Finite fields

In this section we summarize some results on field theory, especially on finite fields. We omit the proofs. See for example [4] ([5] if the reader prefers a Japanese book).

All the rings and fields in this section is assumed to be commutative.

The following lemma is well-known.

LEMMA 5.1   For any prime number $p$ , $\mathbb{Z}/p \mathbb{Z}$ is a field. (We denote it by $\mathbb{F}_p$ .)

Funny things about this field are:

LEMMA 5.2   Let $p$ be a prime number. Let $R$ be a commutative ring which contains $\mathbb{F}_p$ as a subring. Then:

1. $$\underbrace{1+1+\dots+1 }_{\text{$p$-times}}=0$$
   holds in $R$ .
2. For any $x,y\in R$ , we have
   $$(x+y)^p=x^p +y^p$$

We would like to show existence of ``finite fields''. A first thing to do is to know their basic properties.

LEMMA 5.3   Let $F$ be a finite field (that means, a field which has only a finite number of elements.) Then:

1. There exists a prime number $p$ such that $p=0$ holds in $F$ .
2. $F$ contains $\mathbb{F}_p$ as a subfield.
3. $q=\char93 (F)$ is a power of $p$ .
4. For any $x\in F$ , we have $x^q-x=0$ .
5. The multiplicative group $(F_q)^{\times}$ is a cyclic group of order $q-1$ .

The next task is to construct such field. An important tool is the following

LEMMA 5.4   For any field $K$ and for any non zero polynomial $f\in K[X]$ , there exists a field $L$ containing $L$ such that $f$ is decomposed into polynomials of degree $1$ .

To prove it we use the following lemma.

LEMMA 5.5   For any field $K$ and for any irreducible polynomial $f\in K[X]$ of degree $d>0$ , we have the following.

1. $L=K[X]/(f(X))$ is a field.
2. Let $a$ be the class of $X$ in $L$ . Then $a$ satisfies $f(a)=0$ .

LEMMA 5.6   Let $p$ be a prime number. Let $q=p^r$ be a power of $p$ . Let $L$ be a field extension of $\mathbb{F}_p$ such that $X^q-X$ is decomposed into polynomials of degree $1$ in $L$ . Then:

1. $$L_1=\{x \in L; x^q=x\}$$
   is a subfield of $L$ containing $\mathbb{F}_p$ .
2. $L_1$ has exactly $q$ elements.

LEMMA 5.7   Let $p$ be a prime number. Let $r$ be a positive integer. Let $q=p^r$ . Then:

1. There exists a field which has exactly $q$ elements.
2. There exists an irreducible polynomial $f$ of degree $r$ over $\mathbb{F}_p$ .
3. $X^q-X$ is divisible by $f$ .
4. For any field $K$ which has exactly $q$ -elements, there exists an element $a\in K$ such that $f(a)=0$ .

THEOREM 5.8   For any power $q$ of $p$ , there exists a field which has exactly $q$ elements. It is unique up to an isomorphism. (We denote it by $\mathbb{F}_q$ .)

The relation between various $\mathbb{F}_q$ 's is described in the following lemma.

LEMMA 5.9   There exists a homomorphism from $\mathbb{F}_q$ to $\mathbb{F}_{q'}$ if and only if $q'$ is a power of $q$ .

Note: The argument given in previous versions of this note was not good enough - inductive limits were taken for non-cofinal arrows. So we modified it to a corrected version(2006/11/29).

Suppose we are given a power $q$ of a prime number $p$ .

For each positive integer $n$ , we put

$$K_n=\mathbb{F}_{q^{n!}}$$   (a field with $q^n!$ elements)

which is unique up to an isomorphism. Then let us choose for each $n$ a field homomorphism

$$\phi_{n} : K_n \hookrightarrow K_{n+1}.$$

Then we take an inductive limit to define

$$\overline{\mathbb{F}_q}=\varinjlim_{n} (K_n)$$

It is easy to check that the following theorem holds.

THEOREM 5.10   $\overline{\mathbb{F}_q}$ is the algebraic closure of $\mathbb{F}_q$ .

Thus, a fortiori the isomorphism class of the field $\overline{\mathbb{F}_q}$ does not depend of the choice of $\{K_n\}$ or $\{\phi_n\}$ .