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Structure 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.

Structure 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 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.

Proposition

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

Proof

Let \(\phi : {\mathbb Z} \rightarrow F\) be the ring homomorphism defined by \(\phi(n) = n \cdot 1\). Since the characteristic of \(F\) is \(p\), the kernel of \(\phi\) must be \(p {\mathbb Z}\) and the image of \(\phi\) must be a subfield of \(F\) isomorphic to \({\mathbb Z}_p\). We will denote this subfield by \(K\). Since \(F\) is a finite field, it must be a finite extension of \(K\) and, therefore, an algebraic extension of \(K\). Suppose that \([F : K] = n\) is the dimension of \(F\), where \(F\) is a \(K\) vector space. There must exist elements \(\alpha_1, \ldots, \alpha_n \in F\) such that any element \(\alpha\) in \(F\) can be written uniquely in the form \[\begin{aligned}\end{aligned}\], where the \(a_i\)'s are in \(K\). Since there are \(p\) elements in \(K\), there are \(p^n\) possible linear combinations of the \(\alpha_i\)'s. Therefore, the order of \(F\) must be \(p^n\).

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

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

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