# Section22.1Structure of a Finite Field¶ permalink

Recall that a field $F$ has characteristic $p$ if $p$ is the smallest positive integer such that for every nonzero element $\alpha$ in $F$, we have $p \alpha = 0$. If no such integer exists, then $F$ has characteristic 0. From Theorem 16.19 we know that $p$ must be prime. Suppose that $F$ is a finite field with $n$ elements. Then $n \alpha = 0$ for all $\alpha$ in $F$. Consequently, the characteristic of $F$ must be $p$, where $p$ is a prime dividing $n$. This discussion is summarized in the following proposition.

Throughout this chapter we will assume that $p$ is a prime number unless otherwise stated.

Let $F$ be a field. A polynomial $f(x) \in F[x]$ of degree $n$ is separable if it has $n$ distinct roots in the splitting field of $f(x)$; that is, $f(x)$ is separable when it factors into distinct linear factors over the splitting field of $f$. An extension $E$ of $F$ is a separable extension of $F$ if every element in $E$ is the root of a separable polynomial in $F[x]$.

##### Example22.4

The polynomial $x^2 - 2$ is separable over ${\mathbb Q}$ since it factors as $(x - \sqrt{2}\, )(x + \sqrt{2}\, )$. In fact, ${\mathbb Q}(\sqrt{2}\, )$ is a separable extension of ${\mathbb Q}$. Let $\alpha = a + b \sqrt{2}$ be any element in ${\mathbb Q}(\sqrt{2}\, )$. If $b = 0$, then $\alpha$ is a root of $x - a$. If $b \neq 0$, then $\alpha$ is the root of the separable polynomial \begin{equation*}x^2 - 2 a x + a^2 - 2 b^2 = (x - (a + b \sqrt{2}\, ))(x - (a - b \sqrt{2}\, )).\end{equation*}

Fortunately, we have an easy test to determine the separability of any polynomial. Let \begin{equation*}f(x) = a_0 + a_1 x + \cdots + a_n x^n\end{equation*} be any polynomial in $F[x]$. Define the derivative of $f(x)$ to be \begin{equation*}f'(x) = a_1 + 2 a_2 x + \cdots + n a_n x^{n - 1}.\end{equation*}

##### Proof

The unique finite field with $p^n$ elements is called the Galois field of order $p^n$. We will denote this field by $\gf(p^n)$.

##### Example22.8

The lattice of subfields of $\gf(p^{24})$ is given in Figure 22.9.

With each field $F$ we have a multiplicative group of nonzero elements of $F$ which we will denote by $F^*$. The multiplicative group of any finite field is cyclic. This result follows from the more general result that we will prove in the next theorem.

##### Example22.13

The finite field $\gf(2^4)$ is isomorphic to the field ${\mathbb Z}_2/ \langle 1 + x + x^4 \rangle$. Therefore, the elements of $\gf(2^4)$ can be taken to be \begin{equation*}\{ a_0 + a_1 \alpha + a_2 \alpha^2 + a_3 \alpha^3 : a_i \in {\mathbb Z}_2 \text{ and } 1 + \alpha + \alpha^4 = 0 \}.\end{equation*} Remembering that $1 + \alpha +\alpha^4 = 0$, we add and multiply elements of $\gf(2^4)$ exactly as we add and multiply polynomials. The multiplicative group of $\gf(2^4)$ is isomorphic to ${\mathbb Z}_{15}$ with generator $\alpha$: \begin{align*} & \alpha^1 = \alpha & & \alpha^6 = \alpha^2 + \alpha^3 & & \alpha^{11} = \alpha + \alpha^2 + \alpha^3 &\\ & \alpha^2 = \alpha^2 & & \alpha^7 = 1 + \alpha + \alpha^3 & & \alpha^{12} = 1 + \alpha + \alpha^2 + \alpha^3 &\\ & \alpha^3 = \alpha^3 & & \alpha^8 = 1 + \alpha^2 & & \alpha^{13} = 1 + \alpha^2 + \alpha^3 &\\ & \alpha^4 = 1 + \alpha & & \alpha^9 = \alpha + \alpha^3 & & \alpha^{14} = 1 + \alpha^3 &\\ &\alpha^5 = \alpha + \alpha^2 & & \alpha^{10} = 1 + \alpha + \alpha^2 & & \alpha^{15} = 1. & \end{align*}