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Algebraic Number Theory – Lecture 6
Andrew Potter
“It was mentioned on CNN that the new prime number discovered recently is four times bigger then the previous record.”
- Setting
Throughout, let k be a number field of degree n. Recall that Ok is a Dedekind domain, in particular it has unique factorisation of ideals. That is, each ideal a ⊂ Ok factorises uniquely as a product of prime ideals. So, in some sense, “ideals take the place of rational integers”.
Definition. Let a, b ⊂ Ok be ideals. Their greatest common divisor, gcd(a, b), is the ideal g with the properties
(1) g | a and g | b; (2) if g′^ satisfies (1) then g′^ | g.
Similarly, their least common multiple, lcm(a, b), is the ideal l satisfying
(1) a | l and b | l; (2) if l′^ satisfies (1) then l | l′.
We have the useful properties
- gcd(a, b) = a + b
- lcm(a, b) = a ∩ b.
Let a ⊂ Ok be an ideal and b ∈ Ok. We write a | b to mean a | (b), the principal ideal generated by b. Then a | b ⇔ b ∈ a.
This notation is useful because if p is a prime ideal then
p | ab ⇒ p | a or p | b.
What about non-principal ideals?
Theorem 1. Let a 6 = 0 be an ideal of Ok and let β be an element of a. Then there exists α ∈ Ok such that a = (α, β).
1
2
- Norms
Recall that if α ∈ k and σi are the n embeddings k ↪→ C then we define
N (α) =
∏^ n
i=
σi(α).
Definition. The norm of an ideal a ⊂ Ok is
N (a) = |Ok/a|.
This is always a finite number, as seen in Dan’s lecture.
What’s the connexion between the norm of an ideal and that of an element?
Theorem 2. (1) Every ideal a ⊂ Ok, a 6 = 0, has a Z-basis {α 1 ,... , αn}.
(2) N (a) =
∆[α 1 ,... , αn] ∆
1 / 2 where ∆ is the discriminant of k.
Corollary. If a = (a) then N (a) = |N (a)|.
The main, useful property of norms is their multiplicativity: N (ab) = N (a)N (b).
But other interesting properties include:
(1) if N (a) is prime then a is a prime ideal; (2) N (a) ∈ a, i.e. a | N (a); (3) if a is a prime ideal then N (a) = pm^ for some m 6 n. Moreover, a divides exactly one p, so exactly one prime p ∈ Z is in a.
Thus norms are very handy for finding ideal factorisations. They also have several useful finiteness properties:
(1) Every nonzero ideal of Ok has finitely many divisors. (2) A nonzero rational integer belongs to only a finite number of ideals of Ok. (3) Only finitely many ideals of Ok have a given norm.
- Unique factorisation
Let R be a ring. A principal ideal domain is always a unique factorisation domain: PID ⇒ UFD. But, in general, UFD ; PID. However:
Theorem 3. Ok is a UFD if and only if it is a PID.
Proof. (⇐) This implication is always true for rings.
(⇒) Because of unique factorisation of ideals we only need to show that every prime ideal is principal. Let p be a prime ideal. There exists N = N (p) such that p | N. Ok is a UFD by assumption so N = π 1 · · · πs for πi irreducible in Ok. But
