Friday, April 6, 2012

Algebraic Number Theory part :4: Dedekind Domains

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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
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
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).]
(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
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
· · · 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
· · · Pnr
r = Qt1
· · ·Qts
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
⊆ 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,
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
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
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(

3.4 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.
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

−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(

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).
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(

is (3) = P3P
3 for appropriate P3 and P
5. With P3 and P
3 as found in Problem 4, verify that (3) = P3P

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