The next few posts will be me detailing some interesting results in the area of Maths that I hope to specialise in: Algebraic Number Theory. The first result will be the unique prime factorisation of ideals. But first, what is an ideal?

## Ideals

If you’re not familiar with the definition of a **ring** click here, as we’ll need this for the following discussion.

An ideal, *I*, is a subset of a ring *R *if :

- It is closed under addition and has additive inverses (i.e. it is an additive subgroup of (
*R*, +. 0); - If
*a*∈*I*and*b*∈*R*, then*a*·*b*∈*I*.

*I* is a proper ideal if *I *does not equal *R*.

## Ring of Integers

In order to prove the result, I need to introduce the concept of number fields and the ring of integers. A *number field* is a finite field extension over the rationals.

Field Extension:a field extensionFof a fieldE(written F/E) is such that the operations ofEare those ofFrestricted toE,i.e.Eis a subfield ofF.Given such a field extension,

Fis a vector space overE,and the dimension of this vector space is called the degree of the extension. If this degree is finite, we have a finite field extension.

So if *F *is a number field, *E* would be the rationals.

Suppose *F *is a number field. An **algebraic integer **is simply a fancy name for an element *a *of F such that there exists a polynomial *f* with integer coefficients, which is monic (the coefficient of the largest power of *x* is 1), such that *f*(*a*) = 0.

The algebraic integers in a number field *F* form a ring, called the ring of integers, which we denote as *O _{F}*. It turns out that the ring of integers is very important in the study of number fields.

## Prime Ideals

If *P* is a **prime ideal** in a ring *R, *then for all *x, y* in *R*, if *xy* ∈ *P,* then x ∈ *P* or y ∈ *P*. As *O _{F}* is a ring we can consider prime ideals in

*O*.

_{F}## Division = Containment

We want to try and deal with ideals in the same way we deal with numbers, as ideals are easier to deal with (ideals are a sort of abstraction of the concept of numbers). After formalising what it means to be an ideal and proving certain properties of ideals, we can prove that given two ideals *I *and* J, I *dividing *J *(written *I*|*J*) is equivalent to *J* containing *I*.

## Three Key Results

Now, there are three results that we will need in order to prove the prime factorisation of ideals that I will simply state:

- All prime ideals
*P*in*O*are maximal (in other words, there are no other ideals contained between_{F }*P*and*R*). Furthermore, the converse also holds: all maximal ideals in*O*are prime._{F } - Analogously to numbers (elements of a number field
*F*), if*I*,*J*are ideals in*O*with_{F }*J|I,*there exists an ideal*K*contained in*I*, such that*I = JK*. - For prime ideal
*P**,*and ideals*I,J*of*O*,_{F}*P*|*IJ*implies*P*|*I*or*P*|*J*.

## Main Theorem

**Theorem:** Let *I* be a non-zero ideal in *O _{F }*. Then

*I*can be written uniquely as a product of prime ideals.

**Proof:** There are two things we have to prove: the existence of such a factorisation and then its uniqueness.

Existence: If *I *is prime then we are done, so suppose it isn’t. Then it is not maximal (by 1) so there is some ideal *J*, properly contained in *I*. So *J|I, *so (by 2) there is an ideal *K, *contained in *I*, such that *I = JK*. We can continue factoring this way and this must stop eventually (for the curious, I make the technical note that this must stop as we have an infinite chain of strictly ascending ideals, and *O _{F }*is Noetherian).

Uniqueness: If *P _{1} · · · P_{r} = Q_{1} · · · Q_{s},* with

*P*prime, then we know

_{i}, Q_{j}*P*|

_{1}*Q*, which implies

_{1}· · · Q_{s}*P*for some

_{1}| Q_{i}*i*(by 3), and without loss of generality

*i = 1*. So

*Q*is contained in

_{1}*P*. But

_{1}*Q*is prime and hence maximal (by 1). So

_{1}*P*. Simplifying we get

_{1}= Q_{1}*P*. Repeating this we get

_{2}· · · P_{r}= Q_{2}· · · Q_{s}*r = s*and

*P*for all

_{i}= Q_{i}*i*(after renumbering if necessary).

## Why is this important?

For numbers, we ** only** get unique prime factorisation in what is called a unique factorisation domain (UFD). Examples of UFDs are the complex numbers and the rationals. However, the integers mod 10 no longer form a UFD because, for example, 2*2 = 4 = 7*2 (mod 10).

However, we have the unique prime factorisation of ideals in the ring of algebraic integers of ** any** number field. This means that we can prove many cool results by using this unique prime factorisation, which we can then translate into results about numbers in that number field. I will detail some of these in future blog posts.

M x

Thanks that you started with well-known properties of natural numbers to let us get used with terminology and basic concepts. Whatever, my head is aching when I try to imagine rings with extensions with subsets, 3 years of job where you only have S = E x H from one end of the lab to another affect imagination not in the best way possible.

Can you please recommend a book over the subject? Or recommendations will come with last post in the series?

LikeLiked by 1 person

Give Modern Algebra by B. L. van der Waerden a try, specifically chapter 3 for rings and chapter 6 for field extensions!

LikeLiked by 1 person