4.6. Finite fields.

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Let Fp = Z/pZ, the field of p elements. As we noted in §1.2, any other field E of characteristic

p contains a copy of Fp, namely, {m1E | m ∈ Z}. No harm results if we identify Fp with this

subfield of E.

Let E be a field of degree n over Fp. Then E has q = pn elements, and so E× is a group

of order q − 1. Hence the nonzero elements of E are roots Xq−1 − 1, and all elements of E

(including 0) are roots of Xq − X. Hence E is a splitting field for Xq − X, and so any two

fields with q elements are isomorphic.

Now let E be the splitting field of f(X) = Xq − X, q = pn. The derivative f_(X) = −1,

which is relatively prime to f(X) (in fact, to every polynomial), and so f(X) has q distinct

roots in E. Let S be the set of its roots. Then S is obviously closed under multiplication

and the formation of inverses, but it is also closed under subtraction: if aq − a = 0 and

bq − b = 0, then

(a − b)q = aq − bq = a − b.

Hence S is a field, and so S = E. In particular, E has pn elements.

Proposition 4.15. For each power q = pn there is a field Fq with q elements. It is the

splitting field of Xq − X, and hence any two such fields are isomorphic. Moreover, Fq is

Galois over Fp with cyclic Galois group generated by the Frobenius automorphism σ(a) = ap.

Proof. Only the final statement remains to be proved. The field Fq is Galois over Fp

because it is the splitting field of a separable polynomial. We noted in (1.3) that σ = (x → xp)

is an automorphism of Fq . It has order n, and a ∈ Fq is fixed by σ if and only if ap = a. But

Fp consists exactly of such elements, and so the fixed field of < σ > is Fp. This proves that

< σ >= Gal(Fq/Fp).

Corollary 4.16. Let E be a field with pn elements. Then E contains exactly one field

with pm elements for each m|n, m ≥ 0, and E is Galois over that field.

Proof. We know that E is Galois over Fp and that Gal(E/Fp) is the cyclic group of order

n generated by σ. The subgroups of < σ > are the groups < σm > with m|n. The fixed

field of < σm > is Fpm.

Corollary 4.17. Every extension of finite fields is simple.

FIELDS AND GALOIS THEORY 33

Proof. Consider E ⊃ F. Then E× is a finite subgroup of the multiplicative group of a

field, and hence is cyclic (see Exercise 3). If ζ generates E× as a multiplicative group, then

clearly E = Fp[ζ].

Corollary 4.18. Each monic irreducible polynomial of degree d|n in Fp[X] occurs exactly

once as a factor of Xpn − X.

Proof. First, the factors of Xpn−X are distinct because it has no common factor with its

derivative. If f(X) is irreducible of degree d, then f(X) has a root in a field of degree d over

Fp. But the splitting field of Xpn −X contains a copy of every field of degree d over Fp with

d|n. Hence some root of Xpn − X is also a root of f(X), and therefore f(X)|Xpn − X.

Maple factors polynomials modulo p very quickly. The syntax is “Factor(f(X)) mod p;”.

Thus, for example, to obtain a list of all monic polynomials of degree 1, 2, or 4 over F5, ask

Maple to factor X625 − X.

Let F be an algebraic closure of Fp. Then F contains one field Fpn for each integer n ≥ 1—

it consists of all roots of Xpn − X—and Fpm ⊂ Fpn ⇐⇒ m|n. The partially ordered set of

finite subfields of F is isomorphic to the set of integers n ≥ 1 partially ordered by divisibility.

Finite fields were sometimes called Galois fields, and Fq used to be denoted GF(q) (it still

is in Maple). Maple contains a “Galois field package” to do computations in finite fields. For

example, it can find a primitive element for Fq (i.e., a generator for F×

q ). To start it, type:

readlib(GF);.

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