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Section22.1Structure of a Finite Field

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 \(\gf(p^n)\).

Proof
Example22.8

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

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Figure22.9Subfields of \(\gf(p^{24})\)

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.

Proof
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*}