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 _{L }*, is a non-zero prime ideal, then there is a unique prime number

*p*such that

*p*∈

**P**.

For those who are more advanced, this is because the ideal generated by p, namely (p), is the kernel of

Then **P|***pO _{L }*and N(

**P**) = p

*for some*

^{f }*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*∈*O*then (_{L }*𝛼*) = {*x*+ 𝛼*y | 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*K*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) = a*with coefficients in_{n}x^{n }+ a_{n-1}x^{n-1}… + a_{1}x^{1}+ a_{0}*K*. Then its**reduction mod p**is defined as*p(x)*=*a*where_{n}x^{n }+ … + a_{0}*a*_{i}≡*a*(mod p)._{i} - 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**F**._{p}

Okay, now we’re ready for the theorem!

### Theorem: Dedekind’s Criterion

Let 𝛼 ∈ *O _{L }*be 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(

_{L }*x*) be its reduction mod p and factor

where g_{1}(*x*), … , g_{r}(*x*) ∈** F _{𝑝} **[

*x*] are distinct monic irreducible polynomials. Let g

_{i}(

*x*) ∈ Z[

*x*] be any polynomial with g

_{i}(

*x*) (mod 𝑝) = g

_{i}(

*x*), and define

an ideal of *O _{L}*. Let f

_{i }= deg g

_{i}(

*x*).

Then * p_{1} ,…, p_{r }*are disjoint prime ideals of

*O*and

_{L}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), *O _{L}*= . Then, [

*O*: Z[√−11]] = 2 and so we can apply Dedekind’s criterion to 𝛼 = √−11 for

_{L}*p*= 5. Then the minimal polynomial is

*f*(

*x*) =

*x*so 𝑓(

^{2 }+ 11,*x*) =

*f*(

*x*) (mod 5) =

*x*(

^{2 }+ 1 =*x*+

*2*)(

*x*+

*3*) ∈

**F**[

_{5}*x*].

Therefore by Dedekind’s Criterion, 5*O _{L}*=

**P**where

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

and **P, Q** are distinct prime ideals in *O _{L}*. 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!

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