A construction of finite fields

We next turn to the finite field analog of the above section. One advantage we had when constructing $\mathbb{Q}(\sqrt{2})$ (for example), was that ``$\sqrt{2}$'' made sense - it was an element of the larger field $\mathbb{R}$ which contains $\mathbb{Q}$.

Example 2.1.16   Suppose we want to construct the analog of $\mathbb{Q}(\sqrt{2})$ but with $\mathbb{Q}$ replaced by $\mathbb{F}_3$. First, we need to check that $2$ is not a perfect square in $\mathbb{F}_3$ (it isn't but the reader should check this). Second, and more important perhaps, we need to define $\sqrt{2}$.

Let $F$ be a vector space of dimension $2$ over $\mathbb{F}_3$ with vector space basis

$$B=\{e_1=1,e_2=\sqrt{2}\},$$

where $\sqrt{2}$ is a formal symbol for some element which satisfies $e_2^2=2\in \mathbb{F}_3$ and commutes with all the elements of $\mathbb{F}_3$. (We shall see later, thanks to a Kronecker's Theorem 2.6.11, that this makes sense.) In other words, each element of $F$ is of the form

$$x_1e_1+x_2e_2,\ \ \ \ x_i\in \mathbb{F}_3.$$

Note

$$(x_1e_1+x_2e_2)(y_1e_1+y_2e_2) =(x_1y_1+2x_2y_2)e_1+(x_2y_1+x_1y_2)e_2.$$

This implies

$$(x_1e_1+x_2e_2)^{-1}=(x_1e_1-x_2e_2)/(x_1^2-2x_2^2).$$

The field axioms hold for $F$ (the reader should check this), so $F$ is a field of degree $2$ over $\mathbb{F}_3$.

Example 2.1.17   Let $p=5$, so $\mathbb{F}_5= \{{0},{1},{2},{3},{4}\}$ (with addition and multitplication mod $5$). The set of squares is given by

$$\{x^2\ \vert\ x\in \mathbb{F}_5\}=\{{0},{1},{4}\}.$$

In particular, ${2},{3}$ are not squares in this field. Let $e_2=\sqrt{2}$ be a formal symbol for some element which satisfies $e_2^2=2$. This is a root of the polynomial $x^2-2=0$.

The vector space $F$ over $\mathbb{F}_5$ with basis $\{e_1=1,e_2\}$ is 2-dimensional over $\mathbb{F}_5$. Two elements $x_1e_1+x_2e_2=x_1+x_2\sqrt{2}$ and $y_1e_1+y_2e_2=y_1+y_2\sqrt{2}$ are multiplied by the rule

$$(x_1+x_2\sqrt{2}) \cdot (y_1+y_2\sqrt{2})= x_1y_1+2x_2y_2+(x_1y_2+y_1x_2)\sqrt{2}.$$

It is a degree 2 field extension of $\mathbb{F}_5$.

The construction used in the above example may be summarized more generally as follows:

1. Pick an element $m\in \mathbb{F}_p$ which is not the square of another element, if such an element exists.

2. Let $e_1=1$ and $e_2=\sqrt{m}$ be a formal symbol for some element which satisfies $e_2^2=m$.

3. As a set, let $F=\{xe_1+ye_2\ \vert\ x,y\in \mathbb{F}_p\}$. To define $F$ as a field, let $+$ be ``componentwise addition'' mod $p$ and let $\cdot$ be defined by
   $$(x_1+x_2\sqrt{m}) \cdot (y_1+y_2\sqrt{m})= x_1y_1+mx_2y_2+(x_1y_2+y_1x_2)\sqrt{m}.$$

A finite field $F$ constructed in this way is called a quadratic extension of $\mathbb{F}_p$. A finite field $F$ constructed in this way has $p^2$ elements.

Remark 2.1.18   What if every element of $\mathbb{F}_p$ is the square of another element? (This can happen, for example, when $p=2$.) In this case, the above construction does not apply. However, other constructions do work. Details, will be given in the following section and in §2.6.