##### Proposition22.1

If \(F\) is a finite field, then the characteristic of \(F\) is \(p\), where \(p\) is prime.

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.

If \(F\) is a finite field, then the characteristic of \(F\) is \(p\), where \(p\) is prime.

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

If \(F\) is a finite field of characteristic \(p\), then the order of \(F\) is \(p^n\) for some \(n \in {\mathbb N}\).

Let \(p\) be prime and \(D\) be an integral domain of characteristic \(p\). Then \begin{equation*}a^{p^n} + b^{p^n} = (a + b)^{p^n}\end{equation*} for all positive integers \(n\).

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 \(F[x]\).

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*}

Let \(F\) be a field and \(f(x) \in F[x]\). Then \(f(x)\) is separable if and only if \(f(x)\) and \(f'(x)\) are relatively prime.

For every prime \(p\) and every positive integer \(n\), there exists a finite field \(F\) with \(p^n\) elements. Furthermore, any field of order \(p^n\) is isomorphic to the splitting field of \(x^{p^n} -x\) over \({\mathbb Z}_p\).

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

Every subfield of the Galois field \(\gf(p^n)\) has \(p^m\) elements, where \(m\) divides \(n\). Conversely, if \(m \mid n\) for \(m \gt 0\), then there exists a unique subfield of \(\gf(p^n)\) isomorphic to \(\gf(p^m)\).

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.

If \(G\) is a finite subgroup of \(F^\ast\), the multiplicative group of nonzero elements of a field \(F\), then \(G\) is cyclic.

The multiplicative group of all nonzero elements of a finite field is cyclic.

Every finite extension \(E\) of a finite field \(F\) is a simple extension of \(F\).

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*}