Our next goal is to show that, in OK, every ideal can be written as a product of prime ideals uniquely. In fact, this is true in any Dedekind domain.
A fractional ideal A of OK is an OK-module contained in K such that there exists m ∈ Z with mA ⊆ OK. Of course, any ideal of OK is a fractional ideal by taking m = 1.
Exercise 5.3.1 Show that any fractional ideal is finitely generated as an OK -module.
Exercise 5.3.2 Show that the sum and product of two fractional ideals are again fractional ideals.
Lemma 5.3.3 Any proper ideal of OK contains a product of nonzero prime ideals.
Proof. Let S be the set of all proper ideals of OK that do not contain a product of prime ideals. We need to show that S is empty. If not, then since OK is Noetherian, S has a maximal element, say a. Then, a is not prime since a ∈ S, so there exist a, b ∈ OK, with ab ∈ a, a /∈ a, b /∈ a.
Then, (a, a) a, (a, b) a. Thus, (a, a) /∈ S, (a, b) /∈ S, by the maximality of a.
Thus, (a, a) ⊇ ℘1· · · ℘r and (a, b) ⊇ ℘′1· · · ℘′s, with the ℘i and ℘′j non-zero prime ideals. But ab ∈ a, so (a, ab) = a.
Thus, a = (a, ab) ⊇ (a, a)(a, b) ⊇ ℘1· · · ℘r℘′1· · · ℘′s. Therefore a contains a product of prime ideals. This contradicts a being in S, so S must actually be empty.
Thus, any proper ideal of OK contains a product of nonzero prime ideals.
2
Lemma 5.3.4 Let ℘ be a prime ideal of OK. There exists z ∈ K, z /∈ OK, such that z℘ ⊆ OK.
Proof. Take x ∈ ℘. From the previous lemma, (x) = xOK contains a product of prime ideals. Let r be the least integer such that (x) contains a product of r prime ideals, and say (x) ⊇ ℘1· · · ℘r, with the ℘i nonzero prime ideals.
Since ℘ ⊇ ℘1· · · ℘r, ℘ ⊇ ℘i, for some i, from Theorem 5.1.2 (d). With-out loss of generality, we can assume that i = 1, so ℘ ⊇ ℘1. But ℘1 is a nonzero prime ideal of OK, and so is maximal. Thus, ℘ = ℘1.
58 CHAPTER 5. DEDEKIND DOMAINS Now, ℘2· · · ℘r (x), since r was chosen to be minimal. Choose an element b ∈ ℘2· · · ℘r, b /∈ xOK. Then
bx−1℘ = bx−1℘1 ⊆ (℘2· · · ℘r)(x−1℘1)
= x−1(℘1· · · ℘r)
⊆ x−1xOK
= OK.
Put z = bx−1. Then z℘ ⊆ OK. Now, if z were in OK, we would have bx−1 ∈ OK, and so b ∈ xOK. But this is not the case, so z /∈ OK. Thus, we have found z ∈ K, z /∈ OK with z℘ ⊆ OK. 2
Let ℘ be a prime ideal. Define
℘−1= {x ∈ K : x℘ ⊆ OK}.
Lemma 5.3.4 implies, in particular, that ℘−1= OK.
Theorem 5.3.5 Let ℘ be a prime ideal of OK. Then ℘−1 is a fractional ideal and ℘℘−1= OK.
Proof. It is easily seen that ℘−1is an OK-module. Now, ℘ ∩ Z = (0), from Exercise 4.4.1, so let n ∈ ℘ ∩ Z, n = 0. Then, n℘−1 ⊆ ℘℘−1 ⊆ OK, by definition. Thus, ℘−1 is a fractional ideal.
It remains to show that ℘℘−1 = OK. Since 1 ∈ ℘−1, ℘ ⊆ ℘℘−1⊆ OK.
℘℘−1 is an ideal of OK, since it is an OK-module contained in OK. But ℘ is maximal, so either ℘℘−1= OK, in which case we are done, or ℘℘−1= ℘.
Suppose that ℘℘−1= ℘. Then x℘ ⊆ ℘ ∀x ∈ ℘−1. Since ℘ is a finitely generated Z-module (from Exercise 4.4.2), x ∈ OK for all x ∈ ℘−1, by Theorem 5.1.6. Thus, ℘−1 ⊆ OK. But 1 ∈ ℘−1, so ℘−1 = OK. From the comments above, and by the previous lemma, we know this is not true, so
℘℘−1 = ℘. Thus, ℘℘−1= OK. 2
Theorem 5.3.6 Any ideal of OK can be written as a product of prime ideals uniquely.
Proof.
Existence. Let S be the set of ideals of OK that cannot be written as a product of prime ideals. If S is nonempty, then S has a maximal element, since OK is Noetherian. Let a be a maximal element of S. Then a ⊆ ℘ for some maximal ideal ℘, since OK is Noetherian. Recall that every maximal ideal of OK is prime. Since a ∈ S, a = ℘ and therefore a is not prime.
Consider ℘−1a. ℘−1a⊂ ℘−1℘ = OK. Since a ℘,
℘−1a ℘−1℘ = OK,
since for any x ∈ ℘\a,
℘−1x ⊆ ℘−1a ⇒ x ∈ ℘℘−1a= OKa= a,
which is not true. Thus, ℘−1a is a proper ideal of OK, and contains a properly since ℘−1 contains OK properly. Thus, ℘−1a ∈ S, since a is a/ maximal element of S. Thus, ℘−1a = ℘1· · · ℘r, for some prime ideals ℘i. Then, ℘℘−1a = ℘℘1· · · ℘r, so a = ℘℘−1a= ℘℘1· · · ℘r. But then a /∈ S, a contradiction.
Thus, S is empty, so every ideal of OK can be written as a product of prime ideals.
Uniqueness. Suppose that a = ℘1· · · ℘r= ℘′1· · · ℘′sare two factoriza-tions of a as a product of prime ideals.
Then, ℘′1 ⊇ ℘′1· · · ℘′s= ℘1· · · ℘r, so ℘′1⊇ ℘i, for some i, say ℘′1 ⊇ ℘1. But ℘1 is maximal, so ℘′1 = ℘1. Thus, multiplying both sides by (℘′1)−1 and cancelling using (℘′1)−1℘′1= OK, we obtain
℘′2· · · ℘′s= ℘2· · · ℘r.
Thus, continuing in this way, we see that r = s and the primes are unique
up to reordering. 2
It is possible to show that any integral domain in which every non-zero ideal can be factored as a product of prime ideals is necessarily a Dedekind domain. We refer the reader to p. 82 of [Mat] for the details of the proof.
This fact gives us an interesting characterization of Dedekind domains.
When ℘ and ℘′ are prime ideals, we will write ℘/℘′ for (℘′)−1℘. We will also write ℘1℘2· · · ℘r
℘′1℘′2· · · ℘′s
to mean (℘′1)−1(℘′2)−1· · · (℘′s)−1℘1℘2· · · ℘r.
Exercise 5.3.7 Show that any fractional ideal A can be written uniquely in the
form ℘1. . . ℘r
℘′1. . . ℘′s
,
where the ℘i and ℘′j may be repeated, but no ℘i= ℘′j.
Exercise 5.3.8 Show that, given any fractional ideal A = 0 in K, there exists a fractional ideal A−1such that AA−1= OK.
For a and b ideals of OK, we say a divides b (denoted a | b), if a ⊇ b.
Exercise 5.3.9 Show that if a and b are ideals of OK, then b | a if and only if there is an ideal c of OK with a = bc.
Define d to be the greatest common divisor of a, b if:
60 CHAPTER 5. DEDEKIND DOMAINS (i) d | a and d | b; and
(ii) e | a and e | b ⇒ e | d.
Denote d by gcd(a, b).
Similarly, define m to be the least common multiple of a, b if:
(i) a | m and b | m; and (ii) a | n and b | n ⇒ m | n.
Denote m by lcm(a, b).
In the next two exercises, we establish the existence (and uniqueness) of the gcd and lcm of two ideals of OK.
Let a =r
i=1℘eii, b =r
i=1℘fii, with ei, fi∈ Z≥0. Exercise 5.3.10 Show that gcd(a, b) = a + b =r
i=1℘min(ei i,fi). Exercise 5.3.11 Show that lcm(a, b) = a ∩ b =r
i=1℘max(ei i,fi).
Exercise 5.3.12 Suppose a, b, c are ideals of OK. Show that if ab = cg and gcd(a, b) = 1, then a = dg and b = eg for some ideals d and e of OK. (This generalizes Exercise 1.2.1.)
Theorem 5.3.13 (Chinese Remainder Theorem) (a) Let a, b be ide-als so that gcd(a, b) = 1, i.e., a + b = OK. Given a, b ∈ OK, we can solve
x ≡ a (mod a), x ≡ b (mod b).
(b) Let ℘1, . . . , ℘r be r distinct prime ideals in OK. Given ai ∈ OK, ei∈ Z>0, ∃x such that x ≡ ai (mod ℘eii) for all i ∈ {1, . . . , r}.
Proof. (a) Since a + b = OK, ∃x1 ∈ a, x2 ∈ b with x1+ x2 = 1. Let x = bx1+ ax2 ≡ ax2 (mod a). But, x2 = 1 − x1≡ 1 (mod a). Thus, we have found an x such that x ≡ a (mod a). Similarly, x ≡ b (mod b).
(b) We proceed by induction on r. If r = 1, there is nothing to show.
Suppose r > 1, and that we can solve x ≡ ai (mod ℘eii) for i = 1, . . . , r −1, say a ≡ ai (mod ℘eii) for i = 1, . . . , r − 1. From part (a), we can solve
x ≡ a (mod ℘e11· · · ℘er−1r−1), x ≡ ar (mod ℘err).
Then x − ai∈ ℘e11· · · ℘er−1r−1, x ≡ ar (mod ℘err). Thus, x − ai∈ ℘eii ∀i, i.e.,
x ≡ ai (mod ℘eii) ∀i. 2
We define the order of a in ℘ by ord℘(a) = t if ℘t| a and ℘t+1∤ a.
Exercise 5.3.14 Show that ord℘(ab) = ord℘(a) + ord℘(b), where ℘ is a prime ideal.
Exercise 5.3.15 Show that, for α = 0 in OK, N ((α)) = |NK(α)|.
Theorem 5.3.16 (a) If a =r
i=1℘eii, then N (a) =
r
i=1
N (℘eii).
(b) OK/℘ ≃ ℘e−1/℘e, and
N (℘e) = (N (℘))e for any integer e ≥ 0.
Proof. (a) Consider the map
φ : OK −→ ⊕r
i=1(OK/℘eii), x −→ (x1, . . . , xr), where xi ≡ x (mod ℘eii).
The function φ is surjective by the Chinese Remainder Theorem, and φ is a homomorphism since each of the r components x −→ xi (mod ℘eii) is a homomorphism.
Next, we show by induction that r
i=1℘eii =r
i=1℘eii. The base case r = 1 is trivial. Suppose r > 1, and that the result is true for numbers smaller than r.
r
i=1
℘eii = lcm
r−1
i=1
℘eii, ℘err
= lcm
r−1
i=1
℘eii, ℘err
=
r
i=1
℘eii.
Thus, ker(φ) = r
i=1℘eii=r
i=1℘eii, which implies that OK/a ≃ ⊕(OK/℘eii).
Hence, N (a) =r
i=1N (℘eii).
(b) Since ℘e ℘e−1, we can find an element α ∈ ℘e−1/℘e, so that ord℘(α) = e − 1. Then ℘e ⊆ (α) + ℘e ⊆ ℘e−1. So ℘e−1 | (α) + ℘e. But
62 CHAPTER 5. DEDEKIND DOMAINS (α) + ℘e= ℘e, so ℘e−1= (α) + ℘e, by unique factorization. Define the map φ : OK −→ ℘e−1/℘e by φ(γ) = γα + ℘e. This is clearly a homomorphism and is surjective since ℘e−1= (α) + ℘e.
Now,
γ ∈ ker(φ) ⇔ γα ∈ ℘e
⇔ ord℘(γα) ≥ e
⇔ ord℘(γ) + ord℘(α) ≥ e
⇔ ord℘(γ) + e − 1 ≥ e
⇔ ord℘(γ) ≥ 1
⇔ γ ∈ ℘.
Thus, OK/℘ ≃ ℘e−1/℘e. Also,
(OK/℘e)/(℘e−1/℘e) ≃ OK/℘e−1
since the map from OK/℘e to OK/℘e−1 taking x + ℘e to x + ℘e−1 is a surjective homomorphism with kernel ℘e−1/℘e. Thus,
N (℘e) = |OK/℘e| = |OK/℘e−1| |℘e−1/℘e|
= N (℘e−1)N (℘)
= N (℘)e−1N (℘) by the induction hypothesis
= N (℘)e. 2
Thus, the norm function is multiplicative. Also, we can extend the definition of norm to fractional ideals, in the following way. Since any fractional ideal can be written uniquely in the form ab−1 where a, b are ideals of OK, we can put
N (ab−1) =N (a) N (b).
Let OK = Zω1+ · · · + Zωn. Then, if p is a prime number, we have pOK=Zpω1+ · · · + Zpωn, and so N ((p)) = pn, where n = [K :Q].
Exercise 5.3.17 If we write pOK as its prime factorization, pOK = ℘e11· · · ℘egg,
show that N (℘i) is a power of p and that if N (℘i) = pfii, g
i=1eifi= n.