPi, as found above, we have a /∈ IPi. ♣

In (3.3.7), we found that any nonzero ideal is a factor of a principal ideal. We can

sharpen this result as follows.

**3.4.2 Proposition**

Let I be a nonzero ideal of the Dedekind domain R. Then there is a nonzero ideal I such

that II is a principal ideal (a). Moreover, if J is an arbitrary nonzero ideal of R, then I

can be chosen to be relatively prime to J.

Proof. Let P1, . . . ,Pn be the distinct prime divisors of J, and choose a as in (3.4.1). Then

a ∈ I, so (a) ⊆ I. Since divides means contains [see (3.3.5)], I divides (a), so (a) = II

for some nonzero ideal I . If I is divisible by Pi, then I = PiI0 for some nonzero ideal

I0, and (a) = IPiI0. Consequently, a ∈ IPi, contradicting (3.4.1). ♣

**3.4.3 Corollary**

A Dedekind domain with only finitely many prime ideals is a PID.

Proof. Let J be the product of all the nonzero prime ideals. If I is any nonzero ideal,

then by (3.4.2) there is a nonzero ideal I such that II is a principal ideal (a), with I

relatively prime to J. But then the set of prime factors of I is empty, so I = R. Thus

(a) = II = IR = I. ♣

The next result reinforces the idea that a Dedekind domain is not too far away from

a principal ideal domain.

**3.4.4 Corollary**

Let I be a nonzero ideal of the Dedekind domain R, and let a be any nonzero element of

I. Then I can be generated by two elements, one of which is a.

Proof. Since a ∈ I, we have (a) ⊆ I, so I divides (a), say (a) = IJ. By (3.4.2), there is

a nonzero ideal I such that II is a principal ideal (b) and I is relatively prime to J. If

gcd stands for greatest common divisor, then the ideal generated by a and b is

gcd((a), (b)) = gcd(IJ, II

) = I

because gcd(J, I ) = (1). ♣

**3.4.5 The Ideal Class Group**

Let I(R) be the group of nonzero fractional ideals of a Dedekind domain R. If P(R) is

the subset of I(R) consisting of all nonzero principal fractional ideals Rx, x ∈ K, then

P(R) is a subgroup of I(R). To see this, note that (Rx)(Ry)−1 = (Rx)(Ry−1) = Rxy−1,

which belongs to P(R). The quotient group C(R) = I(R)/P(R) is called the ideal class

group of R. Since R is commutative, C(R) is abelian, and we will show later that in the

number field case, C(R) is finite.

**3.4. SOME ARITHMETIC IN DEDEKIND DOMAINS 9**

Let us verify that C(R) is trivial if and only if R is a PID. If C(R) is trivial, then

every integral ideal I of R is a principal fractional ideal Rx, x ∈ K. But I ⊆ R, so x = 1x

must belong to R, proving that R is a PID. Conversely, if R is a PID and I is a nonzero

fractional ideal, then rI ⊆ R for some nonzero r ∈ R. By hypothesis, the integral ideal

rI must be principal, so rI = Ra for some a ∈ R. Thus I = R(a/r) with a/r ∈ K, and

we conclude that every nonzero fractional ideal of R is a principal fractional ideal.

Problems For Section 3.4

We will now go through the factorization of an ideal in a number field. In the next chapter,

we will begin to develop the necessary background, but some of the manipulations are

accessible to us now. By (2.3.11), the ring B of algebraic integers of the number field

Q(

√

−5) is Z[

√

−5]. (Note that −5 ≡ 3 mod 4.) If we wish to factor the ideal (2) = 2B

of B, the idea is to factor x2 + 5 mod 2, and the result is x2 + 5 ≡ (x + 1)2 mod 2.

Identifying x with

√

−5, we form the ideal P2 = (2,1 +

√

−5), which turns out to be

prime. The desired factorization is (2) = P2

2 . This technique works if B = Z[α], where

the number field L is Q(

√

α).

1. Show that 1 −

√

−5 ∈ P2, and conclude that 6 ∈ P2

2 .

2. Show that 2 ∈ P2

2 , hence (2) ⊆ P2

2 .

3. Expand P2

2 = (2,1 +

√

−5)(2,1 +

√

−5), and conclude that P2

2

⊆ (2).

4. Following the technique suggested in the above problems, factor x2 + 5 mod 3, and

conjecture that the prime factorization of (3) in the ring of algebraic integers of Q(

√

−5)

is (3) = P3P

3 for appropriate P3 and P

3.

5. With P3 and P

3 as found in Problem 4, verify that (3) = P3P

3.

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