## 2: Dedekind’s Criterion

In episode 1, I introduced the idea of prime ideals. Today we will extend this idea and prove a really important result in algebraic number theory: Dedekind’s Criterion.

We will use the following fact:

If P, contained in O, is a non-zero prime ideal, then there is a unique prime number p such that pP.

For those who are more advanced, this is because the ideal generated by p, namely (p), is the kernel of Then P|pOand N(P) = pfor some f > 0.

The proof of Dedekind’s Criterion uses a lot of Group Theory and therefore I will not prove it for you. However, it is a really useful tool in algebraic number theory and so I will state it and show how it can be used to factor ideals (remember that in episode 1 we showed that this factorisation is unique).

Before stating the theorem, let me define a few things:

• Let 𝛼 ∈ Othen (𝛼) = { x + 𝛼y | x, y ∈ }
• Let 𝐿/𝐾 be a field extension and let 𝛼 ∈ 𝐿 be algebraic over 𝐾 (i.e. there is a polynomial p with coefficients in such that p(𝛼)=0). We call the minimal polynomial of 𝛼 over 𝐾 the monic polynomial 𝑓 with coefficients in K of the least degree such that f(𝛼) = 0.
• Say we have a polynomial p(x) = anx+ an-1xn-1 … + a1x1 + a0  with coefficients in K. Then its reduction mod p is defined as p(x) = anx+ … + a0 where ai  ≡ ai (mod p).
• In episode 1 we defined the degree of a field extension L/K. We denote this as [L:K].
• Z/pZ is the additive group of the integers mod p. For p prime, this is a finite field. We usually denote this as Fp.

Okay, now we’re ready for the theorem!

### Theorem: Dedekind’s Criterion

Let 𝛼 ∈ Obe such that 𝐿 = Q(𝛼). Let 𝑓(x), with integer coefficients, be its minimal polynomial and let 𝑝 be a prime integer such that 𝑝 does not divide the degree [O∶ Z[𝛼]]. Let 𝑓(x) be its reduction mod p and factor where g1(x), … , gr(x F𝑝 [x] are distinct monic irreducible polynomials. Let gi(x) ∈ Z[x] be any polynomial with gi(x) (mod 𝑝) = gi(x), and define an ideal of OL. Let f= deg gi(x).

Then p1 ,…, pare disjoint prime ideals of OL and If you don’t quite understand the theorem, don’t worry! The first time I read this I was really confused as well. I think the more examples you see and the more you use it the easier it becomes to understand. Because of this, I will give you an example next.

### Example

Let L = (√−11) and p = 5. We will use the following result:

Let d ∈ Z be square-free and not equal to 0 or 1. Let L = (√d). Then As – 11 = 1 (mod 4), OL . Then, [OL: Z[√−11]] = 2 and so we can apply Dedekind’s criterion to 𝛼 = √−11 for p = 5. Then the minimal polynomial is f(x) = x+ 11, so 𝑓(x) = f(x) (mod 5) = x+ 1 = (x+2)(x+3) F5 [x].

Therefore by Dedekind’s Criterion, 5OL= P·where

P = (5, √−11 + 2) and Q = (5, √ −11 + 3)

and P, Q are distinct prime ideals in OL. So we have found how 5 splits in (√−11).

In the next episode I will talk about Dirichlet’s Unit Theorem and then we will be ready to solve some problems in Number Theory!

M x