# Creating new fields

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 $F$. Now, we can look at the ring of polynomials (which we will denote by $F[x]$) with coefficients which are taken from $F$. For example, we can take $F=\mathbb{Q}$, and an element of $\mathbb{Q}[x]$ will be, for example, $f(x)=x^2+\frac{1}{2}x+5$. However, $g(x)=x-\pi\notin \mathbb{Q}[x]$. You can check yourself that $F[x]$ is really a ring, which is not hard at all.

The ring $F[x]$ has a lot of strong properties:

• It is commutative: $f\cdot g = g\cdot f$ with no zero divisors (two elements $f,g\neq 0$ such that $f\cdot g=0$. 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 $I$ as $I=\langle f(x)\rangle$ for some $f\in F[x]$. 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 $f$ is a polynomial with degree $n$, then $F[x]/\langle f(x)\rangle$ is a Vector space with dimension $n$. 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 $F$. It also turns out that if the polynomial $f$ is ‘nice enough’ – the quotient ring is actually a field, and it contains $F$ as a subfield, thus, the process that we will do is going to give us a new field that extends $F$.

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 $g(x)+\langle f(x)\rangle$, I am going to write $\overline{g(x)}$ instead.

In order to prove that $F[x]/\langle f(x) \rangle$ is indeed a vector space of degree $n$, I have to find a basis for it. I am going to pick the set $\mathcal{B}=\{\overline{1}, \overline{x}, \overline{x}^2,\dots,\overline{x}^{n-1}\}$ as my candidate. I am going to to write $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$.

In order to prove that $\mathcal{B}$ 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 $b_i=0$ where $0\leq i \leq n-1$. 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; $f(x) | b_{n-1}\overline{x}^{n-1}+b_{n-2}\overline{x}^{n-2}+\cdots+b_0\cdot 1$, i.e. $f$ divides this polynomial, thus, if $b_i$ are not all zeros, then the degree of $f$ must be smaller or equals the degree of the linear comination. This gives us $n < n-1$ which is absolutely wrong, thus $b_i=0$ for every $i$, as required.

Spanning. I am just going to show how to express $\overline{x}^n$ as a linear combination of element of $\mathcal{B}$, this is enough (think why it that!). We know that in the qutient ring, $f(\overline{x})=0$, 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 $\mathcal{B}$, 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 $\{1,\overline{x},\overline{x}^2\}$. I would like to express the polynomial $\overline{x}^4+\overline{x}^3$ as a linear combination of elements from the basis. Following the steps from the proof, we know that $\overline{x}^3-\overline{x}^2-1=0$, thus $\overline{x}^3=\overline{x}^2+1$. 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 $R$, and a maximal ideal $I$ 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 $I$ is maximal if $I\subseteq J$ where $J$ is also an ideal, then $I = J$ or $J =R$.

Ok then, we have an integral domain $R$ with a maximal ideal $I$ in it. What can we do with this ideal? We can look at the qutient ring $R/I$. The qutient ring is also an integral domain, however, it turns out that it is a little more than that. Let $a+I\in R/I$ be an arbitrary element in the qutient ring, where $a\in R- I$. Let $(a) \vartriangleleft R$ be the ideal generated by $a$. 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 $a$ and some element from $I$. Clearly $I\subsetneq (a) + I$, however, $I$ is maximal, then $(a)+I=R$ , thus, there exists some $r\in R,i\in I$ such that : $r\cdot a + i = 1 \Rightarrow r\cdot a + I = 1 +I$. Then, in the qutient ring: $(r+I)\cdot(a+I)=r\cdot a+I = 1 + I$ and we have found that $a+I$ 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 $F[x]$ 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 $f$ is irreducible if $f = g \cdot h$ implies that $g$ is constant or $h$ 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 $\langle f(x) \rangle$ is maximal, and also suppose that $f(x) = g(x)\cdot h(x)$. Then $f(x) \in \langle g(x) \rangle$ and $f(x) \in \langle h(x) \rangle$. Thus: $\langle f(x) \rangle \subseteq \langle g(x)\rangle$, and also $\langle f(x) \rangle \subseteq \langle h(x)\rangle$. since $\langle f(x)\rangle$ is maximal then $\langle g(x)\rangle = \langle f(x)\rangle$ or $\langle g(x)\rangle = F[x]$, same goes for $\langle h(x) \rangle$ as well. If $\langle g(x)\rangle = \langle f(x)\rangle$ then $g(x) = c\cdot f(x)$ thus $h(x) = \frac{1}{c}$ which is a constant. If $\langle g(x)\rangle = R$ then $g(x)= c$ which is a constant. From both cases we got that one of the polynomials is constant, thus $f$ is irreducible.

On the other hand, suppose that $f$ is irreducible and $\langle f(x) \rangle \subset\langle g(x) \rangle$ where $\langle g(x) \rangle$ is a proper ideal. Then :$f(x) = g(x)h(x)$, and since $\langle f(x) \rangle$ is a proper subset of $\langle g(x) \rangle$, we get a contradiction to $f$ being irreducible, thus $\langle f(x) \rangle$ must be maximal.

## Summary

We now have a way to extend a field. We find an irreducible polynmial $f\in F[x]$ of degree $n$. Then the qutient ring $F[x]/\langle f(x)\rangle$ is a field, which contains $F$ (using the embedding $\varphi : F \to F[x]/\langle f(x)\rangle , \varphi(a)=a + \langle f(x) \rangle$). And is also a vector space above $F$ with dimension $n$. This conclution will give us endless results as we shall see in the future. That’s it for now.