In the last post I have presented what group, rings and fields are. Here I am going to discuss one way to create new fields of out existing one. Before I start, I assume that you are already familiar with **Ideals** and **quotient rings**. If not, I recommend you to come back here when you are familiar with those terms, however, I believe that in the future I am going to write about it here, in this site.

As I said, our goal for this post is to create new fields out of existing one, so let’s pick some arbitrary field . Now, we can look at the **ring** **of polynomials** (which we will denote by ) with coefficients which are taken from . For example, we can take , and an element of will be, for example, . However, . You can check yourself that is really a ring, which is not hard at all.

The ring has a lot of strong properties:

- It is commutative: with no zero divisors (two elements such that . i.e It is an
**integral domain**. - It is an
**euclidian domain**: This basically means you can divide polynomials with a remainder - It is a
**PID**(principal ideal domain) : This means that every ideal in the ring can be generated by a single element.

And a lot more properties. However, those properties are the most improtant one for our needs.

Since It is a PID, we can write every ideal as for some . From that we conclude that the **qutient rings** are of the form

F[x]/\langle f(x)\rangle

And we are now for the first insight. If is a polynomial with degree , then is a **Vector space with dimension **. I find this insight really elegant, we now have a way to create vector spaces out of rings! It gives us a structure that we can consider as two thing: as ring and as a vector space over the field . It also turns out that if the polynomial is ‘nice enough’ – the quotient ring is actually a **field**, and it contains as a subfield, thus, the process that we will do is going to give us a **new field** that **extends **.

First things first though, I still have to prove this insight, so let’s start with that. Small remark first: Instead of writing the elements of the qutient ring as , I am going to write instead.

In order to prove that is indeed a vector space of degree , I have to find a basis for it. I am going to pick the set as my candidate. I am going to to write .

In order to prove that is indeed a basis, we need to show that it is linearly independent, and is a spanning set of the whole space.

**Linearly independent**. Given a linear combination

b_{n-1}\overline{x}^{n-1}+b_{n-2}\overline{x}^{n-2}+\cdots+b_0\cdot1=0

We shall prove that where . since this combination equals zero in the qutient ring, we conclude that :

b_{n-1}\overline{x}^{n-1}+b_{n-2}\overline{x}^{n-2}+\cdots+b_0\cdot1\in\langle f(x) \rangle

Then; , i.e. divides this polynomial, thus, if are not all zeros, then the degree of must be **smaller or equals** the degree of the linear comination. This gives us which is absolutely wrong, thus for every , as required.

**Spanning**. I am just going to show how to express as a linear combination of element of , this is enough (think why it that!). We know that in the qutient ring, , then we get

f(\overline{x})=a_n\overline{x}^n +a_{n-1}\overline{x}^{n-1}+\cdots +a_0 = 0

So we can now write:

\overline{x}^n =-\frac{1}{a_n}(a_{n-1}\overline{x}^{n-1}+\cdots +a_0 )

And that is a valid linear combination of elements of , which completes the proof.

So we have proved this fabulous theorem, before I proceed, I want to show a small example first. Condier the qutient ring:

\mathbb{Q}[x]/\langle x^3-x^2-1\rangle

A basis for this vector space will be . I would like to express the polynomial as a linear combination of elements from the basis. Following the steps from the proof, we know that , thus . We can use that to get:

\overline{x}^4+\overline{x}^3=\overline{x}\cdot\overline{x}^3+\overline{x}^3=\overline{x}\cdot(\overline{x}^2+1) + \overline{x}^2+1

=\overline{x}^3+\overline{x}+\overline{x}^2+1=\overline{x}^2+1+\overline{x}+\overline{x}^2+1=2\overline{x}^2+\overline{x}+2

And we found what we were looking for.

## Maximal ideals and fields

I want to take a little break now from the ring of polynomials and talk about general integral domains, so I’ll start with some integral domain , and a maximal ideal or the ring. A **maximal ideal** is an ideal which is not the whole ring (it is a **proper** ideal), however, it is not contained in any other ideal. In other word, we say that is maximal if where is also an ideal, then or .

Ok then, we have an integral domain with a maximal ideal in it. What can we do with this ideal? We can look at the qutient ring . The qutient ring is also an integral domain, however, it turns out that it is a little more than that. Let be an arbitrary element in the qutient ring, where . Let be the ideal generated by . Consider now the ideal:

(a)+I=\{r\cdot a +i |r\in R, i\in I\}

This is indeed and ideal, and it’s elements are linear combination of and some element from . Clearly , however, is maximal, then , thus, there exists some such that : . Then, in the qutient ring: and we have found that is invertable! Thus, if we look at a qutient ring of a maximal ideal, we get a **field!** Let’s apply this knowledge to our case:

## maximal ideals in the ring of polynomials

The ideals in are all principal ideals, this leads us to think that we need a specific** **property from the polynomial itself who generated the ideal so it would be maximal. Turns out that this property is that the polynomial has to be **irreducible**. We say that is irreducible if implies that is constant or is constant (you can replace ‘constant’ with ‘invertible’ to get a more general defintion which applies to all rings). Let’s prove this statement:

\langle f(x) \rangle \text{ is maximal} \iff f(x)\text{ is irreducible}

Suppose that is maximal, and also suppose that . Then and . Thus: , and also . since is maximal then or , same goes for as well. If then thus which is a constant. If then which is a constant. From both cases we got that one of the polynomials is constant, thus is irreducible.

On the other hand, suppose that is irreducible and where is a proper ideal. Then :, and since is a proper subset of , we get a contradiction to being irreducible, thus must be maximal.

## Summary

We now have a way to extend a field. We find an irreducible polynmial of degree . Then the qutient ring is a field, which contains (using the embedding ). And is also a vector space above with dimension . This conclution will give us endless results as we shall see in the future. That’s it for now.